9-1 Solutions for Chapter 9 Problems 2. Passive Circuit Elements P9.1: Given a 2.0 cm length of AWG20 copper wire, (a) calculate R dc dc, (b) calculate R ac ac at 800 MHz, (c) estimate L.
a
31.96mils 25.4 m
Rdc
2 l
a 2
mil
m 6
10 m 0.02
406x106 m 6 2
5.8 x10 406 x10 7
666
At 800 MHz: Rac
l
2 a
So Rac
, where where 1
f
2.336 x10 6 m 2.336
0.02
5.8 x107 2 406 x10 6 2.336 x10 6
58m
2(0.02) 1 14.4 nH , 6 406 x 10
L 2 x107 0.02 ln so L so L = = 14 nH
P9.2: MATLAB: Repeat MATLAB 9.1 for a typical 200 chip resistor (L L = 0.40 nH, Cx = 50 fF). Compare the resulting Bode plot plot with that of Figure 9.7. %Resistor Equivalent Circuit %P0902: modify ML0901 for typical chip resistor %Enter equivalent circuit values R=200; %resistance in ohms LR=0; %element inductance in nH LLcc=13.4; %lead inductance in nH CRcc=2; %element capacitance, in LLcr=0.4; %lead inductance in nH CRcr=.05; %elem cap, in pF, for
for carbon comp res pF for carb comp res for chip res chip res
f=10e6:10e6:10e11; w=2*pi*f; XLR=complex(0,w*LR*1e-9); XLLcc=complex(0,w*LLcc*1e-9); XCRcc=complex(0,-1./(w*CRcc*1e-12)); XLLcr=complex(0,w*LLcr*1e-9); XCRcr=complex(0,-1./(w*CRcr*1e-12));
9-2
Zcc=XLLcc+parallel(R+XLR,XCRcc); Zcr=XLLcr+parallel(R+XLR,XCRcr); Zmagcc=abs(Zcc); Zmagcr=abs(Zcr); loglog(f,Zmagcc,'--k',f,Zmagcr,'k') legend('carbon composite','chip resistor') grid on xlabel('frequency (Hz)') ylabel('Z magnitude (ohms)')
Fig. P9.2
We see the chip resistor may be operated almost two orders of magnitude higher in frequency than the carbon composite resistor.
P9.3: Recalculate L, Cx and f SRF if the AWG30 wire for the coil of Example 9.2 is replaced with AWG40 wire. For this wire we have t =3.145 mil = 79.9 m. L hasn’t changed from the Example, so L = 1.3 H. d
h Nt N 1
300mil 20(3.145 mil) 19
12.5mils
S t (2 a)( N 1) (3.145mils)(2 )(0.25cm)(19)
1000 mils 2.54cm
37 x10 3 mil 2
9-3
8.854 x10
12
C x
f srf
F m 37 x10 3 mil 2 25.4x106 m
12.5mil 1 2 1.3 x10
6
mil 15
666 x10
666x1015 F
171MHz ,
so f srf = 170 MHz.
P9.4: Estimate L and the SRF if a 99.8% iron core is inserted inside the coil of Example 9.2. From Example 9.2 we had o N 2 a 2 1.3 H . L h Now, we have the iron core with r = 5000, so L (5000)(1.3 H ) 6.5mH . For the self resonance frequency, we use Cx = 5.2 pF as before and find 1 f srf 866kHz , 2 6.5mH 5.2 pF or f srf = 870kHz.
