Consider the circuit shown in Figure 4.28 with circuit parameters V D D = 5 V , R S = 5 k Ω , R 1 = 70.7 k Ω , R 2 = 9.3 k Ω , and R S i = 500 Ω . The transistor parameters are: V T P = − 0.8 V , K p = 0.4 mA/V 2 ,and λ = 0 . Calculate the small−signal voltage gain A υ = υ o / υ i and the output resistance R o seen looking back into the circuit. (Ans. A υ = 0.817 , R o = 0.915 k Ω )

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Answer 1

Textbook Question

Chapter 4, Problem 4.9EP

Consider the circuit shown in Figure 4.28 with circuit parameters $V D D = 5 V$ , $R S = 5 k Ω$ , $R 1 = 70.7 k Ω$ , $R 2 = 9.3 k Ω$ , and $R S i = 500 Ω$ . The transistor parameters are: $V T P = − 0.8 V$ , $K p = 0.4 mA/V 2$ ,and $λ = 0$ . Calculate the small−signal voltage gain $A υ = υ o / υ i$ and the output resistance $R o$ seen looking back into the circuit. (Ans. $A υ = 0.817$ , $R o = 0.915 k Ω$ )

Expert Solution & Answer

To determine

The values of $Av,Ro$ .

Answer to Problem 4.9EP

$Av=0.77Ro=0.914 kΩ$

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 1

$VDD=5 V, RS=5 kΩR1=70.7 kΩ, R2=9.3 kΩRsi=500 Ω, λ=0VTP=−0.8 V, KP=0.4 mAV2$

Calculation:

The coupling capacitor acts like open circuit for DC value calculation.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 2

The value of gate voltage is:

$VG=R2R1+R2×VDDVG=9.370.7+9.3×5VG=9.3×580VG=0.58 V$

From the circuit:

$Vs=VDD−IDRSVs=5−5×103ID$

The value of $VSG$ is:

$VSG=VS−VGVSG=5−5×103ID−0.58VSG=4.42−5×103ID$

Assuming the transistor operates in saturation region:

$ID=KP( V SG+ V TP)2ID=0.4×10−3(4.42−5× 10 3 I D−0.8)2ID=0.4×10−3(3.62−5× 10 3 I D)2104ID2−15.48ID+5.24×10−3=0ID=1.048 mA,0.5 mA$

The value of $VSG$ is:

$VSG=4.42−5×103IDID=1.048 mA,VSG=4.42−5×103×1.048×10−3VSG=−0.82VSG<|VTP|$

Hence, the transistor would be in cutoff mode for $ID=1.048 mA$ .

Plugging $ID=0.5 mA,$

$VSG=4.42−5×103×0.5×10−3VSG=1.92 VVSG>|VTP|$

The value of $VSD$ is:

$VSD=VS−VDVSD=5−5×103ID−0VSD=5−5×103×0.5×10−3VSD=2.5 V$

$VSD>VSG+VTP$

Hence, the assumption is correct and transistor operates in saturation region.

The DC voltage source and coupling capacitor are short-circuited for small-signal equivalent circuit. It is common drain amplifier.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 3

The value of $gm,ro$ is:

$gm=2KPIDgm=20.4× 10 −3×0.5× 10 −3gm=0.8944 mAVro=1λIDro=10×IDro=∞$

The value of output voltage is:

$Vo=(−gmV sg)(Rs||ro)Vo=(−gmV sg)Rs ...(1)$

Applying Kirchhoff’s voltage law from input to output:

$Vin=−Vsg+VoVin=−Vsg−(gmV sg)(Rs)Vsg=−V in1+gm( R s )$

Applying voltage division rule in input:

$Vin=(R1||R2)Rsi+(R1||R2)Vi$

Hence, the value of $Vsg$ is:

$Vsg=−(R1||R2)(1+gmRs)(R si+( R 1 || R 2 ))Vi$

From equation (1):

$Vo=−gm(Rs)−( R 1 || R 2 )( 1+ g m R s )( R si +( R 1 || R 2 ))ViVoVi=gmRs( 1+ g m R s )( R i R si + R i )(Ri=R1||R2)Av=0.8944×51+0.8944×5( 8.22 0.5+8.22)Av=0.77$

The value of output resistance is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 4

Applying Kirchhoff’s current law at output node:

$Ix=gmVsg+VxRs+Vxro$

The value of current in input circuit is zero.

