Eletronica 1 Malvino 7ed Respostas (em ingles)
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Eletronica 1 Malvino 7ed Respostas (em ingles)


DisciplinaEletrônica I6.452 materiais31.651 seguidores
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regulation is needed. For 
high-frequency oscillations, you can try shielding the 
stages, using a single ground point, filter capacitors on each 
stage supply, and ferrite beads on each base or gate lead. 
PROBLEMS 
 23-1. Given: RF = 1 k\u2126 
Solution: The oscillator becomes stable with a lamp 
resistance of 500 \u2126 and from the graph a lamp voltage 
of 3 V rms. 
IL = 3 V/500 \u2126 
IL = 6 mA 
Vout = IL(RF + RL) 
Vout = 6 mA(1 k\u2126 + 500 \u2126) 
Vout = 9 V rms 
Answer: The output voltage is 9 V rms. 
 23-2. Given: 
C = 200 pF 
Rmin = 2 k\u2126 
Rmax = 24 k\u2126 
Solution: 
fr(max) = 1/[2\u3c0RminC] (Eq. 23-4) 
fr(max) = 1/[2\u3c0(2.2 k\u2126)(200 pF)] 
fr(max) = 398 kHz 
fr(min) = 1/[2\u3c0RmaxC] (Eq. 23-4) 
fr(min) = 1/[2\u3c0(2.4 k\u2126)(200 pF)] 
fr(min) = 33.2 kHz 
Answer: The maximum frequency is 398 kHz, and the 
minimum frequency is 33.2 kHz. 
 23-3a. Given: 
C = 0.2 µF 
Rmin = 2 k\u2126 
Rmax = 24 k\u2126 
Solution: 
fr(max) = 1/[2\u3c0RminC] (Eq. 23-4) 
fr(max) = 1/[2\u3c0(2 k\u2126)(0.2 µF)] 
fr(max) = 398 Hz 
fr(min) = 1/[2\u3c0RmaxC] (Eq. 23-4) 
fr(min) = 1/[2\u3c0(24 k\u2126)(0.2 µF)] 
fr(min) = 33.2 Hz 
Answer: The maximum frequency is 398 Hz, and the 
minimum frequency is 33.2 Hz. 
 23-3b. Given: 
C = 0.02 µF 
Rmin = 2 k\u2126 
Rmax = 24 k\u2126 
Solution: 
fr(max) = 1/[2\u3c0RminC] (Eq. 23-4) 
fr(max) = 1/[2\u3c0(2 k\u2126)(0.02 µF)] 
fr(max) = 3.98 kHz 
fr(min) = 1/[2\u3c0RmaxC] (Eq. 23-4) 
fr(min) = 1/[2\u3c0(24 k\u2126)(0.02 µF)] 
fr(min) = 332 Hz 
Answer: The maximum frequency is 3.98 kHz, and the 
minimum frequency is 332 Hz. 
 23-3c. Given: 
C = 0.002 µF 
Rmin = 2 k\u2126 
Rmax = 24 k\u2126 
Solution: 
fr(max)= 1/[2\u3c0RminC] (Eq. 23-4) 
fr(max) = 1/[2\u3c0(2 k\u2126)(0.002 µF)] 
fr(max) = 39.8 kHz 
fr(min) = 1/[2\u3c0RmaxC] (Eq. 23-4) 
fr(min) = 1/[2\u3c0(24 k\u2126)(0.002 µF)] 
fr(min) = 3.32 kHz 
Answer: The maximum frequency is 39.8 kHz, and the 
minimum frequency is 3.32 kHz. 
 23-3d. Given: 
C = 200 pF 
Rmin = 2 k\u2126 
Rmax = 24 k\u2126 
Solution: 
fr(max) = 1/[2\u3c0RminC] (Eq. 23-4) 
fr(max) = 1/[2\u3c0(2 k\u2126)(200 pF)] 
fr(max) = 398 kHz 
fr(min) = 1/[2\u3c0RmaxC] (Eq. 23-4) 
fr(min) = 1/[2\u3c0(24 k\u2126)(200 pF)] 
fr(min) = 33.2 kHz 
Answer: The maximum frequency is 398 kHz, and the 
minimum frequency is 33.2 kHz. 
 23-4. Given: 
Vout = 6 V rms 
RF = 2Rlamp 
Solution: Since the lamp resistance is one-third of the 
total resistance, its voltage will be one-third of the total 
voltage, or 2 V rms. According to the graph, the lamp 
resistance would be 350 \u2126, so the feedback resistor 
would need to be twice that, or 700 \u2126. 
Answer: Change the feedback resistor to 700 \u2126. 
 23-5. Given: Maximum frequency is 398 kHz, from Prob. 
23-3. 
