13. 7 (a) C1 is a coupling capacitor that couples the ac component of vS into the amplifier. C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the drain to the output vO




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13.7 (a) C1 is a coupling capacitor that couples the ac component of vS into the amplifier. C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the drain to the output vO. (b) The signal voltage at the source will be vs = 0.



13.14





13.16





13.18





13.20





13.22





13.24





13.26





13.28





13.30





13.32





13.34 Figure 13.61





13.35 (a) Figure 13.62



13.35 (b) Figure 13.63






13.36 (a) Figure 13.64



13.36 (b) Figure 13.65





13.37 (a) Figure 13.66


13.37 (b) Figure 13.67





13.38 (a) Figure 13.68



13.38 (b) Figure 13.69



13.39 Figure 13.70





13.66





13.105



13.115



14.1 (a) Common-collector Amplifier (emitter-follower)





14.1 (b) Not a useful circuit because the signal is injected into the drain of the transistor.





14.1 (c) Common-emitter Amplifier





14.1 (d) Common-source Amplifier



14.1 (e) Common-gate Amplifier





14.1 (f) Common-collector Amplifier (emitter-follower)





14.1 (g) Common-source Amplifier





14.1 (h) Common-base Amplifier





14.1 (i) Not a useful circuit since the signal is being taken out of the base terminal.





14.1 (j) Common-source Amplifier





14.1 (k) Common-gate Amplifier





14.1 (l) Not a useful circuit because the signal is injected into the drain of the transistor.





14.1 (m) Common-source Amplifier





14.1 (n) Common-emitter Amplifier





14.1 (o) Common-drain Amplifier (Source-follower)





14.8



14.10



14.16



14.17



14.18





14.19



14.20



14.21



14.23



14.32

Some design possibilities are listed in the table below.




RE

IE

VCC

VCC IE

100 

450 mA

53 V

24 W

250 

225 mA

64 V

16 W

360 

179mA

73V

13 W

500 

150 mA

83 V

12 W

750 

125 mA

102 V

13 W

1000 

113mA

120 V

14 W

2000 

93.8 mA

196 V

18 W

Using a result near the minimum-power case in the table: RE = 510  E= 149 mA and VCC = 85 V.

It is very difficult to achieve the required level of linearity!





14.33



14.70

*Problem 14.70 - Common-Collector Amplifier 14.48(a)

VCC 6 0 DC 18

VEE 7 0 DC -18

VI 1 0 AC 1

*For Output Resistance

*VI 1 0 AC 0

*VO 8 0 AC 1

RI 1 2 500

C1 2 3 10UF

R1 3 0 51K

R2 6 3 100K

RE 4 7 4.7K

RC 6 5 2K

C3 5 0 10UF

C2 4 8 47UF

R3 8 0 24K

Q1 5 3 4 NBJT

.OP

.AC LIN 1 10KHZ 10KHZ



.MODEL NBJT NPN IS=1E-16 BF=125 VA=50

.PRINT AC VM(8) VP(8) VM(3) VP(3) IM(VI) IP(VI) IM(C2) IP(C2)

.END

Results: Q-point: (4.67 mA, 4.57 V), Av = 0.982, Rin = 31.1 kRout= 9.12 





14.71

*Problem 14.71 - Common-Emitter Amplifier - 14.48(c)

VCC 6 0 DC 12

VI 1 0 AC 1

*VI 1 0 AC 0

*VO 7 0 AC 1

RI 1 2 1K

C1 2 3 2.2UF

R1 6 3 20K

R2 3 0 62K

RE 6 4 3.9K

C2 6 4 47UF

RC 5 0 8.2K

C3 5 7 10UF

R3 7 0 100K

Q1 5 3 4 PBJT

.OP

.AC LIN 1 5KHZ 5KHZ



.MODEL PBJT PNP IS=1E-16 BF=75 VA=60

.PRINT AC VM(7) VP(7) VM(3) VP(3) IM(VI) IP(VI) IM(C3) IP(C3)

.END

Results: Q-point: (525 A, 5.62V), Av = -110, Rin = 3.16k Rout= 7.69 k




14.72

*Problem 14.72 - Common-Source Amplifier - 14.48(d)

