Chapter 19 Problem Solutions 19-1.
Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus È q!Eg 1 1 ˘ Ï ¸ 14 10 ni(Ti) = Nd = 10 = 10 exp ÍÎ -! 2k !ÌT !-!300˝˙˚ Ó i ˛ Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields Ti = 262 °C or 535 °K.
19-2.
1 1 = 43.5 ohm-cm N-side resistivity rn = q!m !N = -19 (1.6x10 )(1500)(1014) n d 1 1 P-side resistivity rp = q!m !N = = 0.013 ohm-cm -19 p a (1.6x10 )(500)(1018)
19-3.
Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximate formulas given in Chapter 19. 2 ni n = Nd = 1013 cm-3 ; p = N d
19-4.
po =
2 ni [300] Nd
; 2po =
2 2 2 ni [300] = ni [300 + T]
1020 = 1013
= 107 cm-3
2 ni [300!+!T] Nd ;
È q!Eg Ï1 1 ¸˘ 10 10 2x10 = 10 exp ÍÎ -! 2k !ÌT!-!300˝˙˚ Ó
˛
q!Eg!300 Solving for T yields T = (q!E !-!k!300!ln(2)) = 305.2 °K g DT = 305.2 - 300 = 5.2 °K. 19-5.
19-6.
q!V1 I1 = Is exp( k!T
q!V1!+!dV ; 10 I1 = Is exp( k!T ) ;
k!T dV = q ln(10) = 60 mV
(a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
Chapter 19 Problem Solutions 19-1.
Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus È q!Eg 1 1 ˘ Ï ¸ 14 10 ni(Ti) = Nd = 10 = 10 exp ÍÎ -! 2k !ÌT !-!300˝˙˚ Ó i ˛ Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields Ti = 262 °C or 535 °K.
19-2.
1 1 = 43.5 ohm-cm N-side resistivity rn = q!m !N = -19 (1.6x10 )(1500)(1014) n d 1 1 P-side resistivity rp = q!m !N = = 0.013 ohm-cm -19 p a (1.6x10 )(500)(1018)
19-3.
Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximate formulas given in Chapter 19. 2 ni n = Nd = 1013 cm-3 ; p = N d
19-4.
po =
2 ni [300] Nd
; 2po =
2 2 2 ni [300] = ni [300 + T]
1020 = 1013
= 107 cm-3
2 ni [300!+!T] Nd ;
È q!Eg Ï1 1 ¸˘ 10 10 2x10 = 10 exp ÍÎ -! 2k !ÌT!-!300˝˙˚ Ó
˛
q!Eg!300 Solving for T yields T = (q!E !-!k!300!ln(2)) = 305.2 °K g DT = 305.2 - 300 = 5.2 °K. 19-5.
19-6.
q!V1 I1 = Is exp( k!T
q!V1!+!dV ; 10 I1 = Is exp( k!T ) ;
k!T dV = q ln(10) = 60 mV
(a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
Chapter 19 Problem Solutions 19-1.
Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus È q!Eg 1 1 ˘ Ï ¸ 14 10 ni(Ti) = Nd = 10 = 10 exp ÍÎ -! 2k !ÌT !-!300˝˙˚ Ó i ˛ Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields Ti = 262 °C or 535 °K.
19-2.
1 1 = 43.5 ohm-cm N-side resistivity rn = q!m !N = -19 (1.6x10 )(1500)(1014) n d 1 1 P-side resistivity rp = q!m !N = = 0.013 ohm-cm -19 p a (1.6x10 )(500)(1018)
19-3.
Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximate formulas given in Chapter 19. 2 ni n = Nd = 1013 cm-3 ; p = N d
19-4.
po =
2 ni [300] Nd
; 2po =
2 2 2 ni [300] = ni [300 + T]
1020 = 1013
= 107 cm-3
2 ni [300!+!T] Nd ;
È q!Eg Ï1 1 ¸˘ 10 10 2x10 = 10 exp ÍÎ -! 2k !ÌT!-!300˝˙˚ Ó
˛
q!Eg!300 Solving for T yields T = (q!E !-!k!300!ln(2)) = 305.2 °K g DT = 305.2 - 300 = 5.2 °K. 19-5.
19-6.
q!V1 I1 = Is exp( k!T
q!V1!+!dV ; 10 I1 = Is exp( k!T ) ;
k!T dV = q ln(10) = 60 mV
(a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
Chapter 19 Problem Solutions 19-1.
Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus È q!Eg 1 1 ˘ Ï ¸ 14 10 ni(Ti) = Nd = 10 = 10 exp ÍÎ -! 2k !ÌT !-!300˝˙˚ Ó i ˛ Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields Ti = 262 °C or 535 °K.
19-2.
1 1 = 43.5 ohm-cm N-side resistivity rn = q!m !N = -19 (1.6x10 )(1500)(1014) n d 1 1 P-side resistivity rp = q!m !N = = 0.013 ohm-cm -19 p a (1.6x10 )(500)(1018)
19-3.
Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximate formulas given in Chapter 19. 2 ni n = Nd = 1013 cm-3 ; p = N d
19-4.
po =
2 ni [300] Nd
; 2po =
2 2 2 ni [300] = ni [300 + T]
1020 = 1013
= 107 cm-3
2 ni [300!+!T] Nd ;
È q!Eg Ï1 1 ¸˘ 10 10 2x10 = 10 exp ÍÎ -! 2k !ÌT!-!300˝˙˚ Ó
˛
q!Eg!300 Solving for T yields T = (q!E !-!k!300!ln(2)) = 305.2 °K g DT = 305.2 - 300 = 5.2 °K. 19-5.
19-6.
q!V1 I1 = Is exp( k!T
q!V1!+!dV ; 10 I1 = Is exp( k!T ) ;
k!T dV = q ln(10) = 60 mV
(a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
Chapter 19 Problem Solutions 19-1.
Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus È q!Eg 1 1 ˘ Ï ¸ 14 10 ni(Ti) = Nd = 10 = 10 exp ÍÎ -! 2k !ÌT !-!300˝˙˚ Ó i ˛ Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields Ti = 262 °C or 535 °K.
19-2.
1 1 = 43.5 ohm-cm N-side resistivity rn = q!m !N = -19 (1.6x10 )(1500)(1014) n d 1 1 P-side resistivity rp = q!m !N = = 0.013 ohm-cm -19 p a (1.6x10 )(500)(1018)
19-3.
Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximate formulas given in Chapter 19. 2 ni n = Nd = 1013 cm-3 ; p = N d
19-4.
po =
2 ni [300] Nd
; 2po =
2 2 2 ni [300] = ni [300 + T]
1020 = 1013
= 107 cm-3
2 ni [300!+!T] Nd ;
È q!Eg Ï1 1 ¸˘ 10 10 2x10 = 10 exp ÍÎ -! 2k !ÌT!-!300˝˙˚ Ó
˛
q!Eg!300 Solving for T yields T = (q!E !-!k!300!ln(2)) = 305.2 °K g DT = 305.2 - 300 = 5.2 °K. 19-5.
19-6.
q!V1 I1 = Is exp( k!T
q!V1!+!dV ; 10 I1 = Is exp( k!T ) ;
k!T dV = q ln(10) = 60 mV
(a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
Chapter 19 Problem Solutions 19-1.
Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus È q!Eg 1 1 ˘ Ï ¸ 14 10 ni(Ti) = Nd = 10 = 10 exp ÍÎ -! 2k !ÌT !-!300˝˙˚ Ó i ˛ Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields Ti = 262 °C or 535 °K.
19-2.
1 1 = 43.5 ohm-cm N-side resistivity rn = q!m !N = -19 (1.6x10 )(1500)(1014) n d 1 1 P-side resistivity rp = q!m !N = = 0.013 ohm-cm -19 p a (1.6x10 )(500)(1018)
19-3.
Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximate formulas given in Chapter 19. 2 ni n = Nd = 1013 cm-3 ; p = N d
19-4.
po =
2 ni [300] Nd
; 2po =
2 2 2 ni [300] = ni [300 + T]
1020 = 1013
= 107 cm-3
2 ni [300!+!T] Nd ;
È q!Eg Ï1 1 ¸˘ 10 10 2x10 = 10 exp ÍÎ -! 2k !ÌT!-!300˝˙˚ Ó
˛
q!Eg!300 Solving for T yields T = (q!E !-!k!300!ln(2)) = 305.2 °K g DT = 305.2 - 300 = 5.2 °K. 19-5.
19-6.
q!V1 I1 = Is exp( k!T
q!V1!+!dV ; 10 I1 = Is exp( k!T ) ;
k!T dV = q ln(10) = 60 mV
(a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
xn(0) + xp(0) = Wo =
2!e!fc!(Na!+!Nd) ! q!Na!Nd
(1)
È 1014!1015˘ k!T ÈÍ Na!Nd˘˙ ˙ = 0.54 eV fc = q ln Í ! 2 ˙ = 0.026 ln ÍÎ ! 20 ˚ 10 Î ni ˚ Conservation of charge: q Na xp = q Nd xn
(2)
Solving (1) and (2) simultaneously gives using the numerical values given in the problem statement gives: W o = 2.8 microns ; xn(0) = 2.55 microns ; xp(0) = 0.25 microns (b) Electric field profile triangular-shaped as shown in Fig. 19-9b. Maximum electric at zero bias given by 2!fc (2)!(0.54) = 3,900 V/cm Emax = W = (2.8x10-4) o (c) From part a) fc = 0.54 eV (d)
e
C(V) A = Wo
V 1!+!f c
; C(V) = space-charge capacitance at reverse voltage V.
(11.7)(8.9x10-14) = = 3.7x10-9 F/cm2 2.8x10-4 (11.7)(8.9x10-14) C(50) = = 3.8x10-10 F/cm2 A 50 2.8x10-4 1!+!0.54
C(0) A
qV qV 0.7 (e) I = Is exp(kT ) ; exp( kT ) = exp (0.026 ) = 5x1011 È D t Dpt˘˙ Í n Í ! N !t !+! N !t ˙ A Î a d ˚ È -6) ˘˙ Í (38)(10-6) (13)(10 ˙ (2) 2 = (1.6x10-19)(1020) ÍÎ ! 15 !+! 14 -6 -6 (10 )(10 ) (10 )(10 )˚ I = (6.7x10-14 )(5x1011) = 34 mA 2 Is = q ni
Is = 6.7x10-14 A ;
19-7.
r!L L 0.02 Resistance R = A ; A = 0.01 = 2 cm-1 1 1 At 25 °C, Nd = 1014 >> ni so r = q!m !N = -19 n d (1.6x10 )(1500)(1014!) = 41.7 W-cm R(25 °C) = (41.7)(2) = 83.4 ohms È (1.6x10-19)!(1.1) Ï 1 1 ¸˘˙ 10 Í Ì At 250 °C (523 °K), ni[523] = 10 exp Î -! ! !-! ˝ = (2)(1.4x10-23) Ó523 300˛˚ (1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 1014. Thus we should solve Eqs. (19-2) and (19-3) exactly for no and po rather than using equations similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for Nd >> Na yields Nd È no = 2 Í1!+! Î
2˘ 2 4!ni ni 1!+! 2 ˙ and po = n . Putting in numerical values yields o Nd ˚
È 1014 Í no = 2 ÍÎ 1!+!
˘ (4)(7.6x1013)2!˙ ˙ = 1.4x1014 and 1!+! 14 2 ˚ (10 )
5.8x1027 po = = 5.8x1013 14 10 Assuming temperature-independent mobilities (not a valid assumption but no other information is given in text or the problem statement), resistance is r(250!° C)!!L 1 ; r(250 °C) ≈ q!m !n +!q!m A n o! p!po 1 = = 26.2 W-cm ; -19 14 (1.6x10 )(1500)(1.4x10 !)!+!(1.6x10-19)(500)(5.8x1013!)
R(250 °C) =
R(250 °C) ≈ (26.2)(2) = 52.4 ohms
19-8.
2 e!(Na!+!Nd)!EBD (11.7)(8.9x1014)(1015!+!1014)(3x105)2 BVBD = = 2!q!Na!Nd (2)(1.6x10-19)(1015)(1014) = 3,340 volts
19-7.
r!L L 0.02 Resistance R = A ; A = 0.01 = 2 cm-1 1 1 At 25 °C, Nd = 1014 >> ni so r = q!m !N = -19 n d (1.6x10 )(1500)(1014!) = 41.7 W-cm R(25 °C) = (41.7)(2) = 83.4 ohms È (1.6x10-19)!(1.1) Ï 1 1 ¸˘˙ 10 Í Ì At 250 °C (523 °K), ni[523] = 10 exp Î -! ! !-! ˝ = (2)(1.4x10-23) Ó523 300˛˚ (1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 1014. Thus we should solve Eqs. (19-2) and (19-3) exactly for no and po rather than using equations similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for Nd >> Na yields Nd È no = 2 Í1!+! Î
2˘ 2 4!ni ni 1!+! 2 ˙ and po = n . Putting in numerical values yields o Nd ˚
È 1014 Í no = 2 ÍÎ 1!+!
˘ (4)(7.6x1013)2!˙ ˙ = 1.4x1014 and 1!+! 14 2 ˚ (10 )
5.8x1027 po = = 5.8x1013 14 10 Assuming temperature-independent mobilities (not a valid assumption but no other information is given in text or the problem statement), resistance is r(250!° C)!!L 1 ; r(250 °C) ≈ q!m !n +!q!m A n o! p!po 1 = = 26.2 W-cm ; -19 14 (1.6x10 )(1500)(1.4x10 !)!+!(1.6x10-19)(500)(5.8x1013!)
R(250 °C) =
R(250 °C) ≈ (26.2)(2) = 52.4 ohms
19-8.
2 e!(Na!+!Nd)!EBD (11.7)(8.9x1014)(1015!+!1014)(3x105)2 BVBD = = 2!q!Na!Nd (2)(1.6x10-19)(1015)(1014) = 3,340 volts
19-9.
2 4!fc !BVBD 2 2 Emax = EBD ≈ 2 W o!fc
2 4!BVBD Wo ; Eq. (19-13); or f = 2 ! c EBD
2 2 4!BVBD W o!BVBD Wo 2 ; Eq. (19-11) ; Inserting f = and taking W (BVBD) = 2 ! c !fc EBD 2!BVBD the square root yields W (BVBD) ≈ 2 EBD. Dp!t =
(13)(10-6) = 36 microns ; Ln =
19-10.
Lp =
19-11.
Assume a one-sided step junction with Na >> Nd
Dn!t =
(39)(10-6) = 62 microns
Dp!t1 Dp!t2 q!V 2 q!V 2 I1 = q ni A N t exp(k!T ) ; I2 = q ni A N t exp(k!T ) d! 1 d! 2 I2 I1 = 2 =
19-12.
t1 t2 ; Thus 4 t2 = t1
2 ni 2 s = q mp p + q mn n ; np = ni ; Combining yeilds s = q mp n + q mn n 2 mp mn ni ds mn and p = ni mp dn = 0 = - q mp n2 + q mn ; Solving for n yields n = ni p = 1010
1500 10 -3 10 500 = 1.7x10 cm ; n = 10
500 9 -3 1500 = 6x10 cm ;
Thus minimum conductivity realized when silicon is slightly p-type. Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn . Putting in numerical values smin = (2)(1.6x10-19)(1010) (500)(1500) = 2.8x10-6 mhos-cm
19-9.
2 4!fc !BVBD 2 2 Emax = EBD ≈ 2 W o!fc
2 4!BVBD Wo ; Eq. (19-13); or f = 2 ! c EBD
2 2 4!BVBD W o!BVBD Wo 2 ; Eq. (19-11) ; Inserting f = and taking W (BVBD) = 2 ! c !fc EBD 2!BVBD the square root yields W (BVBD) ≈ 2 EBD. Dp!t =
(13)(10-6) = 36 microns ; Ln =
19-10.
Lp =
19-11.
Assume a one-sided step junction with Na >> Nd
Dn!t =
(39)(10-6) = 62 microns
Dp!t1 Dp!t2 q!V 2 q!V 2 I1 = q ni A N t exp(k!T ) ; I2 = q ni A N t exp(k!T ) d! 1 d! 2 I2 I1 = 2 =
19-12.
t1 t2 ; Thus 4 t2 = t1
2 ni 2 s = q mp p + q mn n ; np = ni ; Combining yeilds s = q mp n + q mn n 2 mp mn ni ds mn and p = ni mp dn = 0 = - q mp n2 + q mn ; Solving for n yields n = ni p = 1010
1500 10 -3 10 500 = 1.7x10 cm ; n = 10
500 9 -3 1500 = 6x10 cm ;
Thus minimum conductivity realized when silicon is slightly p-type. Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn . Putting in numerical values smin = (2)(1.6x10-19)(1010) (500)(1500) = 2.8x10-6 mhos-cm
19-9.
2 4!fc !BVBD 2 2 Emax = EBD ≈ 2 W o!fc
2 4!BVBD Wo ; Eq. (19-13); or f = 2 ! c EBD
2 2 4!BVBD W o!BVBD Wo 2 ; Eq. (19-11) ; Inserting f = and taking W (BVBD) = 2 ! c !fc EBD 2!BVBD the square root yields W (BVBD) ≈ 2 EBD. Dp!t =
(13)(10-6) = 36 microns ; Ln =
19-10.
Lp =
19-11.
Assume a one-sided step junction with Na >> Nd
Dn!t =
(39)(10-6) = 62 microns
Dp!t1 Dp!t2 q!V 2 q!V 2 I1 = q ni A N t exp(k!T ) ; I2 = q ni A N t exp(k!T ) d! 1 d! 2 I2 I1 = 2 =
19-12.
t1 t2 ; Thus 4 t2 = t1
2 ni 2 s = q mp p + q mn n ; np = ni ; Combining yeilds s = q mp n + q mn n 2 mp mn ni ds mn and p = ni mp dn = 0 = - q mp n2 + q mn ; Solving for n yields n = ni p = 1010
1500 10 -3 10 500 = 1.7x10 cm ; n = 10
500 9 -3 1500 = 6x10 cm ;
Thus minimum conductivity realized when silicon is slightly p-type. Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn . Putting in numerical values smin = (2)(1.6x10-19)(1010) (500)(1500) = 2.8x10-6 mhos-cm
19-9.
2 4!fc !BVBD 2 2 Emax = EBD ≈ 2 W o!fc
2 4!BVBD Wo ; Eq. (19-13); or f = 2 ! c EBD
2 2 4!BVBD W o!BVBD Wo 2 ; Eq. (19-11) ; Inserting f = and taking W (BVBD) = 2 ! c !fc EBD 2!BVBD the square root yields W (BVBD) ≈ 2 EBD. Dp!t =
(13)(10-6) = 36 microns ; Ln =
19-10.
Lp =
19-11.
Assume a one-sided step junction with Na >> Nd
Dn!t =
(39)(10-6) = 62 microns
Dp!t1 Dp!t2 q!V 2 q!V 2 I1 = q ni A N t exp(k!T ) ; I2 = q ni A N t exp(k!T ) d! 1 d! 2 I2 I1 = 2 =
19-12.
t1 t2 ; Thus 4 t2 = t1
2 ni 2 s = q mp p + q mn n ; np = ni ; Combining yeilds s = q mp n + q mn n 2 mp mn ni ds mn and p = ni mp dn = 0 = - q mp n2 + q mn ; Solving for n yields n = ni p = 1010
1500 10 -3 10 500 = 1.7x10 cm ; n = 10
500 9 -3 1500 = 6x10 cm ;
Thus minimum conductivity realized when silicon is slightly p-type. Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn . Putting in numerical values smin = (2)(1.6x10-19)(1010) (500)(1500) = 2.8x10-6 mhos-cm
Chapter 20 Problem Solutions
20-1.
1.3x1017 1.3x1017 Nd = BV ! = 2500 = 5x1013 cm-3 ; W(2500 V) = (10-5)(2500) = 250 BD
microns 20-2.
Drift region length of 50 microns is much less than the 250 microns found in the previous problem (20-1) for the same drift region doping density. Hence this must be a punchthrough structure and Eq. (20-9) applies. (1.6x10-19)(5x1013)(5x10-3)2 = 900 V BVBD = (2x105)(5x10-3) (2)(11.7)(8.9x10-14)
20-3.
2 q!A!ni !Lp ; For one-sided step junction Is = !N !t ; d o
k!T È I ˘ Von = Vj + Vdrift ; Vj = q ln ÍÎ !I ˙˚ s (1.6x10-19)(2)(1010)2! (13)(2x10-6) Evaluating Is yields = 1.6x10-9 A !(5x1013)(2x10-6)
Vd = K1 I + K2 (I)2/3 Eq. (20-16) with I = forward bias current through the diode. Wd 5x10-3 K1 = q!m !A!n = o b (1.6x10-19)(900)(2)(1017) 3 K2 = = 7.5x10-4
= 1.7x10-4
4 3 Wd (5x10-3)4 = 3 2 (1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6) q2!mo!nb!A2!to
Chapter 20 Problem Solutions
20-1.
1.3x1017 1.3x1017 Nd = BV ! = 2500 = 5x1013 cm-3 ; W(2500 V) = (10-5)(2500) = 250 BD
microns 20-2.
Drift region length of 50 microns is much less than the 250 microns found in the previous problem (20-1) for the same drift region doping density. Hence this must be a punchthrough structure and Eq. (20-9) applies. (1.6x10-19)(5x1013)(5x10-3)2 = 900 V BVBD = (2x105)(5x10-3) (2)(11.7)(8.9x10-14)
20-3.
2 q!A!ni !Lp ; For one-sided step junction Is = !N !t ; d o
k!T È I ˘ Von = Vj + Vdrift ; Vj = q ln ÍÎ !I ˙˚ s (1.6x10-19)(2)(1010)2! (13)(2x10-6) Evaluating Is yields = 1.6x10-9 A !(5x1013)(2x10-6)
Vd = K1 I + K2 (I)2/3 Eq. (20-16) with I = forward bias current through the diode. Wd 5x10-3 K1 = q!m !A!n = o b (1.6x10-19)(900)(2)(1017) 3 K2 = = 7.5x10-4
= 1.7x10-4
4 3 Wd (5x10-3)4 = 3 2 (1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6) q2!mo!nb!A2!to
Chapter 20 Problem Solutions
20-1.
1.3x1017 1.3x1017 Nd = BV ! = 2500 = 5x1013 cm-3 ; W(2500 V) = (10-5)(2500) = 250 BD
microns 20-2.
Drift region length of 50 microns is much less than the 250 microns found in the previous problem (20-1) for the same drift region doping density. Hence this must be a punchthrough structure and Eq. (20-9) applies. (1.6x10-19)(5x1013)(5x10-3)2 = 900 V BVBD = (2x105)(5x10-3) (2)(11.7)(8.9x10-14)
20-3.
