For the circuit in Figure P4.39, R S = 1 k Ω and the quiescent drain current is I D Q = 5 mA . The transistor parameters are V T N = − 2 V , k ′ n = 100 μ A/V 2 , and λ = 0.01 V − 1 . (a) Determine the transistor width−to− length ratio. (b) Using the results of part (a), find the small−signal voltage gain for R L = ∞ . (c) Find the small−signal output resistance R o . (d) Using the results of part (a), find A υ for R L = 2 k Ω .

Question

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Answer 1

Textbook Question

Chapter 4, Problem 4.40P

For the circuit in Figure P4.39, R S = 1 k Ω and the quiescent drain current is I D Q = 5 mA . The transistor parameters are V T N = − 2 V , k ′ n = 100 μ A/V 2 , and λ = 0.01 V − 1 . (a) Determine the transistor width−to− length ratio. (b) Using the results of part (a), find the small−signal voltage gain for R L = ∞ . (c) Find the small−signal output resistance R o . (d) Using the results of part (a), find A υ for R L = 2 k Ω .

(a)

Expert Solution

To determine

The width to length ratio of the transistor in the given circuit.

Answer to Problem 4.40P

The width to length ratio of the transistors is given by (WL)=23.8

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

Consider the given circuit in Figure 1 which consists a source follower depletion NMOS circuit .

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.40P

The gate terminal gets grounded when we consider the DC equivalent of the above circuit. The gate current is zero and gate source voltage is expressed as,

VGSQ=VGQ−VSQ=0−(−VDD+IDQRS)=−(−5+5×10−3×1×103)=0 V

The drain to source voltage corresponding to operating point is,

VDSQ=VDQ−VSQ=5−0=5 V

Assuming the transistor is in saturation, the drain current at quiescent condition is given by,

IDQ=k'n2WL((VGSQ−VTN)2(1+λVDSQ))=k'n2WL((−VTN)2(1+λVDSQ))

Therefore the above equation can be rearranged to get the W/L ratio as,

WL=2IDQk'n[(−VTN)2(1+λVDSQ)]

Substituting the values of quiescent drain current ,gate-source voltage , threshold voltage and k'n=100 μA/V2

WL=2×5×10−3100×10−6[(−(−2))2(1+0.01×5)]=23.8

(b)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=∞ is given by Av=0.819

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

Here, gm represents the transconductance of the circuit given by,

gm=2k'n2WL(VGSQ−VTN) /

Substituting the value of width to length ratio from part (a) along with the voltages and k'n , the transconductance is obtained as,

gm=100×10−6×23.8(0−(−2))=4.76 mA/V

The transistor output resistance ro is given by,

ro≅1λIDQ=10.01×5×10−3=20 kΩ

Substituting these parameters along with the source resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103)1+4.76×10−3(1×103||20×103)=4.76×10−3×952.381+4.76×10−3×952.38=0.819

(c)

Expert Solution

To determine

The small signal output resistance

Answer to Problem 4.40P

The output voltage is given by Vo=0.297 V

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the small signal output resistance is given by,

Ro=1gm||RS||ro

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ

Substituting these parameters along with the source resistance, small signal output resistance can be calculated as,

Ro=1gm||RS||ro=14.76×10−3||1×103||20×103=172.117 Ω≃172 Ω

(d)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=2 kΩ is given by Av=0.754

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ .

Now, when there is a finite load resistance which is given here as RL=2 kΩ ,the expression for voltage gain gets modified as,

Av=gm(Rs||ro||RL)1+gm(Rs||ro||RL)

Substituting these parameters along with the load resistance resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103||2×103)1+4.76×10−3(1×103||20×103||2×103)=4.76×10−3×645.1611+4.76×10−3×645.161=0.754

It can be observed that, with a finite load resistance, the gain is less compared to the output with infinite load.

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Sketch VL and IL for the given half-wave rectifier.