P9.5: Consider a 99.8% iron toroidal core of inner diameter 0.50 cm and outer diameter 1.0 cm wrapped with 20 turns of evenly spaced AWG26 copper wire. Estimate inductance and the self-resonance frequency of this toroidal inductor. For inductance, from P3.56 we found: L
I
N 2
2b
b a
2
where b = (a+c)/2 = 0.375 cm. For the capacitance we have S C , where S = t (2ac)( N -1), and ac = (c-a)/2 = 0.125 cm, and t = 405 m. d Also, h Nt d for a toroid, where h = core length = 2b. N Plugging in the numbers we have 2
2
5000 4 x10 7 20 0.00375 0.0025 L 524 H 2 0.00375 S 405x106 2 0.0012519 60.4x106 m2
9-4 2 0.00375 20 405 x10 6
d
20
773x106 m
8.854 x10 60.4 x10 0.69 fF C 773 x10 12
6
6
f srf
1
1
2 LC
2
6
15
524 x10 0.69 x10
8.4MHz .
P9.6: Calculate the self-resonance frequency for a 47 nF mica capacitor with a pair of 1.0 cm long AWG 24 copper leads. We can apply the equivalent circuit of Figure 9.13(b), where we can neglect the very small resistance R x. Then, H 2l L x 2 x10 7 l ln 1 , where each lead is l = 1 cm, and m a
25.4 x10 6 m 6 a 20.1mils 255 x10 m, so 2 mil 2 0.01 H L x 2 x10 7 0.01 ln 1 6.7 nH . 6 m 255x10 1
With the two leads in series we then have Ltot = 13.4 nH, and then 1 1 f srf 6.3MHz , 2 Ltot C 2 13.4 x10 9 47 x10 9 so f srf = 6 MHz.
P9.7: A thin film capacitor is made by sandwiching a 0.10 m thick layer of Teflon between copper conductive layers. Determine the capacitance per unit area and the maximum voltage that can be applied across such a capacitor. 6
For Teflon, from the appendix we have r = 2.1 and E br = 60x10 V/m. So, C
2.18.854 x10 12
186
F
, S d 0.1x10 6 m2 and V Ebr max , or Vmax Ebr d 60 x10 6 0.1x10 6 6V . d
9-5 2
P9.8: If the 2.2 nF capacitor of Example 9.3 has an area of 20. mm , what thickness mica is used? What is the maximum voltage that can be applied across this capacitor? 6
For mica, from the appendix we have r = 5.4 and E br = 200x10 V/m. F 5.4 8.854 x10 12 20 mm2 2 r o S S m 1m C , d 435nm d C 2.2 x10 9 F 1000 mm
Vmax Ebr d 200 x106 435 x109 87V .
P9.9: (JustAsk): Suppose a standard 300. twin-lead T-line is constructed with AWG 24 wire separated by a center-to-center spacing of 0.800 cm. If this line is terminated in a short circuit realized using the shortest possible length of AWG 24 wire, calculate the reflection coefficient looking into this “short” at 100 MHz, 1 GHz and 10 GHz. AWG24 has a radius a = 255 m. Using Equation (9.4),
2 0.008 H 2l 7 ln 1 2 x 10 0.008 1 5.02 nH l ln 6 m a 255x10
L x 2 x10 7
The impedance looking into the line, neglecting the very small resistance, is Z L = j L. At 100 MHz, we then have Z L j 2 100 x106 5.02 x109 j3.16. The reflection coefficient is found by Z Z o j 3.16 300 L L 1e j179 . Z L Z o j 3.16 300 Repeating the calculation at 1 GHz and at 10 GHz, we have: Frequency 100 MHz 1 GHz 10 GHz
Z L( ) j3.16 j31.5 j315
L 1e 1e 1e
3. Digital Signals P9.10: (JustAsk): What is the spectral bandwidth for a 4.0 ns risetime signal, using equation (9.13)? What risetime is required to achieve a 1 GHz bandwidth?
(a) The bandwidth is approximated by Equation (9.13), 1 1 BW 250MHz . tr 4 x10 9 s (b) Modifying Equation (9.13), 1 1 tr 1ns. BW 1x109
9-6
P9.11: Suppose a 1.0 GHz clock rate is assumed. bandwidth calculated using Eqn. (9.13).