VDD 4 0 DC 18

VI 1 0 AC 1

*For output resistance

*VI 1 0 AC 0

*VO 7 0 AC 1

RI 1 2 1K

C1 2 3 2.2U

R1 3 0 500K

R2 4 3 1.4MEG

R4 6 0 27K

C2 6 0 47U

RD 4 5 75K

C3 5 7 10U

R3 7 0 470K

M1 5 3 6 6 NMOSFET

.OP


.AC LIN 1 5KHZ 5KHZ

.MODEL NMOSFET NMOS VTO=1 KP=250U LAMBDA=0.02

.PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VI) IP(VI) IM(C3) IP(C3)

.END


Results: Q-point: (106A, 7.14V), Av = -14.2, Rin = 369kRout= 65.8 k





14.73 *Problem 14.73 - Common-Gate Amplifier - 14.48(e)

VSS 5 0 DC 12

VDD 6 0 DC -12

VI 1 0 AC 1

*For Output Resistance

*VI 1 0 AC 0

*VO 7 0 AC 1

RI 1 2 500

C1 2 3 10U

R1 5 3 33K

RD 6 4 24K

C2 4 7 47U

R3 7 0 100K

M1 4 0 3 3 PFET

.OP

.AC LIN 1 50KHZ 50KHZ



.MODEL PFET PMOS KP=200U VTO=-1 LAMBDA=0.02

.PRINT AC VM(7) VP(7) VM(3) VP(3) IM(VI) IP(VI) IM(C2) IP(C2)

.END

Results: Q-point: (286 A, 7.72V), Av = +5.50, Rin = 2.73 kRout= 21.8 k





14.74

*Problem 14.74 - Common-Collector Amplifier - 14.1(f)

VCC 7 0 DC 5

VEE 8 0 DC -5

VI 1 0 AC 1

*For Output Resistance

*VI 1 0 AC 0

*VO 6 0 AC 1

RI 1 2 500

C1 2 3 10UF

R1 3 5 500K

R2 5 0 500K

RE 4 8 430K

C2 4 5 47UF

C3 4 6 10UF

R3 6 0 500K

Q1 7 3 4 NBJT

.OP


.AC LIN 1 10KHZ 10KHZ

.MODEL NBJT NPN IS=1E-16 BF=100 VA=60

.PRINT AC VM(6) VP(6) VM(3) VP(3) IM(VI) IP(VI) IM(C3) IP(C3)

.END


Results: Q-point: (9.81 A, 5.74V), Av = 0.983, Rin = 11.0 MRout= 2.58 k



14.75

*Problem 14.75 - Common-Source Amplifier - 14.48(g)

VDD 6 0 DC 20

VI 1 0 AC 1

*For output resistance

*VI 1 0 AC 0

*VO 7 0 AC 1

RI 1 2 500

C1 2 3 2.2UF

RG 3 0 1MEG

R4 4 0 11K

C2 5 7 47UF

RD 6 5 39K

R3 7 0 500K

J1 5 3 4 NJFET

.OP


.AC LIN 1 4KHZ 4KHZ

.MODEL NJFET NJF BETA=1.25M VTO=-4 LAMBDA=0.02

.PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VI) IP(VI) IM(C2) IP(C2)

.END


Results: Q-point: (319 A, 4.03V), Av = -3.02, Rin = 1.00 MRout= 38.4 k



14.76

*Problem 14.76 - Common-Base Amplifier - 14.48(h)

VCC 7 0 DC -9

VEE 4 0 DC 9

VI 1 0 AC 1

*For output resistance

*VI 1 0 AC 0

*VO 8 0 AC 1

RI 1 2 500

C2 2 3 47UF

RB 5 0 100K

RE 3 4 82K

C1 5 0 4.7UF

RC 7 6 39K

C3 6 8 10UF

R3 8 0 100K

Q1 6 5 3 PBJT

.OP


.AC LIN 1 12KHZ 12KHZ

.MODEL PBJT PNP IS=1E-16 BF=50 VA=50

.PRINT AC VM(3) VP(3) VM(8) VP(8) IM(VI) IP(VI) IM(C3) IP(C3)

.END


Results: Q-point: (97.2 A, 6.10V), Av = 35.5, Rin = 273 Rout= 38.1 k



14.77

*Problem 14.77 - Common-Source Amplifier 14.48(j)