Answer 2

Textbook Question

Chapter 4, Problem 4.40P

For the circuit in Figure P4.39, R S = 1 k Ω and the quiescent drain current is I D Q = 5 mA . The transistor parameters are V T N = − 2 V , k ′ n = 100 μ A/V 2 , and λ = 0.01 V − 1 . (a) Determine the transistor width−to− length ratio. (b) Using the results of part (a), find the small−signal voltage gain for R L = ∞ . (c) Find the small−signal output resistance R o . (d) Using the results of part (a), find A υ for R L = 2 k Ω .

(a)

Expert Solution

To determine

The width to length ratio of the transistor in the given circuit.

Answer to Problem 4.40P

The width to length ratio of the transistors is given by (WL)=23.8

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

Consider the given circuit in Figure 1 which consists a source follower depletion NMOS circuit .

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.40P

The gate terminal gets grounded when we consider the DC equivalent of the above circuit. The gate current is zero and gate source voltage is expressed as,

VGSQ=VGQ−VSQ=0−(−VDD+IDQRS)=−(−5+5×10−3×1×103)=0 V

The drain to source voltage corresponding to operating point is,

VDSQ=VDQ−VSQ=5−0=5 V

Assuming the transistor is in saturation, the drain current at quiescent condition is given by,

IDQ=k'n2WL((VGSQ−VTN)2(1+λVDSQ))=k'n2WL((−VTN)2(1+λVDSQ))

Therefore the above equation can be rearranged to get the W/L ratio as,

WL=2IDQk'n[(−VTN)2(1+λVDSQ)]

Substituting the values of quiescent drain current ,gate-source voltage , threshold voltage and k'n=100 μA/V2

WL=2×5×10−3100×10−6[(−(−2))2(1+0.01×5)]=23.8

(b)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=∞ is given by Av=0.819

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

Here, gm represents the transconductance of the circuit given by,

gm=2k'n2WL(VGSQ−VTN) /

Substituting the value of width to length ratio from part (a) along with the voltages and k'n , the transconductance is obtained as,

gm=100×10−6×23.8(0−(−2))=4.76 mA/V

The transistor output resistance ro is given by,

ro≅1λIDQ=10.01×5×10−3=20 kΩ

Substituting these parameters along with the source resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103)1+4.76×10−3(1×103||20×103)=4.76×10−3×952.381+4.76×10−3×952.38=0.819

(c)

Expert Solution

To determine

The small signal output resistance

Answer to Problem 4.40P

The output voltage is given by Vo=0.297 V

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the small signal output resistance is given by,

Ro=1gm||RS||ro

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ

Substituting these parameters along with the source resistance, small signal output resistance can be calculated as,

Ro=1gm||RS||ro=14.76×10−3||1×103||20×103=172.117 Ω≃172 Ω

(d)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=2 kΩ is given by Av=0.754

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ .

Now, when there is a finite load resistance which is given here as RL=2 kΩ ,the expression for voltage gain gets modified as,

Av=gm(Rs||ro||RL)1+gm(Rs||ro||RL)

Substituting these parameters along with the load resistance resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103||2×103)1+4.76×10−3(1×103||20×103||2×103)=4.76×10−3×645.1611+4.76×10−3×645.161=0.754

It can be observed that, with a finite load resistance, the gain is less compared to the output with infinite load.

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Answer 3

Textbook Question

Chapter 4, Problem 4.40P

For the circuit in Figure P4.39, R S = 1 k Ω and the quiescent drain current is I D Q = 5 mA . The transistor parameters are V T N = − 2 V , k ′ n = 100 μ A/V 2 , and λ = 0.01 V − 1 . (a) Determine the transistor width−to− length ratio. (b) Using the results of part (a), find the small−signal voltage gain for R L = ∞ . (c) Find the small−signal output resistance R o . (d) Using the results of part (a), find A υ for R L = 2 k Ω .

(a)

Expert Solution

To determine

The width to length ratio of the transistor in the given circuit.