What is the minimum spectral
The minimum bandwidth occurs for the maximum possible rise time, or when the signal is a sawtooth function. At 1 GHz, the period is T = 1/ f = 1 ns. For a sawtooth wave, then, the risetime would be half the period, or t r = 0.5 ns. Then we have 1 1 2GHz . BW tr 0.5x10 9 s
P9.12: MATLAB: Modify MATLAB 9.3 to look at a 1.0 GHz clock rate signal. Minimize the rise and fall times by letting the signal be a sawtooth wave. % MLP0912 % Modify ML0903 to look at a 1 GHz % clock rate signal, with minimum % rise and fall times of a sawtooth wave. % clc %clears the command window clear %clears variables % Initialize variables N=40; fo=1e9; wo=2*pi*fo; To=1/fo; Vo=1; t1=0; t2=0.50*To; tf=t2-t1; % Determine the coefficients a0=(Vo/To)*(t1+t2); for i=1:N n(i)=i; a(i)=((2*Vo)/(pi*wo*tf*i^2))*(cos(i*wo*t1)cos(i*wo*t2)); end % Determine the function components for j=1:1000 t(j)=j*To/1000; for i=1:N
9-7 Fn(i)=a(i)*cos(i*wo*t(j)); end F(j)=a0+sum(Fn); end % Generate the plot subplot(2,1,1) plot(t,F) axis([0 1e-9 0 1]) xlabel('time(sec)') ylabel('voltage(V)') grid on subplot(2,1,2) m=1:40; b=a(m); bar(m,b,'-k') axis([0 40 0 0.5]) xlabel('n') ylabel('an')
Fig. P9.12
4. Grounds P9.13: Repeat Example 9.4 using AWG22 wire and 200 MHz current.
AWG22 wire has a
1 2
25.4 m 322 m, so we can use equation (9.4) to mil
25.35mils
find
2 0.04 H 2l 7 1 36 nH l ln 1 2 x 10 0.04 ln 6 322 x 10 m a
L x 2 x10 7
At 200 MHz, the impedance is
Z j L j 2 200 x106 36 x109 j 45 Figure P9.13 indicates our situation. By circuit analysis we then have V A 3mA 45 135mV , V B 2mA 45 135mV 225mV , VC 1mA 45 225mV 270 mV .
9-8
Fig. P9.13
P9.14: Repeat Example 9.5 using AWG22 wire and 200 MHz current. As in P9.13, we have for AWG22 wire a = 322 m, and for a 4 cm length equation (9.4) gives
2 0.04 H 2l 7 1 36 nH ln 1 2 x 10 0.04 l ln 6 322 x 10 m a
L x 2 x10 7
At 200 MHz, the impedance of the 4 cm length impedance is
Z j L j 2 200 x106 36 x109 j 45 . Likewise, for the 8 cm length wire we have L x = 83nH and Z = j105. For the 12 cm length wire we have L x = 135nH and Z = j170. Now, V A 1mA 45 45mV , V B 1mA105 105mV , VC 1mA170 170mV .
5. Shields P9.15: The field within a shielded enclosure is 12 kV/m. What shielding effectiveness is required such that the field outside the shield is no more than 1.0 nV/m?
E ns E s
SE 20log
12 kV m 20log 262 dB 1 nV m
P9.16: (JustAsk): Compare the attenuation in dB at 1.0 GHz for 20 m thick layers of (a) copper, (b) aluminum and (c) nickel.
1
f 1x109 r 4 x10 7
9-9
For copper, 1 x109 1 4 x10 7 5.8 x10 7 479 x10 3 Np m,
dB Np 6 20 x 10 m 8.686 , which for copper is m Np Atten(dB) 479 x103 20 x106 m 8.686 83dB.