Solution: 1 decade above 398 kHz is 3.98 MHz. 
Answer: The cutoff frequency is 3.98 MHz. 
 23-6. Given: 
R = 10 k\u2126 
C = 0.01 µF 
Solution: 
fr = 1/[2\u3c0RC] (Eq. 23-4) 
fr = 1/[2\u3c0(10 k\u2126)(0.01 µF)] 
fr = 1.59 kHz 
Answer: The resonant frequency is 1.59 kHz. 
 23-7. Given: 
R = 20 k\u2126 
C = 0.02 µF 
Solution: 
fr = 1/[2\u3c0RC] (Eq. 23-4) 
fr = 1/[2\u3c0(20 k\u2126)(0.02 µF)] 
fr = 398 Hz 
1-117 
Answer: The resonant frequency is 398 Hz. 
 23-8. Given: 
R1 = 10 k\u2126 
R2 = 5 k\u2126 
RE = 1 k\u2126 
VBE = 0.7 V 
VCC = 12 V 
Solution: 
VB = [R2/(R1 + R2)]VCC (Eq. 8-1) 
VB = [5 k\u2126/(10 k\u2126 + 5 k\u2126)]12 V 
VB = 4 V 
VE = VB \u2013 VBE (Eq. 8-2) 
VE = 4 V \u2013 0.7 V 
VE = 3.3 V 
IE = VE/RE (Eq. 8-3) 
IE = 3.3 V/1 k\u2126 
IE = 3.3 mA 
Since the RF choke is a short to direct current, the 
collector voltage is 12 V. 
VCE = VC \u2013 VE (Eq. 8-6) 
VCE = 12 V \u2013 3.3 V 
VCE = 8.7 V 
Answer: The emitter current is 3.3 mA, and the 
collector-to-emitter voltage is 8.7 V. 
 23-9. Given: 
C1 = 0.001 µF 
C2 = 0.01 µF 
L = 10 µH 
Solution: 
C = C1C2/(C1 + C2) (Eq. 23-6) 
C = (0.001 µF)(0.01 µF)/(0.001 µF + 0.01 µF) 
C = 909 pF ( )1/ 2\u3c0rf LC= (Eq. 23-5) ( )1/ 2\u3c0 (10 µH)(909 pF)rf = 
fr = 1.67 MHz 
B = C1/C2 (Eq. 23-7) 
B = 0.001 µF/0.01 µF 
B = 0.10 
Av(min) = C2/C1 (Eq. 23-8) 
Av(min) = 0.001 µF/0.01 µF 
Av(min) = 10 
Answer: The frequency is 1.67 MHz, the feedback 
fraction is 0.10, and the minimum gain is 10. 
 23-10. Given: 
C1 = 0.001 µF 
C2 = 0.01 µF 
Solution: 
B = C1/(C1 + C2) 
B = 0.001 µF/(0.001 µF + 0.01 µF) 
B = 0.091 
Answer: The feedback fraction is 0.091. 
 23-11. Given: 
C1 = 0.001 µF 
C2 = 0.01 µF 
L = 20 µH 
Solution: 
C = C1C2/(C1 + C2) (Eq. 23-6) 
C = (0.001 µF)(0.01 µF)/(0.001 µF + 0.01 µF) 
C = 909 pF 
1/(2\u3c0 )rf LC= (Eq. 23-5) 
=1/(2\u3c0 (20 µH)(909 pF))rf 
fr = 1.18 MHz 
Answer: The frequency is 1.18 MHz. 
 23-12. Answer: Reduce the inductance by a factor of 4 (since 
there is a square root in the denominator). 
 23-13. Given: 
C1 = 0.001 µF 
C2 = 0.01 µF 
C3 = 47 pF 
L = 10 µH 
Solution: 
31/(2\u3c0 ) rf LC= (Eq. 23-18) 
1/(2\u3c0 (10 µH)(47 pF))rf = 
fr = 7.34 MHz 
Answer: The frequency is 7.34 MHz. 
 23-14. Given: 
L1 = 1 µH 
L2 = 0.2 µH 
C = 1000 pF 
Solution: 
B = L2L1 (Eq. 23-16) 
B = 0.2 µH/1 µH 
B = 0.2 
L = L1 + L2 
L = 1 µH + 0.2 µH 
L = 1.2 µH 
 1/[2\u3c0 ] rf LC= (Eq. 23-5) ( )( ) 1/[2\u3c0 1.2 µH 1000 pF ]rf = 
fr = 4.59 MHz 
Av(min) = L1/L2 
Av(min) = 1 µH/0.2 µH 
Av(min) = 5 
Answer: The frequency is 4.59 MHz, the feedback 
fraction is 0.2, and the minimum gain is 5. 
