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Eletronica 1 Malvino 7ed Respostas (em ingles)

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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
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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