Answer to Problem 4.40P

The width to length ratio of the transistors is given by (WL)=23.8

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

Consider the given circuit in Figure 1 which consists a source follower depletion NMOS circuit .

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.40P

The gate terminal gets grounded when we consider the DC equivalent of the above circuit. The gate current is zero and gate source voltage is expressed as,

VGSQ=VGQ−VSQ=0−(−VDD+IDQRS)=−(−5+5×10−3×1×103)=0 V

The drain to source voltage corresponding to operating point is,

VDSQ=VDQ−VSQ=5−0=5 V

Assuming the transistor is in saturation, the drain current at quiescent condition is given by,

IDQ=k'n2WL((VGSQ−VTN)2(1+λVDSQ))=k'n2WL((−VTN)2(1+λVDSQ))

Therefore the above equation can be rearranged to get the W/L ratio as,

WL=2IDQk'n[(−VTN)2(1+λVDSQ)]

Substituting the values of quiescent drain current ,gate-source voltage , threshold voltage and k'n=100 μA/V2

WL=2×5×10−3100×10−6[(−(−2))2(1+0.01×5)]=23.8

(b)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=∞ is given by Av=0.819

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

Here, gm represents the transconductance of the circuit given by,

gm=2k'n2WL(VGSQ−VTN) /

Substituting the value of width to length ratio from part (a) along with the voltages and k'n , the transconductance is obtained as,

gm=100×10−6×23.8(0−(−2))=4.76 mA/V

The transistor output resistance ro is given by,

ro≅1λIDQ=10.01×5×10−3=20 kΩ

Substituting these parameters along with the source resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103)1+4.76×10−3(1×103||20×103)=4.76×10−3×952.381+4.76×10−3×952.38=0.819

(c)

Expert Solution

To determine

The small signal output resistance

Answer to Problem 4.40P

The output voltage is given by Vo=0.297 V

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the small signal output resistance is given by,

Ro=1gm||RS||ro

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ

Substituting these parameters along with the source resistance, small signal output resistance can be calculated as,

Ro=1gm||RS||ro=14.76×10−3||1×103||20×103=172.117 Ω≃172 Ω

(d)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=2 kΩ is given by Av=0.754

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ .

Now, when there is a finite load resistance which is given here as RL=2 kΩ ,the expression for voltage gain gets modified as,

Av=gm(Rs||ro||RL)1+gm(Rs||ro||RL)

Substituting these parameters along with the load resistance resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103||2×103)1+4.76×10−3(1×103||20×103||2×103)=4.76×10−3×645.1611+4.76×10−3×645.161=0.754

It can be observed that, with a finite load resistance, the gain is less compared to the output with infinite load.

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Answer 4

Textbook Question

Chapter 4, Problem 4.40P

For the circuit in Figure P4.39, R S = 1 k Ω and the quiescent drain current is I D Q = 5 mA . The transistor parameters are V T N = − 2 V , k ′ n = 100 μ A/V 2 , and λ = 0.01 V − 1 . (a) Determine the transistor width−to− length ratio. (b) Using the results of part (a), find the small−signal voltage gain for R L = ∞ . (c) Find the small−signal output resistance R o . (d) Using the results of part (a), find A υ for R L = 2 k Ω .

(a)

Expert Solution

To determine

The width to length ratio of the transistor in the given circuit.

Answer to Problem 4.40P

The width to length ratio of the transistors is given by (WL)=23.8

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

Consider the given circuit in Figure 1 which consists a source follower depletion NMOS circuit .

