Consider the circuit shown in Figure 4.14. Assume transistor parameters of V T N = 0.8 V , K n = 0.20 mA/V 2 , and λ = 0 . Let V D D = 5 V , R i = R 1 ∥ R 2 = 200 k Ω , and R S i = 0 . Design the circuit such that I D Q = 0.5 mA and the Q −point is in the center of the saturation region. Find the small−signal voltage gain. (Ans. R D = 2.76 k Ω , R 1 = 420 k Ω , R 2 = 382 k Ω , A υ = − 1.75 )

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Answer 1

Textbook Question

Chapter 4, Problem 4.4EP

Consider the circuit shown in Figure 4.14. Assume transistor parameters of V T N = 0.8 V , K n = 0.20 mA/V 2 , and λ = 0 . Let V D D = 5 V , R i = R 1 ∥ R 2 = 200 k Ω , and R S i = 0 . Design the circuit such that I D Q = 0.5 mA and the Q−point is in the center of the saturation region. Find the small−signal voltage gain. (Ans. R D = 2.76 k Ω , R 1 = 420 k Ω , R 2 = 382 k Ω , A υ = − 1.75 )

Expert Solution & Answer

To determine

The design parameters of the circuit. The small-signal voltage gain.

Answer to Problem 4.4EP

Design parameters are:

RD=2.76 kΩR1=420 kΩR2=382 kΩ

Small-signal voltage gain is Av=−1.74 .

Explanation of Solution

Given Information:

The given circuit diagram is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 1

VTN=0.8 VKn=0.2 mAV2λ=0VDD=5 VRi=R1||R2=200 kΩRSi=0 IDQ=0.5 mA

The Q -point is in the center of the saturation region.

Calculation:

If the Q -point is to be in the middle of saturation region, the current at the transition point must be 2IDQ that is 1 mA.

IDt=2IDQ =2×0.5 mA =1 mA

Calculating VDS(sat) at the transition point.

IDt=Kn( V GSt− V TN)21=0.2( V GSt− V TN)2VGSt−VTN=5VGSt=5+0.8VGSt=3.036 VVDSt=VGSt−VTN =5 =2.236 V

If the Q -point is to be in the middle of saturation region, the value of VDSQ is:

VDSQ=VDSt+V DD−V DSt2VDSQ=2.236+5−2.2362VDSQ=3.618 V

It would yield a 2.76 V peak to peak symmetrical output voltage.

The value of RD is:

VDSQ=VDD−IDQRDRD=V DD−V DSQI DQRD=5−3.6180.5 mARD=2.76 kΩ

The value of VGSQ is determined as follows:

IDQ=Kn( V GSQ− V TN)20.5=0.2( V GSQ−0.8)2( V GSQ−0.8)2=2.5VGSQ=2.5+0.8VGSQ=2.38 V

The figure of load line is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 2

The value of R1,R2 is determined as follows:

VGSQ=( R 2 R 1 + R 2 )(V DD)2.38 =1R1( R 1 R 2 R 1 + R 2 )(V DD)2.38=RiR1(V DD)2.38=200R1×5R1=10002.38R1=420 kΩ

The value of R2 is:

Ri=R1||R2200=R1R2R1+R2200=420R2420+R284000+200R2=420R2R2=84000220R2=382 kΩ

The value of transconductance is:

gm=2KnI DQ=20.2×0.5=2×0.1 mAV=0.632 mAV

The small-signal output resistance is:

ro=1λI DQ=10×I DQ=∞

The small-signal equivalent circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 3

The output voltage is:

Vo=Vds=−(RD||ro)gmVgs ...(1)

The value of Vgs is:

Applying voltage division rule:

Vgs=(R1||R2( R 1 || R 2 )+R Si)Vi

Putting the value of Vgs in equation 1:

Vo=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )ViVoVi=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )VoVi=−(RD)gm( R 1 || R 2 ( R 1 || R 2 )) ( r o =∞ R Si =0)VoVi=−gmRD