2 q!A!ni !Lp ; For one-sided step junction Is = !N !t ; d o
k!T È I ˘ Von = Vj + Vdrift ; Vj = q ln ÍÎ !I ˙˚ s (1.6x10-19)(2)(1010)2! (13)(2x10-6) Evaluating Is yields = 1.6x10-9 A !(5x1013)(2x10-6)
Vd = K1 I + K2 (I)2/3 Eq. (20-16) with I = forward bias current through the diode. Wd 5x10-3 K1 = q!m !A!n = o b (1.6x10-19)(900)(2)(1017) 3 K2 = = 7.5x10-4
= 1.7x10-4
4 3 Wd (5x10-3)4 = 3 2 (1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6) q2!mo!nb!A2!to
Von in volts 1.6 •
1.4 I 0A 1 10 100 1000 3000
Vj 0V 0.53 0.59 0.65 0.71 0.74
Vdrift 0V 0.001 0.005 0.033 0.25 0.67
Von 0V 0.53 0.59 0.68 0.96 1.41
1.2 1
•
0.8 0.6
•
•
•
0.4 0.2 0 1
10
100
1000
10000
Forward current in amperes 20-4.
L 1 Von(t) = Rdrift I(t) >> Vj ≈ 1 V ; Rdrift = r A ; r = q!m !N n d -3 L 5x10 1 = 85 ohm-cm ; = 2 = 2.5x10-3 r= A -19 13 (1.6x10 )(1500)(5x10 )
a)
Rdrift = (85)(2.5x10-3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds Von(t) = (0.21)(2.5x108 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds Von(4 ms) = (5.3x107)(4x10-6) = 212 volts Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x107 t] Von(t) = {0.21[1 - 2.5x107 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds
b)
Von in volts 1.6 •
1.4 I 0A 1 10 100 1000 3000
Vj 0V 0.53 0.59 0.65 0.71 0.74
Vdrift 0V 0.001 0.005 0.033 0.25 0.67
Von 0V 0.53 0.59 0.68 0.96 1.41
1.2 1
•
0.8 0.6
•
•
•
0.4 0.2 0 1
10
100
1000
10000
Forward current in amperes 20-4.
L 1 Von(t) = Rdrift I(t) >> Vj ≈ 1 V ; Rdrift = r A ; r = q!m !N n d -3 L 5x10 1 = 85 ohm-cm ; = 2 = 2.5x10-3 r= A -19 13 (1.6x10 )(1500)(5x10 )
a)
Rdrift = (85)(2.5x10-3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds Von(t) = (0.21)(2.5x108 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds Von(4 ms) = (5.3x107)(4x10-6) = 212 volts Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x107 t] Von(t) = {0.21[1 - 2.5x107 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds
b)
250
200 Von in volts
No carrier injection
150
100
50 With carrier injection 0 0
20-5.
1
1.5
2
2.5
3
3.5
4
IF 2000 toff = trr + t3 = trr + di /dt = trr + = trr + 8 ms R 2.5x108 2!t!IF -12 2 = 4x10-12(2000)2 = 16 ms trr = diR/dt ; t = 4x10 (BVBD)
trr = 20-6.
0.5
(2)(1.6x10-5)(2x103) = 16 ms ; toff = 8 ms + 16 ms = 24 ms 2.5x108
Assume a non-punch-through structure for the Schottky diode. 1.3x1017 Nd = BV BD
1.3x1017 = 150
= 8.7x1014 cm-3
W d = 10-5 BVBD = (10-5) (150) = 15 microns
20-7.
Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 =
A =
2x10-3 = 0.42 cm2 -19 15 -2 !(1.6x10 )(1500)(10 )(2x10 )
1 L q!mn!Nd A
250
200 Von in volts
No carrier injection
150
100
50 With carrier injection 0 0
20-5.
1
1.5
2
2.5
3
3.5
4
IF 2000 toff = trr + t3 = trr + di /dt = trr + = trr + 8 ms R 2.5x108 2!t!IF -12 2 = 4x10-12(2000)2 = 16 ms trr = diR/dt ; t = 4x10 (BVBD)
trr = 20-6.
0.5
(2)(1.6x10-5)(2x103) = 16 ms ; toff = 8 ms + 16 ms = 24 ms 2.5x108
Assume a non-punch-through structure for the Schottky diode. 1.3x1017 Nd = BV BD
1.3x1017 = 150
= 8.7x1014 cm-3
W d = 10-5 BVBD = (10-5) (150) = 15 microns
20-7.
Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 =
A =
2x10-3 = 0.42 cm2 -19 15 -2 !(1.6x10 )(1500)(10 )(2x10 )
1 L q!mn!Nd A
250
200 Von in volts
No carrier injection
150
100
50 With carrier injection 0 0
20-5.
1
1.5
2
2.5
3
3.5
4
IF 2000 toff = trr + t3 = trr + di /dt = trr + = trr + 8 ms R 2.5x108 2!t!IF -12 2 = 4x10-12(2000)2 = 16 ms trr = diR/dt ; t = 4x10 (BVBD)
trr = 20-6.
0.5
(2)(1.6x10-5)(2x103) = 16 ms ; toff = 8 ms + 16 ms = 24 ms 2.5x108
Assume a non-punch-through structure for the Schottky diode. 1.3x1017 Nd = BV BD
1.3x1017 = 150
= 8.7x1014 cm-3
W d = 10-5 BVBD = (10-5) (150) = 15 microns
20-7.
Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 =
A =
2x10-3 = 0.42 cm2 -19 15 -2 !(1.6x10 )(1500)(10 )(2x10 )
1 L q!mn!Nd A
250
200 Von in volts
No carrier injection
150
100
50 With carrier injection 0 0
20-5.
1
1.5
2
2.5
3
3.5
4
IF 2000 toff = trr + t3 = trr + di /dt = trr + = trr + 8 ms R 2.5x108 2!t!IF -12 2 = 4x10-12(2000)2 = 16 ms trr = diR/dt ; t = 4x10 (BVBD)
trr = 20-6.
0.5
(2)(1.6x10-5)(2x103) = 16 ms ; toff = 8 ms + 16 ms = 24 ms 2.5x108
Assume a non-punch-through structure for the Schottky diode. 1.3x1017 Nd = BV BD
1.3x1017 = 150
= 8.7x1014 cm-3
W d = 10-5 BVBD = (10-5) (150) = 15 microns
20-7.
Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 =
A =
2x10-3 = 0.42 cm2 -19 15 -2 !(1.6x10 )(1500)(10 )(2x10 )
1 L q!mn!Nd A
20-8.
20-9.
È 2!e ˘ ˙ Use Eq. (20-9) to solve for Nd ; Nd = [EBD Wd - BVBD] ÍÍ 2˙ Î q!Wd˚ (2)(11.7)(8.9x10-14) 5 -3 = Nd = 3.4x1014 cm-3 = [(2x10 )(2x10 ) - 300] -19 -3 2 (1.6x10 )(2x10 ) W d(npt) R A npt = q!mn!Nnpt W d(pt) R = A pt q!mn!Npt
; Nnpt = Nd of non-punch-throuth (npt) diode
;
Npt = Nd of punch-through (pt) diode
e!EBD W d(npt) = !q!N ; Derived from Eqs. (19-11), (19-12), and (19-13) npt e!EBD W d(pt) = !q!N pt
È Í _! Í 1!+ Î
2!q!Npt!BVBD˘˙ 1!-! ˙ 2 ˚ e!EBD
q!Npt Nnpt 2!BVBD q!Nnpt 2!BVBD Npt 2!q!Npt!BVBD = e!E = e!E 2 BD Nnpt EBD BD EBD Nnpt e!EBD Npt Npt 1 = W (npt) Wd(npt) N =x= N d npt npt
1 ; Wd(pt) = Wd(npt) x [ 1!+ _ ! 1!-!x]
If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10) 1 _ ! 1!-!x] as x approaches is infinite for the plus root and 0.5 for the Limit of x [ 1!+ minus root. Hence the minus root is the correct choice to use. 1 W (npt)! x![ 1!-!! 1!-!x] R d A pt = q!mn!Npt
=
1 W d(npt)!x![ 1!-!! 1!-!x] Nnpt !Nnpt q!mn!Npt
W d(npt) 1 R 1 = !q!m !N [1 - 1!-!x ] = A [1 - 1!-!x ] 2 n npt x npt x2 d dx
ÈÍ 1 ˘˙ -2 1 Î !x2!! 1!-!x!˚ = 0 = x3 [1 - 1!-!x ] + 2!x2! 1!-!x
8 8 Solving for x yields x = 9 i.e. Npt = 9 Nnpt ; Wd(pt) = 0.75 Wd(npt)
20-8.
20-9.
È 2!e ˘ ˙ Use Eq. (20-9) to solve for Nd ; Nd = [EBD Wd - BVBD] ÍÍ 2˙ Î q!Wd˚ (2)(11.7)(8.9x10-14) 5 -3 = Nd = 3.4x1014 cm-3 = [(2x10 )(2x10 ) - 300] -19 -3 2 (1.6x10 )(2x10 ) W d(npt) R A npt = q!mn!Nnpt W d(pt) R = A pt q!mn!Npt
; Nnpt = Nd of non-punch-throuth (npt) diode
;
Npt = Nd of punch-through (pt) diode
e!EBD W d(npt) = !q!N ; Derived from Eqs. (19-11), (19-12), and (19-13) npt e!EBD W d(pt) = !q!N pt
È Í _! Í 1!+ Î
2!q!Npt!BVBD˘˙ 1!-! ˙ 2 ˚ e!EBD
q!Npt Nnpt 2!BVBD q!Nnpt 2!BVBD Npt 2!q!Npt!BVBD = e!E = e!E 2 BD Nnpt EBD BD EBD Nnpt e!EBD Npt Npt 1 = W (npt) Wd(npt) N =x= N d npt npt
1 ; Wd(pt) = Wd(npt) x [ 1!+ _ ! 1!-!x]
If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10) 1 _ ! 1!-!x] as x approaches is infinite for the plus root and 0.5 for the Limit of x [ 1!+ minus root. Hence the minus root is the correct choice to use. 1 W (npt)! x![ 1!-!! 1!-!x] R d A pt = q!mn!Npt
=
1 W d(npt)!x![ 1!-!! 1!-!x] Nnpt !Nnpt q!mn!Npt
W d(npt) 1 R 1 = !q!m !N [1 - 1!-!x ] = A [1 - 1!-!x ] 2 n npt x npt x2 d dx
ÈÍ 1 ˘˙ -2 1 Î !x2!! 1!-!x!˚ = 0 = x3 [1 - 1!-!x ] + 2!x2! 1!-!x
8 8 Solving for x yields x = 9 i.e. Npt = 9 Nnpt ; Wd(pt) = 0.75 Wd(npt)
R R A npt = 0.84 A npt
20-10.
eA Cjo = W (0) ; Area A determined by on-state voltage and current. Depletion-layer d width Wd(0) set by breakdown voltage. Wd(0) same for both diodes. 150 1!+! 0.7
W d(150) = Wd(0) W d(0) =
15 (150)/(0.7)
= (10-5)(150) = 15 mm ;
≈ 1 mm
W d(150) Schottky diode area ; Vdrift = 2 volts = q!m !N !A I n d Schottky F Drift region doping density Nd same for both diodes. Nd = = 8.7x1014 cm-3
1.3x1017 150
(1.5x10-3)(300) ASchottky = = 1.07 cm2 -19 14 (2)(1.6x10 )(1500)(8.7x10 ) PN junction diode area: Vdrift = 2 volts = K1 IF + K2 (IF)2/3 ; Eq. (20-16) W d(150) 1.5x10-3 K1 = q!m !A !n = o pn b (1.6x10-19)(900)(Apn)(1017) 3 K2 =
4 W d(150) = 3 2 2 q2!mo!nb!Apn!to
3
10-4 = A pn
(1.5x10-3)4 2 (1.6x10-19)2!(900)3!(1017)2!(Apn)!(2x10-6)
10-4 K2 = 2.4x10-4 (Apn)-0.67 ; 2 volts = A (300) + 2.4x10-4 (Apn)-0.67 (300)0.67 pn 0.333 Apn = 0.015 + 0.00154 (Apn) ; Solve by successive approximation ; Apn ≈ 0.017 cm2 Schottky diode Cjo =
(11.7)(8.9x10-14)(1.07) 10-4
≈ 11 nF = 0.011 mF
R R A npt = 0.84 A npt
20-10.
eA Cjo = W (0) ; Area A determined by on-state voltage and current. Depletion-layer d width Wd(0) set by breakdown voltage. Wd(0) same for both diodes. 150 1!+! 0.7
W d(150) = Wd(0) W d(0) =
15 (150)/(0.7)
= (10-5)(150) = 15 mm ;
≈ 1 mm
W d(150) Schottky diode area ; Vdrift = 2 volts = q!m !N !A I n d Schottky F Drift region doping density Nd same for both diodes. Nd = = 8.7x1014 cm-3
1.3x1017 150
(1.5x10-3)(300) ASchottky = = 1.07 cm2 -19 14 (2)(1.6x10 )(1500)(8.7x10 ) PN junction diode area: Vdrift = 2 volts = K1 IF + K2 (IF)2/3 ; Eq. (20-16) W d(150) 1.5x10-3 K1 = q!m !A !n = o pn b (1.6x10-19)(900)(Apn)(1017) 3 K2 =
4 W d(150) = 3 2 2 q2!mo!nb!Apn!to
3
10-4 = A pn
(1.5x10-3)4 2 (1.6x10-19)2!(900)3!(1017)2!(Apn)!(2x10-6)
10-4 K2 = 2.4x10-4 (Apn)-0.67 ; 2 volts = A (300) + 2.4x10-4 (Apn)-0.67 (300)0.67 pn 0.333 Apn = 0.015 + 0.00154 (Apn) ; Solve by successive approximation ; Apn ≈ 0.017 cm2 Schottky diode Cjo =
(11.7)(8.9x10-14)(1.07) 10-4
≈ 11 nF = 0.011 mF
PN junction diode Cjo =
20-11.
BVcyl BVp
(11.7)(8.9x10-14)(0.017) ≈ 180 pF 10-4
1 1 = 2 r2 (1 + r ) ln (1 + r ) - 2r
BVcyl / BVp 1 0.9
•
0.8 0.7
•
0.6 0.5
•
•
•
• •
•
•
0.4 0.3 0.2 0.1 0 0.1
1
10
r = R/(2W n ) 20-12.
BVcyl BVp
950 R = 1000 = 0.95 ; From graph in problem 20-11, r ≈ 6 = 2!W n
R ≈ 12 Wn
PN junction diode Cjo =
20-11.
BVcyl BVp
(11.7)(8.9x10-14)(0.017) ≈ 180 pF 10-4
1 1 = 2 r2 (1 + r ) ln (1 + r ) - 2r
BVcyl / BVp 1 0.9
•
0.8 0.7
•
0.6 0.5
•
•
•
• •
•
•
0.4 0.3 0.2 0.1 0 0.1
1
10
r = R/(2W n ) 20-12.
BVcyl BVp
950 R = 1000 = 0.95 ; From graph in problem 20-11, r ≈ 6 = 2!W n
R ≈ 12 Wn
PN junction diode Cjo =
20-11.
BVcyl BVp
(11.7)(8.9x10-14)(0.017) ≈ 180 pF 10-4
1 1 = 2 r2 (1 + r ) ln (1 + r ) - 2r
BVcyl / BVp 1 0.9
•
0.8 0.7
•
0.6 0.5
•
•
•
• •
•
•
0.4 0.3 0.2 0.1 0 0.1
1
10
r = R/(2W n ) 20-12.
BVcyl BVp
950 R = 1000 = 0.95 ; From graph in problem 20-11, r ≈ 6 = 2!W n
R ≈ 12 Wn
Chapter 21 Problem Solutions
21-1.
NPN BJT ; BVCEO =
BVCBO BVCBO ; PNP BJT ; BV = CEO b 1/6 b 1/4
1 JB • *
0.9 0.8 0.7 BVCEO BVCBO
0.6
• *
• *
• • •• •• * * * ** *
0.5
• *
0.4 0.3
PNP • * NPN
• • • •• • * * * ** *
0.2 0.1 0 1 21-2.
10
100
The flow of large reverse base currents when emitter current is still flowing in the forward direction will lead to emitter current crowding towards the center of the emitter. This accenuates the possibility of second breakdown. See Fig. 21-8b. When emitter-open switching is used, there is no emitter current and thus no emitter current crowding and second breakdown is much less likely to occur.
21-3.
a)
Idealized BJT current and voltage waveforms in step-down converter
Chapter 21 Problem Solutions
21-1.
NPN BJT ; BVCEO =
BVCBO BVCBO ; PNP BJT ; BV = CEO b 1/6 b 1/4
1 JB • *
0.9 0.8 0.7 BVCEO BVCBO
0.6
• *
• *
• • •• •• * * * ** *
0.5
• *
0.4 0.3
PNP • * NPN
• • • •• • * * * ** *
0.2 0.1 0 1 21-2.
10
100
The flow of large reverse base currents when emitter current is still flowing in the forward direction will lead to emitter current crowding towards the center of the emitter. This accenuates the possibility of second breakdown. See Fig. 21-8b. When emitter-open switching is used, there is no emitter current and thus no emitter current crowding and second breakdown is much less likely to occur.
21-3.
a)
Idealized BJT current and voltage waveforms in step-down converter
Chapter 21 Problem Solutions
21-1.
NPN BJT ; BVCEO =
BVCBO BVCBO ; PNP BJT ; BV = CEO b 1/6 b 1/4
1 JB • *
0.9 0.8 0.7 BVCEO BVCBO
0.6
• *
• *
• • •• •• * * * ** *
0.5
• *
0.4 0.3
PNP • * NPN
• • • •• • * * * ** *
0.2 0.1 0 1 21-2.
10
100
The flow of large reverse base currents when emitter current is still flowing in the forward direction will lead to emitter current crowding towards the center of the emitter. This accenuates the possibility of second breakdown. See Fig. 21-8b. When emitter-open switching is used, there is no emitter current and thus no emitter current crowding and second breakdown is much less likely to occur.
21-3.
a)
Idealized BJT current and voltage waveforms in step-down converter
T/2 V
i (t) C
V
I d
t d,on
t ri
t fv
(t) CE
o
t d,off
t rv
t fi
1 T 1 BJT power dissipation Pc = T Û ı!vCE(t)!iC(t)!dt = {Eon + Esw }fs ; fs = T 0 T 1 40 Eon = VCE,on Io {2 + td,off - td,on} ≈ (2)(40) 2!f = f Joules s s Esw = Eri + Efv + Erv + Efi Eri =
Vd!Io!tri 2
=
(100)(40)(2x10-7) = 4x10-4 Joules 2
Vd!Io!tfv Efv = 2
(100)(40)(1x10-7) = = 2x10-4 Joules 2
Vd!Io!trv Erv = 2
(100)(40)(1x10-7) = = 2x10-4 Joules 2
Efi =
Vd!Io!tfi 2
(100)(40)(2x10-7) = = 4x10-4 Joules 2
40 Esw = 1.2x10-3 Joules ; Pc = [ f + 1.2x10-3] fs = 40 + 1.2x10-3 fs s
P c 80 [watts]
40 f
0 33 kHz b)
Tj = Rqja Pc + Ta ; Pc,max =
(125!-!40) fs,max = 1.2x10-3 21-4.
150!-!25 1
s
= 125 watts = 40 + 1.2x10-3 fs,max
= 71 kHz
From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as 0.4 Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = (125!-!25) = 4x10-3 110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5.
Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC =
q!Dn!Na!A (1.6x10-19)(38)(1016)(1) = = 200 A !Wb (3x10-4)
21-6.
The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7.
The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the baseside protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
P c 80 [watts]
40 f
0 33 kHz b)
Tj = Rqja Pc + Ta ; Pc,max =
(125!-!40) fs,max = 1.2x10-3 21-4.
150!-!25 1
s
= 125 watts = 40 + 1.2x10-3 fs,max
= 71 kHz
From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as 0.4 Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = (125!-!25) = 4x10-3 110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5.
Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC =
q!Dn!Na!A (1.6x10-19)(38)(1016)(1) = = 200 A !Wb (3x10-4)
21-6.
The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7.
The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the baseside protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
P c 80 [watts]
40 f
0 33 kHz b)
Tj = Rqja Pc + Ta ; Pc,max =
(125!-!40) fs,max = 1.2x10-3 21-4.
150!-!25 1
s
= 125 watts = 40 + 1.2x10-3 fs,max
= 71 kHz
From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as 0.4 Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = (125!-!25) = 4x10-3 110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5.
Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC =
q!Dn!Na!A (1.6x10-19)(38)(1016)(1) = = 200 A !Wb (3x10-4)
21-6.
The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7.
The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the baseside protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
P c 80 [watts]
40 f
0 33 kHz b)
Tj = Rqja Pc + Ta ; Pc,max =
(125!-!40) fs,max = 1.2x10-3 21-4.
150!-!25 1
s
= 125 watts = 40 + 1.2x10-3 fs,max
= 71 kHz
From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as 0.4 Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = (125!-!25) = 4x10-3 110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5.
Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC =
q!Dn!Na!A (1.6x10-19)(38)(1016)(1) = = 200 A !Wb (3x10-4)
21-6.
The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7.
The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the baseside protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
P c 80 [watts]
40 f
0 33 kHz b)
Tj = Rqja Pc + Ta ; Pc,max =
(125!-!40) fs,max = 1.2x10-3 21-4.
150!-!25 1
s
= 125 watts = 40 + 1.2x10-3 fs,max
= 71 kHz
From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as 0.4 Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = (125!-!25) = 4x10-3 110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5.
Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC =
q!Dn!Na!A (1.6x10-19)(38)(1016)(1) = = 200 A !Wb (3x10-4)
21-6.
The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7.
The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the baseside protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
2!e!fc!(NaB!+!NdE) È NaB!NdE˘ k!T Í ˙ W o,EB = ; fcE = q ln Í ! !q!NaB!NdE 2 ˙ Î ˚ ni È 1034˘ fcE = 0.26 ln ÍÎ ! 20˙˚ = 0.84 V 10
W o,EB =
(2)(11.7)(8.9x10-14)(0.84)(1019!+!1015) = 0.33 microns (1.6x10-19)(1019)(1015)
Estimate of collector-base base-side protusion of WCB,depl. V 1!+!f = xp + xn ; c W(V) xp = protrusion of CB depletion layer into p-type base region. xp = 11 using xp Na xn Nd (charge neutrality). CB depletion layer thickness W(V) = Wo,CB
=
2!e!fcC!(NaB!+!NdC) È NaB!NdC˘ k!T Í ˙ ; fcC = q ln Í ! W o,CB = !q!NaB!NdC 2 ˙ Î ˚ ni È 1029˘ fcC = 0.26 ln ÍÎ ! 20˙˚ = 0.54 V 10
W o,CB =
(2)(11.7)(8.9x10-14)(0.54)(1014!+!1015) = 2.8 microns (1.6x10-19)(1014)(1015)
V {(3 - 0.33)x10-4}(11) = 2.8x10-4 1!+!0.54 : Solving for V yields V = 59 volts. Reach-through voltage of 59 V is much less than the expected value of BVBD = 1000 V.
21-8.