$Vsg=Vx$

Plugging the value:

$Ix=gmVx+VxRs+VxroIx=Vx(gm+1 R s +1 r o )IxVx=gm+1Rs+1ro1Ro=gm+1Rs+1roRo=1gm||Rs||ro$

Plugging the values:

$Ro=1gm||Rs||roRo=10.8944||5 (ro=∞)Ro=0.914 kΩ$

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Answer 2

Textbook Question

Chapter 4, Problem 4.9EP

Consider the circuit shown in Figure 4.28 with circuit parameters $V D D = 5 V$ , $R S = 5 k Ω$ , $R 1 = 70.7 k Ω$ , $R 2 = 9.3 k Ω$ , and $R S i = 500 Ω$ . The transistor parameters are: $V T P = − 0.8 V$ , $K p = 0.4 mA/V 2$ ,and $λ = 0$ . Calculate the small−signal voltage gain $A υ = υ o / υ i$ and the output resistance $R o$ seen looking back into the circuit. (Ans. $A υ = 0.817$ , $R o = 0.915 k Ω$ )

Expert Solution & Answer

To determine

The values of $Av,Ro$ .

Answer to Problem 4.9EP

$Av=0.77Ro=0.914 kΩ$

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 1

$VDD=5 V, RS=5 kΩR1=70.7 kΩ, R2=9.3 kΩRsi=500 Ω, λ=0VTP=−0.8 V, KP=0.4 mAV2$

Calculation:

The coupling capacitor acts like open circuit for DC value calculation.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 2

The value of gate voltage is:

$VG=R2R1+R2×VDDVG=9.370.7+9.3×5VG=9.3×580VG=0.58 V$

From the circuit:

$Vs=VDD−IDRSVs=5−5×103ID$

The value of $VSG$ is:

$VSG=VS−VGVSG=5−5×103ID−0.58VSG=4.42−5×103ID$

Assuming the transistor operates in saturation region:

$ID=KP( V SG+ V TP)2ID=0.4×10−3(4.42−5× 10 3 I D−0.8)2ID=0.4×10−3(3.62−5× 10 3 I D)2104ID2−15.48ID+5.24×10−3=0ID=1.048 mA,0.5 mA$

The value of $VSG$ is:

$VSG=4.42−5×103IDID=1.048 mA,VSG=4.42−5×103×1.048×10−3VSG=−0.82VSG<|VTP|$

Hence, the transistor would be in cutoff mode for $ID=1.048 mA$ .

Plugging $ID=0.5 mA,$

$VSG=4.42−5×103×0.5×10−3VSG=1.92 VVSG>|VTP|$

The value of $VSD$ is:

$VSD=VS−VDVSD=5−5×103ID−0VSD=5−5×103×0.5×10−3VSD=2.5 V$

$VSD>VSG+VTP$

Hence, the assumption is correct and transistor operates in saturation region.

The DC voltage source and coupling capacitor are short-circuited for small-signal equivalent circuit. It is common drain amplifier.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 3

The value of $gm,ro$ is:

$gm=2KPIDgm=20.4× 10 −3×0.5× 10 −3gm=0.8944 mAVro=1λIDro=10×IDro=∞$

The value of output voltage is:

$Vo=(−gmV sg)(Rs||ro)Vo=(−gmV sg)Rs ...(1)$

Applying Kirchhoff’s voltage law from input to output:

$Vin=−Vsg+VoVin=−Vsg−(gmV sg)(Rs)Vsg=−V in1+gm( R s )$

Applying voltage division rule in input:

$Vin=(R1||R2)Rsi+(R1||R2)Vi$

Hence, the value of $Vsg$ is:

$Vsg=−(R1||R2)(1+gmRs)(R si+( R 1 || R 2 ))Vi$

From equation (1):

$Vo=−gm(Rs)−( R 1 || R 2 )( 1+ g m R s )( R si +( R 1 || R 2 ))ViVoVi=gmRs( 1+ g m R s )( R i R si + R i )(Ri=R1||R2)Av=0.8944×51+0.8944×5( 8.22 0.5+8.22)Av=0.77$

The value of output resistance is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 4

Applying Kirchhoff’s current law at output node:

$Ix=gmVsg+VxRs+Vxro$

The value of current in input circuit is zero.