 23-15. Given: 
M = 0.1 µH 
L = 3.3 µH 
Solution: 
B = M/L (Eq. 23-14) 
B = 0.1 µH/3.3 µH 
B = 0.030 
Av(min) = L/M 
Av(min) = 3.3 µH/0.1 µH 
Av(min) = 33 
Answer: The feedback fraction is 0.03, and the 
minimum gain is 33. 
 23-16. Given: f = 5 MHz 
Answer: The first overtone is 10 MHz, the second 
overtone is 15 MHz, and the third overtone is 20 MHz. 
 23-17. Answer: Since the frequency is inversely proportional to 
thickness, if thickness is reduced by 1% the frequency 
will increase by 1%. 
1-119 
Answer: The period is 100 µs, the quiescent pulse width 
is 5.61 µs, the maximum pulse width is 8.66 µs, the 
minimum pulse width is 3.71 µs, the maximum duty 
cycle is 0.0866, and the minimum duty cycle is 0.0371. 
 23-24. Given: 
VCC = 10 V 
R1 = 1.2 k\u2126 
R2 = 1.5 k\u2126 
C = 4.7 nF 
vmod = 1.5 V 
Solution: 
W = 0.693(R1 + R2)C (Eq. 23-26) 
W = 0.693(1.2 k\u2126 + 1.5 k\u2126)(4.7 nF) 
W = 8.79 µs 
T = 0.693(R1 + 2R2)C (Eq. 23-27) 
T = 0.693[1.2 k\u2126 + 2(1.5 k\u2126)](4.7 nF) 
T = 13.68 µs 
UTPmax = 2VCC/3 + vmod (Eq. 23-34) 
UTPmax = 2(10 V)/3 + 1.5 V 
UTPmax = 8.17 V 
UTPmin = 2VCC/3 \u2013 vmod (Eq. 23-34) 
UTPmin = 2(10 V)/3 \u2013 1.5 V 
UTPmin = 5.17 V 
Wmax = \u2013 (R1 + R2)C1n[(VCC \u2013 UTPmax)/(VCC \u2013 0.5UTPmax)] 
(Eq. 23-35) 
Wmax = \u2013 {[(1.2 k\u2126 + 1.5 k\u2126)(4.7 nF)]ln[(10 \u2013 8.17 V)/ 
(10 V \u2013 0.5(8.17 V)]} 
Wmax = 14.89 µs 
Wmin = \u2013 (R1 + R2)C1n[(VCC \u2013 UTPmin)/(VCC \u2013 0.5UTPmin)] 
(Eq. 23-35) 
Wmin = \u2013 {[(1.2 k\u2126 + 1.5 k\u2126)(4.7 nF)]ln[(10 \u2013 5.17 V)/ 
(10 V \u2013 0.5(5.17 V)]} 
Wmin = 5.44 µs 
Space = 0.693R2C 
Space = 0.693(1.5 k\u2126)(4.7 nF) 
Space = 4.89 µs 
Answer: The quiescent pulse width is 8.79 µs, the 
quiescent period is 13.69 µs, the maximum pulse width 
is 14.89 µs, the minimum pulse width is 5.44 µs, and the 
space between pulses is 4.89 µs. 
 23-25. Given: 
IC = 0.5 mA 
VCC = 10 V 
C = 47 nF 
Solution: 
S = IC/C (Eq. 23-39) 
S = 0.5 mA/47 nF 
S = 10.6 V/mS 
V = 2VCC/3 (Eq. 23-40) 
V = 2(10 V)/3 
V = 6.67 V 
T = 2VCC/3S (Eq. 23-41) 
T = 2(10 V)/3(10.6 V/mS) 
T = 0.629 mS 
Answer: The slope is 10.6 V/mS, the peak value is 
6.67 V, and the duration is 0.629 mS. 
 23-26. Given: 
S1 = Closed 
R = 20 k\u2126 
R3 = 40 k\u2126 
C = 0.1 µF 
Solution: 
The waveform is a sine wave. 
f = 1.1RC 
f = 1/(20 k\u2126)(0.1 µF) 
f = 500 Hz 
Amplitude = 2.4 Vp 
Amplitude = 4.8 Vpp 
Answer: The output is a sine wave at a frequency of 500 
Hz and a peak voltage of 2.4 V. 
 23-27. Given: 
S1 = Open 
R = 10 k\u2126 
R3 = 40 k\u2126 
C = 0.01 µF 
Solution: 
The waveform is a triangle wave. 
f = 1.1RC 
f = 1/(10 k\u2126)(0.01 µF) 
f = 10 kHz 
Amplitude = 5 Vp 
Amplitude = 10 Vpp 
Answer: The output is a triangle wave at a frequency of 
10 kHz and a peak voltage of 5 V. 
 23-28. Given: 
R1 = 2
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