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# Eletronica 1 Malvino 7ed Respostas (em ingles)

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```13-17. Given:
VDD = 15 V
VEE = \u20139 V
RD = 7.5 k\u2126
RE = 8.2 k\u2126
VBE = 0.7 V
Solution:
ID = (VEE \u2013 VBE)/RE (Eq. 13-13)
ID = (9 V \u2013 0.7 V)/8.2 k\u2126
ID = 1.01 mA
VD = VDD \u2013 IDRD (Eq. 13-4)
VD = 15 V \u2013 (1.01 mA)(7.5 k\u2126)
VD = 7.43 V
Answer: The drain voltage is 7.43 V, and the drain
current is 1.01 mA.
13-18. Given:
VDD = 15 V
VEE = \u20139 V
RD = 4.7 k\u2126
RE = 8.2 k\u2126
VBE = 0.7 V
1-64
Solution:
ID = (VEE \u2013 VBE)/RE (Eq. 13-13)
ID = (9 V \u2013 0.7 V)/8.2 k\u2126
ID = 1.01 mA
VD = VDD \u2013 IDRD (Eq. 13-4)
VD = 15 V \u2013 (1.01 mA)(4.7 k\u2126)
VD = 10.25 V
Answer: The drain voltage is 10.25 V, and the drain
current is 1.01 mA.
13-19. Given:
VDD = 25 V
RD = 8.2 k\u2126
RS = 1 k\u2126
ID = 1.5 mA
Solution:
VGS = \u2013IDRS (Eq. 13-7)
VGS = \u2013(1.5 mA)(1 k\u2126)
VGS = \u20131.5 V
VD = VDD \u2013 IDRD \u2013 IDRS
VD = 25 V \u2013 (1.5 mA)(8.2 k\u2126) \u2013 (1.5 mA)(1 k\u2126)
VD = 11.2 V
Answer: The gate-source voltage is \u20131.5 V, and the
drain-source voltage is 11.2 V.
13-20. Given:
VDD = 25 V
RD = 8.2 k\u2126
RS = 1 k\u2126
VS = 1.5 V
Solution:
ID = VS/RS
ID = 1.5 V/1 k\u2126
ID = 1.5 mA
VD = VDD \u2013 IDRD
VD = 25 V \u2013 (1.5 mA)(8.2 k\u2126)
VD = 12.7 V
Answer: The drain voltage is 12.7 V.
13-21. Given:
VDD = 25 V
RD = 10 k\u2126
RS = 22 k\u2126
R1 = 1.5 M\u2126
R2 = 1 M\u2126
Answer: The gate-source voltage is \u20132.5 V, and the drain
current is 0.55 mA.
13-22. Given:
VDD = 15 V
RG = 2.2 M\u2126
RE = 8.2 k\u2126
VEE = \u20139 V
Answer: The gate-source voltage is \u20132.0 V, and the drain
voltage is 7.5 V.
13-23. Given:
VDD = 25 V
RG = 1.5 M\u2126
RS = 1 k\u2126
Answer: The gate-source voltage is \u20132.0 V, and the drain
current is 1.5 mA.
13-24. Given:
VDD = 25 V
RG = 1.5 M\u2126
RS = 2 k\u2126
Answer: The gate-source voltage is \u20135.0 V, and the drain
current is 1 mA and the drain-source voltage is 14.8 V.
13-25. Given:
gm0 = 4000 \u3bcS
IDSS = 10 mA
Solution:
VGS(off) = \u20132I/DSS/gm0 (Eq. 13-15)
VGS(off) = \u20132(10 mA)/4000 \u3bcS
VGS(off) = \u20135 V
gm = gm0 [1 \u2013 (VGS/VGS(off))] (Eq.13-16)
gm = 4000 \u3bcS[1 \u2013 (\u20131 V/\u20135V)]
gm = 3200 \u3bcS
Answer: The gate-source cutoff voltage is \u20135 V, and the
gm0 for VGS = \u20131 V is 3200 \u3bcS.
13-26. Given:
gm0 = 1500 \u3bcS
IDSS = 2.5 mA
VGS = \u20131 V
Solution:
VGS(off) = \u20132IDSS/gm0 (Eq. 13-15)
VGS(off) = \u20132(2.5 mA)/1500 \u3bcS
VGS(off) = \u20133.33 V
gm = gm0 [1 \u2013 (VGS/VGS(off))] (Eq. 13-16)
gm = 1500 \u3bcS[1 \u2013 (\u20131 V/\u20133.3 V)]
gm = 1045 \u3bcS
Answer: The gm for VGS = \u20131 V is 1045 \u3bcS.