BVEBO = 10 V =
1.3x1017 ; NaB = acceptor doping density in base = 1.3x1016 cm-3 !NaB
1.3x1017 1/4 1/4 BVCBO = b BVCEO = (5) (1000) = ≈ 1500 Volts ≈ !N dC 13 -3 NdC = collector drift region donor density = 8.7x10 cn
2!e!fc!(NaB!+!NdE) È NaB!NdE˘ k!T Í ˙ W o,EB = ; fcE = q ln Í ! !q!NaB!NdE 2 ˙ Î ˚ ni È 1034˘ fcE = 0.26 ln ÍÎ ! 20˙˚ = 0.84 V 10
W o,EB =
(2)(11.7)(8.9x10-14)(0.84)(1019!+!1015) = 0.33 microns (1.6x10-19)(1019)(1015)
Estimate of collector-base base-side protusion of WCB,depl. V 1!+!f = xp + xn ; c W(V) xp = protrusion of CB depletion layer into p-type base region. xp = 11 using xp Na xn Nd (charge neutrality). CB depletion layer thickness W(V) = Wo,CB
=
2!e!fcC!(NaB!+!NdC) È NaB!NdC˘ k!T Í ˙ ; fcC = q ln Í ! W o,CB = !q!NaB!NdC 2 ˙ Î ˚ ni È 1029˘ fcC = 0.26 ln ÍÎ ! 20˙˚ = 0.54 V 10
W o,CB =
(2)(11.7)(8.9x10-14)(0.54)(1014!+!1015) = 2.8 microns (1.6x10-19)(1014)(1015)
V {(3 - 0.33)x10-4}(11) = 2.8x10-4 1!+!0.54 : Solving for V yields V = 59 volts. Reach-through voltage of 59 V is much less than the expected value of BVBD = 1000 V.
21-8.
BVEBO = 10 V =
1.3x1017 ; NaB = acceptor doping density in base = 1.3x1016 cm-3 !NaB
1.3x1017 1/4 1/4 BVCBO = b BVCEO = (5) (1000) = ≈ 1500 Volts ≈ !N dC 13 -3 NdC = collector drift region donor density = 8.7x10 cn
Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO) Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10-5)(1500) = 150 mm xn(BVCBO)!NdC 8.7x1013 = xn(BVCBO) = xn(BVCBO) 6.7x10-3 xp(BVCBO) = 16 NaB 1.3x10 xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm Hence xp(BVCBO) = (1.5x10-2 cm)(6.7x10-3) = 1 micron = Wbase Beta = 150 = bD bM + b D + b M = 20 bM + 20 + b M
21-9.
bM = 21-10.
150!-!20 = 6.2 21
Must first ascertain the operating states of the two transistors.Two likely choices including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT both saturated. Initially assume driver BJT saturated and main BJT active. C
B
0.02 W 0.6 V +
10 I
B,M
+ 0.8 V
I B,M
+ 0.8 V E
10 IB,M + IB,M = 100 A IB,M = IC,D = 9.1A IC,M = 91 A VCE,D = 0.2 +(.02)(9.1) = 0.382 V VCE,M = VCE,D + 0.8 V = 1.18 v But a saturated main BJT with I C,M = 91 A going through 0.02 ohms generates a voltage drop of 1.8 V which is > 1.18 V. Hence main BJT must be saturated.
Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO) Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10-5)(1500) = 150 mm xn(BVCBO)!NdC 8.7x1013 = xn(BVCBO) = xn(BVCBO) 6.7x10-3 xp(BVCBO) = 16 NaB 1.3x10 xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm Hence xp(BVCBO) = (1.5x10-2 cm)(6.7x10-3) = 1 micron = Wbase Beta = 150 = bD bM + b D + b M = 20 bM + 20 + b M
21-9.
bM = 21-10.
150!-!20 = 6.2 21
Must first ascertain the operating states of the two transistors.Two likely choices including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT both saturated. Initially assume driver BJT saturated and main BJT active. C
B
0.02 W 0.6 V +
10 I
B,M
+ 0.8 V
I B,M
+ 0.8 V E
10 IB,M + IB,M = 100 A IB,M = IC,D = 9.1A IC,M = 91 A VCE,D = 0.2 +(.02)(9.1) = 0.382 V VCE,M = VCE,D + 0.8 V = 1.18 v But a saturated main BJT with I C,M = 91 A going through 0.02 ohms generates a voltage drop of 1.8 V which is > 1.18 V. Hence main BJT must be saturated.
Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO) Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10-5)(1500) = 150 mm xn(BVCBO)!NdC 8.7x1013 = xn(BVCBO) = xn(BVCBO) 6.7x10-3 xp(BVCBO) = 16 NaB 1.3x10 xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm Hence xp(BVCBO) = (1.5x10-2 cm)(6.7x10-3) = 1 micron = Wbase Beta = 150 = bD bM + b D + b M = 20 bM + 20 + b M
21-9.
bM = 21-10.
150!-!20 = 6.2 21
Must first ascertain the operating states of the two transistors.Two likely choices including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT both saturated. Initially assume driver BJT saturated and main BJT active. C
B
0.02 W 0.6 V +
10 I
B,M
+ 0.8 V
I B,M
+ 0.8 V E
10 IB,M + IB,M = 100 A IB,M = IC,D = 9.1A IC,M = 91 A VCE,D = 0.2 +(.02)(9.1) = 0.382 V VCE,M = VCE,D + 0.8 V = 1.18 v But a saturated main BJT with I C,M = 91 A going through 0.02 ohms generates a voltage drop of 1.8 V which is > 1.18 V. Hence main BJT must be saturated.
C
B
0.02 W 0.6 V +
0.02 W 0.6 V +
+ 0.8 V
I
-
C,M
I B,M
+ 0.8 V -
Neglecting IB,D IC,M + IC,D = 100 A (.02)I C,M -0.6 = (.02) I C,D +0.2 IC,D = 30 A ; IC,M = 70 A VCE,M = 0.2 + (70)(.02) = 1.6 V PDarl = VCE,M [IC,M + IC,D ] PDarl = (1.6 V)(100 A) = 160 W
E 21-11.
e!AE CEBO = W EBO
;
e!AC CCBO = W CBO
W EBO = zero-bias emitter-base depletion layer thickness W CBO = zero-bias collector-base depletion layer thickness 2!e!fcE!(NaB!+!NdE) È NaB!NdE˘ k!T Í ˙ ; fcE = q ln Í ! W EBO = !q!NaB!NdE 2 ˙ Î ˚ ni È (1019)(1016)˘ ˙ = 0.89 V ; fcE = 0.026 ln ÍÎ ! 20 ˚ 10 W EBO =
CEBO =
(2)(11.7)(8.9x10-14)(0.89)(1019!+!1016) = 0.34 microns (1.6x10-19)(1019)(1016)
(11.7)(8.9x10-14)(0.3) = 9.2 nF 3.4x10-5
2!e!fcC!(NaB!+!NdC) È NaB!!NdC˘ k!T Í ˙ ; f = ln q !q!NaB!NdC Í! 2 ˙ cC Î ˚ ni È (1.3x1016)(8.7x1013)˘ ˙ = 0.6 V ; fcC = 0.026 ln ÍÎ ! ˚ 1020
W CBO =
C
B
0.02 W 0.6 V +
0.02 W 0.6 V +
+ 0.8 V
I
-
C,M
I B,M
+ 0.8 V -
Neglecting IB,D IC,M + IC,D = 100 A (.02)I C,M -0.6 = (.02) I C,D +0.2 IC,D = 30 A ; IC,M = 70 A VCE,M = 0.2 + (70)(.02) = 1.6 V PDarl = VCE,M [IC,M + IC,D ] PDarl = (1.6 V)(100 A) = 160 W
E 21-11.
e!AE CEBO = W EBO
;
e!AC CCBO = W CBO
W EBO = zero-bias emitter-base depletion layer thickness W CBO = zero-bias collector-base depletion layer thickness 2!e!fcE!(NaB!+!NdE) È NaB!NdE˘ k!T Í ˙ ; fcE = q ln Í ! W EBO = !q!NaB!NdE 2 ˙ Î ˚ ni È (1019)(1016)˘ ˙ = 0.89 V ; fcE = 0.026 ln ÍÎ ! 20 ˚ 10 W EBO =
CEBO =
(2)(11.7)(8.9x10-14)(0.89)(1019!+!1016) = 0.34 microns (1.6x10-19)(1019)(1016)
(11.7)(8.9x10-14)(0.3) = 9.2 nF 3.4x10-5
2!e!fcC!(NaB!+!NdC) È NaB!!NdC˘ k!T Í ˙ ; f = ln q !q!NaB!NdC Í! 2 ˙ cC Î ˚ ni È (1.3x1016)(8.7x1013)˘ ˙ = 0.6 V ; fcC = 0.026 ln ÍÎ ! ˚ 1020
W CBO =
W CBO =
CCBO = 21-12.
(2)(11.7)(8.9x10-14)(0.6)(1.3x1016!+!8.7x1013) = 2.1 microns (1.6x10-19)(1.3x1016)(8.7x1013)
(11.7)(8.9x10-14)(3) = 14.8 nF 2.1x10-4
Equivalent circuit for turn-on delay time, td,on, calculation. V
10 W
C CB +
V
in
C
+ -
BE
-
in
8V
VBE
-8 V
t
È -t ˘ VBE(t) = 8 - 16 exp ÍÎ !t˙˚
; t = (10 W)(CBE + CCB) È 16 ˘ At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln ÍÎ 7.3˙˚ The space-charge capacitances are nonlinear functions of the voltages across them. Need to find an average value for each of the two capacitors. During the td,on interval, the voltage VCB changes from 108 V to 100 volts and thus will be considered a constant. Hence CCB will be given by CCBO
CCB = !
VCB 1!+! f cC
1.5x10-8 = 100 1!+! 0.6
= 1.2 nF
The voltage VBE changes from -8 V to 0.7 volts during the same interval. We must find the average value of CBE.
W CBO =
CCBO = 21-12.
(2)(11.7)(8.9x10-14)(0.6)(1.3x1016!+!8.7x1013) = 2.1 microns (1.6x10-19)(1.3x1016)(8.7x1013)
(11.7)(8.9x10-14)(3) = 14.8 nF 2.1x10-4
Equivalent circuit for turn-on delay time, td,on, calculation. V
10 W
C CB +
V
in
C
+ -
BE
-
in
8V
VBE
-8 V
t
È -t ˘ VBE(t) = 8 - 16 exp ÍÎ !t˙˚
; t = (10 W)(CBE + CCB) È 16 ˘ At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln ÍÎ 7.3˙˚ The space-charge capacitances are nonlinear functions of the voltages across them. Need to find an average value for each of the two capacitors. During the td,on interval, the voltage VCB changes from 108 V to 100 volts and thus will be considered a constant. Hence CCB will be given by CCBO
CCB = !
VCB 1!+! f cC
1.5x10-8 = 100 1!+! 0.6
= 1.2 nF
The voltage VBE changes from -8 V to 0.7 volts during the same interval. We must find the average value of CBE.
0
Û Ù! Ù! ı CBE =
CBE =
CEBO
!dV VEB EB 1!+! f cE
8 0 Û ıdVEB 8 (9.2x10-9)(0.89) [ 8
=
CEBO!fcE È Í (-8) Î
0 1!+!f !!-!! cE
8 ˘˙ 1!+!f ˚ cE
8 1!+!0.89 - 1] = 2.2 nF
È 16 ˘ td,on = (10) [2.2x10-9 + 1.2x10-9] ln ÍÎ 7.3˙˚ = (3.4x10-8)(0.78) ≈ 27 nanoseconds
Chapter 22 Problem Solutions 22-1.
The capacitance of the gate-source terminals can be modeled as two capacitors connected electrically in series. Cox is the capacitance of the oxide layer and is a constant independent of vGS. Cdepl(vGS) is the capacitance of the depletion layer which increases in thickness as VGS increases as is shown in Fig. 22-6. Any increase in the depletion layer thickness reduces the value of Cdepl(vGS) and hence the atotal gatesource capacitance Cgs. However once vGS becomes equal to or greater than VGS(th), the depletion layer thickness becomes constant because the formation of the inversion layer jillustrated in Fig. 22-6c shields the depletion layer from further increases in vGS (any additional increase in vGS is dropped across the oxide layer). Thus both components of Cgs are constant for vGS > VGS(th).
22-2.
a) Idealized MOSFET waveforms shown below. To dimensions the waveforms, we need the numerical values of the various waveform parameters. The voltage and current amplitude parameters are given in the problem statement as: Vd = 300 V, VGS(th) = 4 V, VGS,Io = 7 V, Io = 10 A. To complete the dimensioning, we must calculate the various switching times.
Chapter 22 Problem Solutions 22-1.
The capacitance of the gate-source terminals can be modeled as two capacitors connected electrically in series. Cox is the capacitance of the oxide layer and is a constant independent of vGS. Cdepl(vGS) is the capacitance of the depletion layer which increases in thickness as VGS increases as is shown in Fig. 22-6. Any increase in the depletion layer thickness reduces the value of Cdepl(vGS) and hence the atotal gatesource capacitance Cgs. However once vGS becomes equal to or greater than VGS(th), the depletion layer thickness becomes constant because the formation of the inversion layer jillustrated in Fig. 22-6c shields the depletion layer from further increases in vGS (any additional increase in vGS is dropped across the oxide layer). Thus both components of Cgs are constant for vGS > VGS(th).
22-2.
a) Idealized MOSFET waveforms shown below. To dimensions the waveforms, we need the numerical values of the various waveform parameters. The voltage and current amplitude parameters are given in the problem statement as: Vd = 300 V, VGS(th) = 4 V, VGS,Io = 7 V, Io = 10 A. To complete the dimensioning, we must calculate the various switching times.
V
T/2
(t) CE
i (t) C
V
I d
t d,on V
t ri
t fv
o
V (t) GS
t d,off
GG
V GS,Io V GS(th) 0
-V GG td,on estimate - Use equivalent circuit of Fig. 22-12a. dvGS vGS VGG Governing equation is dt + R (C !+!C ) = R (C !+!C ) ; G gs gd G gs gd Boundary condition vGS(0) = - VGG Solution is vGS(t) = VGG - 2 VGG e-t/t ; t = RG(Cgs + Cgd) ; At t = td,on , vGS = VGS(th). Solving for td,on yields
t rv
t fi
È 2!VGG ˘ ˙ td,on = RG(Cgs + Cgd) ln ÍÎ !!V !-!V GG GS(th)˚ È (2)(15) ˘ td,on = (50) (1.15x10-9) ln ÍÎ !(15!-!4)˙˚ = 58 ns tri estimate - Use equivalent circuit of Fig. 22-12b. vGS(t) still given by governing equation given above in td,on estimate. Changing time origin to when vGS = VGS(th) yields; vGS(t) = VGG + [VGG - VGS(th)] e-t/t . The drain current is given by d(Vd!-!vGS) 10 iD(t) = Cgd + gm[vGS(t) - VGs(th)] ; gm = 7!-!4 = 3.3 mhos dt At t = tri, iD = Io. Substituting vGS(t) into iD(t) and solving for tri yields Cgd È (V !-!V ˘ ){gm!+ !R (C !+!C )} GG GS(th) Í G gs gd ˙ tri = RG(Cgs + Cgd) ln Î! ˚ gm(VGG!-!VGS(th))!-!Io È -10 ˘ Í (15!-!4)(3.3!+! 1.5x10 ˙ -9))˙ Í (50)(1.15x10 ˚ = 21 ns tri = (50)(1.15x10-9) ln Î ! (3.3)(15!-!4)!-!10 tfv estimate - Use equivalent circuit of Fig. 22-12c. Io vGS approximately constant at VGS,Io = g + VGS(th) during this interval. m
Governing equation is Cgd
dvDS =dt
Io ˘ È V !-!V Í GG GS(th)!-!gm˙ Î! ˚ RG
with vDS(0) = Vd.
Solution is given by È Io ˘ t vDS(t) = Vd - ÍÎ VGG!-!VGS(th)!-!g ˙˚ R C ; At t = tfv, vDS = 0. G gd m Solving for tfv yields
tfv =
RG!Cgd!Vd
300 -10 Io = (50)(1.5x10 ) (15!-!4!-!3) VGG!-!VGS(th)!-!g m
= 300 ns
td,off estimate - use equivalent circuit of Fig. 22-12d with the input voltagge VGG reversed. vGS(t) = - VGG + 2 VGG e-t/t ; At t = td,off, vGS = VGS,Io. Solving for td,off td,off = RG(Cgs + Cgd) ln
2!VGG È! ˘ I Í o˙ V +!V !+! Î GG! GS(th) gm˚
È (2)(15) ˘ ˙ = 18 ns = (50)(1.15x10-9) ln Í ! 10 Í ˙ Î 3.3!+!4!+!15˚
trv estimate - Use equivalent circuit of Fig. 22-12c with the input voltage VGG reversed. vGS approximately constant at VGS,Io as in previous of tfv. Governing equation is VGG!+!VGS,Io d{vDS!-!VGS,Io} = with vDS(0) = 0. Solution given by Cgd RG dt VGG!+!VGS,Io t . At t = trv , vDS = Vd . Solving for trv yields vDS(t) = RG!Cgd Vd!RG!Cgd (300)(50)(1.5x10-10) = = 100 ns trv = V !+!V (15!+!7) GG GS,Io tfi estimate - use equivalent circuit of Fig. 22-12b with the input voltage VGG reversed. Governing equation the same as in previous calculation of tri. At t = 0, vGS(0) = VGS,Io. Solution in this caae is given by vGS(t) = - VGG + [VGS,Io + VGG] e-t/t ; At t = tfi, vGS = VGS(th). Solving for tfi È VGG!+!VGS,Io ˘ È ˘ ˙ = (50)(1.15x10-9)ln Í !15!+!7˙ = 9 ns tfi = RG(Cgs + Cgd) ln ÍÎ !V +!V Î ˚ 15!+!4 GG GS(th)˚ b) Estimate the power dissipated in the MOSFET in the same manner as was done for the BJT in problem 21-3.Waveforms for the MOSFET are the same as for the BJT except for appropriate re-labeling of the currents and voltages.
Eri = (0.5)(300)(10)(2.1x10-8) = 3x10-5 Joules Efv = (0.5)(300)(10)(3x10-7) = 4.5x10-4 Joules Eon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 V T >> td,on and td,off Eon = (10)(300)(0.5)(5x10-5) = 1.25x10-3 Joules Erv = (0.5)(300)(10)(10-7) = 1.5x10-4 Joules Efi = (0.5)(300)(10)(9x10-9) = 1.5x10-5 Joules Pc = (1.95x10-3)(2x104) = 39 watts 22-3.
Use test conditons to estimate Cgd. Then estimate the switching times in the circuit with the 150 ohm load. Test circuit waveforms are shown below.
Eri = (0.5)(300)(10)(2.1x10-8) = 3x10-5 Joules Efv = (0.5)(300)(10)(3x10-7) = 4.5x10-4 Joules Eon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 V T >> td,on and td,off Eon = (10)(300)(0.5)(5x10-5) = 1.25x10-3 Joules Erv = (0.5)(300)(10)(10-7) = 1.5x10-4 Joules Efi = (0.5)(300)(10)(9x10-9) = 1.5x10-5 Joules Pc = (1.95x10-3)(2x104) = 39 watts 22-3.
Use test conditons to estimate Cgd. Then estimate the switching times in the circuit with the 150 ohm load. Test circuit waveforms are shown below.
V
GG
V (t) G
t VGS,Io V GS(th)
V
V (t) GS
GG t
t d,off
t d,on
t ri = t
t f i = tr v
fv
i (t) D
V (t) DS
V d
Io
t Equivalent circuit during voltage and current rise and fall intervals: RG
V (t) G
+ -
C gd
R
D
C gs
V d
g (V - V ) GS(th) m GS Governing equation using Miller capacitance approximation:
vGS !VG(t) dvGS + = ; t = RG [Cgs + Cgd{1 + gmRD}] ; t t dt During tri = tfv interval, VG(t) = VGG. Solution is Vd vGS(t) = VGG + {VGS(th) - VGG} e-t/t ; At t = tri, VGS = VGS(th) + g R ; m D Solving for tri = tfv yields tri = tfv = t ln
È! VGG!-!VGS(th) ˘ Vd ˙ Í Î VGG!-!VGS(th)!-!!!gm!RD˚
During trv = tfi, vG(t) = 0 and solution is vGS(t) = VGS,Io e-t/t . At t = trv, vGS(t) = VGS(th). Solving for trv yields
trv = t ln
Cgd =
Vd ˘ ÈV Í GS(th)!+!!gm!RD˙ Î! VGS(th) ˚ .
Invert equation for tri to find Cgd. Result is
tri 1 ÏÔ ¸Ï ¸ !!-!!RG!CgsÔ ÌR (1!+!g R )˝ V !-!V GG GS(th) m D˛ ˘ Ì!ln!ÈÍ! ˝Ó G V ˙ ! d ÔÓ Î V !-!V Ô˛ !-!! ˚ !g R GG GS(th) m! D
ÏÔ ¸ ¸ 1 3x10-8 -9Ô ÏÌ ˝ = 2.3x10-9 F = 2.3 nF Cgd = Ì È !!-!!5x10 ˝ 5(1!+!25) ˘ 15!-!4 Ó ˛ ˙ ÔÓ!ln!ÍÎ ! Ô˛ 15!-!4!-!1˚ Solving for switching times in circuit with RD = 150 ohms. t = [10-9 + 2.3x10-9 {1 + 150}][100] = 35 ms È 15!-!4 ˘ È 4!+!2˘ tri = 3.5x10-5 ln ÍÎ !15!-!4!-!2˙˚ = 7 ms ; trv = 3.5x10-5 ln ÍÎ ! 4 ˙˚ 22-4.
= 14 ms
Waveforms for vDS(t) and iD(t) shown in previous problem. Power dissipation in MOSFET given by 1 T
= [Eon + Esw] fs ; fs = T ; Eon = [ID]2 rDS,on(Tj) 2 ;
vGS !VG(t) dvGS + = ; t = RG [Cgs + Cgd{1 + gmRD}] ; t t dt During tri = tfv interval, VG(t) = VGG. Solution is Vd vGS(t) = VGG + {VGS(th) - VGG} e-t/t ; At t = tri, VGS = VGS(th) + g R ; m D Solving for tri = tfv yields tri = tfv = t ln
È! VGG!-!VGS(th) ˘ Vd ˙ Í Î VGG!-!VGS(th)!-!!!gm!RD˚
During trv = tfi, vG(t) = 0 and solution is vGS(t) = VGS,Io e-t/t . At t = trv, vGS(t) = VGS(th). Solving for trv yields
trv = t ln
Cgd =
Vd ˘ ÈV Í GS(th)!+!!gm!RD˙ Î! VGS(th) ˚ .
