The small-signal voltage gain for given R L .

Question

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Answer 1

Question

Chapter 10, Problem 10.81P

(a)

To determine

The small-signal voltage gain for given $RL$ .

(a)

Expert Solution

Answer to Problem 10.81P

$Av=−1978$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 1

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=∞$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1∞)Av=−(1.964×10−3)(9.928×10−7+0)Av=−(1.964×10−3)(9.928×10−7)$

$Av=−1978$

Conclusion:

$Av=−1978$

(b)

To determine

The small-signal voltage gain for given $RL$ .

(b)

Expert Solution

Answer to Problem 10.81P

$Av=−454$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 2

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=300kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1300×103)Av=−(1.964×10−3)(9.928×10−7+3.33×10−6)Av=−(1.964×10−3)(4.326×10−6)$

$Av=−454$

Conclusion:

$Av=−454$

(c)

To determine

The small-signal voltage gain for given $RL$ .

(c)

Expert Solution

Answer to Problem 10.81P

$Av=−256$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 3

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=150kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1150×103)Av=−(1.964×10−3)(9.928×10−7+6.66×10−6)Av=−(1.964×10−3)(7.659×10−6)$

$Av=−256$

Conclusion:

$Av=−256$

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Answer 2

Question

Chapter 10, Problem 10.81P

(a)

To determine

The small-signal voltage gain for given $RL$ .

(a)

Expert Solution

Answer to Problem 10.81P

$Av=−1978$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 1

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=∞$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1∞)Av=−(1.964×10−3)(9.928×10−7+0)Av=−(1.964×10−3)(9.928×10−7)$

$Av=−1978$

Conclusion:

$Av=−1978$

(b)

To determine

The small-signal voltage gain for given $RL$ .

(b)

Expert Solution

Answer to Problem 10.81P

$Av=−454$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 2

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=300kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1300×103)Av=−(1.964×10−3)(9.928×10−7+3.33×10−6)Av=−(1.964×10−3)(4.326×10−6)$

$Av=−454$

Conclusion:

$Av=−454$

(c)

To determine

The small-signal voltage gain for given $RL$ .

(c)

Expert Solution

Answer to Problem 10.81P

$Av=−256$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 3

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=150kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1150×103)Av=−(1.964×10−3)(9.928×10−7+6.66×10−6)Av=−(1.964×10−3)(7.659×10−6)$

$Av=−256$

Conclusion:

$Av=−256$

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Answer 3

Question

Chapter 10, Problem 10.81P

(a)

To determine

The small-signal voltage gain for given $RL$ .

(a)

Expert Solution

Answer to Problem 10.81P

$Av=−1978$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 1

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=∞$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1∞)Av=−(1.964×10−3)(9.928×10−7+0)Av=−(1.964×10−3)(9.928×10−7)$

$Av=−1978$

Conclusion:

$Av=−1978$

(b)

To determine

The small-signal voltage gain for given $RL$ .

(b)

Expert Solution

Answer to Problem 10.81P

$Av=−454$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 2

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=300kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1300×103)Av=−(1.964×10−3)(9.928×10−7+3.33×10−6)Av=−(1.964×10−3)(4.326×10−6)$

$Av=−454$

Conclusion:

$Av=−454$

(c)

To determine

The small-signal voltage gain for given $RL$ .

(c)

Expert Solution

Answer to Problem 10.81P

$Av=−256$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 3

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=150kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1150×103)Av=−(1.964×10−3)(9.928×10−7+6.66×10−6)Av=−(1.964×10−3)(7.659×10−6)$

$Av=−256$

Conclusion:

$Av=−256$

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Answer 4

Question

Chapter 10, Problem 10.81P

(a)

To determine

The small-signal voltage gain for given $RL$ .

(a)

Expert Solution

Answer to Problem 10.81P

$Av=−1978$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 1

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=∞$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1∞)Av=−(1.964×10−3)(9.928×10−7+0)Av=−(1.964×10−3)(9.928×10−7)$

$Av=−1978$

Conclusion:

$Av=−1978$

(b)

To determine

The small-signal voltage gain for given $RL$ .

(b)

Expert Solution

Answer to Problem 10.81P

$Av=−454$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 2

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=300kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1300×103)Av=−(1.964×10−3)(9.928×10−7+3.33×10−6)Av=−(1.964×10−3)(4.326×10−6)$

$Av=−454$

Conclusion:

$Av=−454$

(c)

To determine

The small-signal voltage gain for given $RL$ .

(c)

Expert Solution

Answer to Problem 10.81P

$Av=−256$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 3

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=150kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1150×103)Av=−(1.964×10−3)(9.928×10−7+6.66×10−6)Av=−(1.964×10−3)(7.659×10−6)$

$Av=−256$

Conclusion:

$Av=−256$

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Answer 5

Chapter 10, Problem 10.81P

(a)

To determine

The small-signal voltage gain for given $RL$ .