The value of small signal voltage gain is:

AV=VoVi =−gmRD =−0.632 ( mAV)×2.76 (kΩ) =−1.74

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Answer 2

Textbook Question

Chapter 4, Problem 4.4EP

Consider the circuit shown in Figure 4.14. Assume transistor parameters of V T N = 0.8 V , K n = 0.20 mA/V 2 , and λ = 0 . Let V D D = 5 V , R i = R 1 ∥ R 2 = 200 k Ω , and R S i = 0 . Design the circuit such that I D Q = 0.5 mA and the Q−point is in the center of the saturation region. Find the small−signal voltage gain. (Ans. R D = 2.76 k Ω , R 1 = 420 k Ω , R 2 = 382 k Ω , A υ = − 1.75 )

Expert Solution & Answer

To determine

The design parameters of the circuit. The small-signal voltage gain.

Answer to Problem 4.4EP

Design parameters are:

RD=2.76 kΩR1=420 kΩR2=382 kΩ

Small-signal voltage gain is Av=−1.74 .

Explanation of Solution

Given Information:

The given circuit diagram is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 1

VTN=0.8 VKn=0.2 mAV2λ=0VDD=5 VRi=R1||R2=200 kΩRSi=0 IDQ=0.5 mA

The Q -point is in the center of the saturation region.

Calculation:

If the Q -point is to be in the middle of saturation region, the current at the transition point must be 2IDQ that is 1 mA.

IDt=2IDQ =2×0.5 mA =1 mA

Calculating VDS(sat) at the transition point.

IDt=Kn( V GSt− V TN)21=0.2( V GSt− V TN)2VGSt−VTN=5VGSt=5+0.8VGSt=3.036 VVDSt=VGSt−VTN =5 =2.236 V

If the Q -point is to be in the middle of saturation region, the value of VDSQ is:

VDSQ=VDSt+V DD−V DSt2VDSQ=2.236+5−2.2362VDSQ=3.618 V

It would yield a 2.76 V peak to peak symmetrical output voltage.

The value of RD is:

VDSQ=VDD−IDQRDRD=V DD−V DSQI DQRD=5−3.6180.5 mARD=2.76 kΩ

The value of VGSQ is determined as follows:

IDQ=Kn( V GSQ− V TN)20.5=0.2( V GSQ−0.8)2( V GSQ−0.8)2=2.5VGSQ=2.5+0.8VGSQ=2.38 V

The figure of load line is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 2

The value of R1,R2 is determined as follows:

VGSQ=( R 2 R 1 + R 2 )(V DD)2.38 =1R1( R 1 R 2 R 1 + R 2 )(V DD)2.38=RiR1(V DD)2.38=200R1×5R1=10002.38R1=420 kΩ

The value of R2 is:

Ri=R1||R2200=R1R2R1+R2200=420R2420+R284000+200R2=420R2R2=84000220R2=382 kΩ

The value of transconductance is:

gm=2KnI DQ=20.2×0.5=2×0.1 mAV=0.632 mAV

The small-signal output resistance is:

ro=1λI DQ=10×I DQ=∞

The small-signal equivalent circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 3

The output voltage is:

Vo=Vds=−(RD||ro)gmVgs ...(1)

The value of Vgs is:

Applying voltage division rule:

Vgs=(R1||R2( R 1 || R 2 )+R Si)Vi

Putting the value of Vgs in equation 1:

Vo=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )ViVoVi=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )VoVi=−(RD)gm( R 1 || R 2 ( R 1 || R 2 )) ( r o =∞ R Si =0)VoVi=−gmRD

The value of small signal voltage gain is:

AV=VoVi =−gmRD =−0.632 ( mAV)×2.76 (kΩ) =−1.74

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Answer 3

Textbook Question

Chapter 4, Problem 4.4EP

Consider the circuit shown in Figure 4.14. Assume transistor parameters of V T N = 0.8 V , K n = 0.20 mA/V 2 , and λ = 0 . Let V D D = 5 V , R i = R 1 ∥ R 2 = 200 k Ω , and R S i = 0 . Design the circuit such that I D Q = 0.5 mA and the Q−point is in the center of the saturation region. Find the small−signal voltage gain. (Ans. R D = 2.76 k Ω , R 1 = 420 k Ω , R 2 = 382 k Ω , A υ = − 1.75 )

Expert Solution & Answer

To determine

The design parameters of the circuit. The small-signal voltage gain.