Invert equation for tri to find Cgd. Result is
tri 1 ÏÔ ¸Ï ¸ !!-!!RG!CgsÔ ÌR (1!+!g R )˝ V !-!V GG GS(th) m D˛ ˘ Ì!ln!ÈÍ! ˝Ó G V ˙ ! d ÔÓ Î V !-!V Ô˛ !-!! ˚ !g R GG GS(th) m! D
ÏÔ ¸ ¸ 1 3x10-8 -9Ô ÏÌ ˝ = 2.3x10-9 F = 2.3 nF Cgd = Ì È !!-!!5x10 ˝ 5(1!+!25) ˘ 15!-!4 Ó ˛ ˙ ÔÓ!ln!ÍÎ ! Ô˛ 15!-!4!-!1˚ Solving for switching times in circuit with RD = 150 ohms. t = [10-9 + 2.3x10-9 {1 + 150}][100] = 35 ms È 15!-!4 ˘ È 4!+!2˘ tri = 3.5x10-5 ln ÍÎ !15!-!4!-!2˙˚ = 7 ms ; trv = 3.5x10-5 ln ÍÎ ! 4 ˙˚ 22-4.
= 14 ms
Waveforms for vDS(t) and iD(t) shown in previous problem. Power dissipation in MOSFET given by 1 T = [Eon + Esw] fs ; fs = T ; Eon = [ID]2 rDS,on(Tj) 2 ;
Vd ÈÍ Tj!-!25˘˙ ÈÍ Tj ˘˙ 300 ID = R = 150 = 2 A ; rDS,on(Tj) = 2 Î 1!+! 150 ˚ = 2 Î 0.833!+!150˚ D ÈÍ Tj ˘˙ 1 1 Eon = (4)(2) Î 0.833!+!150˚ 2f = {3.32 + 0.027 Tj} f s s tri 1 Û t t Esw = T Ù ıVd!ID(1!-!t )(t )dt + ri ri 0 Esw =
tfi V !I 1 Û ÙV !I (1!-! t )( t )dt = d D [t + t ] T ı d D 6 tfi tfi ri fi 0
(300)(2) [7x10-6 + 14x10-6] = 2.1x10-3 joules 6
1 = {3.32 + 0.027 Tj} f fs + 2.1x10-3 fs s = {3.32 + 0.027 Tj} +[ 2.1x10-3][104] = 24.3 + 0.027 Tj PMOSFET 30 25
B
B
B
B
B
20 Watts
15 10 5 0 0
10
20
30
40
50
60
70
Temperature [ °K] 22-5.
Von = on-state voltage of three MOSFETs in parallel = Io reff
80
90 100
Vd ÈÍ Tj!-!25˘˙ ÈÍ Tj ˘˙ 300 ID = R = 150 = 2 A ; rDS,on(Tj) = 2 Î 1!+! 150 ˚ = 2 Î 0.833!+!150˚ D ÈÍ Tj ˘˙ 1 1 Eon = (4)(2) Î 0.833!+!150˚ 2f = {3.32 + 0.027 Tj} f s s tri 1 Û t t Esw = T Ù ıVd!ID(1!-!t )(t )dt + ri ri 0 Esw =
tfi V !I 1 Û ÙV !I (1!-! t )( t )dt = d D [t + t ] T ı d D 6 tfi tfi ri fi 0
(300)(2) [7x10-6 + 14x10-6] = 2.1x10-3 joules 6
1 = {3.32 + 0.027 Tj} f fs + 2.1x10-3 fs s = {3.32 + 0.027 Tj} +[ 2.1x10-3][104] = 24.3 + 0.027 Tj PMOSFET 30 25
B
B
B
B
B
20 Watts
15 10 5 0 0
10
20
30
40
50
60
70
Temperature [ °K] 22-5.
Von = on-state voltage of three MOSFETs in parallel = Io reff
80
90 100
r1!r2!r3 reff = r !r +!r !r +!r r ; r1 etc. = on-state resistance of MOSFET #1 etc. 1 2! 2 3! 3! 1 ÈÍ Tj!-!25˘˙ r1(Tj) = r1(25 °C) Î 1!+!0.8!! 100 ˚ ; r1(105 °C) = (1.64) r1(25 °C) etc. r1(105 °C) = 2.95 W ; r2(105 °C) = 3.28 W ; r3(105 °C) = 3.61 W (2.95)(3.28)(3.61) reff(105 °C) = [(2.95)(3.28)!+!(3.28)(3.61)!+!(3.61)(2.95)] = 1.09 ohms Von2 Io2!reff2 For the ith MOSFET, Pi = 2!r = 2!r ; Assume a 50% duty cycle and ignore i i switching losses. (5)2(1.09)2 (5)2(1.09)2 (5)2(1.09)2 P1 = (2)(2.95) = 5 W ; P2 = (2)(3.28) = 4.5 W ; P3 = (2)(3.61) = 4.1 W 22-6.
Hybrid switch would combine the low on-state losses of the BJT and the faster switching of the MOSFET. In order to obtain these advantages, The MOSFET would be turned on before the BJT and turned off after the BJT. The waveforms shown below indicate the relative timing. The switch blocks Vd volts in the off-state and conducts Io amps in the on-state.
r1!r2!r3 reff = r !r +!r !r +!r r ; r1 etc. = on-state resistance of MOSFET #1 etc. 1 2! 2 3! 3! 1 ÈÍ Tj!-!25˘˙ r1(Tj) = r1(25 °C) Î 1!+!0.8!! 100 ˚ ; r1(105 °C) = (1.64) r1(25 °C) etc. r1(105 °C) = 2.95 W ; r2(105 °C) = 3.28 W ; r3(105 °C) = 3.61 W (2.95)(3.28)(3.61) reff(105 °C) = [(2.95)(3.28)!+!(3.28)(3.61)!+!(3.61)(2.95)] = 1.09 ohms Von2 Io2!reff2 For the ith MOSFET, Pi = 2!r = 2!r ; Assume a 50% duty cycle and ignore i i switching losses. (5)2(1.09)2 (5)2(1.09)2 (5)2(1.09)2 P1 = (2)(2.95) = 5 W ; P2 = (2)(3.28) = 4.5 W ; P3 = (2)(3.61) = 4.1 W 22-6.
Hybrid switch would combine the low on-state losses of the BJT and the faster switching of the MOSFET. In order to obtain these advantages, The MOSFET would be turned on before the BJT and turned off after the BJT. The waveforms shown below indicate the relative timing. The switch blocks Vd volts in the off-state and conducts Io amps in the on-state.
VDS,on
V CE,on
v =v CE DS r I o; r < 1
Io
iD
(1 - r)I o
i
v
C
GS
v
22-7.
BE 1.3x1017 BVDSS ≈ N = 750 volts ; Ndrift = 1.7x1014 cm-3 drift -5 W drift ≈ (10 )(750) = 75 microns ; W d,body = protrusion of drain depletion layer into body region W drift!Ndrift (75)(1.7x1014) ≈ = ≈ 0.3 microns Nbody 5x1016 Even though body-source junction is shorted, there is a depletion layer associated with it which is contained entirely on the body side of the junction. This must be included in the estimate of the required length of the body region.
W s,body ≈
2!e!fc k!T ÈÍ Na!Nd˘˙ q!Na,body ; fc = q ln ÍÎ ! n 2 ˙˚ ; i
VDS,on
V CE,on
v =v CE DS r I o; r < 1
Io
iD
(1 - r)I o
i
v
C
GS
v
22-7.
BE 1.3x1017 BVDSS ≈ N = 750 volts ; Ndrift = 1.7x1014 cm-3 drift -5 W drift ≈ (10 )(750) = 75 microns ; W d,body = protrusion of drain depletion layer into body region W drift!Ndrift (75)(1.7x1014) ≈ = ≈ 0.3 microns Nbody 5x1016 Even though body-source junction is shorted, there is a depletion layer associated with it which is contained entirely on the body side of the junction. This must be included in the estimate of the required length of the body region.
W s,body ≈
2!e!fc k!T ÈÍ Na!Nd˘˙ q!Na,body ; fc = q ln ÍÎ ! n 2 ˙˚ ; i
È (1019)(5x1016)˘ ˙ = 0.94 fc = 0.026 ln ÍÎ ! 20 ˚ (10 ) W s,body ≈
(2)(11.7)(8.9x10-14)(0.94) ≈ 0.16 microns (1.6x10-19)(5x1016)
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8.
dvDS dvGD ≈ Cgd dt ; vDS ≈ vGD>> vGS Displacement current = Cgd dt BJT will turn on if Rbody Cgd
dvDS dt = 0.7 V
dvDS 0.7 dt > Rbody!Cgd will turn on the BJT.
22-9.
VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10.
a) iD =
mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L
(11.7)(8.9x10-14) Cox = t = = 1.04x10-7 F/cm2 10-5 ox e
2!iD!L N = m !C !W !(v !-!V n ox cell GS GS(th)) !(2)(100)(10-4) ≈ 5,800 cells N = (1500)(1.04x10-7)(2x10-3)(15!-!4) 100 b) Icell = 5800 = 17 milliamps 22-11.
Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms W drift Ron = q!m !N !A n d
: Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
È (1019)(5x1016)˘ ˙ = 0.94 fc = 0.026 ln ÍÎ ! 20 ˚ (10 ) W s,body ≈
(2)(11.7)(8.9x10-14)(0.94) ≈ 0.16 microns (1.6x10-19)(5x1016)
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8.
dvDS dvGD ≈ Cgd dt ; vDS ≈ vGD>> vGS Displacement current = Cgd dt BJT will turn on if Rbody Cgd
dvDS dt = 0.7 V
dvDS 0.7 dt > Rbody!Cgd will turn on the BJT.
22-9.
VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10.
a) iD =
mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L
(11.7)(8.9x10-14) Cox = t = = 1.04x10-7 F/cm2 10-5 ox e
2!iD!L N = m !C !W !(v !-!V n ox cell GS GS(th)) !(2)(100)(10-4) ≈ 5,800 cells N = (1500)(1.04x10-7)(2x10-3)(15!-!4) 100 b) Icell = 5800 = 17 milliamps 22-11.
Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms W drift Ron = q!m !N !A n d
: Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
È (1019)(5x1016)˘ ˙ = 0.94 fc = 0.026 ln ÍÎ ! 20 ˚ (10 ) W s,body ≈
(2)(11.7)(8.9x10-14)(0.94) ≈ 0.16 microns (1.6x10-19)(5x1016)
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8.
dvDS dvGD ≈ Cgd dt ; vDS ≈ vGD>> vGS Displacement current = Cgd dt BJT will turn on if Rbody Cgd
dvDS dt = 0.7 V
dvDS 0.7 dt > Rbody!Cgd will turn on the BJT.
22-9.
VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10.
a) iD =
mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L
(11.7)(8.9x10-14) Cox = t = = 1.04x10-7 F/cm2 10-5 ox e
2!iD!L N = m !C !W !(v !-!V n ox cell GS GS(th)) !(2)(100)(10-4) ≈ 5,800 cells N = (1500)(1.04x10-7)(2x10-3)(15!-!4) 100 b) Icell = 5800 = 17 milliamps 22-11.
Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms W drift Ron = q!m !N !A n d
: Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
È (1019)(5x1016)˘ ˙ = 0.94 fc = 0.026 ln ÍÎ ! 20 ˚ (10 ) W s,body ≈
(2)(11.7)(8.9x10-14)(0.94) ≈ 0.16 microns (1.6x10-19)(5x1016)
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8.
dvDS dvGD ≈ Cgd dt ; vDS ≈ vGD>> vGS Displacement current = Cgd dt BJT will turn on if Rbody Cgd
dvDS dt = 0.7 V
dvDS 0.7 dt > Rbody!Cgd will turn on the BJT.
22-9.
VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10.
a) iD =
mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L
(11.7)(8.9x10-14) Cox = t = = 1.04x10-7 F/cm2 10-5 ox e
2!iD!L N = m !C !W !(v !-!V n ox cell GS GS(th)) !(2)(100)(10-4) ≈ 5,800 cells N = (1500)(1.04x10-7)(2x10-3)(15!-!4) 100 b) Icell = 5800 = 17 milliamps 22-11.
Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms W drift Ron = q!m !N !A n d
: Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
È (1019)(5x1016)˘ ˙ = 0.94 fc = 0.026 ln ÍÎ ! 20 ˚ (10 ) W s,body ≈
(2)(11.7)(8.9x10-14)(0.94) ≈ 0.16 microns (1.6x10-19)(5x1016)
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8.
dvDS dvGD ≈ Cgd dt ; vDS ≈ vGD>> vGS Displacement current = Cgd dt BJT will turn on if Rbody Cgd
dvDS dt = 0.7 V
dvDS 0.7 dt > Rbody!Cgd will turn on the BJT.
22-9.
VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10.
a) iD =
mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L
(11.7)(8.9x10-14) Cox = t = = 1.04x10-7 F/cm2 10-5 ox e
2!iD!L N = m !C !W !(v !-!V n ox cell GS GS(th)) !(2)(100)(10-4) ≈ 5,800 cells N = (1500)(1.04x10-7)(2x10-3)(15!-!4) 100 b) Icell = 5800 = 17 milliamps 22-11.
Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms W drift Ron = q!m !N !A n d
: Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
1.3x1017 Nd = BV DSS
=
1.3x1017 800
≈ 1.6x1014 cm-3
8x10-3 A= ≈ 0.5 cm2 (1.6x10-19)(1500)(1.6x1014)(0.4) 10!A A A = 20 2 << the allowable maximum of 200 , so estimate is alright. 2 0.5!cm cm cm2
22-12.
Cgs ≈ Cox N Wcell L = (1.04x10-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13.
Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first. È 100 ˘˙ di = 300 V > BVDSS = 150 V VDS(turn-off) = Vd + L dt = 100 + (10-7) ÍÎ ! 5x10-8˚ Check for excessive power dissipation. Pallowed =
Tj,max!-!Ta 150!-!50 = = 100 watts ; Pdissipated = [Eon + Esw] fs 1 Rq,j-a
Io2!rDS(on) (100)2(0.01) = = 50 watts Eonfs = 2 2 Esw =
Vd!Io (100)(100) [(2)(5x10-8) + (2)(2x10-7)] 2 [tri + tfi + trv +tfv] = 2
Esw = 2.5x10-3 joules ; Eswfs = (2.5x10-3)(3x104) = 75 watts Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14.
Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have:
1.3x1017 Nd = BV DSS
=
1.3x1017 800
≈ 1.6x1014 cm-3
8x10-3 A= ≈ 0.5 cm2 (1.6x10-19)(1500)(1.6x1014)(0.4) 10!A A A = 20 2 << the allowable maximum of 200 , so estimate is alright. 2 0.5!cm cm cm2
22-12.
Cgs ≈ Cox N Wcell L = (1.04x10-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13.
Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first. È 100 ˘˙ di = 300 V > BVDSS = 150 V VDS(turn-off) = Vd + L dt = 100 + (10-7) ÍÎ ! 5x10-8˚ Check for excessive power dissipation. Pallowed =
Tj,max!-!Ta 150!-!50 = = 100 watts ; Pdissipated = [Eon + Esw] fs 1 Rq,j-a
Io2!rDS(on) (100)2(0.01) = = 50 watts Eonfs = 2 2 Esw =
Vd!Io (100)(100) [(2)(5x10-8) + (2)(2x10-7)] 2 [tri + tfi + trv +tfv] = 2
Esw = 2.5x10-3 joules ; Eswfs = (2.5x10-3)(3x104) = 75 watts Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14.
Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have:
1.3x1017 Nd = BV DSS
=
1.3x1017 800
≈ 1.6x1014 cm-3
8x10-3 A= ≈ 0.5 cm2 (1.6x10-19)(1500)(1.6x1014)(0.4) 10!A A A = 20 2 << the allowable maximum of 200 , so estimate is alright. 2 0.5!cm cm cm2
22-12.
Cgs ≈ Cox N Wcell L = (1.04x10-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13.
Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first. È 100 ˘˙ di = 300 V > BVDSS = 150 V VDS(turn-off) = Vd + L dt = 100 + (10-7) ÍÎ ! 5x10-8˚ Check for excessive power dissipation. Pallowed =
Tj,max!-!Ta 150!-!50 = = 100 watts ; Pdissipated = [Eon + Esw] fs 1 Rq,j-a
Io2!rDS(on) (100)2(0.01) = = 50 watts Eonfs = 2 2 Esw =
Vd!Io (100)(100) [(2)(5x10-8) + (2)(2x10-7)] 2 [tri + tfi + trv +tfv] = 2
Esw = 2.5x10-3 joules ; Eswfs = (2.5x10-3)(3x104) = 75 watts Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14.
Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have:
1.3x1017 Nd = BV DSS
=
1.3x1017 800
≈ 1.6x1014 cm-3
8x10-3 A= ≈ 0.5 cm2 (1.6x10-19)(1500)(1.6x1014)(0.4) 10!A A A = 20 2 << the allowable maximum of 200 , so estimate is alright. 2 0.5!cm cm cm2
22-12.
Cgs ≈ Cox N Wcell L = (1.04x10-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13.
Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first. È 100 ˘˙ di = 300 V > BVDSS = 150 V VDS(turn-off) = Vd + L dt = 100 + (10-7) ÍÎ ! 5x10-8˚ Check for excessive power dissipation. Pallowed =
Tj,max!-!Ta 150!-!50 = = 100 watts ; Pdissipated = [Eon + Esw] fs 1 Rq,j-a
Io2!rDS(on) (100)2(0.01) = = 50 watts Eonfs = 2 2 Esw =
Vd!Io (100)(100) [(2)(5x10-8) + (2)(2x10-7)] 2 [tri + tfi + trv +tfv] = 2
Esw = 2.5x10-3 joules ; Eswfs = (2.5x10-3)(3x104) = 75 watts Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14.
Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have:
VGG Current rise/fall times proportional to [Cgs + Cgd] I G Vd Voltage rise/fall times proportional to Cgd I G Cgs roughly the same size as Cgd and Vd >> VGG Hence voltage switching times much greater than current switching times.
Chapter 23 Problem Solutions
23-1.
vs(t) ; a < wt < p vs(t) = 2 Vs sin(wt) ; iL(t) = R L p 1 Û = 2p ı[(1)iL(wt)!+!{iL(wt)}2Ron]!d(wt) a Vs 1 {1 + cos(a)} + π = 2!pRL
23-2.
È Vs ˘ 2 Í ! ˙ R [π - a + sin(2a) ] 2 Î RL˚ on
Tj,max!-!Ta,max 120 °F = 49 °C ; Tj,max = 125 °C ; |max = Rqja |max =
125!-!49 = 760 Watts 0.1
Check at a = 0 1 220 [1 + cos(0)] + π = 2π!(1)
ÈÍ 220˘˙ 2 sin(0) -3 Î ! 1 ˚ (2x10 )[π - 0 + 2 ] = 107 watts
= 107 watts less than the allowable 760 watts. Hence trigger angle of zero where maximum load power is delivered is permissible. π 1 Û Average load power = 2π ı{iL(wt)}2RL!d(wt) ; iL(wt) = 2 (220) sin(wt) a π sin(2a) 1 Û (220)2 2 2 ı = 2π { 2!(220)} sin (wt)!d(wt) = 6.28 [3.14 - a + ] 2 a For a = 0 = 24.2 kW 23-3.
PSCR(t) = instantaneous power dissipated in the SCR during turn-on. t dI PSCR(t) = vAK(t) iA(t) = VAK {1 - t } dt t during tf f P(t) = power density =
PSCR(t) = watts per cm2 ; A(t) = conducting area of SCR A(t)
Chapter 23 Problem Solutions
23-1.
vs(t) ; a < wt < p vs(t) = 2 Vs sin(wt) ; iL(t) = R L p 1 Û = 2p ı[(1)iL(wt)!+!{iL(wt)}2Ron]!d(wt) a Vs 1 {1 + cos(a)} + π = 2!pRL
23-2.
È Vs ˘ 2 Í ! ˙ R [π - a + sin(2a) ] 2 Î RL˚ on
Tj,max!-!Ta,max 120 °F = 49 °C ; Tj,max = 125 °C ; |max = Rqja |max =
125!-!49 = 760 Watts 0.1
Check at a = 0 1 220 [1 + cos(0)] + π = 2π!(1)
ÈÍ 220˘˙ 2 sin(0) -3 Î ! 1 ˚ (2x10 )[π - 0 + 2 ] = 107 watts
= 107 watts less than the allowable 760 watts. Hence trigger angle of zero where maximum load power is delivered is permissible. π 1 Û Average load power = 2π ı{iL(wt)}2RL!d(wt) ; iL(wt) = 2 (220) sin(wt) a π sin(2a) 1 Û (220)2 2 2 ı = 2π { 2!(220)} sin (wt)!d(wt) = 6.28 [3.14 - a + ] 2 a For a = 0 = 24.2 kW 23-3.
PSCR(t) = instantaneous power dissipated in the SCR during turn-on. t dI PSCR(t) = vAK(t) iA(t) = VAK {1 - t } dt t during tf f P(t) = power density =
PSCR(t) = watts per cm2 ; A(t) = conducting area of SCR A(t)
Chapter 23 Problem Solutions
23-1.
vs(t) ; a < wt < p vs(t) = 2 Vs sin(wt) ; iL(t) = R L p 1 Û = 2p ı[(1)iL(wt)!+!{iL(wt)}2Ron]!d(wt) a Vs 1 {1 + cos(a)} + π = 2!pRL
23-2.
È Vs ˘ 2 Í ! ˙ R [π - a + sin(2a) ] 2 Î RL˚ on
Tj,max!-!Ta,max 120 °F = 49 °C ; Tj,max = 125 °C ; |max = Rqja |max =
125!-!49 = 760 Watts 0.1
Check at a = 0 1 220 [1 + cos(0)] + π = 2π!(1)
ÈÍ 220˘˙ 2 sin(0) -3 Î ! 1 ˚ (2x10 )[π - 0 + 2 ] = 107 watts
= 107 watts less than the allowable 760 watts. Hence trigger angle of zero where maximum load power is delivered is permissible. π 1 Û Average load power = 2π ı{iL(wt)}2RL!d(wt) ; iL(wt) = 2 (220) sin(wt) a π sin(2a) 1 Û (220)2 2 2 ı = 2π { 2!(220)} sin (wt)!d(wt) = 6.28 [3.14 - a + ] 2 a For a = 0 = 24.2 kW 23-3.
PSCR(t) = instantaneous power dissipated in the SCR during turn-on. t dI PSCR(t) = vAK(t) iA(t) = VAK {1 - t } dt t during tf f P(t) = power density =
PSCR(t) = watts per cm2 ; A(t) = conducting area of SCR A(t)
A(t) = π [ro + us t]2 - π ro2 = π [2 ro us t + (us t)2] tf
Û V !{1!-!!tt !}!dI tf !!t! f dt 1 Ù AK 1 Û !dt dTj = C ıP(t)!dt = C v ıπ![2!ro!us!t!+!(us!t)2] v 0 0 tf
1 VAK dI dTj = C 2πr u dt o s v
Û È 1!-!!tt ˘ ÙÍ f ˙ ! ! !dt u ÙÍ st ˙ ı Î 1!+!2ro˚
us -1 ; Let a = a' = 1, b = t , and b' = 2r f o
0 tf Û Ù È a!+!bt ˘ Integral becomes ı!ÍÎ !a'!+!b't˙˚ !dt ; Using integral tables 0 tf btf [ab'!-!ab] Û Ù ÈÍ a!+!bt ˘˙ ı!Î !a'!+!b't˚ !dt = b' + ln[a' + b' tf] ; [b']2 0 1 b= 2x10-5
104 4 -1 = - 5x10 sec ; b' = (2)(0.5)
= 104 sec-1
Evaluating the integral yields tf È Û -1 5x104˘˙ Ù ÈÍ a!+!bt ˘˙ Í -4 ı!Î !a'!+!b't˚ !dt = 4 + Î 10 !+!! ln[1 + (104)(2x10-5)] = 9.4x10-6 sec 8 ˚ 10 10 0 With Cv = 1.75 Joule/(°C-cm3), the expression for dTj becomes (103)(!9.4x10-6) dI dI dTj = = 125 - 25 = 100 °C ; Solving for dt yields 4 dt (2π)(1.75)(10 )(0.5) dI 100 dt = 1.7x10-7 = 590 A/ms
23-4.