(a)

Expert Solution

Answer to Problem 10.81P

$Av=−1978$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 1

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=∞$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1∞)Av=−(1.964×10−3)(9.928×10−7+0)Av=−(1.964×10−3)(9.928×10−7)$

$Av=−1978$

Conclusion:

$Av=−1978$

(b)

To determine

The small-signal voltage gain for given $RL$ .

(b)

Expert Solution

Answer to Problem 10.81P

$Av=−454$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 2

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=300kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1300×103)Av=−(1.964×10−3)(9.928×10−7+3.33×10−6)Av=−(1.964×10−3)(4.326×10−6)$

$Av=−454$

Conclusion:

$Av=−454$

(c)

To determine

The small-signal voltage gain for given $RL$ .

(c)

Expert Solution

Answer to Problem 10.81P

$Av=−256$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 3

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=150kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1150×103)Av=−(1.964×10−3)(9.928×10−7+6.66×10−6)Av=−(1.964×10−3)(7.659×10−6)$

$Av=−256$

Conclusion:

$Av=−256$

Answer 6

Chapter 10, Problem 10.81P

(a)

To determine

The small-signal voltage gain for given $RL$ .

(a)

Expert Solution

Answer to Problem 10.81P

$Av=−1978$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 1

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=∞$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1∞)Av=−(1.964×10−3)(9.928×10−7+0)Av=−(1.964×10−3)(9.928×10−7)$

$Av=−1978$

Conclusion:

$Av=−1978$

Answer 7

Chapter 10, Problem 10.81P

(a)

To determine

The small-signal voltage gain for given $RL$ .

Answer 8

(a)

Expert Solution

Answer to Problem 10.81P

$Av=−1978$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 1

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=∞$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1∞)Av=−(1.964×10−3)(9.928×10−7+0)Av=−(1.964×10−3)(9.928×10−7)$

$Av=−1978$

Conclusion:

$Av=−1978$

Answer 13

(b)

To determine

The small-signal voltage gain for given $RL$ .

(b)

Expert Solution

Answer to Problem 10.81P

$Av=−454$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 2

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=300kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1300×103)Av=−(1.964×10−3)(9.928×10−7+3.33×10−6)Av=−(1.964×10−3)(4.326×10−6)$

$Av=−454$

Conclusion:

$Av=−454$

Answer 14

(b)

To determine

The small-signal voltage gain for given $RL$ .

Answer 15

(b)

Expert Solution

Answer to Problem 10.81P

$Av=−454$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 2

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=300kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1300×103)Av=−(1.964×10−3)(9.928×10−7+3.33×10−6)Av=−(1.964×10−3)(4.326×10−6)$

$Av=−454$

Conclusion:

$Av=−454$

Answer 20

(c)

To determine

The small-signal voltage gain for given $RL$ .

(c)

Expert Solution

Answer to Problem 10.81P

$Av=−256$

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 3

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=150kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1150×103)Av=−(1.964×10−3)(9.928×10−7+6.66×10−6)Av=−(1.964×10−3)(7.659×10−6)$

$Av=−256$

Conclusion:

$Av=−256$

Answer 21

(c)

To determine

The small-signal voltage gain for given $RL$ .

Answer 22

(c)

Expert Solution

Answer to Problem 10.81P

$Av=−256$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

$R1=47kΩVAN=120VVAP=90VV+=3VVEB(on)=0.6V$

Calculation:

The given circuit is,

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.81P , additional homework tip 3

The transistor $Q1$ and $Q2$ are matched so expression for reference current will be,

$IREF=V+−VEB(on)R1$

$IREF=3−0.647×103IREF=2.447×103IREF=51.06μA$

Now calculate the small-signal voltage gain,

$Av=−(IREFVT)(IREFVAN+1RL+IREFVAP)$

Substitute the given values,

$Av=−(51.06×10−60.026)(51.06×10−6120+1RL+51.06×10−690)Av=−(1.964×10−3)(4.255×10−7+1RL+5.673×10−7)Av=−(1.964×10−3)(9.928×10−7+1RL) (1)$

Substitute $RL=150kΩ$ in equation (1)

$Av=−(1.964×10−3)(9.928×10−7+1150×103)Av=−(1.964×10−3)(9.928×10−7+6.66×10−6)Av=−(1.964×10−3)(7.659×10−6)$

$Av=−256$

Conclusion:

$Av=−256$

Answer 27

Ch. 10 - The circuit parameters for the two-transistor...Ch. 10 - Consider the circuit shown in Figure 10.3. The...Ch. 10 - The parameters of the circuit shown in Figure 10.5...Ch. 10 - Consider the Widlar current source in Figure 10.9....Ch. 10 - Consider the circuit in Figure 10.10. Assume the...Ch. 10 - A Widlar current source is shown in Figure 10.9....Ch. 10 - Figure 10.12 shows the N-output current mirror....Ch. 10 - Prob. 10.1TYU Ch. 10 - Prob. 10.2TYU Ch. 10 - For the Wilson current source in Figure 10.8, the...

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