Answer to Problem 4.4EP

Design parameters are:

RD=2.76 kΩR1=420 kΩR2=382 kΩ

Small-signal voltage gain is Av=−1.74 .

Explanation of Solution

Given Information:

The given circuit diagram is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 1

VTN=0.8 VKn=0.2 mAV2λ=0VDD=5 VRi=R1||R2=200 kΩRSi=0 IDQ=0.5 mA

The Q -point is in the center of the saturation region.

Calculation:

If the Q -point is to be in the middle of saturation region, the current at the transition point must be 2IDQ that is 1 mA.

IDt=2IDQ =2×0.5 mA =1 mA

Calculating VDS(sat) at the transition point.

IDt=Kn( V GSt− V TN)21=0.2( V GSt− V TN)2VGSt−VTN=5VGSt=5+0.8VGSt=3.036 VVDSt=VGSt−VTN =5 =2.236 V

If the Q -point is to be in the middle of saturation region, the value of VDSQ is:

VDSQ=VDSt+V DD−V DSt2VDSQ=2.236+5−2.2362VDSQ=3.618 V

It would yield a 2.76 V peak to peak symmetrical output voltage.

The value of RD is:

VDSQ=VDD−IDQRDRD=V DD−V DSQI DQRD=5−3.6180.5 mARD=2.76 kΩ

The value of VGSQ is determined as follows:

IDQ=Kn( V GSQ− V TN)20.5=0.2( V GSQ−0.8)2( V GSQ−0.8)2=2.5VGSQ=2.5+0.8VGSQ=2.38 V

The figure of load line is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 2

The value of R1,R2 is determined as follows:

VGSQ=( R 2 R 1 + R 2 )(V DD)2.38 =1R1( R 1 R 2 R 1 + R 2 )(V DD)2.38=RiR1(V DD)2.38=200R1×5R1=10002.38R1=420 kΩ

The value of R2 is:

Ri=R1||R2200=R1R2R1+R2200=420R2420+R284000+200R2=420R2R2=84000220R2=382 kΩ

The value of transconductance is:

gm=2KnI DQ=20.2×0.5=2×0.1 mAV=0.632 mAV

The small-signal output resistance is:

ro=1λI DQ=10×I DQ=∞

The small-signal equivalent circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 3

The output voltage is:

Vo=Vds=−(RD||ro)gmVgs ...(1)

The value of Vgs is:

Applying voltage division rule:

Vgs=(R1||R2( R 1 || R 2 )+R Si)Vi

Putting the value of Vgs in equation 1:

Vo=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )ViVoVi=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )VoVi=−(RD)gm( R 1 || R 2 ( R 1 || R 2 )) ( r o =∞ R Si =0)VoVi=−gmRD

The value of small signal voltage gain is:

AV=VoVi =−gmRD =−0.632 ( mAV)×2.76 (kΩ) =−1.74

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Answer 4

Textbook Question

Chapter 4, Problem 4.4EP

Consider the circuit shown in Figure 4.14. Assume transistor parameters of V T N = 0.8 V , K n = 0.20 mA/V 2 , and λ = 0 . Let V D D = 5 V , R i = R 1 ∥ R 2 = 200 k Ω , and R S i = 0 . Design the circuit such that I D Q = 0.5 mA and the Q−point is in the center of the saturation region. Find the small−signal voltage gain. (Ans. R D = 2.76 k Ω , R 1 = 420 k Ω , R 2 = 382 k Ω , A υ = − 1.75 )

Expert Solution & Answer

To determine

The design parameters of the circuit. The small-signal voltage gain.

