BRAC University ECE230&EEE209 (Semiconductor Materials and Devices) Summer 2011 Date Due: 29.06.11
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Assignment 2 Solution 1. Consider a sample of silicon at T = 300 K doped at an impurity concentration of: Nd= 1015 cm 3. Assume electron and hole mobilities are 1350 and 480 cm2/V.s, respectively. Calculate the conductivity of the of the silicon. If an If an electric field is E = 35 V/cm is applied, calculate the drift current density. ‐
J Drift
= q n0 μ n + p 0 μ p E
n0 ≅ N d = 10 J Drift
15
-3
cm ; p0
= 1.6 × 10
−19
=
ni2 n
(1.5 × 10 ) =
10 2
10
15
= 2.25 × 10 5 cm -3
× (10 × 1350 + 2.25 × 10 5 × 480)× 35 = 7.56 A/cm 2 15
Note:
1. The unit of the of the current density (A/cm2)
Current Density = Charge(q ) • Carrier Density(n) • Mobility( ) • Electric Field(E ) cm 2 = Coulomb • 3 • V.s cm 1
•
V cm
=
Coulomb s cm
2
=
A cm 2
2. The drift current is quite high even with a small electric field and moderate doping density. 3. Since electron concentration n0 is much much larger than hole concentration p0, we can write,
J Drift = q(n0 μ n + p 0 μ p ) E ≈ qn0 μ n E = 1.6 ×10
19
×1015 ×1350 × 35 = 7.56 A
cm 2 !
which is exactly the same as that calculated considering both electron and hole concentrations. For extrinsic semiconductors, Drift current Drift current is is determined by determined by the the majority carriers only, contribution of minority of minority carriers carriers is negligible. negligible. 2. Consider a compensated GaAs semiconductor at T = 300 K. Nd= 5 x 1015 cm 3 and Na = 2 x 1016 cm 3. Calculate the conductivity of the semiconductor. If an electric field is E = 100 V/cm is applied, calculate the drift current density. Assume electron and hole mobilities are 1350 and 480 cm2/V.s, respectively. ‐
‐
Conductivity of a of a semiconductor is given by,
σ
= q nμ n + pμ p
p
= p 0 = N a − N d = 2 ×1016 − 5 ×1015 = 1.5 ×1016 cm −3 2
n = n0 Q p
=
ni
p 0
>> n,
σ
(2.16 ×10 ) =
6 2
1.5 × 10
16
= 3.1×10 −4 cm −3
= q(nμ n + pμ p ) ≈ qpμ p = 1.6 ×10 −19 ×1.5 ×1016 × 480 = 0.768 S/cm
= σ • E = 0.768 • 100 = 76.8 A/cm 2
J Drift
3. It is required to have an n‐type semiconductor with resistivity 0.1 ohm‐cm. Find the doping density needed to achieve this resistivity. If the semiconductor has a dimension of 1 μm х2 μm х 10μm, calculate the resistance of the material. ρ
=
1 σ
⇒n=
=
q (nμ n
1 qμ n ρ
=
1
+ pμ p )
≅
1 qnμ n
(Assuming n >> p )
1 1.6 ×10
−19
×1350 × 0.1
= 4.63 ×1016 cm -3
2 10 2 ⎡ ⎤ ni 1.5 ×10 ) ( 4860 , = ⎢Note : p = = ⎥ 16 n 4 . 63 10 × ⎢ ⎥ ⎢⎣so our original assumption n >> p is correct.⎥⎦
Resistance of the material, L 10μ m R = ρ = (0.1Ω.cm )× A 1× 2 μ m 2
= (0.1Ω.cm )×
10 ×10 − 4 cm 1×10 − 4 × 2 ×10 − 4 cm 2
= 5000Ω = 5 k Ω
[Note: Units of dimensions must be converted into cm before the calculation to match with that of resistivity ] 4. A bar of Si semiconductor with length 10 μm has electron concentration at one end 1016 cm 3 and 8x1016 cm 3 at the other end. Calculate the diffusion current density if the electron density varies linearly through the bar. Assume electron mobility 1350 cm2/V.s and T = 300 K. ‐
‐
The diffuson current density is gven by, J n, Diff .