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.40P

The gate terminal gets grounded when we consider the DC equivalent of the above circuit. The gate current is zero and gate source voltage is expressed as,

VGSQ=VGQ−VSQ=0−(−VDD+IDQRS)=−(−5+5×10−3×1×103)=0 V

The drain to source voltage corresponding to operating point is,

VDSQ=VDQ−VSQ=5−0=5 V

Assuming the transistor is in saturation, the drain current at quiescent condition is given by,

IDQ=k'n2WL((VGSQ−VTN)2(1+λVDSQ))=k'n2WL((−VTN)2(1+λVDSQ))

Therefore the above equation can be rearranged to get the W/L ratio as,

WL=2IDQk'n[(−VTN)2(1+λVDSQ)]

Substituting the values of quiescent drain current ,gate-source voltage , threshold voltage and k'n=100 μA/V2

WL=2×5×10−3100×10−6[(−(−2))2(1+0.01×5)]=23.8

(b)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=∞ is given by Av=0.819

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

Here, gm represents the transconductance of the circuit given by,

gm=2k'n2WL(VGSQ−VTN) /

Substituting the value of width to length ratio from part (a) along with the voltages and k'n , the transconductance is obtained as,

gm=100×10−6×23.8(0−(−2))=4.76 mA/V

The transistor output resistance ro is given by,

ro≅1λIDQ=10.01×5×10−3=20 kΩ

Substituting these parameters along with the source resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103)1+4.76×10−3(1×103||20×103)=4.76×10−3×952.381+4.76×10−3×952.38=0.819

(c)

Expert Solution

To determine

The small signal output resistance

Answer to Problem 4.40P

The output voltage is given by Vo=0.297 V

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the small signal output resistance is given by,

Ro=1gm||RS||ro

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ

Substituting these parameters along with the source resistance, small signal output resistance can be calculated as,

Ro=1gm||RS||ro=14.76×10−3||1×103||20×103=172.117 Ω≃172 Ω

(d)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=2 kΩ is given by Av=0.754

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ .

Now, when there is a finite load resistance which is given here as RL=2 kΩ ,the expression for voltage gain gets modified as,

Av=gm(Rs||ro||RL)1+gm(Rs||ro||RL)

Substituting these parameters along with the load resistance resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103||2×103)1+4.76×10−3(1×103||20×103||2×103)=4.76×10−3×645.1611+4.76×10−3×645.161=0.754

It can be observed that, with a finite load resistance, the gain is less compared to the output with infinite load.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The width to length ratio of the transistor in the given circuit.

Answer to Problem 4.40P

The width to length ratio of the transistors is given by (WL)=23.8

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

Consider the given circuit in Figure 1 which consists a source follower depletion NMOS circuit .

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.40P

The gate terminal gets grounded when we consider the DC equivalent of the above circuit. The gate current is zero and gate source voltage is expressed as,

VGSQ=VGQ−VSQ=0−(−VDD+IDQRS)=−(−5+5×10−3×1×103)=0 V

The drain to source voltage corresponding to operating point is,

VDSQ=VDQ−VSQ=5−0=5 V

Assuming the transistor is in saturation, the drain current at quiescent condition is given by,

IDQ=k'n2WL((VGSQ−VTN)2(1+λVDSQ))=k'n2WL((−VTN)2(1+λVDSQ))

Therefore the above equation can be rearranged to get the W/L ratio as,

WL=2IDQk'n[(−VTN)2(1+λVDSQ)]

Substituting the values of quiescent drain current ,gate-source voltage , threshold voltage and k'n=100 μA/V2

WL=2×5×10−3100×10−6[(−(−2))2(1+0.01×5)]=23.8

(b)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=∞ is given by Av=0.819

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

Here, gm represents the transconductance of the circuit given by,

gm=2k'n2WL(VGSQ−VTN) /

Substituting the value of width to length ratio from part (a) along with the voltages and k'n , the transconductance is obtained as,

gm=100×10−6×23.8(0−(−2))=4.76 mA/V

The transistor output resistance ro is given by,

ro≅1λIDQ=10.01×5×10−3=20 kΩ

Substituting these parameters along with the source resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103)1+4.76×10−3(1×103||20×103)=4.76×10−3×952.381+4.76×10−3×952.38=0.819

(c)

Expert Solution

To determine

The small signal output resistance

Answer to Problem 4.40P

The output voltage is given by Vo=0.297 V

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the small signal output resistance is given by,

Ro=1gm||RS||ro

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ

Substituting these parameters along with the source resistance, small signal output resistance can be calculated as,

Ro=1gm||RS||ro=14.76×10−3||1×103||20×103=172.117 Ω≃172 Ω

(d)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=2 kΩ is given by Av=0.754

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ .

Now, when there is a finite load resistance which is given here as RL=2 kΩ ,the expression for voltage gain gets modified as,

Av=gm(Rs||ro||RL)1+gm(Rs||ro||RL)

Substituting these parameters along with the load resistance resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103||2×103)1+4.76×10−3(1×103||20×103||2×103)=4.76×10−3×645.1611+4.76×10−3×645.161=0.754

It can be observed that, with a finite load resistance, the gain is less compared to the output with infinite load.

Answer 8

(a)

Expert Solution

To determine

The width to length ratio of the transistor in the given circuit.

Answer to Problem 4.40P

The width to length ratio of the transistors is given by (WL)=23.8

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

Consider the given circuit in Figure 1 which consists a source follower depletion NMOS circuit .

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.40P

The gate terminal gets grounded when we consider the DC equivalent of the above circuit. The gate current is zero and gate source voltage is expressed as,

VGSQ=VGQ−VSQ=0−(−VDD+IDQRS)=−(−5+5×10−3×1×103)=0 V

The drain to source voltage corresponding to operating point is,

VDSQ=VDQ−VSQ=5−0=5 V

Assuming the transistor is in saturation, the drain current at quiescent condition is given by,

IDQ=k'n2WL((VGSQ−VTN)2(1+λVDSQ))=k'n2WL((−VTN)2(1+λVDSQ))

Therefore the above equation can be rearranged to get the W/L ratio as,

WL=2IDQk'n[(−VTN)2(1+λVDSQ)]

Substituting the values of quiescent drain current ,gate-source voltage , threshold voltage and k'n=100 μA/V2

WL=2×5×10−3100×10−6[(−(−2))2(1+0.01×5)]=23.8

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The width to length ratio of the transistor in the given circuit.

Answer 13

Answer to Problem 4.40P

The width to length ratio of the transistors is given by (WL)=23.8

Answer 14

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

Consider the given circuit in Figure 1 which consists a source follower depletion NMOS circuit .

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.40P

The gate terminal gets grounded when we consider the DC equivalent of the above circuit. The gate current is zero and gate source voltage is expressed as,

VGSQ=VGQ−VSQ=0−(−VDD+IDQRS)=−(−5+5×10−3×1×103)=0 V

The drain to source voltage corresponding to operating point is,

VDSQ=VDQ−VSQ=5−0=5 V

Assuming the transistor is in saturation, the drain current at quiescent condition is given by,

IDQ=k'n2WL((VGSQ−VTN)2(1+λVDSQ))=k'n2WL((−VTN)2(1+λVDSQ))

Therefore the above equation can be rearranged to get the W/L ratio as,

WL=2IDQk'n[(−VTN)2(1+λVDSQ)]

Substituting the values of quiescent drain current ,gate-source voltage , threshold voltage and k'n=100 μA/V2

WL=2×5×10−3100×10−6[(−(−2))2(1+0.01×5)]=23.8

Answer 15

(b)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=∞ is given by Av=0.819

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

Here, gm represents the transconductance of the circuit given by,

gm=2k'n2WL(VGSQ−VTN) /

Substituting the value of width to length ratio from part (a) along with the voltages and k'n , the transconductance is obtained as,

gm=100×10−6×23.8(0−(−2))=4.76 mA/V

The transistor output resistance ro is given by,

ro≅1λIDQ=10.01×5×10−3=20 kΩ

Substituting these parameters along with the source resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103)1+4.76×10−3(1×103||20×103)=4.76×10−3×952.381+4.76×10−3×952.38=0.819

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The small signal voltage gain for a load resistance.