Advantages - Shorter N- region length Wd means shorter carrier lifetimes can be accomodated and thus faster switching times. Disadvantages - Junction J1 is now a P+ - N+ junction with a low breakdown voltage. Since J3 already has a low breakdown voltage, the modified thyristor has no significant reverse blocking capability. (4!-!0.5)!cm us
=
3.5 = 350 microseconds 104
23-5.
ton =
23-6.
Lateral voltage drops caused by base currents cause current density nonuniformities. At large currents, these nonuniformities become severe and the increasing possibility of second breakdown limit the total current that the BJT can safely conduct. In the thyristor, no significant gate current is needed to keep the thyristor on and there is consequently no lateral current flow and thus lateral voltage drop. The current density is uniform across the entire cross-sectional area of the thyristor and there is much less likelyhood of second breakdown.
23-7.
a) Breakover in a thyristor is not due to impact ionization. However in a well-designed thyristor, the value of the breakover voltage is an appreciable fraction of the actual avalanche breakdown voltage. Thus an estimate of the n1 thickness and doping level based avalanche breakdown would be a reasonable first attempt. Nd =
1.3x1017 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns 3 2x10
qWd2 (1.6x10-19)(2x10-2)2 b) t = kT(m !+!m ) = = 17 microseconds (1.4x10-23)(300)(900) n p Used (mn + m p) = 900 cm2/V-sec which is value appropriate to large excess carrier densities (approaching 1017 cm-3) c) Von = I Rdrift ; Ignore I2/3 contribution as it is usually small compared to the linear term. Wd 2 0.02 ; Rdrift ≈ q!(m !+!m )!n !A ; 2000 = 10-3 = -19 (1.6x10 )(900)(1017)!A n p b Solving for A gives A =
0.02 = 1.4 cm2 -19 17 -3 (1.6x10 )(900)(10 )(10 )
23-4.
Advantages - Shorter N- region length Wd means shorter carrier lifetimes can be accomodated and thus faster switching times. Disadvantages - Junction J1 is now a P+ - N+ junction with a low breakdown voltage. Since J3 already has a low breakdown voltage, the modified thyristor has no significant reverse blocking capability. (4!-!0.5)!cm us
=
3.5 = 350 microseconds 104
23-5.
ton =
23-6.
Lateral voltage drops caused by base currents cause current density nonuniformities. At large currents, these nonuniformities become severe and the increasing possibility of second breakdown limit the total current that the BJT can safely conduct. In the thyristor, no significant gate current is needed to keep the thyristor on and there is consequently no lateral current flow and thus lateral voltage drop. The current density is uniform across the entire cross-sectional area of the thyristor and there is much less likelyhood of second breakdown.
23-7.
a) Breakover in a thyristor is not due to impact ionization. However in a well-designed thyristor, the value of the breakover voltage is an appreciable fraction of the actual avalanche breakdown voltage. Thus an estimate of the n1 thickness and doping level based avalanche breakdown would be a reasonable first attempt. Nd =
1.3x1017 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns 3 2x10
qWd2 (1.6x10-19)(2x10-2)2 b) t = kT(m !+!m ) = = 17 microseconds (1.4x10-23)(300)(900) n p Used (mn + m p) = 900 cm2/V-sec which is value appropriate to large excess carrier densities (approaching 1017 cm-3) c) Von = I Rdrift ; Ignore I2/3 contribution as it is usually small compared to the linear term. Wd 2 0.02 ; Rdrift ≈ q!(m !+!m )!n !A ; 2000 = 10-3 = -19 (1.6x10 )(900)(1017)!A n p b Solving for A gives A =
0.02 = 1.4 cm2 -19 17 -3 (1.6x10 )(900)(10 )(10 )
23-4.
Advantages - Shorter N- region length Wd means shorter carrier lifetimes can be accomodated and thus faster switching times. Disadvantages - Junction J1 is now a P+ - N+ junction with a low breakdown voltage. Since J3 already has a low breakdown voltage, the modified thyristor has no significant reverse blocking capability. (4!-!0.5)!cm us
=
3.5 = 350 microseconds 104
23-5.
ton =
23-6.
Lateral voltage drops caused by base currents cause current density nonuniformities. At large currents, these nonuniformities become severe and the increasing possibility of second breakdown limit the total current that the BJT can safely conduct. In the thyristor, no significant gate current is needed to keep the thyristor on and there is consequently no lateral current flow and thus lateral voltage drop. The current density is uniform across the entire cross-sectional area of the thyristor and there is much less likelyhood of second breakdown.
23-7.
a) Breakover in a thyristor is not due to impact ionization. However in a well-designed thyristor, the value of the breakover voltage is an appreciable fraction of the actual avalanche breakdown voltage. Thus an estimate of the n1 thickness and doping level based avalanche breakdown would be a reasonable first attempt. Nd =
1.3x1017 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns 3 2x10
qWd2 (1.6x10-19)(2x10-2)2 b) t = kT(m !+!m ) = = 17 microseconds (1.4x10-23)(300)(900) n p Used (mn + m p) = 900 cm2/V-sec which is value appropriate to large excess carrier densities (approaching 1017 cm-3) c) Von = I Rdrift ; Ignore I2/3 contribution as it is usually small compared to the linear term. Wd 2 0.02 ; Rdrift ≈ q!(m !+!m )!n !A ; 2000 = 10-3 = -19 (1.6x10 )(900)(1017)!A n p b Solving for A gives A =
0.02 = 1.4 cm2 -19 17 -3 (1.6x10 )(900)(10 )(10 )
23-4.
Advantages - Shorter N- region length Wd means shorter carrier lifetimes can be accomodated and thus faster switching times. Disadvantages - Junction J1 is now a P+ - N+ junction with a low breakdown voltage. Since J3 already has a low breakdown voltage, the modified thyristor has no significant reverse blocking capability. (4!-!0.5)!cm us
=
3.5 = 350 microseconds 104
23-5.
ton =
23-6.
Lateral voltage drops caused by base currents cause current density nonuniformities. At large currents, these nonuniformities become severe and the increasing possibility of second breakdown limit the total current that the BJT can safely conduct. In the thyristor, no significant gate current is needed to keep the thyristor on and there is consequently no lateral current flow and thus lateral voltage drop. The current density is uniform across the entire cross-sectional area of the thyristor and there is much less likelyhood of second breakdown.
23-7.
a) Breakover in a thyristor is not due to impact ionization. However in a well-designed thyristor, the value of the breakover voltage is an appreciable fraction of the actual avalanche breakdown voltage. Thus an estimate of the n1 thickness and doping level based avalanche breakdown would be a reasonable first attempt. Nd =
1.3x1017 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns 3 2x10
qWd2 (1.6x10-19)(2x10-2)2 b) t = kT(m !+!m ) = = 17 microseconds (1.4x10-23)(300)(900) n p Used (mn + m p) = 900 cm2/V-sec which is value appropriate to large excess carrier densities (approaching 1017 cm-3) c) Von = I Rdrift ; Ignore I2/3 contribution as it is usually small compared to the linear term. Wd 2 0.02 ; Rdrift ≈ q!(m !+!m )!n !A ; 2000 = 10-3 = -19 (1.6x10 )(900)(1017)!A n p b Solving for A gives A =
0.02 = 1.4 cm2 -19 17 -3 (1.6x10 )(900)(10 )(10 )
2000 Resulting current density is 1.4 = 1430 A/cm2 which is excessively large. Probably should use nb ≈ 1016 cm-3 which would give a current density of 140 A/cm2, a more realistic value.
23-8.
15 3000 Cathode area = 200 = 15 cm2 = 0.65 Asi ; Asi = 0.65 = 23 cm2 23 cm2 = π Rsi2 ; Rsi =
23-9.
23 π = 2.7 cm
dvAK IBO dt max ≈ Cj2(0) ; Cj2(0) = zero bias value of junction J2 space charge capacitance
|
e!A
Cj2(0) ≈ W depl(0)
; Wdepl(0) ≈
2!e!fj2 q!Nd
NaNd kT ; fj2 = q ln[ 2 ] ni
(1014)(1017) fj2 = 0.026 ln[ ] = 0.66 V ; (1020)
W depl(0) ≈
(2)(11.7)(8.9x10-14)(0.66) = 2.9 microns (1.6x10-19)(1014)
(11.7)(8.9x10-14)(10) = 36 nF Cj2(0) = 2.9x10-4 dvAK 0.05 6 dt max = 3.6x10-8 = 1.4x10 V/sec or 1.4 V per microsecond
|
2000 Resulting current density is 1.4 = 1430 A/cm2 which is excessively large. Probably should use nb ≈ 1016 cm-3 which would give a current density of 140 A/cm2, a more realistic value.
23-8.
15 3000 Cathode area = 200 = 15 cm2 = 0.65 Asi ; Asi = 0.65 = 23 cm2 23 cm2 = π Rsi2 ; Rsi =
23-9.
23 π = 2.7 cm
dvAK IBO dt max ≈ Cj2(0) ; Cj2(0) = zero bias value of junction J2 space charge capacitance
|
e!A
Cj2(0) ≈ W depl(0)
; Wdepl(0) ≈
2!e!fj2 q!Nd
NaNd kT ; fj2 = q ln[ 2 ] ni
(1014)(1017) fj2 = 0.026 ln[ ] = 0.66 V ; (1020)
W depl(0) ≈
(2)(11.7)(8.9x10-14)(0.66) = 2.9 microns (1.6x10-19)(1014)
(11.7)(8.9x10-14)(10) = 36 nF Cj2(0) = 2.9x10-4 dvAK 0.05 6 dt max = 3.6x10-8 = 1.4x10 V/sec or 1.4 V per microsecond
|
2000 Resulting current density is 1.4 = 1430 A/cm2 which is excessively large. Probably should use nb ≈ 1016 cm-3 which would give a current density of 140 A/cm2, a more realistic value.
23-8.
15 3000 Cathode area = 200 = 15 cm2 = 0.65 Asi ; Asi = 0.65 = 23 cm2 23 cm2 = π Rsi2 ; Rsi =
23-9.
23 π = 2.7 cm
dvAK IBO dt max ≈ Cj2(0) ; Cj2(0) = zero bias value of junction J2 space charge capacitance
|
e!A
Cj2(0) ≈ W depl(0)
; Wdepl(0) ≈
2!e!fj2 q!Nd
NaNd kT ; fj2 = q ln[ 2 ] ni
(1014)(1017) fj2 = 0.026 ln[ ] = 0.66 V ; (1020)
W depl(0) ≈
(2)(11.7)(8.9x10-14)(0.66) = 2.9 microns (1.6x10-19)(1014)
(11.7)(8.9x10-14)(10) = 36 nF Cj2(0) = 2.9x10-4 dvAK 0.05 6 dt max = 3.6x10-8 = 1.4x10 V/sec or 1.4 V per microsecond
|
Chapter 24 Problem Solutions 24-1.
Cross-sectional view of GTO gate-cathode area with reverse gate current flowing.
Gate Cathode -i N
2 R
R P 2
-
v
G
GK +
+ vGK -
t
Center Line The negative gate current, -iG, flowing through the P2 layher beneath the N2 cathode layer develops a lateral voltage drop vGK as indicated. Maximum negative vGK = BVJ3 BVJ3 |vGK| = |IG,max| RGK < BVJ3 in order to avoid breakdown or IG,max < R GK r p2!W R RGK = 2!N = 4!N!L!t ; N = number of cathode islands in parallel. BVJ3 IG,max = R GK
IA,max = b : Solving for IA,max yields off
4!N!L!t!boff!BVJ3 IA,max = r p2!W 24-2.
Assume that the current is communtated from the GTO to the turn-off snubber and associated stray inductance linearly as a function of time. That is the inductor current
Chapter 24 Problem Solutions 24-1.
Cross-sectional view of GTO gate-cathode area with reverse gate current flowing.
Gate Cathode -i N
2 R
R P 2
-
v
G
GK +
+ vGK -
t
Center Line The negative gate current, -iG, flowing through the P2 layher beneath the N2 cathode layer develops a lateral voltage drop vGK as indicated. Maximum negative vGK = BVJ3 BVJ3 |vGK| = |IG,max| RGK < BVJ3 in order to avoid breakdown or IG,max < R GK r p2!W R RGK = 2!N = 4!N!L!t ; N = number of cathode islands in parallel. BVJ3 IG,max = R GK
IA,max = b : Solving for IA,max yields off
4!N!L!t!boff!BVJ3 IA,max = r p2!W 24-2.
Assume that the current is communtated from the GTO to the turn-off snubber and associated stray inductance linearly as a function of time. That is the inductor current
t iLs = Io t ; Assume that just prior to the end of the current fall time interval, the fi voltage across the snubber capacitor has built up to approximately Vd. diLs Io vAK,max = 1.5 Vd = Ls dt + vcap = Ls t + Vd ; Solving for Ls yields fi Vd!tfi Ls = 2!I o 24-3.
Equivalent circuit during tgq shown below. P N P
i (t) G L
N
G V GG+
J3 forward biased during t gq
-!VGGIo diG LG dt = - VGG- ; iG(t) = L t ; At t = tgq want iG = - b ; Solve for LG off G b off!VGG-!tgq LG = Io
=
(5)(15)(5x10-6) (500)
Equivalent circuit during tw2 interval.
= 0.75 microhenries
t iLs = Io t ; Assume that just prior to the end of the current fall time interval, the fi voltage across the snubber capacitor has built up to approximately Vd. diLs Io vAK,max = 1.5 Vd = Ls dt + vcap = Ls t + Vd ; Solving for Ls yields fi Vd!tfi Ls = 2!I o 24-3.
Equivalent circuit during tgq shown below. P N P
i (t) G L
N
G V GG+
J3 forward biased during t gq
-!VGGIo diG LG dt = - VGG- ; iG(t) = L t ; At t = tgq want iG = - b ; Solve for LG off G b off!VGG-!tgq LG = Io
=
(5)(15)(5x10-6) (500)
Equivalent circuit during tw2 interval.
= 0.75 microhenries
i (t) G L
G
-
V GG+
+
BV J3
Io Io !BVJ3!-!VGG-! diG LG dt = BVJ3 - VGG- ; iG(0) = - b ; iG(t) = - b + t LG off off At t = tw2 , iG = 0 ; solving for tw2 yields Io!LG tw2 = b ![BV !-!V off J3 GG-]
(500)(7.5x10-7) = = 7.5 microseconds (5)(25!-!15)
Chapter 25 Problem Solutions
25-1.
Ron(MOS) 1 proprotional to m ; mn = 3 mp ; Hence A majority Ron(n-channel) Ron(p-channel) = 3 A A Ron(IGBT) 1 proportional to dn(m +!m ) ; dn = excess carrier density A n! p dn = dp so p-channel IGBTs have the same Ron as n-channel IGBTs
25-2.
Turn-off waveforms of short versus long lifetime IGBTs i
D long lifetime
I
(long) BJT
I
(short) BJT
short lifetime
t
Long lifetime IGBT a. BJT portion of the device has a larger beta and thus the BJT section carries the largest fraction of the IGBT current. Thus IBJT(long) > IBJT(short). b. Longer lifetime leads to longer BJT turn-off times. Short lifetime IGBT a. BJT beta smaller. MOSFET section of the device carries most of the current. b. Shorter lifetime means less stored charge in the BJT section and thus faster turn-off.
Chapter 25 Problem Solutions
25-1.
Ron(MOS) 1 proprotional to m ; mn = 3 mp ; Hence A majority Ron(n-channel) Ron(p-channel) = 3 A A Ron(IGBT) 1 proportional to dn(m +!m ) ; dn = excess carrier density A n! p dn = dp so p-channel IGBTs have the same Ron as n-channel IGBTs
25-2.
Turn-off waveforms of short versus long lifetime IGBTs i
D long lifetime
I
(long) BJT
I
(short) BJT
short lifetime
t
Long lifetime IGBT a. BJT portion of the device has a larger beta and thus the BJT section carries the largest fraction of the IGBT current. Thus IBJT(long) > IBJT(short). b. Longer lifetime leads to longer BJT turn-off times. Short lifetime IGBT a. BJT beta smaller. MOSFET section of the device carries most of the current. b. Shorter lifetime means less stored charge in the BJT section and thus faster turn-off.
25-3. Collector junction of pnp BJT section of IGBT Body region of MOSFET section of IGBT
Base of pnp BJT Emitter of pnp BJT
N-
P
V V
DS2
P >V
+ Drain of IGBT
DS1
DS1
Effective base width Significant encroachment intoa the base of the PNP BJT section by the depletion layer of the blocking junction. The effective base width is thus lowered and the beta increases as vDS increases. This is base width modulation and it results in a lower output resistance ro (steeper slope in the active region of the iD-vDS characteristics). Depletion layer
Collector junction of pnp BJT section of IGBT
Base of pnp BJT Emitter of pnp BJT
Body region of MOSFET section of IGBT
N-
P
N V
+
P
+
DS
Drain of IGBT
Effective base width independent of V DS
Depletion layer
Depletion encroaches into the N- layer but the advance is halted at moderate vDS values by the N+ buffer layer. The PNP base width becomes constant and so the effective resistance ro remains large. 25-4.
One dimensional model of n-channel IGBT 25 m m
Source N
+
19 10
P 17 10
N+
N 14 10
19 10
P
+
19 10
Drain
Reverse blocking junction is the P+ - N+ junction because of body-source short. BVRB ≈
1.3x1017 < 1 volt. No reverse blocking capability. 1019
Forward breakdown - limited by P - N- junction.
Collector junction of pnp BJT section of IGBT
Base of pnp BJT Emitter of pnp BJT
Body region of MOSFET section of IGBT
N-
P
N V
+
P
+
DS
Drain of IGBT
Effective base width independent of V DS
Depletion layer
Depletion encroaches into the N- layer but the advance is halted at moderate vDS values by the N+ buffer layer. The PNP base width becomes constant and so the effective resistance ro remains large. 25-4.
One dimensional model of n-channel IGBT 25 m m
Source N
+
19 10
P 17 10
N+
N 14 10
19 10
P
+
19 10
Drain
Reverse blocking junction is the P+ - N+ junction because of body-source short. BVRB ≈
1.3x1017 < 1 volt. No reverse blocking capability. 1019
Forward breakdown - limited by P - N- junction.
BVFB =
1.3x1017 ≈ 1300 volts ; But Wdepl(1300 V) = (10-5)(1300) = 130 microns 14 10
130 microns > 25 micron drift region length. Hence forward blocking limited by punch-through. (1.6x10-19)(1014)(2.5x10-3)2 = 453 volts BVFB = (2x105)(2.5x10-3) (2)(11.7)(8.9x10-14) 25-5.
IGBT current - Ion,IGBT
; Von(IGBT) = Vj + Ion,IGBT Ron,IGBT
Assume Vj ≈ 0.8 V ; Exact value not critical to an approximate estimate of Ion,IGBT. Ion,IGBT ≈
Wd 3!-!0.8 ; R ≈ Ron,IGBT! q!(mn!+!m p)!nb!A on,IGBT
7.5x10-3 = 2.6x10-3 W W d = (10-5)(750) = 75 mm ; Ron,IGBT = (1.6x10-19)(900)(1016)(2) 2.2 Ion,IGBT ≈ ≈ 850 amps 2.6x10-3 MOSFET current - Ion,MOS ; Von(MOS) = Ion,MOS Ron,MOS Von(MOS) ; Ron,MOS = Ion,MOS = R on,MOS Nd =
Wd q!mn!!Nd!A
; Wd = 75 mm
1.3x1017 = 1.7x1014 cm-3 750
7.5x10-3 Ron,MOS = = 0.09 ohms (1.6x10-19)(1.5x103)(1.7x1014)(2) 3 Ion,MOS = 0.09 = 33 amps 25-6.
Von(PT) = Vj,PT + Ion,PT Ron,PT = Vj,NPT + Ion,NPT Ron,NPT Ion,PT Ion,NPT
≈
Ron,NPT Ron,PT since Vj,NPT ≈ Vj,PT
BVFB =
1.3x1017 ≈ 1300 volts ; But Wdepl(1300 V) = (10-5)(1300) = 130 microns 14 10
130 microns > 25 micron drift region length. Hence forward blocking limited by punch-through. (1.6x10-19)(1014)(2.5x10-3)2 = 453 volts BVFB = (2x105)(2.5x10-3) (2)(11.7)(8.9x10-14) 25-5.
IGBT current - Ion,IGBT
; Von(IGBT) = Vj + Ion,IGBT Ron,IGBT
Assume Vj ≈ 0.8 V ; Exact value not critical to an approximate estimate of Ion,IGBT. Ion,IGBT ≈
Wd 3!-!0.8 ; R ≈ Ron,IGBT! q!(mn!+!m p)!nb!A on,IGBT
7.5x10-3 = 2.6x10-3 W W d = (10-5)(750) = 75 mm ; Ron,IGBT = (1.6x10-19)(900)(1016)(2) 2.2 Ion,IGBT ≈ ≈ 850 amps 2.6x10-3 MOSFET current - Ion,MOS ; Von(MOS) = Ion,MOS Ron,MOS Von(MOS) ; Ron,MOS = Ion,MOS = R on,MOS Nd =
Wd q!mn!!Nd!A
; Wd = 75 mm
1.3x1017 = 1.7x1014 cm-3 750
7.5x10-3 Ron,MOS = = 0.09 ohms (1.6x10-19)(1.5x103)(1.7x1014)(2) 3 Ion,MOS = 0.09 = 33 amps 25-6.
Von(PT) = Vj,PT + Ion,PT Ron,PT = Vj,NPT + Ion,NPT Ron,NPT Ion,PT Ion,NPT
≈
Ron,NPT Ron,PT since Vj,NPT ≈ Vj,PT
BVFB =
1.3x1017 ≈ 1300 volts ; But Wdepl(1300 V) = (10-5)(1300) = 130 microns 14 10
130 microns > 25 micron drift region length. Hence forward blocking limited by punch-through. (1.6x10-19)(1014)(2.5x10-3)2 = 453 volts BVFB = (2x105)(2.5x10-3) (2)(11.7)(8.9x10-14) 25-5.