= qDn
D n
kT
dn
, D n dx From Einstein relation, μ n
=
q
⇒ Dn = μ n dn dx
∴
J n, Diff .
(10 =
kT
16
q
→ Diffusion constant of electrons
= (1350 cm 2 / V.s)× (0.0259 V) = 34.96 cm 2 / s
− 8 ×1015 ) cm -3 10μ m
=
2 × 1015 cm -3 −4
10 × 10 cm
= 2 ×1018 cm -4
⎛ cm 2 ⎞ ⎟⎟ × (2 ×1018 cm -4 ) = 11.19 A/cm 2 = (1.6 ×10 −19 C)× ⎜⎜ 34.96 s ⎠ ⎝
[Note: Relatively small change in carrier concentration can cause large diffusion current ]
5. Consider a semiconductor that is nonuniformly doped with acceptor impurity atoms as shown in the figure below. If the semiconductor is in thermal equilibrium, draw the energy band diagram showing the intrinsic and the Fermi energy levels through the crystal. Derive an expression for the induced electric field and show its direction in the diagram. If the nonuniformly doped semiconductor is in thermal equilibrium and has no external connections, the individual electron and hole currents must be zero.
J p
= 0 = qp( x )μ p E x − eD p
dp ( x ) dx
Ex is the induced electric field EInduced . Solving for Ex
E x
=
kT 1
=
kT 1
=
1 dE i
dp ( x )
q p( x ) dx q
⎡ p( x ) dE i ⎤ p( x ) ⎢⎣ kT dx ⎥⎦
− E F ⎤ ⎣ kT ⎥⎦ dp ( x ) ⎡ E − E F ⎤ d ⎛ E i − E F ⎞ = ni exp ⎢ i ⎜ ⎟ ⎥ dx kT dx kT ⎣ ⎦ ⎝ ⎠ 144 244 3 ⎡ E p ( x ) = ni exp ⎢ i
n ( x )
q dx
= =
p ( x ) ⎛ dE i kT
− ⎜ ⎝ dx
p ( x ) dE i kT dx
dE F ⎞ dx
⎟ ⎠
,
⎛ dE F = 0, ⎜Q ⎝ dx
⎞ ⎠
as the current is zero ⎟
E Induced
[Note: It is not clear from the final expression of Ex = (1/q)(dEx/dx) that if Ex is positive or negative. But from the 1st line where Ex is expressed in terms of carrier density gradient (dp/dx), it can be said that Ex is negative since (dp/dx) is negative. So Ex will point towards negative x‐axis and the energy band diagram will have a negative slope.]
6. An intrinsic Si sample is doped with donors from one side such that Nd = N0exp(‐ax). (i) Find an expression for E(x) at equilibrium over the range for which Nd >> ni. (ii) Evaluate E when a = 1 (μm) 1. (iii) Sketch a band diagram showing the intrinsic and the Fermi energy levels through the crystal and indicate the direction of E. ‐
kT 1 dN d ( x ) q N d ( x ) dx kT 1 d [ N exp(− ax )] =− q N d ( x ) dx 0 kT 1 (− a ) N 0 exp(− ax ) =− q N d ( x ) kT 1 = aN ( x ) q N d ( x ) d kT =a q
E ( x ) = −
−1
For a = 1 (μ m ) , kT = 1 (μ m )−1 × 0.0259 V E = a q −1
= 1 (10 −4 cm) × 0.0259 V = 259 V/cm
[Note: • Electric field is expressed in units of V/cm. So the unit of a is converted into cm; • The induced Electric field is independent of the position x for the exponential distribution of the carrier density ] E = a
kT 1 dE i = q q dx
⇒
dE i dx
= akT
The energy band diagram will have a constant slope of +qakT . Since the semiconductor is at thermal equilibrium, Fermi level will be flat, i.e., (dE F dx ) = 0 .