And the attenuation is Atten(dB)
Using this approach the following table is generated: r (S/m) (a) copper 1 5.8x10 (b) aluminum 1 3.8x10 (c) nickel 600 1.5 x10
(Np/m) 479x10 387x10 5.96x10
Atten(dB) 83 67 1040
P9.17: A particular silver-filled paint is to be used as an absorptive layer. It has = 6 1.0x10 S/m with r and r assumed equal to one. Calculate the attenuation of a 100 MHz wave propagating through a 50. m thick layer and compare with the attenuation through a pure silver layer of the same thickness. Atten = t , where f . For the silver paint we have paint 100 x10 6 4 x10
7
10 19.9 x10 6
3
Np m
.
Then,
Atten paint 19.9 x103
dB 50 x10 6 m 8.686 8.6 dB. m Np
Np
Repeating these calculations for pure silver, we find 3 silver = 156x10 Np/m and Atten silver = 68dB.
P9.18: Shielding low frequency magnetic fields often requires a magnetic shield. What thickness of 99.8% iron is required to give 20 dB attenuation of a 1.0 kHz magnetic field? 7
For 99.8% iron we have r = 5000 and = 1x10 S/m. So we have Np f 1000 5000 4 x10 7 1 x10 7 14 x10 3 . m In terms of dB, we have Np dB 3 dB (dB ) 14 x103 . 8.686 122 x10 m Np m Since atten = t , then we have Atten 20dB t 163 m. 122 x103 dB m
9-10 P9.19: Find the shielding effectiveness for the silver-filled paint shield of P9.17 and compare the result with that of pure silver. Using the program ML0904, we find SE paint = 80 dB SE silver = 156 dB
P9.20: MATLAB: Consider a 10.0 m thick copper shield. Plot the contributions to shielding effectiveness (and the total shielding effectiveness) from each of the reflective terms and from the absorption term from 1 MHz up to 1 GHz. Repeat for the same thickness nickel shield. % % % % % % % % % % % % % % % % % % % % % % % % % % % %
used for P9.20 Modify ML0904 Here we want to plot the SE contributions vs frequency. Wentworth, 2/12/03 Variables d shield thickness (m) s shield conductivity (S/m) ur rel permeability uo free space permeability er rel permittivity eo free space permittivity f,w freq. and ang. freq. (1/s) c speed of light (m/s) Zo free space impedance (ohms) A,B,C calculation variables prop propagation constant (1/m) Z1 impedance (ohms) taud transmission coeff at z = -d tao0 trans coeff at z = 0 ratio power ratio SErefd SE from reflection at front face SEabs SE from atten in shield SEref0 SE from reflection at back face SEtot total shielding effectiveness
clc clear
%clears the command window %clears variables
9-11
% Initialize variables d=10e-6; s=5.8e7; ur=1; er=1; f=100e6; eo=8.854e-12; uo=pi*4e-7; c=2.998e8; Zo=120*pi; % Perform calculations for i=1:1:4 f(i)=10^(i+5); w=2*pi*f(i); A=i*w*ur*uo; B=s+i*w*er*eo; prop=sqrt(A*B); Z1=sqrt(A/B); C=tanh(prop*d); Zin=(Z1*(Zo+Z1*C))/(Z1+Zo*C); taud=2*Zin/(Zin+Zo); tau0=2*Zo/(Zo+Z1); ratio=abs(taud*tau0*exp(-prop*d)); SErefd(i)=-20*log10(abs(taud)); SEabs(i)=-20*log10(abs(exp(-prop*d))); SEref0(i)=-20*log10(abs(tau0)); SEtot(i)=-20*log10(ratio); end semilogx(f,SErefd,'-o',f,SEabs,'-+',f,SEref0,'*',f,SEtot,'-^') legend('front face reflection','absorption','back reflection','total') title('10 micron thick copper shield') xlabel('frequency (Hz)') ylabel('loss (dB)') grid on
This is repeated for nickel and the results plotted in Figure P9.20.