IGBT current - Ion,IGBT
; Von(IGBT) = Vj + Ion,IGBT Ron,IGBT
Assume Vj ≈ 0.8 V ; Exact value not critical to an approximate estimate of Ion,IGBT. Ion,IGBT ≈
Wd 3!-!0.8 ; R ≈ Ron,IGBT! q!(mn!+!m p)!nb!A on,IGBT
7.5x10-3 = 2.6x10-3 W W d = (10-5)(750) = 75 mm ; Ron,IGBT = (1.6x10-19)(900)(1016)(2) 2.2 Ion,IGBT ≈ ≈ 850 amps 2.6x10-3 MOSFET current - Ion,MOS ; Von(MOS) = Ion,MOS Ron,MOS Von(MOS) ; Ron,MOS = Ion,MOS = R on,MOS Nd =
Wd q!mn!!Nd!A
; Wd = 75 mm
1.3x1017 = 1.7x1014 cm-3 750
7.5x10-3 Ron,MOS = = 0.09 ohms (1.6x10-19)(1.5x103)(1.7x1014)(2) 3 Ion,MOS = 0.09 = 33 amps 25-6.
Von(PT) = Vj,PT + Ion,PT Ron,PT = Vj,NPT + Ion,NPT Ron,NPT Ion,PT Ion,NPT
≈
Ron,NPT Ron,PT since Vj,NPT ≈ Vj,PT
Ron,NPT W d,NPT = Ron,PT !Wd,PT ≈ 2 assuming doping level in PT drift region is much less than the doping level in the NPT drift region. Ion,PT Hence I on,NPT
25-7.
≈ 2
P Cv dT = dQ ; dQ = V dt ; P = power dissipated in IGBT during overcurrent transient. V = volume in IGBT where power is dissipated.Duration of transient = dt. P = Iov2 Ron ; Iov =
!V!Cv!dT dt!Ron ; V ≈ A Wdrift
W drift Ron = q!(m !+!m )!n !A ; Iov = n p b
Iov =
q!(mn!+!m p)!nb!A2!Cv!dT dt
(1.6x10-19)(900)(1016)(0.5)2(1.75)(100) ≈ 2.5x103 amps -5 (10 )
Estimate is overly optimistic because it ignores any other ohmic losses in the device such as channel resistance or resistance of the heavily doped source and drain diffusions. However IGBTs nominally rated at 100 A have repeatedly been tested at 1000 A for 10 microseconds or less and survived. 25-8.
25-9.
The IGBT has the smaller values of Cgd and Cgs because its effective cross-sectional area is smaller than that of the MOSFET. The IGBT has a smaller area even though its current rating is identical to the MOSFET's rating because the IGBT utilizes conductivity modulation of the drift region to significantly reduce the specific on-state resistance. dvDS Vd dvDS = !t ; trv < 0.75 microseconds Check dt at turn-off ; dt rv Vd dvDS dt > !toff
=
700 = 2330 V/ms > 800 V/ms limit. !3x10-7
Device is overstressed by an overly large
dvDS dt .
Ron,NPT W d,NPT = Ron,PT !Wd,PT ≈ 2 assuming doping level in PT drift region is much less than the doping level in the NPT drift region. Ion,PT Hence I on,NPT
25-7.
≈ 2
P Cv dT = dQ ; dQ = V dt ; P = power dissipated in IGBT during overcurrent transient. V = volume in IGBT where power is dissipated.Duration of transient = dt. P = Iov2 Ron ; Iov =
!V!Cv!dT dt!Ron ; V ≈ A Wdrift
W drift Ron = q!(m !+!m )!n !A ; Iov = n p b
Iov =
q!(mn!+!m p)!nb!A2!Cv!dT dt
(1.6x10-19)(900)(1016)(0.5)2(1.75)(100) ≈ 2.5x103 amps -5 (10 )
Estimate is overly optimistic because it ignores any other ohmic losses in the device such as channel resistance or resistance of the heavily doped source and drain diffusions. However IGBTs nominally rated at 100 A have repeatedly been tested at 1000 A for 10 microseconds or less and survived. 25-8.
25-9.
The IGBT has the smaller values of Cgd and Cgs because its effective cross-sectional area is smaller than that of the MOSFET. The IGBT has a smaller area even though its current rating is identical to the MOSFET's rating because the IGBT utilizes conductivity modulation of the drift region to significantly reduce the specific on-state resistance. dvDS Vd dvDS = !t ; trv < 0.75 microseconds Check dt at turn-off ; dt rv Vd dvDS dt > !toff
=
700 = 2330 V/ms > 800 V/ms limit. !3x10-7
Device is overstressed by an overly large
dvDS dt .
Ron,NPT W d,NPT = Ron,PT !Wd,PT ≈ 2 assuming doping level in PT drift region is much less than the doping level in the NPT drift region. Ion,PT Hence I on,NPT
25-7.
≈ 2
P Cv dT = dQ ; dQ = V dt ; P = power dissipated in IGBT during overcurrent transient. V = volume in IGBT where power is dissipated.Duration of transient = dt. P = Iov2 Ron ; Iov =
!V!Cv!dT dt!Ron ; V ≈ A Wdrift
W drift Ron = q!(m !+!m )!n !A ; Iov = n p b
Iov =
q!(mn!+!m p)!nb!A2!Cv!dT dt
(1.6x10-19)(900)(1016)(0.5)2(1.75)(100) ≈ 2.5x103 amps -5 (10 )
Estimate is overly optimistic because it ignores any other ohmic losses in the device such as channel resistance or resistance of the heavily doped source and drain diffusions. However IGBTs nominally rated at 100 A have repeatedly been tested at 1000 A for 10 microseconds or less and survived. 25-8.
25-9.
The IGBT has the smaller values of Cgd and Cgs because its effective cross-sectional area is smaller than that of the MOSFET. The IGBT has a smaller area even though its current rating is identical to the MOSFET's rating because the IGBT utilizes conductivity modulation of the drift region to significantly reduce the specific on-state resistance. dvDS Vd dvDS = !t ; trv < 0.75 microseconds Check dt at turn-off ; dt rv Vd dvDS dt > !toff
=
700 = 2330 V/ms > 800 V/ms limit. !3x10-7
Device is overstressed by an overly large
dvDS dt .
Ron,NPT W d,NPT = Ron,PT !Wd,PT ≈ 2 assuming doping level in PT drift region is much less than the doping level in the NPT drift region. Ion,PT Hence I on,NPT
25-7.
≈ 2
P Cv dT = dQ ; dQ = V dt ; P = power dissipated in IGBT during overcurrent transient. V = volume in IGBT where power is dissipated.Duration of transient = dt. P = Iov2 Ron ; Iov =
!V!Cv!dT dt!Ron ; V ≈ A Wdrift
W drift Ron = q!(m !+!m )!n !A ; Iov = n p b
Iov =
q!(mn!+!m p)!nb!A2!Cv!dT dt
(1.6x10-19)(900)(1016)(0.5)2(1.75)(100) ≈ 2.5x103 amps -5 (10 )
Estimate is overly optimistic because it ignores any other ohmic losses in the device such as channel resistance or resistance of the heavily doped source and drain diffusions. However IGBTs nominally rated at 100 A have repeatedly been tested at 1000 A for 10 microseconds or less and survived. 25-8.
25-9.
The IGBT has the smaller values of Cgd and Cgs because its effective cross-sectional area is smaller than that of the MOSFET. The IGBT has a smaller area even though its current rating is identical to the MOSFET's rating because the IGBT utilizes conductivity modulation of the drift region to significantly reduce the specific on-state resistance. dvDS Vd dvDS = !t ; trv < 0.75 microseconds Check dt at turn-off ; dt rv Vd dvDS dt > !toff
=
700 = 2330 V/ms > 800 V/ms limit. !3x10-7
Device is overstressed by an overly large
dvDS dt .
!tri!+!trv!+!!tfi!+!tfv Check switching losses Psw = Vd Io fs 2 (3x10-7!+!7.5x10-7) (2.5x104) = 875 W Psw = (700)(100) 2 Tj,max!-!Ta 150!-!25 Allowable power loss = = = 250 watts Rqja 0.5 Switching losses exceed allowable losses. Module is overstressed by too much power dissipation.
Chapter 26 Problem Solutions 26-1.
Equivalent circuit for JFET in active region.
+ v
C
GS
+
ro
C GD +
GS
-
Equivelent circuit for JFET in the blocking state.
v
mv GS
D 0
C GD
v
C
GS
-V
GS1
+
GS
-
v
DS
0
26-2.
-
Linearized I-V characteristics i
+
DS
V
DS1
V DS2
Drive circuit configuration V DD
R1 R
RL
MOSFET off V R2 VDS = VKG = - VGK = RDD! 1!+!R2 Negative enough to insure that the FCT is off.
2
MOSFET on + V drive -
VDS = VKG = - VGK = 0 and FCT is on.
-V
GS2
Chapter 26 Problem Solutions 26-1.
Equivalent circuit for JFET in active region.
+ v
C
GS
+
ro
C GD +
GS
-
Equivelent circuit for JFET in the blocking state.
v
mv GS
D 0
C GD
v
C
GS
-V
GS1
+
GS
-
v
DS
0
26-2.
-
Linearized I-V characteristics i
+
DS
V
DS1
V DS2
Drive circuit configuration V DD
R1 R
RL
MOSFET off V R2 VDS = VKG = - VGK = RDD! 1!+!R2 Negative enough to insure that the FCT is off.
2
MOSFET on + V drive -
VDS = VKG = - VGK = 0 and FCT is on.
-V
GS2
MOSFET characteirstics: - High current sinking capability - Low Ron ; Low BVDss
26-3.
Wd2 EBD!Wd Vdrift = (m !+!!m )!t ; BVBD = ; Vdrift,GaAs = Vdrift,Si 2 n p Wd2(Si) (mn!+!!mp)|Si!tSi =
Wd2(GaAs) (mn!+!!mp)|GaAs!tGaAs
(mn!+!!mp)|Si È EBD(Si) ˘ 2 tGaA Í ˙ tSi = (mn!+!!mp)|GaAs ÎE (GaAs)˚ BD (mn + mp)|Si = 2000 cm2/V-sec ; (m n + mp)|GaAs = 9000 cm2/V-sec EBD(Si) = 300 kV/cm tGaA tSi 26-4.
; EBD(GaAs) = 400 kV/cm
2 È3˘ 2 = 9 ÍÎ4˙˚ = 0.125 ; GaAs has the shorter lifetime.
EBD = 107 V/cm ; BVBD = EBD tox 103 = 10-4 cm = 1 micron tox = 7 10
26-5.
IA,max = (105)(1.5x10-2) = 1500 amperes
26-6.
P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7.
Assume an n-type drift region since m n > mp.
MOSFET characteirstics: - High current sinking capability - Low Ron ; Low BVDss
26-3.
Wd2 EBD!Wd Vdrift = (m !+!!m )!t ; BVBD = ; Vdrift,GaAs = Vdrift,Si 2 n p Wd2(Si) (mn!+!!mp)|Si!tSi =
Wd2(GaAs) (mn!+!!mp)|GaAs!tGaAs
(mn!+!!mp)|Si È EBD(Si) ˘ 2 tGaA Í ˙ tSi = (mn!+!!mp)|GaAs ÎE (GaAs)˚ BD (mn + mp)|Si = 2000 cm2/V-sec ; (m n + mp)|GaAs = 9000 cm2/V-sec EBD(Si) = 300 kV/cm tGaA tSi 26-4.
; EBD(GaAs) = 400 kV/cm
2 È3˘ 2 = 9 ÍÎ4˙˚ = 0.125 ; GaAs has the shorter lifetime.
EBD = 107 V/cm ; BVBD = EBD tox 103 = 10-4 cm = 1 micron tox = 7 10
26-5.
IA,max = (105)(1.5x10-2) = 1500 amperes
26-6.
P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7.
Assume an n-type drift region since m n > mp.
MOSFET characteirstics: - High current sinking capability - Low Ron ; Low BVDss
26-3.
Wd2 EBD!Wd Vdrift = (m !+!!m )!t ; BVBD = ; Vdrift,GaAs = Vdrift,Si 2 n p Wd2(Si) (mn!+!!mp)|Si!tSi =
Wd2(GaAs) (mn!+!!mp)|GaAs!tGaAs
(mn!+!!mp)|Si È EBD(Si) ˘ 2 tGaA Í ˙ tSi = (mn!+!!mp)|GaAs ÎE (GaAs)˚ BD (mn + mp)|Si = 2000 cm2/V-sec ; (m n + mp)|GaAs = 9000 cm2/V-sec EBD(Si) = 300 kV/cm tGaA tSi 26-4.
; EBD(GaAs) = 400 kV/cm
2 È3˘ 2 = 9 ÍÎ4˙˚ = 0.125 ; GaAs has the shorter lifetime.
EBD = 107 V/cm ; BVBD = EBD tox 103 = 10-4 cm = 1 micron tox = 7 10
26-5.
IA,max = (105)(1.5x10-2) = 1500 amperes
26-6.
P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7.
Assume an n-type drift region since m n > mp.
MOSFET characteirstics: - High current sinking capability - Low Ron ; Low BVDss
26-3.
Wd2 EBD!Wd Vdrift = (m !+!!m )!t ; BVBD = ; Vdrift,GaAs = Vdrift,Si 2 n p Wd2(Si) (mn!+!!mp)|Si!tSi =
Wd2(GaAs) (mn!+!!mp)|GaAs!tGaAs
(mn!+!!mp)|Si È EBD(Si) ˘ 2 tGaA Í ˙ tSi = (mn!+!!mp)|GaAs ÎE (GaAs)˚ BD (mn + mp)|Si = 2000 cm2/V-sec ; (m n + mp)|GaAs = 9000 cm2/V-sec EBD(Si) = 300 kV/cm tGaA tSi 26-4.
; EBD(GaAs) = 400 kV/cm
2 È3˘ 2 = 9 ÍÎ4˙˚ = 0.125 ; GaAs has the shorter lifetime.
EBD = 107 V/cm ; BVBD = EBD tox 103 = 10-4 cm = 1 micron tox = 7 10
26-5.
IA,max = (105)(1.5x10-2) = 1500 amperes
26-6.
P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7.
Assume an n-type drift region since m n > mp.
MOSFET characteirstics: - High current sinking capability - Low Ron ; Low BVDss
26-3.
Wd2 EBD!Wd Vdrift = (m !+!!m )!t ; BVBD = ; Vdrift,GaAs = Vdrift,Si 2 n p Wd2(Si) (mn!+!!mp)|Si!tSi =
Wd2(GaAs) (mn!+!!mp)|GaAs!tGaAs
(mn!+!!mp)|Si È EBD(Si) ˘ 2 tGaA Í ˙ tSi = (mn!+!!mp)|GaAs ÎE (GaAs)˚ BD (mn + mp)|Si = 2000 cm2/V-sec ; (m n + mp)|GaAs = 9000 cm2/V-sec EBD(Si) = 300 kV/cm tGaA tSi 26-4.
; EBD(GaAs) = 400 kV/cm
2 È3˘ 2 = 9 ÍÎ4˙˚ = 0.125 ; GaAs has the shorter lifetime.
EBD = 107 V/cm ; BVBD = EBD tox 103 = 10-4 cm = 1 micron tox = 7 10
26-5.
IA,max = (105)(1.5x10-2) = 1500 amperes
26-6.
P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7.
Assume an n-type drift region since m n > mp.
MOSFET characteirstics: - High current sinking capability - Low Ron ; Low BVDss
26-3.
Wd2 EBD!Wd Vdrift = (m !+!!m )!t ; BVBD = ; Vdrift,GaAs = Vdrift,Si 2 n p Wd2(Si) (mn!+!!mp)|Si!tSi =
Wd2(GaAs) (mn!+!!mp)|GaAs!tGaAs
(mn!+!!mp)|Si È EBD(Si) ˘ 2 tGaA Í ˙ tSi = (mn!+!!mp)|GaAs ÎE (GaAs)˚ BD (mn + mp)|Si = 2000 cm2/V-sec ; (m n + mp)|GaAs = 9000 cm2/V-sec EBD(Si) = 300 kV/cm tGaA tSi 26-4.
; EBD(GaAs) = 400 kV/cm
2 È3˘ 2 = 9 ÍÎ4˙˚ = 0.125 ; GaAs has the shorter lifetime.
EBD = 107 V/cm ; BVBD = EBD tox 103 = 10-4 cm = 1 micron tox = 7 10
26-5.
IA,max = (105)(1.5x10-2) = 1500 amperes
26-6.
P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7.
Assume an n-type drift region since m n > mp.
Wd Rdrift = q!m !N !A n d
Wd ; Rdrift,sp = Rdrift A = q!m !N n d
e!EBD2 Using Eq. (20-1) Nd = 2!q!BV BD
; Using Eq. (20-3) Wd =
2!!BVBD EBD
Substituting into the expression for Rdrift,sp yields 2!!BVBD 1 2!q!BVBD Rdrift,sp = E BD q!e e!EBD2
26-8.
!!!4!!(BVBD)2 = e!mn!(EBD)3
(4)(500)2 Silicon : Rdrift,sp = = 0.024 ohms-cm2 -14 5 3 (11.7)(1500)(8.9x10 )(3x10 ) GaAs: Rdrift,sp =
(4)(500)2 = 0.0016 ohms-cm2 (12.8)(8500)(8.9x10-14)(4x105)3
(4)(500)2 6H-SiC: Rdrift,sp = = 2.3x10-4 ohms-cm2 (10)(600)(8.9x10-14)(2x106)3 (4)(500)2 Diamond: Rdrift,sp = = 9.3x10-7 ohms-cm2 -14 7 3 (5.5)(2200)(8.9x10 )(10 ) 26-9.
26-10.
Diamond is the most suitable material for high temperature operation. It has the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier density at any given temperature. This statement presumes that the phase change listed for diamond in the table of material properties exceeds the sublimation temperature of SiC (1800 °C). e!EBD2 Eq. (20-1): Nd = 2!q!BV BD 5.7x1017 (12.8)(8.9x10-14)(4x105)2 = BV For GaAs: Nd = BD (2)(1.6x10-19)(BVBD) (10)(8.9x10-14)(2x106)2 1.1x1019 For 6H-SiC: Nd = = BV (2)(1.6x10-19)(BVBD) BD
Wd Rdrift = q!m !N !A n d
Wd ; Rdrift,sp = Rdrift A = q!m !N n d
e!EBD2 Using Eq. (20-1) Nd = 2!q!BV BD
; Using Eq. (20-3) Wd =
2!!BVBD EBD
Substituting into the expression for Rdrift,sp yields 2!!BVBD 1 2!q!BVBD Rdrift,sp = E BD q!e e!EBD2
26-8.
!!!4!!(BVBD)2 = e!mn!(EBD)3
(4)(500)2 Silicon : Rdrift,sp = = 0.024 ohms-cm2 -14 5 3 (11.7)(1500)(8.9x10 )(3x10 ) GaAs: Rdrift,sp =
(4)(500)2 = 0.0016 ohms-cm2 (12.8)(8500)(8.9x10-14)(4x105)3
(4)(500)2 6H-SiC: Rdrift,sp = = 2.3x10-4 ohms-cm2 (10)(600)(8.9x10-14)(2x106)3 (4)(500)2 Diamond: Rdrift,sp = = 9.3x10-7 ohms-cm2 -14 7 3 (5.5)(2200)(8.9x10 )(10 ) 26-9.
26-10.
Diamond is the most suitable material for high temperature operation. It has the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier density at any given temperature. This statement presumes that the phase change listed for diamond in the table of material properties exceeds the sublimation temperature of SiC (1800 °C). e!EBD2 Eq. (20-1): Nd = 2!q!BV BD 5.7x1017 (12.8)(8.9x10-14)(4x105)2 = BV For GaAs: Nd = BD (2)(1.6x10-19)(BVBD) (10)(8.9x10-14)(2x106)2 1.1x1019 For 6H-SiC: Nd = = BV (2)(1.6x10-19)(BVBD) BD
Wd Rdrift = q!m !N !A n d
Wd ; Rdrift,sp = Rdrift A = q!m !N n d
e!EBD2 Using Eq. (20-1) Nd = 2!q!BV BD
; Using Eq. (20-3) Wd =
2!!BVBD EBD
Substituting into the expression for Rdrift,sp yields 2!!BVBD 1 2!q!BVBD Rdrift,sp = E BD q!e e!EBD2
26-8.
!!!4!!(BVBD)2 = e!mn!(EBD)3
(4)(500)2 Silicon : Rdrift,sp = = 0.024 ohms-cm2 -14 5 3 (11.7)(1500)(8.9x10 )(3x10 ) GaAs: Rdrift,sp =
(4)(500)2 = 0.0016 ohms-cm2 (12.8)(8500)(8.9x10-14)(4x105)3
(4)(500)2 6H-SiC: Rdrift,sp = = 2.3x10-4 ohms-cm2 (10)(600)(8.9x10-14)(2x106)3 (4)(500)2 Diamond: Rdrift,sp = = 9.3x10-7 ohms-cm2 -14 7 3 (5.5)(2200)(8.9x10 )(10 ) 26-9.
26-10.
Diamond is the most suitable material for high temperature operation. It has the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier density at any given temperature. This statement presumes that the phase change listed for diamond in the table of material properties exceeds the sublimation temperature of SiC (1800 °C). e!EBD2 Eq. (20-1): Nd = 2!q!BV BD 5.7x1017 (12.8)(8.9x10-14)(4x105)2 = BV For GaAs: Nd = BD (2)(1.6x10-19)(BVBD) (10)(8.9x10-14)(2x106)2 1.1x1019 For 6H-SiC: Nd = = BV (2)(1.6x10-19)(BVBD) BD
Wd Rdrift = q!m !N !A n d
Wd ; Rdrift,sp = Rdrift A = q!m !N n d
e!EBD2 Using Eq. (20-1) Nd = 2!q!BV BD
; Using Eq. (20-3) Wd =
2!!BVBD EBD
Substituting into the expression for Rdrift,sp yields 2!!BVBD 1 2!q!BVBD Rdrift,sp = E BD q!e e!EBD2
26-8.
!!!4!!(BVBD)2 = e!mn!(EBD)3
(4)(500)2 Silicon : Rdrift,sp = = 0.024 ohms-cm2 -14 5 3 (11.7)(1500)(8.9x10 )(3x10 ) GaAs: Rdrift,sp =
(4)(500)2 = 0.0016 ohms-cm2 (12.8)(8500)(8.9x10-14)(4x105)3
(4)(500)2 6H-SiC: Rdrift,sp = = 2.3x10-4 ohms-cm2 (10)(600)(8.9x10-14)(2x106)3 (4)(500)2 Diamond: Rdrift,sp = = 9.3x10-7 ohms-cm2 -14 7 3 (5.5)(2200)(8.9x10 )(10 ) 26-9.
26-10.
