Eletronica 1 Malvino 7ed Respostas (em ingles)
100 pág.

Eletronica 1 Malvino 7ed Respostas (em ingles)


DisciplinaEletrônica I5.950 materiais31.342 seguidores
Pré-visualização32 páginas
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 
 13-34. Answer: 
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. 
 13-38. Answer: 
 a. Multiply
Estudante
Estudante fez um comentário
falatando pagina vergonha
0 aprovações
Marcus
Marcus fez um comentário
tá faltando páginas véi....
0 aprovações
naruto
naruto fez um comentário
alguien me podria enviar el archivo a mi email por favor 5542sp@gmail.com
0 aprovações
Carregar mais