Answer to Problem 4.4EP

Design parameters are:

RD=2.76 kΩR1=420 kΩR2=382 kΩ

Small-signal voltage gain is Av=−1.74 .

Explanation of Solution

Given Information:

The given circuit diagram is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 1

VTN=0.8 VKn=0.2 mAV2λ=0VDD=5 VRi=R1||R2=200 kΩRSi=0 IDQ=0.5 mA

The Q -point is in the center of the saturation region.

Calculation:

If the Q -point is to be in the middle of saturation region, the current at the transition point must be 2IDQ that is 1 mA.

IDt=2IDQ =2×0.5 mA =1 mA

Calculating VDS(sat) at the transition point.

IDt=Kn( V GSt− V TN)21=0.2( V GSt− V TN)2VGSt−VTN=5VGSt=5+0.8VGSt=3.036 VVDSt=VGSt−VTN =5 =2.236 V

If the Q -point is to be in the middle of saturation region, the value of VDSQ is:

VDSQ=VDSt+V DD−V DSt2VDSQ=2.236+5−2.2362VDSQ=3.618 V

It would yield a 2.76 V peak to peak symmetrical output voltage.

The value of RD is:

VDSQ=VDD−IDQRDRD=V DD−V DSQI DQRD=5−3.6180.5 mARD=2.76 kΩ

The value of VGSQ is determined as follows:

IDQ=Kn( V GSQ− V TN)20.5=0.2( V GSQ−0.8)2( V GSQ−0.8)2=2.5VGSQ=2.5+0.8VGSQ=2.38 V

The figure of load line is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 2

The value of R1,R2 is determined as follows:

VGSQ=( R 2 R 1 + R 2 )(V DD)2.38 =1R1( R 1 R 2 R 1 + R 2 )(V DD)2.38=RiR1(V DD)2.38=200R1×5R1=10002.38R1=420 kΩ

The value of R2 is:

Ri=R1||R2200=R1R2R1+R2200=420R2420+R284000+200R2=420R2R2=84000220R2=382 kΩ

The value of transconductance is:

gm=2KnI DQ=20.2×0.5=2×0.1 mAV=0.632 mAV

The small-signal output resistance is:

ro=1λI DQ=10×I DQ=∞

The small-signal equivalent circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 3

The output voltage is:

Vo=Vds=−(RD||ro)gmVgs ...(1)

The value of Vgs is:

Applying voltage division rule:

Vgs=(R1||R2( R 1 || R 2 )+R Si)Vi

Putting the value of Vgs in equation 1:

Vo=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )ViVoVi=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )VoVi=−(RD)gm( R 1 || R 2 ( R 1 || R 2 )) ( r o =∞ R Si =0)VoVi=−gmRD

The value of small signal voltage gain is:

AV=VoVi =−gmRD =−0.632 ( mAV)×2.76 (kΩ) =−1.74

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The design parameters of the circuit. The small-signal voltage gain.

Answer to Problem 4.4EP

Design parameters are:

RD=2.76 kΩR1=420 kΩR2=382 kΩ

Small-signal voltage gain is Av=−1.74 .

Explanation of Solution

Given Information:

The given circuit diagram is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 1

VTN=0.8 VKn=0.2 mAV2λ=0VDD=5 VRi=R1||R2=200 kΩRSi=0 IDQ=0.5 mA

The Q -point is in the center of the saturation region.

Calculation:

If the Q -point is to be in the middle of saturation region, the current at the transition point must be 2IDQ that is 1 mA.

IDt=2IDQ =2×0.5 mA =1 mA

Calculating VDS(sat) at the transition point.

IDt=Kn( V GSt− V TN)21=0.2( V GSt− V TN)2VGSt−VTN=5VGSt=5+0.8VGSt=3.036 VVDSt=VGSt−VTN =5 =2.236 V

If the Q -point is to be in the middle of saturation region, the value of VDSQ is:

VDSQ=VDSt+V DD−V DSt2VDSQ=2.236+5−2.2362VDSQ=3.618 V

It would yield a 2.76 V peak to peak symmetrical output voltage.