Diamond is the most suitable material for high temperature operation. It has the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier density at any given temperature. This statement presumes that the phase change listed for diamond in the table of material properties exceeds the sublimation temperature of SiC (1800 °C). e!EBD2 Eq. (20-1): Nd = 2!q!BV BD 5.7x1017 (12.8)(8.9x10-14)(4x105)2 = BV For GaAs: Nd = BD (2)(1.6x10-19)(BVBD) (10)(8.9x10-14)(2x106)2 1.1x1019 For 6H-SiC: Nd = = BV (2)(1.6x10-19)(BVBD) BD
For diamond: Nd =
(5.5)(8.9x10-14)(107)2 1.5x1020 = BVBD (2)(1.6x10-19)(BVBD)
Eq. (20-3): Wd =
2!!BVBD EBD
For GaAs: Wd =
2!!BVBD = 5x10-6 BVBD [cm] 4x105
2!!BVBD = 10-6 BVBD For 6H-SiC: Wd = 6 2x10
[cm]
2!!BVBD = 2x10-7 BVBD For diamond: Wd = 7 10 26-11.
26-12.
[cm]
Use equations from problem 26-10. Material
Nd
Wd
GaAs
2.9x1015 cm-3
10-2 cm
6H-SiC
5.5x1016 cm-3
2x10-3 cm
Diamond
7.5x1017 cm-3
4x10-5 cm
Tj = Rqjc Pdiode + Tcase : Rqjc = C.• (k) -1 k = thermal conductivity and C = constant (Tj!!-!!Tcase)!k Using silicon diode data: C = Pdiode
=
(150!-!50)(!1.5) = 0.75 cm-1 200
0.75 0.75 Rqjc(GaAs) = 0.5 = 1.5 °C/W : R qjc(SiC) = 5 = 0.15°C/W 0.75 Rqjc(diamond) = 20 = 0.038°C/W Tj(GaAs) = (1.5)(200) + 50 = 350 °C : Tj(SiC) = (0.15)(200) + 50 = 80 °C
For diamond: Nd =
(5.5)(8.9x10-14)(107)2 1.5x1020 = BVBD (2)(1.6x10-19)(BVBD)
Eq. (20-3): Wd =
2!!BVBD EBD
For GaAs: Wd =
2!!BVBD = 5x10-6 BVBD [cm] 4x105
2!!BVBD = 10-6 BVBD For 6H-SiC: Wd = 6 2x10
[cm]
2!!BVBD = 2x10-7 BVBD For diamond: Wd = 7 10 26-11.
26-12.
[cm]
Use equations from problem 26-10. Material
Nd
Wd
GaAs
2.9x1015 cm-3
10-2 cm
6H-SiC
5.5x1016 cm-3
2x10-3 cm
Diamond
7.5x1017 cm-3
4x10-5 cm
Tj = Rqjc Pdiode + Tcase : Rqjc = C.• (k) -1 k = thermal conductivity and C = constant (Tj!!-!!Tcase)!k Using silicon diode data: C = Pdiode
=
(150!-!50)(!1.5) = 0.75 cm-1 200
0.75 0.75 Rqjc(GaAs) = 0.5 = 1.5 °C/W : R qjc(SiC) = 5 = 0.15°C/W 0.75 Rqjc(diamond) = 20 = 0.038°C/W Tj(GaAs) = (1.5)(200) + 50 = 350 °C : Tj(SiC) = (0.15)(200) + 50 = 80 °C
For diamond: Nd =
(5.5)(8.9x10-14)(107)2 1.5x1020 = BVBD (2)(1.6x10-19)(BVBD)
Eq. (20-3): Wd =
2!!BVBD EBD
For GaAs: Wd =
2!!BVBD = 5x10-6 BVBD [cm] 4x105
2!!BVBD = 10-6 BVBD For 6H-SiC: Wd = 6 2x10
[cm]
2!!BVBD = 2x10-7 BVBD For diamond: Wd = 7 10 26-11.
26-12.
[cm]
Use equations from problem 26-10. Material
Nd
Wd
GaAs
2.9x1015 cm-3
10-2 cm
6H-SiC
5.5x1016 cm-3
2x10-3 cm
Diamond
7.5x1017 cm-3
4x10-5 cm
Tj = Rqjc Pdiode + Tcase : Rqjc = C.• (k) -1 k = thermal conductivity and C = constant (Tj!!-!!Tcase)!k Using silicon diode data: C = Pdiode
=
(150!-!50)(!1.5) = 0.75 cm-1 200
0.75 0.75 Rqjc(GaAs) = 0.5 = 1.5 °C/W : R qjc(SiC) = 5 = 0.15°C/W 0.75 Rqjc(diamond) = 20 = 0.038°C/W Tj(GaAs) = (1.5)(200) + 50 = 350 °C : Tj(SiC) = (0.15)(200) + 50 = 80 °C
Tj(diamond) = (0.038)(200) + 50 = 57.4 °C
26-13.
Pinch-off of the channel occurs when the depletion region of the gate-channel (P+N-) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-source voltage of -Vp. The other half of the channel is depleted by the depletion region from the gate-channel junction on the other side of the channel. See Figs. 26-1 and 26-3. W 2 = Wo
Vp 1!+! f c
;
ÈNa!Nd˘ kT Í ˙ fc = q ln Í 2 ˙ Î ni ˚
Solving for Vp yields Vp =
;
Wo =
2!e!fc q!Nd
È W ˘ 2 fc ÍÎ2!W ˙˚ - fc o
È(1019)(2x1014)˘ ˙ = 0.8 V fc = 0.026 ln ÍÎ ˚ 1010
Wo =
(2)(11.7)(8.9x10-14)(0.8) (1.6x10-19)(2x1014)
= 2.3 microns
È 10 ˘ 2 Vp = (0.8) ÍÎ(2)(2.3)˙˚ - 0.8 = 3.8 - 0.8 = 3 V 26-14.
Single cell of the multi-cell JFET shown below. See Fig. 26-1 for a fuller picture of the multi-cell nature of the JFET. The diagram on the left indicates the various contributions to the on-state resistance and the figure on the right shows the various geometrical factors that determine the resistance. Each cell is d centimeters deep in the direction perpendicular to the plane (page) of the diagram. The gate-source voltage is set at zero.
Tj(diamond) = (0.038)(200) + 50 = 57.4 °C
26-13.
Pinch-off of the channel occurs when the depletion region of the gate-channel (P+N-) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-source voltage of -Vp. The other half of the channel is depleted by the depletion region from the gate-channel junction on the other side of the channel. See Figs. 26-1 and 26-3. W 2 = Wo
Vp 1!+! f c
;
ÈNa!Nd˘ kT Í ˙ fc = q ln Í 2 ˙ Î ni ˚
Solving for Vp yields Vp =
;
Wo =
2!e!fc q!Nd
È W ˘ 2 fc ÍÎ2!W ˙˚ - fc o
È(1019)(2x1014)˘ ˙ = 0.8 V fc = 0.026 ln ÍÎ ˚ 1010
Wo =
(2)(11.7)(8.9x10-14)(0.8) (1.6x10-19)(2x1014)
= 2.3 microns
È 10 ˘ 2 Vp = (0.8) ÍÎ(2)(2.3)˙˚ - 0.8 = 3.8 - 0.8 = 3 V 26-14.
Single cell of the multi-cell JFET shown below. See Fig. 26-1 for a fuller picture of the multi-cell nature of the JFET. The diagram on the left indicates the various contributions to the on-state resistance and the figure on the right shows the various geometrical factors that determine the resistance. Each cell is d centimeters deep in the direction perpendicular to the plane (page) of the diagram. The gate-source voltage is set at zero.
Tj(diamond) = (0.038)(200) + 50 = 57.4 °C
26-13.
Pinch-off of the channel occurs when the depletion region of the gate-channel (P+N-) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-source voltage of -Vp. The other half of the channel is depleted by the depletion region from the gate-channel junction on the other side of the channel. See Figs. 26-1 and 26-3. W 2 = Wo
Vp 1!+! f c
;
ÈNa!Nd˘ kT Í ˙ fc = q ln Í 2 ˙ Î ni ˚
Solving for Vp yields Vp =
;
Wo =
2!e!fc q!Nd
È W ˘ 2 fc ÍÎ2!W ˙˚ - fc o
È(1019)(2x1014)˘ ˙ = 0.8 V fc = 0.026 ln ÍÎ ˚ 1010
Wo =
(2)(11.7)(8.9x10-14)(0.8) (1.6x10-19)(2x1014)
= 2.3 microns
È 10 ˘ 2 Vp = (0.8) ÍÎ(2)(2.3)˙˚ - 0.8 = 3.8 - 0.8 = 3 V 26-14.
Single cell of the multi-cell JFET shown below. See Fig. 26-1 for a fuller picture of the multi-cell nature of the JFET. The diagram on the left indicates the various contributions to the on-state resistance and the figure on the right shows the various geometrical factors that determine the resistance. Each cell is d centimeters deep in the direction perpendicular to the plane (page) of the diagram. The gate-source voltage is set at zero.
source
Rs P
+
W g
P
Rc R
+
P
+
W
W - 2W o Wo + Wg /2
t
R
l gs
P
+
l gd - W o - W g/2
W + Wg
d
Rs
drain lgs 10-3 = q!m !N !d!W = = 298 ohms n d (1.6x10-19)(1500)(2x1014)(0.07)(10-3)
Rc
lc = q!m !N !d!(W!-!2W ) n d o 10-3 = 552 ohms = (1.6x10-19)(1500)(2x1014)(0.07)(10-3!-!4.6x10-4)
Rt estimate. Treat the region of thickness Wo + Wg/2 as though it has an average width (W!-!2!Wo)!+!(W!+!Wg) given by = W + Wg/2 - Wo. Rt now approximately given by 2 Rt Rt
Wo!+!Wg/2 = q!m !N !d!(W!+!W /2!-!W ) n d g o (10-3!+!5x10-4) = 351 ohms = (1.6x10-19)(1500)(2x1014)(0.07)(10-3!+!5x10-4!-!2.3x10-4)
(lgd!-!Wo!-Wg/2) Rd = q!m !N !d!(W!+!W ) = n d g -4! (35x10 -!2.3x10-4!-!5x10-4) = 412 ohms Rd = (1.6x10-19)(1500)(2x1014)(0.07)(10-3!+10-3) Total resistance of a single cell is Rcell = Rs + Rc + Rt + Rd
lc
Rcell = 298 + 552 + 351 + 412 = 1613 ohms There are N = 28 cells in parallel so the the net on-state resistance is Rcell 1613 = 28 = 58 ohms Ron = N 26-15.
As the drain-source voltage increases, the reverse-bias on the gate-drain pn junction increases. The depletion region of the two adjacent P+ regions merge and then grow towards the drain. The drift region of length lgd and doping Nd must contain this depletion region and will determine the breakdown voltage. The short length of the drift region suggests that punch-through will limit the breakdown voltage. Check for this possibility first. Non-punch-through estimate: 1.3x1017 = 650 V ; Wd > (10-5)(6.5x102) = 65 microns > 35 microns BV = 14 2x10 Hence this is a punch-through structure. Use Eq. (21-21) and EBD = 2x105 V/cm (1.6x10-19)(2x1014)(3.5x10-3)2 5 -3 BV = (2x10 )(3.5x10 ) (2)(11.7)(8.9x10-14) BV = 511 V
= 700 - 189
Rcell = 298 + 552 + 351 + 412 = 1613 ohms There are N = 28 cells in parallel so the the net on-state resistance is Rcell 1613 = 28 = 58 ohms Ron = N 26-15.
As the drain-source voltage increases, the reverse-bias on the gate-drain pn junction increases. The depletion region of the two adjacent P+ regions merge and then grow towards the drain. The drift region of length lgd and doping Nd must contain this depletion region and will determine the breakdown voltage. The short length of the drift region suggests that punch-through will limit the breakdown voltage. Check for this possibility first. Non-punch-through estimate: 1.3x1017 = 650 V ; Wd > (10-5)(6.5x102) = 65 microns > 35 microns BV = 14 2x10 Hence this is a punch-through structure. Use Eq. (21-21) and EBD = 2x105 V/cm (1.6x10-19)(2x1014)(3.5x10-3)2 5 -3 BV = (2x10 )(3.5x10 ) (2)(11.7)(8.9x10-14) BV = 511 V
= 700 - 189
Chapter 27 Problem Solutions 27-1.
a.
During turn-off of the GTO, Io communtates linearly to Cs. dvC dvC dvAK Io!t t Cs dt = Io t ; dt = = C !t < 5x107 V/s dt fi s fi Maximum
Cs >
dvAK occurs at tfi. Solving for Cs yields dt
ÈÍ dvAK˘˙ -1 Io Î! dt ˚ =
500 5x107
= 10 microfarads
Rs chosen on basis of limiting discharge current from Cs to safe level when GTO turns on. ICs,max = IAM - Io - Irr . Assume irr = 0.2 Io. Then ICs,max = 1000 - 500 -100 = 400 A !500 Rs = !400 ≈ 1.3 ohms Snubber recovery time = 2.3 Rs Cs = (2.3)(10-5)(1.3) = 30 microseconds.
b.
fsw!Cs!Vd2 Power dissipated in snubber PRs ≈ 2 PRs = (0.5)(103)(10-5)(500)2 = 1.25 kW
27-2.
diA 500 L s dt max = Vd ; Ls ≈ ≈ 1.7 microhenries. 3x108 Voltage across GTO at turn-off = Vd + Io Rs : Assume Io Rs = 0.2 Vd Rs =
27-3.
(0.2)(500) = 0.2 ohms. (500)
vCs(t) = Vd - Vd cos(wot) + Vd
Cbase Cs sin(wot) = Vd + K sin(wot - f)
vCs,max = Vd + K ; K sin(wot - f) = K sin(wot) cos(f) - Kcos(wot) sin(f)
Chapter 27 Problem Solutions 27-1.
a.
During turn-off of the GTO, Io communtates linearly to Cs. dvC dvC dvAK Io!t t Cs dt = Io t ; dt = = C !t < 5x107 V/s dt fi s fi Maximum
Cs >
dvAK occurs at tfi. Solving for Cs yields dt
ÈÍ dvAK˘˙ -1 Io Î! dt ˚ =
500 5x107
= 10 microfarads
Rs chosen on basis of limiting discharge current from Cs to safe level when GTO turns on. ICs,max = IAM - Io - Irr . Assume irr = 0.2 Io. Then ICs,max = 1000 - 500 -100 = 400 A !500 Rs = !400 ≈ 1.3 ohms Snubber recovery time = 2.3 Rs Cs = (2.3)(10-5)(1.3) = 30 microseconds.
b.
fsw!Cs!Vd2 Power dissipated in snubber PRs ≈ 2 PRs = (0.5)(103)(10-5)(500)2 = 1.25 kW
27-2.
diA 500 L s dt max = Vd ; Ls ≈ ≈ 1.7 microhenries. 3x108 Voltage across GTO at turn-off = Vd + Io Rs : Assume Io Rs = 0.2 Vd Rs =
27-3.
(0.2)(500) = 0.2 ohms. (500)
vCs(t) = Vd - Vd cos(wot) + Vd
Cbase Cs sin(wot) = Vd + K sin(wot - f)
vCs,max = Vd + K ; K sin(wot - f) = K sin(wot) cos(f) - Kcos(wot) sin(f)
Chapter 27 Problem Solutions 27-1.
a.
During turn-off of the GTO, Io communtates linearly to Cs. dvC dvC dvAK Io!t t Cs dt = Io t ; dt = = C !t < 5x107 V/s dt fi s fi Maximum
Cs >
dvAK occurs at tfi. Solving for Cs yields dt
ÈÍ dvAK˘˙ -1 Io Î! dt ˚ =
500 5x107
= 10 microfarads
Rs chosen on basis of limiting discharge current from Cs to safe level when GTO turns on. ICs,max = IAM - Io - Irr . Assume irr = 0.2 Io. Then ICs,max = 1000 - 500 -100 = 400 A !500 Rs = !400 ≈ 1.3 ohms Snubber recovery time = 2.3 Rs Cs = (2.3)(10-5)(1.3) = 30 microseconds.
b.
fsw!Cs!Vd2 Power dissipated in snubber PRs ≈ 2 PRs = (0.5)(103)(10-5)(500)2 = 1.25 kW
27-2.
diA 500 L s dt max = Vd ; Ls ≈ ≈ 1.7 microhenries. 3x108 Voltage across GTO at turn-off = Vd + Io Rs : Assume Io Rs = 0.2 Vd Rs =
27-3.
(0.2)(500) = 0.2 ohms. (500)
vCs(t) = Vd - Vd cos(wot) + Vd
Cbase Cs sin(wot) = Vd + K sin(wot - f)
vCs,max = Vd + K ; K sin(wot - f) = K sin(wot) cos(f) - Kcos(wot) sin(f)
Cbase Cs sin(wot) - Vd cos(wot)
K sin(wot) cos(f) - K cos(wot) sin(f) = Vd
K cos(f) = Vd
Cbase Cs
and K sin(f) = Vd ;
C 2 + K!sin(f) 2 = K2 = V 2 base + V 2 K!cos(f) [ ] [ ] d Cs d
vCs,max = Vd + K = Vd + Vd 27-4.
a.
Cbase 1!+! C s
Equivalent circuit after diode reverse recovery. L = 10 mH + 200 V -
i
L
Rs Cs
diR + iL(0 ) = Irr ; During reverse recovery L dt = 200 V diR Irr 200 7 7 -7 dt = trr = 10-5 = 2x10 A/sec ; Irr = (2x10 )(3x10 ) = 6 A
b.
vCs,max = 500 V = 200 + 200
Cbase 1!+! C s
Cbase Cbase 1!+! C = 1.5 ; C = 1.5 ≈ 1.25 s s È 62 ˘ ˙ = 9 nF Cbase = (10-5) ÍÎ 2 (200) ˚ 9!nF Cs = 1.25 ≈ 7 nF
Cbase Cs sin(wot) - Vd cos(wot)
K sin(wot) cos(f) - K cos(wot) sin(f) = Vd
K cos(f) = Vd
Cbase Cs
and K sin(f) = Vd ;
C 2 + K!sin(f) 2 = K2 = V 2 base + V 2 K!cos(f) [ ] [ ] d Cs d
vCs,max = Vd + K = Vd + Vd 27-4.
a.
Cbase 1!+! C s
Equivalent circuit after diode reverse recovery. L = 10 mH + 200 V -
i
L
Rs Cs
diR + iL(0 ) = Irr ; During reverse recovery L dt = 200 V diR Irr 200 7 7 -7 dt = trr = 10-5 = 2x10 A/sec ; Irr = (2x10 )(3x10 ) = 6 A
b.
vCs,max = 500 V = 200 + 200
Cbase 1!+! C s
Cbase Cbase 1!+! C = 1.5 ; C = 1.5 ≈ 1.25 s s È 62 ˘ ˙ = 9 nF Cbase = (10-5) ÍÎ 2 (200) ˚ 9!nF Cs = 1.25 ≈ 7 nF
27-5.
Use the circuit shown in problem 27-4. È 62 ˘ ˙ = 9 nF Cs = Cbase = (10-5) ÍÎ 2 (200) ˚ È200˘ Rs = 1.3 Rbase = (1.3) ÍÎ 6 ˙˚ = 43 ohms vCs,max = (1.5)(200) = 300 V
27-6.
ÈL !I 2!+!C !V 2˘ Í s rr s in ˙ ˚ f P = WR fsw = Î 2 sw WR = (0.5)(10-5)(6)2 + (0.5)(9x10-9)(200)2 = 3.6x10-4 Joules P = WR fsw = (3.6x10-4 )(2x104) = 7.2 watts
27-7.
a.
BJT waveforms (trv assumed to be zero for Cs = 0) i
Cs
Io 0 v
t
tfi
CE Cs = 0 Cs > 0 0
V d t
Vd!Io!tfi Power dissipation for Cs = 0 is Pc = fsw 2
27-5.
Use the circuit shown in problem 27-4. È 62 ˘ ˙ = 9 nF Cs = Cbase = (10-5) ÍÎ 2 (200) ˚ È200˘ Rs = 1.3 Rbase = (1.3) ÍÎ 6 ˙˚ = 43 ohms vCs,max = (1.5)(200) = 300 V
27-6.
ÈL !I 2!+!C !V 2˘ Í s rr s in ˙ ˚ f P = WR fsw = Î 2 sw WR = (0.5)(10-5)(6)2 + (0.5)(9x10-9)(200)2 = 3.6x10-4 Joules P = WR fsw = (3.6x10-4 )(2x104) = 7.2 watts
27-7.
a.
BJT waveforms (trv assumed to be zero for Cs = 0) i
Cs
Io 0 v
t
tfi
CE Cs = 0 Cs > 0 0
V d t
Vd!Io!tfi Power dissipation for Cs = 0 is Pc = fsw 2
27-5.
Use the circuit shown in problem 27-4. È 62 ˘ ˙ = 9 nF Cs = Cbase = (10-5) ÍÎ 2 (200) ˚ È200˘ Rs = 1.3 Rbase = (1.3) ÍÎ 6 ˙˚ = 43 ohms vCs,max = (1.5)(200) = 300 V
27-6.
ÈL !I 2!+!C !V 2˘ Í s rr s in ˙ ˚ f P = WR fsw = Î 2 sw WR = (0.5)(10-5)(6)2 + (0.5)(9x10-9)(200)2 = 3.6x10-4 Joules P = WR fsw = (3.6x10-4 )(2x104) = 7.2 watts
27-7.
a.
BJT waveforms (trv assumed to be zero for Cs = 0) i
Cs
Io 0 v
t
tfi
CE Cs = 0 Cs > 0 0
V d t
Vd!Io!tfi Power dissipation for Cs = 0 is Pc = fsw 2
Pc =
(200)(25)(4x10-7) (2x104) = 20 W 2
Power dissipation for Cs = Cs1 tfi Û Io t2 Vd!Io!tfi t Ù Pc = Wc fsw ; Wc = ıIo(1!-!t )!!2!C !t !!d!t = 12 s1 fi fi 0 (200)(25)(4x10-7)(2x104) = 3.3 watts Pc = 12 Factor of six reduction in the turn-off losses. b.
BJT losses increase at turn-on only becaue of energy stored in Cs being dissipated in the BJT, but also because the time to complete turn-on is extended as shown in Fig. 27-14a. This extended duration of traversal of the active region also increases the turn-on losses. During the turn-on interval, the collector-emitter voltage is given by (assuming that the external circuit dominates the transient) diC diC t2 dvCE Cs1 dt = - dt t - Irr ; vCE(t) = Vd - dt 2!C s1
t - Irr C s1
Seting the expression for vCE(t) equal to zero and solving for the time DT = t2 - (tri + trr) (see Fig. 27-14a) required for vCE to reach zero yields Irr DT = - di /dt + C
È Irr ˘2 2!Vd!Cs1 Í!! ˙ Î diC/dt˚ !+! diC/dt !