Answer 20

Answer to Problem 4.40P

The small signal voltage gain for RL=∞ is given by Av=0.819

Answer 21

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

Here, gm represents the transconductance of the circuit given by,

gm=2k'n2WL(VGSQ−VTN) /

Substituting the value of width to length ratio from part (a) along with the voltages and k'n , the transconductance is obtained as,

gm=100×10−6×23.8(0−(−2))=4.76 mA/V

The transistor output resistance ro is given by,

ro≅1λIDQ=10.01×5×10−3=20 kΩ

Substituting these parameters along with the source resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103)1+4.76×10−3(1×103||20×103)=4.76×10−3×952.381+4.76×10−3×952.38=0.819

Answer 22

(c)

Expert Solution

To determine

The small signal output resistance

Answer to Problem 4.40P

The output voltage is given by Vo=0.297 V

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the small signal output resistance is given by,

Ro=1gm||RS||ro

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ

Substituting these parameters along with the source resistance, small signal output resistance can be calculated as,

Ro=1gm||RS||ro=14.76×10−3||1×103||20×103=172.117 Ω≃172 Ω

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The small signal output resistance

Answer 27

Answer to Problem 4.40P

The output voltage is given by Vo=0.297 V

Answer 28

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the small signal output resistance is given by,

Ro=1gm||RS||ro

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ

Substituting these parameters along with the source resistance, small signal output resistance can be calculated as,

Ro=1gm||RS||ro=14.76×10−3||1×103||20×103=172.117 Ω≃172 Ω

Answer 29

(d)

Expert Solution

To determine

The small signal voltage gain for a load resistance.

Answer to Problem 4.40P

The small signal voltage gain for RL=2 kΩ is given by Av=0.754

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ .

Now, when there is a finite load resistance which is given here as RL=2 kΩ ,the expression for voltage gain gets modified as,

Av=gm(Rs||ro||RL)1+gm(Rs||ro||RL)

Substituting these parameters along with the load resistance resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103||2×103)1+4.76×10−3(1×103||20×103||2×103)=4.76×10−3×645.1611+4.76×10−3×645.161=0.754

It can be observed that, with a finite load resistance, the gain is less compared to the output with infinite load.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The small signal voltage gain for a load resistance.

Answer 34

Answer to Problem 4.40P

The small signal voltage gain for RL=2 kΩ is given by Av=0.754

Answer 35

Explanation of Solution

Given Information:

A source follower depletion NMOS circuit with transistor parameters VTN=−2 V , k'n=100 μA/V2 and λ=0.01 V−1 . The quiescent drain current is IDQ=5 mA

Calculation:

For the circuit in Figure 1, the voltage gain is given by,

Av=gm(Rs||ro)1+gm(Rs||ro)

The transconductance of the circuit was obtained in previous part as gm=4.76 mA/V and the transistor output resistance ro is given by ro=20 kΩ .

Now, when there is a finite load resistance which is given here as RL=2 kΩ ,the expression for voltage gain gets modified as,

Av=gm(Rs||ro||RL)1+gm(Rs||ro||RL)

Substituting these parameters along with the load resistance resistance, the voltage gain can be calculated as,

Av=4.76×10−3(1×103||20×103||2×103)1+4.76×10−3(1×103||20×103||2×103)=4.76×10−3×645.1611+4.76×10−3×645.161=0.754

It can be observed that, with a finite load resistance, the gain is less compared to the output with infinite load.

Answer 36

Ch. 4 - Prob. 4.1EP Ch. 4 - For the circuit shown in Figure 4.1, VDD=3.3V and...Ch. 4 - Prob. 4.1TYU Ch. 4 - For the circuit shown in Figure 4.1, VDD=3.3V and...Ch. 4 - For the circuit in Figure 4.1, the circuit and...Ch. 4 - The parameters for the circuit in Figure 4.8 are...Ch. 4 - A transistor has the same parameters as those...Ch. 4 - The parameters of the circuit shown in Figure 4.14...Ch. 4 - Consider the circuit shown in Figure 4.14. Assume...Ch. 4 - For the circuit shown in Figure 4.19, the...

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