face
9-12
Fig. P9.20
6. Filters P9.21: Derive the insertion loss expression Equation (9.20) for the series inductor circuit Figure 9.25(b).
Removing the inductor from the circuit of Figure 9.25(b) (that is, replacing it with a short), we have the voltage across the load without the filter element 1 v L vS . 2 Now, with the filter in place, we have R v Lf vS . 2 R j L
9-13 Therefore, v Lf 2R v L
2R j L
1
1 j L
1 . j fL
1 2 R R We find the magnitude of this ratio, v Lf 1 . 2 v L fL 1 R The insertion loss is then
2 v v Lf fL 2 fL L 20log 1 IL 20log 10log 1 . v Lf vL R R
P9.22: (JustAsk): Suppose an L = 100. nH inductor is used in the series inductance filter of Figure 9.25(b). Determine the insertion loss at 200 MHz if (a) R = 10. and (b )R = 10. k .
fL 2 We apply the equation IL 10log 1 . R 2 200 x106 100 x10 9 (a) IL 10log 1 16 dB. 10 200 x106 100 x10 9 2 (b) IL 10log 1 0.17 x10 3 dB. 3 10 x10
P9.23: Suppose a C = 47. pF capacitor is used in the shunt capacitance filter of Figure 9.25(a). Determine the insertion loss at 200. MHz if (a) R = 10. and (b) R = 10. k .
2
We apply the equation IL 10log 1 fRC .
0.36 dB. (b) IL 10log 1 200 x10 10 x10 47 x10 49 dB. (a) IL 10log 1 200 x10 6 10 47 x10 12 6
3
2
12
2
P9.24: Determine the insertion loss at 1.0 GHz for a T-filter inserted between a 10. source impedance and a 10. load impedance. Consider L = 10. nH and C = 47. pF.
9-14
Fig. P9.24 (a) We can analyzer the circuit shown in Figure 9.24(a) by using the models in Figures 9.24 (b) & (c): Z1 R j L, and
j Z 2 Z 1
.
C
Then we have Z 2 R R v2 vS , vLf v2 v2 , Z1 Z 2 R j L Z1 v L
1 2
Fig. 9.24(b) & (c)
vLf
R
Z 2
Z1 Z1 Z 2
vS .
vS ,
v and we then wish to calculate insertion loss as IL 20 log L . v Lf This is calculated using MLP0924. % MLP0924 % % A T-Filter Problem % clc clear % Variables R=10; L=10e-9; C=47e-12; f=1e9; % Run Program w=2*pi*f; XL=i*w*L; XC=-i/(w*C);
9-15 Z1=R+XL; Z2=parallel(Z1,XC); VLF=R*Z2/(Z1*(Z1+Z2)); ratio=1/(2*abs(VLF)); IL=20*Log10(ratio)
Now we run the program: IL = 34.5662
So the insertion loss is IL = 34.6 dB.
P9.25: Determine the insertion loss at 40 MHz for a -filter inserted between a 10. k source impedance and a 10. k load impedance. Consider L = 10. nH and C = 47. pF. Referring to Figure P9.25, we have j Z1 R L , C and Z2 j L Z 1. Then, Z3 Z 2
j
. C Using these values we find Z 3 v1 vS , Z1 R s v2
v
Z 1 Z1 j L
Lf
v1
v Lf ,
Z 1
Z 3
Z1 j L Z1 R s
vS
and v L
R L
R L RS Finally,
.
v IL 20 log L . v Lf This is calculated using MLP0925. % MLP0925 % % This is a pi filter. %
Fig. P9.25
9-16 clc clear % enter variables RS=10e3; RL=10e3; L=10e-9; C=47e-12; f=40e6; % Perform calculations w=2*pi*f; XC=-i/(w*C); XL=i*w*L; Z1=parallel(RL,XC); Z2=XL+Z1; Z3=parallel(XC,Z2); vLf=(Z1*Z3)/(Z2*(Z3+RS)); vL=RL/(RL+RS); IL=20*Log10(vL/abs(vLf))
Now running this program: IL = 41.3
So IL = 41 dB.