Note that DT = 0 if Cs1 = 0 which is consistent with the assumption that the external circuit and not the BJT that dominates the turn-on transient. Extra energy disspated in the BJT at turn-on due to Cs1 is thus ÈVd diC [Io!+!Irr]!Irr˘ DT 2 Í ! !!!-! ˙ Û v (t)!i (t)!dt = V I DT + ı CE C 2!Cs1 ˚ DT d rr Î 2 dt 0 (Io!+!3!Irr) diC ÈÍdiC˘˙ 2 DT4 3 2!Cs1 dt DT - Î dt ˚ 8!Cs1
ÈDT ˘ Í Û The increase in the BJT loss is Í ıvCE(t)!iC(t)!dt˙˙ fs where fs is the switching Î0 ˚ frequency. Numerical evaluation of DT gives DT = 0.29 ms (I rr = 10 A and Cs1 = 25 nF). Evaluation of the loss gives 1.3x10-3 fs = 26.5 watts 27-8.
a. Ds shorts out the snubber resistance during the BJT turn-off. Hence Cs1 is directly across the BJT as in problem 27-7a. Thus the loss reduction is the same as in problem 27-7a. b. Equivalent circuit during BJT turn-off after free-wheeling diode reverse recovery is shown below. Rs + v CE -
I r r+
di C dt
t
Cs
+
v C -
v (0+ ) = V C d
dvC diC dvC vCE = CsRs dt + vC ;; Cs dt = - Irr - dt t Combining equations and solving for vCE(t) yields diC t2 È Irr diC˘ vCE(t) = Vd - Irr Rs - ÍÎ!C !+!Rs! dt ˙˚ t - dt 2!C s s At t = D T, vCE = 0 and turn-on is completed.
diC Irr!+!Rs!Cs!! dt DT= + diC ! dt !
in
diC˘ È ÍIrr!+!Rs!Cs!! dt ˙2 2!Cs diC Í ˙ !+!! diC ![Vd!-!Irr!Rs] Î ! dt ! ˚ dt
Vd D T goes to zero when Rs = I . hence there is no increase in power dissipation rr the BJT due to the presence of Cs.
ÈDT ˘ Í Û The increase in the BJT loss is Í ıvCE(t)!iC(t)!dt˙˙ fs where fs is the switching Î0 ˚ frequency. Numerical evaluation of DT gives DT = 0.29 ms (I rr = 10 A and Cs1 = 25 nF). Evaluation of the loss gives 1.3x10-3 fs = 26.5 watts 27-8.
a. Ds shorts out the snubber resistance during the BJT turn-off. Hence Cs1 is directly across the BJT as in problem 27-7a. Thus the loss reduction is the same as in problem 27-7a. b. Equivalent circuit during BJT turn-off after free-wheeling diode reverse recovery is shown below. Rs + v CE -
I r r+
di C dt
t
Cs
+
v C -
v (0+ ) = V C d
dvC diC dvC vCE = CsRs dt + vC ;; Cs dt = - Irr - dt t Combining equations and solving for vCE(t) yields diC t2 È Irr diC˘ vCE(t) = Vd - Irr Rs - ÍÎ!C !+!Rs! dt ˙˚ t - dt 2!C s s At t = D T, vCE = 0 and turn-on is completed.
diC Irr!+!Rs!Cs!! dt DT= + diC ! dt !
in
diC˘ È ÍIrr!+!Rs!Cs!! dt ˙2 2!Cs diC Í ˙ !+!! diC ![Vd!-!Irr!Rs] Î ! dt ! ˚ dt
Vd D T goes to zero when Rs = I . hence there is no increase in power dissipation rr the BJT due to the presence of Cs.
27-9.
a.
Proposed snubber circuit configuration shown below. Ls
2 Vs
1
Rs
4 Io
3
Cs
2
Equivalent circuit swith SCRs 3 & 4 on and 1&2 going off or vice-versa. Continuous flow of load current formces SCRs 1 & 2 to remain on past the time of natural commutation (when vs(t) goes through zero and becomes negative). 3 or 4
Ls Irr 2 Vs
Rs 1 or 2 Cs
With 3 & 4 on, 1 & 2 are off, and effectively in parallel with the R s-Cs snubber. same is true when 1 & 2 are on and 3 & 4 are off. Thus the Rs-Cs snubber functions as a turn-off snubber. 0.05!Vs ; worst case situation (maximum reverse voltage across SCR Ia1 which is turning off) occurs when SCR which is turning on is triggered with a di delay angle of 90°. During reverse recovery of SCR1, L s dt = 2 Vs and b.
w Ls =
Irr di = dt trr . Solving for Irr yields È Irr ˘ 2 2!w!trr!Ia1 Í ˙ Irr = ; C = L ! Í !0.05 base s Î ! 2!V ˙˚ = s
!w!!Ia1!trr2 !0.05!Vs
Smallest overvoltage occurs when Cs = Cbase. Putting in numerical values Cs = 0.9933 Ia1 mF !0.05!Vs 2!Vs = w!t !I Rbase = I rr a1 rr Rs,opt = 1.3 Rbase = c.
4000 Ia1
=
(0.05)!(230) 3050 = I -5 (377)!(10 )!Ia1 a1
ohms
ohms
Peak line voltage = 2 (230) = 322 V Smallest overvoltage = (1.5)(322) = 483 V which occurs when Cs = Cbase and Rs = 1.3 Rbase. For Ia1 = 100 A 4000 Rs,opt = 1.3 Rbase = 100 = 40 ohms ; Cs = (0.0033)(100) = 0.33 mF
27-10.
The resistor in the BJT/MOSFET snubber must be shorted out during the device turn-off so that the snubber capacitance is in parallel with the device. The uncharged capacitor delays the build-up of the large Vd voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forward biased during turn-off, thus providing the shorting of the snubber resistor as required. The turn-off of the thyristor limits overvoltages arising from the interruption of current through the stray inductance in series with the thyristor. Lowest overvoltages are obtained when Rs = 1.3 Rbase and Cs = Cbase. Overvoltages are 70-% larger if Rs is zero. Hence a diode in parallel with the resistor is not desirable.
27-11.
a.
dvDS Vd dvDS ; trv < 0.75 microseconds Check dt at turn-off; dt = t rv dvDS 700 dt > 7.5x10-7 = 930 V/ms > 800 V/ms limit so snubber is needed. Not enough information available to check on power dissipation or overcurrents.
Irr di = dt trr . Solving for Irr yields È Irr ˘ 2 2!w!trr!Ia1 Í ˙ Irr = ; C = L ! Í !0.05 base s Î ! 2!V ˙˚ = s
!w!!Ia1!trr2 !0.05!Vs
Smallest overvoltage occurs when Cs = Cbase. Putting in numerical values Cs = 0.9933 Ia1 mF !0.05!Vs 2!Vs = w!t !I Rbase = I rr a1 rr Rs,opt = 1.3 Rbase = c.
4000 Ia1
=
(0.05)!(230) 3050 = I -5 (377)!(10 )!Ia1 a1
ohms
ohms
Peak line voltage = 2 (230) = 322 V Smallest overvoltage = (1.5)(322) = 483 V which occurs when Cs = Cbase and Rs = 1.3 Rbase. For Ia1 = 100 A 4000 Rs,opt = 1.3 Rbase = 100 = 40 ohms ; Cs = (0.0033)(100) = 0.33 mF
27-10.
The resistor in the BJT/MOSFET snubber must be shorted out during the device turn-off so that the snubber capacitance is in parallel with the device. The uncharged capacitor delays the build-up of the large Vd voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forward biased during turn-off, thus providing the shorting of the snubber resistor as required. The turn-off of the thyristor limits overvoltages arising from the interruption of current through the stray inductance in series with the thyristor. Lowest overvoltages are obtained when Rs = 1.3 Rbase and Cs = Cbase. Overvoltages are 70-% larger if Rs is zero. Hence a diode in parallel with the resistor is not desirable.
27-11.
a.
dvDS Vd dvDS ; trv < 0.75 microseconds Check dt at turn-off; dt = t rv dvDS 700 dt > 7.5x10-7 = 930 V/ms > 800 V/ms limit so snubber is needed. Not enough information available to check on power dissipation or overcurrents.
Irr di = dt trr . Solving for Irr yields È Irr ˘ 2 2!w!trr!Ia1 Í ˙ Irr = ; C = L ! Í !0.05 base s Î ! 2!V ˙˚ = s
!w!!Ia1!trr2 !0.05!Vs
Smallest overvoltage occurs when Cs = Cbase. Putting in numerical values Cs = 0.9933 Ia1 mF !0.05!Vs 2!Vs = w!t !I Rbase = I rr a1 rr Rs,opt = 1.3 Rbase = c.
4000 Ia1
=
(0.05)!(230) 3050 = I -5 (377)!(10 )!Ia1 a1
ohms
ohms
Peak line voltage = 2 (230) = 322 V Smallest overvoltage = (1.5)(322) = 483 V which occurs when Cs = Cbase and Rs = 1.3 Rbase. For Ia1 = 100 A 4000 Rs,opt = 1.3 Rbase = 100 = 40 ohms ; Cs = (0.0033)(100) = 0.33 mF
27-10.
The resistor in the BJT/MOSFET snubber must be shorted out during the device turn-off so that the snubber capacitance is in parallel with the device. The uncharged capacitor delays the build-up of the large Vd voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forward biased during turn-off, thus providing the shorting of the snubber resistor as required. The turn-off of the thyristor limits overvoltages arising from the interruption of current through the stray inductance in series with the thyristor. Lowest overvoltages are obtained when Rs = 1.3 Rbase and Cs = Cbase. Overvoltages are 70-% larger if Rs is zero. Hence a diode in parallel with the resistor is not desirable.
27-11.
a.
dvDS Vd dvDS ; trv < 0.75 microseconds Check dt at turn-off; dt = t rv dvDS 700 dt > 7.5x10-7 = 930 V/ms > 800 V/ms limit so snubber is needed. Not enough information available to check on power dissipation or overcurrents.
b.
dvDS Io 100 = dt Cs = 400 V/ms ; Cs = 4x108 = 0.25 mF Choose Rs to limit total current ID to less than 150 A 700 700 150 A = 100 + R ; Rs = !150!-!100 = 14 ohms s Check snubber recovery time = 2.3 Rs Cs = (2.3)(14)(2.5x10-7) = 8 ms
Off time of the IGBT is 10 microseconds which is greater than the snubber recovery time. Hence choice of Rs is fine.
Chapater 28 Problem Solutions 28-1.
Schematic of drive circuit shown below.
V
100 V GG+
R
100 A +
G
V
DS
V
GG-
vDS(t) waveform same as in problem 22-2. dvDS Vd During MOSFET turn-on dt = t < 500 V/ms fv
Vd From problem 22-2, t fv
ÈV !-!V !-!! Io ˘ Í GG+ GSth gm˙ =Î ˚ RG!Cgd
dvDS Vd During MOSFET turn-off, = t dt rv
Vd From problem 23-2, t rv
< 500 V/ms
ÈV !+!V !+!! Io ˘ Í GG- GSth gm˙ =Î ˚ RG!Cgd
Io 60 gm = V !-!V = 7!-!4 = 20 A/V GS GSth Estimate of RG for MOSFET turn-on:
5x108 V/sec >
100 VGG+!-!4!-! 20 (RG)(4x10-10)
Io 100 ; VGG+,min = VGSth + g = 4 + 20 = 9 V m
Choose VGG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses. RG >
15!-!9 = 30 ohms -10 (4x10 )(5x108)
Estimate of RG at MOSFET turn-off:
5x108 V/sec >
100 VGG-!+!4!+! 20 (RG)(4x10-10)
Choose VGG- = -15 V to insure that MOSFET is held in off-state and to minimize turn-off times. 15!+!9 = 115 ohms RG > (4x10-10)(5x108) Satisfy both turn-on and turn-joff requirements by choosing VGG+ = VGG= 15 V and RG > 115 ohms. 28-2.
a.
Circuit diagram shown below.
+
R
G1
Tsw
V = 1000 V d
-
R
I o = 200 A
Df
G2
Qs
RG2 1000 When FCT is off we need VKG = (1.25) 40 = 31.25 V = (1000) R !+!!R G1 G2 Now R G1 + RG2 = 10-6 ohms so 31.25 = (1000) (10-6) RG2 RG2 = 31.25 kW and RG1 = 969 kW
Choose VGG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses. RG >
15!-!9 = 30 ohms -10 (4x10 )(5x108)
Estimate of RG at MOSFET turn-off:
5x108 V/sec >
100 VGG-!+!4!+! 20 (RG)(4x10-10)
Choose VGG- = -15 V to insure that MOSFET is held in off-state and to minimize turn-off times. 15!+!9 = 115 ohms RG > (4x10-10)(5x108) Satisfy both turn-on and turn-joff requirements by choosing VGG+ = VGG= 15 V and RG > 115 ohms. 28-2.
a.
Circuit diagram shown below.
+
R
G1
Tsw
V = 1000 V d
-
R
I o = 200 A
Df
G2
Qs
RG2 1000 When FCT is off we need VKG = (1.25) 40 = 31.25 V = (1000) R !+!!R G1 G2 Now R G1 + RG2 = 10-6 ohms so 31.25 = (1000) (10-6) RG2 RG2 = 31.25 kW and RG1 = 969 kW
b.
MOSFET characteristics - large on-state current capability - low Ron - low BVDSS (BVDSS of 50-100 V should work)
Chapter 29 Problem Solutions 29-1.
Assume Ts = 120 °C and T a = 20 °C. 0.12 ; A in m2 ; Eq. (29-18) Rq,rad ≈ A Èd ˘ 1 Í vert˙ 1/4 Rq,conv ≈ (1.34)(A) Î DT ˚ ; Eq. (29-20) Rq,conv ≈
0.24 1/4 for DT = 100 °C. A [dvert]
A cube having a side of length dvert has a surface area A = 6 [dvert]2 Rq,conv ≈
0.4 [dvert]1.75
0.02 ; Rq,rad ≈ [dvert]2 Rq,conv!Rq,rad!
Rq,sa = net surface-to-ambient thermal resistance = R q,conv!!+!Rq,rad Rq,sa =
0.04 1.75 ![dvert] !+!2![dvert]2
Heat Sink #
1
2
3
5
6
Volume [m]3
7.6x10-5
10-4
1.8x10-4
2x10-4
3x10-4
dv = (vol.)1/3 [m]
0.042
0.046
0.057
0.058
0.067
A = 6 [dv]2 [m]2
0.011
0.013
0.019
0.002
0.027
dv1.75
0.004
0.046
0.0066
0.0069
0.0088
dv2
0.0018
0.0021
0.0032
0.0034
0.0045
Rq,sa [°C/W]
5.3
4.5
3.1
2.9
2.3
Rq,sa (measured)
3.2
2.3
2.2
2.1
1.7
Heat Sink #
7
8
9
10
11
12
Volume [m]3
4.4x10-4
6.810-4
6.1x10-4
6.3x10-4
7x10-4
1.4x10-3
dv = (vol.)1/3 [m]
0.076
0.088
0.085
0.086
0.088
0.11
A = 6 [dv]2 [m]2
0.034
0.046
0.043
0.044
0.047
0.072
dv1.75
0.011
0.014
0.013
0.014
0.014
0.021
dv2
0.0058
0.0078
0.0071
0.0073
0.0078
0.012
Rq,sa [°C/W]
1.8
1.4
1.5
1.4
1.3
0.9
Rq,sa (measured)
1.3
1.3
1.25
1.2
0.8
0.65
Heat sink #9 is relatively large and cubical in shape with only a few cooling fins. Heat sink #9 is small and flat with much more surface area compared to its volume. Large surface-to-volume ratios give smaller values of Rq,sa.
29-2.
Rq,conv ≈
0.24 1/4 for DT = 100 °C ; From problem 29-1 A [dvert]
Rq,conv ≈ 24 0 [d vert]1/4 for DT = 100 °C and A = 10 cm2 = 10-3 m2 dvert
Rq,conv
1 cm
76 °C/W
5 cm
113 °C/W
12 cm
141 °C/W
20 cm
160 °C/W
Volume [m]3
4.4x10-4
6.810-4
6.1x10-4
6.3x10-4
7x10-4
1.4x10-3
dv = (vol.)1/3 [m]
0.076
0.088
0.085
0.086
0.088
0.11
A = 6 [dv]2 [m]2
0.034
0.046
0.043
0.044
0.047
0.072
dv1.75
0.011
0.014
0.013
0.014
0.014
0.021
dv2
0.0058
0.0078
0.0071
0.0073
0.0078
0.012
Rq,sa [°C/W]
1.8
1.4
1.5
1.4
1.3
0.9
Rq,sa (measured)
1.3
1.3
1.25
1.2
0.8
0.65
Heat sink #9 is relatively large and cubical in shape with only a few cooling fins. Heat sink #9 is small and flat with much more surface area compared to its volume. Large surface-to-volume ratios give smaller values of Rq,sa.
29-2.
Rq,conv ≈
0.24 1/4 for DT = 100 °C ; From problem 29-1 A [dvert]
Rq,conv ≈ 24 0 [d vert]1/4 for DT = 100 °C and A = 10 cm2 = 10-3 m2 dvert
Rq,conv
1 cm
76 °C/W
5 cm
113 °C/W
12 cm
141 °C/W
20 cm
160 °C/W
B
160 B
140 120 R q ,conv °C/W
B
100 80
B
60 40 20 0 0
2
4
6
8 10 12 14 16 18 20 dvert [cm]
29-3.
1 Rq,conv ≈ (1.34)(A)
Èd ˘ Í vert˙ 1/4 ; Eq. (29-20) Î DT ˚
Rq,conv ≈ 353 [DT]-.25
A = 10 cm2 and dvert = 5 cm DT
Rq,conv
60 °C
127 °C/W
80 °C
118 °C/W
100 °C
112 °C/W
120 °C
107 °C/W
B
160 B
140 120 R q ,conv °C/W
B
100 80
B
60 40 20 0 0
2
4
6
8 10 12 14 16 18 20 dvert [cm]
29-3.
1 Rq,conv ≈ (1.34)(A)
Èd ˘ Í vert˙ 1/4 ; Eq. (29-20) Î DT ˚
Rq,conv ≈ 353 [DT]-.25
A = 10 cm2 and dvert = 5 cm DT
Rq,conv
60 °C
127 °C/W
80 °C
118 °C/W
100 °C
112 °C/W
120 °C
107 °C/W
140 120
B
B
B
B
100 Rq ,conv °C/W
80 60 40 20 0 60
70
80
90
100
110
120
DT [°C]
29-4.
Rq,rad =
!DT ÊÁ Ê Ts ˆ4 ÊÁ Ta ˆ˜4 ˆ˜ Á ˜ 5.1!A!Ë!Ë100¯ !-!Ë100¯ !¯
120!-!Ta(° C) Rq,rad = 196 È ÈÍTa(° K)˘˙4!˘˙ Í !Î239!-!Î 100 ˚ ˚
; Eq. (29-17)
; A = 10 cm2 and Ts = 120 °C
Ta
Rq,rad
0 °C
128 °C/W
10 °C
123 °C/W
20 °C
119 °C/W
40 °C
110 °C/W
140 120
B
B
B
B
100 Rq ,conv °C/W
80 60 40 20 0 60
70
80
90
100
110
120
DT [°C]
29-4.
Rq,rad =
!DT ÊÁ Ê Ts ˆ4 ÊÁ Ta ˆ˜4 ˆ˜ Á ˜ 5.1!A!Ë!Ë100¯ !-!Ë100¯ !¯
120!-!Ta(° C) Rq,rad = 196 È ÈÍTa(° K)˘˙4!˘˙ Í !Î239!-!Î 100 ˚ ˚
; Eq. (29-17)
; A = 10 cm2 and Ts = 120 °C
Ta
Rq,rad
0 °C
128 °C/W
10 °C
123 °C/W
20 °C
119 °C/W
40 °C
110 °C/W
140 120 R
B
B
B
B
100 q ,rad 80
[ °C / W ]
60 40 20 0 0
5
10
15
20
25
30
35
Ta [°C ]
29-5.
Rq,rad =
!DT ÊÁ Ê Ts ˆ4 ÊÁ Ta ˆ˜4 ˆ˜ Á ˜ 5.1!A!Ë!Ë100¯ !-!Ë100¯ !¯
Ts(° C)!-!40 Rq,rad = 196 ÈÈT (° K)˘4! ˘˙ ÍÍ s ˙ Î Î ˚ ! 100 -!96˚
; Eq. (29-17)
; A = 10 cm2 and Ta= 40 °C
Ts
Rq,rad
80 °C
114 °C/W
100 °C
120 °C/W
120 °C
110 °C/W
140 °C
101 °C/W
40
140 120 R
B
B
B
B
100 q ,rad 80
[ °C / W ]
60 40 20 0 0
5
10
15
20
25
30
35
Ta [°C ]
29-5.
Rq,rad =
!DT ÊÁ Ê Ts ˆ4 ÊÁ Ta ˆ˜4 ˆ˜ Á ˜ 5.1!A!Ë!Ë100¯ !-!Ë100¯ !¯
Ts(° C)!-!40 Rq,rad = 196 ÈÈT (° K)˘4! ˘˙ ÍÍ s ˙ Î Î ˚ ! 100 -!96˚
; Eq. (29-17)
; A = 10 cm2 and Ta= 40 °C
Ts
Rq,rad
80 °C
114 °C/W
100 °C
120 °C/W
120 °C
110 °C/W
140 °C
101 °C/W
40
120
B
B
B B
100 R q,rad
80
[ °C / W ]
60 40 20 0 80
90
100
110
120
130
140
T s [ °C ]
29-6.
PMOSFET,max =
150!° C!!-!50!° C! !1!° C/W
= 100 W ; 100 W = 50 + 10-3 fs
Solving for fs yields fs = 50 kHz
29-7.
PMOSFET = 50 + 10-3 • 2.5x104 = 75 W Rq,ja = Rq,jc + Rq,ca =
150!° C!!-!35!° C! = 1.53 °C/W !75!W
Rq,ca = 1.53 - 1.00 = 0.53 °C/W
120
B
B
B B
100 R q,rad
80
[ °C / W ]
60 40 20 0 80
90
100
110
120
130
140
T s [ °C ]
29-6.
PMOSFET,max =
150!° C!!-!50!° C! !1!° C/W
= 100 W ; 100 W = 50 + 10-3 fs
Solving for fs yields fs = 50 kHz
29-7.
PMOSFET = 50 + 10-3 • 2.5x104 = 75 W Rq,ja = Rq,jc + Rq,ca =
150!° C!!-!35!° C! = 1.53 °C/W !75!W
Rq,ca = 1.53 - 1.00 = 0.53 °C/W
120
B
B
B B
100 R q,rad
80
[ °C / W ]
60 40 20 0 80
90
100
110
120
130
140
T s [ °C ]
29-6.
PMOSFET,max =
150!° C!!-!50!° C! !1!° C/W
= 100 W ; 100 W = 50 + 10-3 fs
Solving for fs yields fs = 50 kHz
29-7.
PMOSFET = 50 + 10-3 • 2.5x104 = 75 W Rq,ja = Rq,jc + Rq,ca =
150!° C!!-!35!° C! = 1.53 °C/W !75!W
Rq,ca = 1.53 - 1.00 = 0.53 °C/W