VDD 6 0 DC 18

VI 1 0 AC 1

*For output resistance

*VI 1 0 AC 0

*VO 7 0 AC 1

RI 1 2 1K

C1 2 3 2.2UF

R2 6 3 2.2MEG

R1 3 0 2.2MEG

RS 6 4 22K

C2 6 4 47UF

RD 5 0 18K

C3 5 7 10UF

R3 7 0 470K

M1 5 3 4 4 PFET

.OP


.AC LIN 1 7.5KHZ 7.5KHZ

.MODEL PFET PMOS KP=400U VTO=-1 LAMBDA=0.02

.PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VI) IP(VI) IM(C3) IP(C3)

.END


Results: Q-point: (268A, 8.60V), Av = -8.29, Rin = 1.10 MRout= 16.4 k





14.78

*Problem 14.78 - Common-Gate Amplifier - 14.48(k)

VDD 5 0 DC 15

VI 1 0 AC 1

*For output resistance

*VI 1 0 AC 0

*VO 6 0 AC 1

RI 1 2 1K

C1 2 3 2.2U

R1 3 0 3.9K

RD 4 5 20K

C2 4 6 47U

RS 6 0 51K

M1 4 0 3 3 NMOSFET

.OP

.AC LIN 1 20KHZ 20KHZ



.MODEL NMOSFET NMOS VTO=-2 KP=500U LAMBDA=0.02

.PRINT AC VM(3) VP(3) VM(6) VP(6) IM(VI) IP(VI) IM(C2) IP(C2)

.END

Results: Q-point: (268A, 8.60V), Av = 4.26, Rin = 1.27 kRout= 18.8 k





14.79

*Problem 14.79 - JFET Common-Source Amplifier - 14.48(m)

VDD 5 0 DC -16

VI 1 0 AC 1

*For output resistance

*VI 1 0 AC 0

*VO 6 0 AC 1

RI 1 2 5K

C1 2 3 2.2U

R1 3 0 10MEG

RD 4 5 1.8K

C2 4 6 10U

R3 6 0 36K

J1 4 3 0 PJFET

.OP

.AC LIN 1 3KHZ 3KHZ



.MODEL PJFET PJF BETA=200U VTO=-5 LAMBDA=0.02

.PRINT AC VM(3) VP(3) VM(6) VP(6) IM(VI) IP(VI) IM(C2) IP(C2)

.END

Results: Q-point: (5.59 A, 5.93 V), Av = -3.27, Rin = 10.0 M Rout= 1.53 k





14.80

*Problem 14.69 - Common-Emitter Amplifier - 14.48(n)

VCC 7 0 DC 10

VEE 6 0 DC -10

VI 1 0 AC 1

*For output resistance

*VI 1 0 AC 0

*VO 8 0 AC 1

RI 1 2 250

C1 2 3 4.7UF

RB 3 0 20K

R4 4 6 9.1K

C2 4 0 100UF

L 7 5 1H

C3 5 8 1UF

R3 8 0 1MEG

Q1 5 3 4 NBJT

.OP

.AC LIN 1 500KHZ 500KHZ



.MODEL NBJT NPN IS=1E-16 BF=80 VA=100

.PRINT AC VM(3) VP(3) VM(8) VP(8) IM(VI) IP(VI) IM(C3) IP(C3)

.END

Results: Q-point: (979 A, 11.0 V), Av = -3420, Rin = 2.09 kRout= 113 k





14.81

*Problem 14.81 - Source Follower - 14.48(o)

VDD 5 0 DC 5

VSS 6 0 DC -5

VI 1 0 AC 1

*For output resistance

*VI 1 0 AC 0

*VO 7 0 AC 1

RI 1 2 10K

C1 2 3 2.2U

R1 3 0 1MEG

L 4 6 100mH

C3 4 7 4.7U

R3 7 0 100K

M1 5 3 4 4 NMOSFET

.OP


.AC LIN 1 100KHZ 100KHZ

.MODEL NMOSFET NMOS VTO=1 KP=400U LAMBDA=0.02

.PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VI) IP(VI) IM(C3) IP(C3)

.END


Results: Q-point: (3.84 mA, 10.0 V), Av = 0.953, Rin = 1.00 M Rout= 504 




14-


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