13-27. Given:
gm0 = 6000 \u3bcS
IDSS = 12 mA
VGS = \u20132 V
Solution:
VGS(off) = \u20132IDSS/gm0 (Eq. 13-15)
VGS(off) = \u20132(12 mA)/6000 \u3bcS
VGS(off) = \u20134 V
Since the ratio of VGS to VGS(off) is one-half, the
following equation can be used:
ID/IDSS = 1/4
ID = 1/4(IDSS)
ID = 1/4(12 mA)
ID = 3 mA
gm = gm0 [1 \u2013 (VGS/VGS(off))] (Eq. 13-16)
gm = 6000 \u3bcS[1 \u2013 (\u20132 V/\u20134 V)]
gm = 3000 \u3bcS
Answer: The drain current is 3 mA, and the transcon-
ductance is 3000 \u3bcS.
13-28. Given:
VDD = 30 V
R1 = 20 M\u2126
R2 = 10 M\u2126
RD = 1 k\u2126
RS = 2 k\u2126
RL = 10 k\u2126
vin = 2 mV
gm = 3000 \u3bcS
Solution:
rd = RD||RL
rd = 1 k\u2126||10 k\u2126
rd = 909 \u2126
1-65
Av = gmrd (Eq. 13-17)
Av = (3000 \u3bcS)(909 \u2126)
Av = 2.73
vout = Av(vin)
vout = (2.73)(2 mV)
vout = 5.46 mV
Answer: The output voltage is 5.46 mV.
13-29. Given:
VDD = 30 V
R1 = 20 M\u2126
R2 = 10 M\u2126
RD = 1 k\u2126
RS = 2 k\u2126
RL = 10 k\u2126
vin = 2 mV
IDSS = 12 mA (from the graph)
VGS(off) = \u20134 V (from the graph)
Solution:
rd = RD||RL
rd = 1 k\u2126||10 k\u2126
rd = 909 \u2126
VGS(off) = \u20132IDSS/gm0 (Eq. 13-15)
gm0 = -2IDSS/VGS(off)
gm0 = \u20132(12 mA)/\u20134 V
gm0 = 6000 \u3bcS
VG = [R2/(R1 + R2)]VDD
VG = [10 M\u2126/(20 M\u2126 + 10 M\u2126)] 30 V
VG = 10 V
ID = VG/RS (Eq. 13-10)
ID = 10 V/2 k\u2126
ID = 5 mA
From the graph, VGS is approximately \u20131.4 V when ID is
5 mA.
With Eq. 13-16, gm0 = 3900 \u3bcS. Then:
Av = gmrd (Eq. 13-17)
Av = (3900 \u3bcS)(909 \u2126)
Av = 3.54
vout = Av(vin)
vout = 3.54(2 mV)
vout = 7.09 mV
Answer: The output voltage is 7.09 mV.
13-30. Given:
VDD = 30 V
R1 = 20 M\u2126
R2 = 10 M\u2126
RS = 3.3 k\u2126
RL = 1 k\u2126
vin = 5 mV
gm = 2000 \u3bcS
Solution:
rS = RS||RL
rS = 3.3 k\u2126||1 k\u2126
rS = 767 \u2126
Av = (gmrs)/(1 + gmrs) (Eq. 13-18)
Av = (2000 \u3bcS)(767 \u2126)/[1 + (2000 \u3bcS)(767 \u2126)]
Av = 0.605
vout = Av(vin)
vout = (0.605)(5 mV)
vout = 3.03 mV
Answer: The output voltage is 3.03 mV.
13-31. Given:
VDD = 30 V
R1 = 20 M\u2126
R2 = 10 M\u2126
RS = 3.3 k\u2126
RL = 1 k\u2126
vin = 5 mV
IDSS = 6 mA (from the graph)
VGS(off) = \u20134 V (from the graph)
rS = 767 \u2126 (from Prob. 13-30)
Solution:
VGS(off) = \u20132IDSS/gm0 (Eq. 13-15)
gm0 = \u20132IDSS/VGS(off)
gm0 = \u20132(6 mA)/ \u20134 V
gm0 = 3000 \u3bcS
VG = [R2/(R1 + R2)]VDD
VG = [10 M\u2126/(20 M\u2126 + 10 M\u2126)] 30 V
VG = 10 V
ID = VG/RS (Eq. 13-7)
ID = 10 V/3.3 k\u2126
ID \u2245 3 mA
From the graph, VGS is roughly \u20131.25 V when ID = 3
mA. With Eq. (13-16), gm = 2060 \u3bcS.
With Eq. (13-18):
gmrS = (2060 \u3bcS)(767 \u2126) = 1.58
Av = 1.58/(1 + 1.58) = 0.612
vout = Avin
vout = (0.612)(5 mV)
vout = 3.06 mV
Answer: The output voltage is 3.06 mV.