The value of RD is:

VDSQ=VDD−IDQRDRD=V DD−V DSQI DQRD=5−3.6180.5 mARD=2.76 kΩ

The value of VGSQ is determined as follows:

IDQ=Kn( V GSQ− V TN)20.5=0.2( V GSQ−0.8)2( V GSQ−0.8)2=2.5VGSQ=2.5+0.8VGSQ=2.38 V

The figure of load line is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 2

The value of R1,R2 is determined as follows:

VGSQ=( R 2 R 1 + R 2 )(V DD)2.38 =1R1( R 1 R 2 R 1 + R 2 )(V DD)2.38=RiR1(V DD)2.38=200R1×5R1=10002.38R1=420 kΩ

The value of R2 is:

Ri=R1||R2200=R1R2R1+R2200=420R2420+R284000+200R2=420R2R2=84000220R2=382 kΩ

The value of transconductance is:

gm=2KnI DQ=20.2×0.5=2×0.1 mAV=0.632 mAV

The small-signal output resistance is:

ro=1λI DQ=10×I DQ=∞

The small-signal equivalent circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 3

The output voltage is:

Vo=Vds=−(RD||ro)gmVgs ...(1)

The value of Vgs is:

Applying voltage division rule:

Vgs=(R1||R2( R 1 || R 2 )+R Si)Vi

Putting the value of Vgs in equation 1:

Vo=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )ViVoVi=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )VoVi=−(RD)gm( R 1 || R 2 ( R 1 || R 2 )) ( r o =∞ R Si =0)VoVi=−gmRD

The value of small signal voltage gain is:

AV=VoVi =−gmRD =−0.632 ( mAV)×2.76 (kΩ) =−1.74

Answer 8

Expert Solution & Answer

To determine

The design parameters of the circuit. The small-signal voltage gain.

Answer to Problem 4.4EP

Design parameters are:

RD=2.76 kΩR1=420 kΩR2=382 kΩ

Small-signal voltage gain is Av=−1.74 .

Explanation of Solution

Given Information:

The given circuit diagram is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 1

VTN=0.8 VKn=0.2 mAV2λ=0VDD=5 VRi=R1||R2=200 kΩRSi=0 IDQ=0.5 mA

The Q -point is in the center of the saturation region.

Calculation:

If the Q -point is to be in the middle of saturation region, the current at the transition point must be 2IDQ that is 1 mA.

IDt=2IDQ =2×0.5 mA =1 mA

Calculating VDS(sat) at the transition point.

IDt=Kn( V GSt− V TN)21=0.2( V GSt− V TN)2VGSt−VTN=5VGSt=5+0.8VGSt=3.036 VVDSt=VGSt−VTN =5 =2.236 V

If the Q -point is to be in the middle of saturation region, the value of VDSQ is:

VDSQ=VDSt+V DD−V DSt2VDSQ=2.236+5−2.2362VDSQ=3.618 V

It would yield a 2.76 V peak to peak symmetrical output voltage.

The value of RD is:

VDSQ=VDD−IDQRDRD=V DD−V DSQI DQRD=5−3.6180.5 mARD=2.76 kΩ

The value of VGSQ is determined as follows:

IDQ=Kn( V GSQ− V TN)20.5=0.2( V GSQ−0.8)2( V GSQ−0.8)2=2.5VGSQ=2.5+0.8VGSQ=2.38 V

The figure of load line is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 2

The value of R1,R2 is determined as follows:

VGSQ=( R 2 R 1 + R 2 )(V DD)2.38 =1R1( R 1 R 2 R 1 + R 2 )(V DD)2.38=RiR1(V DD)2.38=200R1×5R1=10002.38R1=420 kΩ

The value of R2 is:

Ri=R1||R2200=R1R2R1+R2200=420R2420+R284000+200R2=420R2R2=84000220R2=382 kΩ

The value of transconductance is:

gm=2KnI DQ=20.2×0.5=2×0.1 mAV=0.632 mAV

The small-signal output resistance is:

ro=1λI DQ=10×I DQ=∞

The small-signal equivalent circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 3

The output voltage is:

Vo=Vds=−(RD||ro)gmVgs ...(1)

The value of Vgs is:

Applying voltage division rule:

Vgs=(R1||R2( R 1 || R 2 )+R Si)Vi

Putting the value of Vgs in equation 1:

Vo=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )ViVoVi=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )VoVi=−(RD)gm( R 1 || R 2 ( R 1 || R 2 )) ( r o =∞ R Si =0)VoVi=−gmRD

The value of small signal voltage gain is:

AV=VoVi =−gmRD =−0.632 ( mAV)×2.76 (kΩ) =−1.74

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The design parameters of the circuit. The small-signal voltage gain.

Answer 12

Answer to Problem 4.4EP

Design parameters are:

RD=2.76 kΩR1=420 kΩR2=382 kΩ

Small-signal voltage gain is Av=−1.74 .

Answer 13

Explanation of Solution

Given Information:

The given circuit diagram is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 1

VTN=0.8 VKn=0.2 mAV2λ=0VDD=5 VRi=R1||R2=200 kΩRSi=0 IDQ=0.5 mA

The Q -point is in the center of the saturation region.

Calculation:

If the Q -point is to be in the middle of saturation region, the current at the transition point must be 2IDQ that is 1 mA.

IDt=2IDQ =2×0.5 mA =1 mA

Calculating VDS(sat) at the transition point.

IDt=Kn( V GSt− V TN)21=0.2( V GSt− V TN)2VGSt−VTN=5VGSt=5+0.8VGSt=3.036 VVDSt=VGSt−VTN =5 =2.236 V

If the Q -point is to be in the middle of saturation region, the value of VDSQ is:

VDSQ=VDSt+V DD−V DSt2VDSQ=2.236+5−2.2362VDSQ=3.618 V

It would yield a 2.76 V peak to peak symmetrical output voltage.

The value of RD is:

VDSQ=VDD−IDQRDRD=V DD−V DSQI DQRD=5−3.6180.5 mARD=2.76 kΩ

The value of VGSQ is determined as follows:

IDQ=Kn( V GSQ− V TN)20.5=0.2( V GSQ−0.8)2( V GSQ−0.8)2=2.5VGSQ=2.5+0.8VGSQ=2.38 V

The figure of load line is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 2

The value of R1,R2 is determined as follows:

VGSQ=( R 2 R 1 + R 2 )(V DD)2.38 =1R1( R 1 R 2 R 1 + R 2 )(V DD)2.38=RiR1(V DD)2.38=200R1×5R1=10002.38R1=420 kΩ

The value of R2 is:

Ri=R1||R2200=R1R2R1+R2200=420R2420+R284000+200R2=420R2R2=84000220R2=382 kΩ

The value of transconductance is:

gm=2KnI DQ=20.2×0.5=2×0.1 mAV=0.632 mAV

The small-signal output resistance is:

ro=1λI DQ=10×I DQ=∞

The small-signal equivalent circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.4EP , additional homework tip 3

The output voltage is:

Vo=Vds=−(RD||ro)gmVgs ...(1)

The value of Vgs is:

Applying voltage division rule:

Vgs=(R1||R2( R 1 || R 2 )+R Si)Vi

Putting the value of Vgs in equation 1:

Vo=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )ViVoVi=−(RD||ro)gm( R 1 || R 2 ( R 1 || R 2 )+ R Si )VoVi=−(RD)gm( R 1 || R 2 ( R 1 || R 2 )) ( r o =∞ R Si =0)VoVi=−gmRD

The value of small signal voltage gain is:

AV=VoVi =−gmRD =−0.632 ( mAV)×2.76 (kΩ) =−1.74

Answer 14

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Answer to Problem 4.4EP

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