13-32. Given:
Rin = 22 k\u2126
vin = 50 mV pp
IDSS = 10 mA
VP = 2 V
Solution:
RDS = VP/IDSS (Eq. 13-1)
RDS = 2 V/10 mA
RDS = 200 \u2126
With VGS at \u201310 V the JFET is cut off and appears as an
open; thus vout = vin = 50 mV pp.
With VGS at 0 V, the JFET is conducting and a voltage
divider is created with the input resistance.
vout = [RDS/(RDS + Rin)]vin
vout = [200 \u2126/(200 \u2126 + 22 k\u2126)]50 mV pp
vout = 0.45 mV pp
On-off ratio = vout(max)/vout(min) (Eq. 13-19)
On-off ratio = 50 mV pp/0.45 mV pp
On-off ratio =111
Answer: The output voltage at a VGS of \u201310 V is 50 mV
pp, the output voltage at a VGS of 0 V is 0.45 mV pp, and
the on-off ratio is 111.
13-33. Given:
Rout = 33 k\u2126
vin = 25 mV pp
IDSS = 5 mA
VP = 3 V
Solution:
RDS = VP/IDSS (Eq. 13-1)
1-66
RDS = 3 V/5 mA
RDS = 600 \u2126
With VGS at \u201310 V the JFET is cut off and appears as an
open; thus vin = 0 mV pp.
With VGS at 0 V, the JFET is conducting and a voltage
divider is created with the output resistance.
vout = [Rout/(RDS + Rout)]vin
vout = [33 k\u2126/(600 \u2126 + 33 k\u2126)]25 mV pp
vout = 24.55 mV pp
On-off ratio = vout(max)/vout(min) (Eq. 13-19)
On-off ratio = 24.55 mV pp/0 mV pp
On-off ratio = \u221e
Answer: The output voltage at a VGS of \u201310 V is 0 mV
pp, the output voltage at a VGS of 0 V is 24.55 mV pp,
and the on-off ratio is \u221e.
CRITICAL THINKING
IDSS = 20 mA
VDS(max) = 5 V for the ohmic region
VDS = 5 to 30 V in the active range
13-35. Given:
VGS(off) = \u20138 V (from the graph)
IDSS = 32 mA (from the graph)
VGS(1) = \u20134 V
VGS(2) = \u20132 V
Solution:
VGS(off) = \u20132IDSS/gm0 (Eq. 13-15)
gm0 = \u20132IDSS/VGS(off)
gm0 = \u20132(32 mA)/\u20138 V
gm0 = 8000 \u3bcS
gm = gm0 [1 \u2013 (VGS/VGS(off))] (Eq. 13-16)
gm = 8000 \u3bcS[1 \u2013 (VGS/\u20138 V)]
ID = IDSS[1 \u2013 (VGS(1)/VGS(off))]2 (Eq. 13-3)
ID = 32 mA[1 \u2013 (\u20134 V/\u20138V)]2
ID = 8 mA
ID = IDSS[1 \u2013 (VGS(2)/VGS(off))]2 (Eq. 13-3)
ID = 32 mA[1 \u2013 (\u20132 V/\u20138V)]2
ID = 8 mA
Answer: The transconductance equation is gm = 8000 \u3bcS
[1 \u2013 (VGS/\u20138 V)], the drain current at \u20134 V is 8 mA, and
the drain current at \u20132 V is 18 mA.
13-36. Given:
VGS(off) = \u20135 V (from the graph)
IDSS = 12 mA (from the graph)
VGS(1) = \u20131 V
Solution:
ID = IDSS[1 \u2013 (VGS(1)/VGS(off))]2 (Eq. 13-3)
ID = 12 mA[1 \u2013 (\u20131 V/\u20135V)]2
ID = 7.68 mA
Answer: The drain current is 7.68 mA.
13-37. Given:
VDD = 15 V
VEE = \u201310 V
RD = 3.3 k\u2126
RE = 4.7 k\u2126
VBE = 0.7 V
gm = 2000 \u3bcS
vin = 3 mV
Solution:
ID = (VEE \u2013 VBE)/RE (Eq. 13-13)
ID = (10 V \u2013 0.7 V)/4.7 k\u2126
ID = 2 mA
VD = VDD \u2013 IDRD (Eq. 13-4)
VD = 15 V \u2013 (2 mA)(3.3 k\u2126)
VD = 8.4 V
rd = RD||RL
rd = 3.3 k\u2126||15 k\u2126
rd = 2.7 k\u2126
Av = gmrd (Eq. 13-17)
Av = (2000 \u3bcS)(2.7 k\u2126)
Av = 5.4
vout = Av(vin)
vout = (5.4)(3 mV)
vout = 16.2 mV
Answer: The drain voltage is 8.4 V, and the output
voltage is 16.2 mV.