$Vsg=Vx$

Plugging the value:

$Ix=gmVx+VxRs+VxroIx=Vx(gm+1 R s +1 r o )IxVx=gm+1Rs+1ro1Ro=gm+1Rs+1roRo=1gm||Rs||ro$

Plugging the values:

$Ro=1gm||Rs||roRo=10.8944||5 (ro=∞)Ro=0.914 kΩ$

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Answer 3

Textbook Question

Chapter 4, Problem 4.9EP

Consider the circuit shown in Figure 4.28 with circuit parameters $V D D = 5 V$ , $R S = 5 k Ω$ , $R 1 = 70.7 k Ω$ , $R 2 = 9.3 k Ω$ , and $R S i = 500 Ω$ . The transistor parameters are: $V T P = − 0.8 V$ , $K p = 0.4 mA/V 2$ ,and $λ = 0$ . Calculate the small−signal voltage gain $A υ = υ o / υ i$ and the output resistance $R o$ seen looking back into the circuit. (Ans. $A υ = 0.817$ , $R o = 0.915 k Ω$ )

Expert Solution & Answer

To determine

The values of $Av,Ro$ .

Answer to Problem 4.9EP

$Av=0.77Ro=0.914 kΩ$

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 1

$VDD=5 V, RS=5 kΩR1=70.7 kΩ, R2=9.3 kΩRsi=500 Ω, λ=0VTP=−0.8 V, KP=0.4 mAV2$

Calculation:

The coupling capacitor acts like open circuit for DC value calculation.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 2

The value of gate voltage is:

$VG=R2R1+R2×VDDVG=9.370.7+9.3×5VG=9.3×580VG=0.58 V$

From the circuit:

$Vs=VDD−IDRSVs=5−5×103ID$

The value of $VSG$ is:

$VSG=VS−VGVSG=5−5×103ID−0.58VSG=4.42−5×103ID$

Assuming the transistor operates in saturation region:

$ID=KP( V SG+ V TP)2ID=0.4×10−3(4.42−5× 10 3 I D−0.8)2ID=0.4×10−3(3.62−5× 10 3 I D)2104ID2−15.48ID+5.24×10−3=0ID=1.048 mA,0.5 mA$

The value of $VSG$ is:

$VSG=4.42−5×103IDID=1.048 mA,VSG=4.42−5×103×1.048×10−3VSG=−0.82VSG<|VTP|$

Hence, the transistor would be in cutoff mode for $ID=1.048 mA$ .

Plugging $ID=0.5 mA,$

$VSG=4.42−5×103×0.5×10−3VSG=1.92 VVSG>|VTP|$

The value of $VSD$ is:

$VSD=VS−VDVSD=5−5×103ID−0VSD=5−5×103×0.5×10−3VSD=2.5 V$

$VSD>VSG+VTP$

Hence, the assumption is correct and transistor operates in saturation region.

The DC voltage source and coupling capacitor are short-circuited for small-signal equivalent circuit. It is common drain amplifier.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 3

The value of $gm,ro$ is:

$gm=2KPIDgm=20.4× 10 −3×0.5× 10 −3gm=0.8944 mAVro=1λIDro=10×IDro=∞$

The value of output voltage is:

$Vo=(−gmV sg)(Rs||ro)Vo=(−gmV sg)Rs ...(1)$

Applying Kirchhoff’s voltage law from input to output:

$Vin=−Vsg+VoVin=−Vsg−(gmV sg)(Rs)Vsg=−V in1+gm( R s )$

Applying voltage division rule in input:

$Vin=(R1||R2)Rsi+(R1||R2)Vi$

Hence, the value of $Vsg$ is:

$Vsg=−(R1||R2)(1+gmRs)(R si+( R 1 || R 2 ))Vi$

From equation (1):

$Vo=−gm(Rs)−( R 1 || R 2 )( 1+ g m R s )( R si +( R 1 || R 2 ))ViVoVi=gmRs( 1+ g m R s )( R i R si + R i )(Ri=R1||R2)Av=0.8944×51+0.8944×5( 8.22 0.5+8.22)Av=0.77$

The value of output resistance is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 4

Applying Kirchhoff’s current law at output node:

$Ix=gmVsg+VxRs+Vxro$

The value of current in input circuit is zero.

$Vsg=Vx$

Plugging the value:

$Ix=gmVx+VxRs+VxroIx=Vx(gm+1 R s +1 r o )IxVx=gm+1Rs+1ro1Ro=gm+1Rs+1roRo=1gm||Rs||ro$

Plugging the values:

$Ro=1gm||Rs||roRo=10.8944||5 (ro=∞)Ro=0.914 kΩ$

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Answer 4

Textbook Question

Chapter 4, Problem 4.9EP

Consider the circuit shown in Figure 4.28 with circuit parameters $V D D = 5 V$ , $R S = 5 k Ω$ , $R 1 = 70.7 k Ω$ , $R 2 = 9.3 k Ω$ , and $R S i = 500 Ω$ . The transistor parameters are: $V T P = − 0.8 V$ , $K p = 0.4 mA/V 2$ ,and $λ = 0$ . Calculate the small−signal voltage gain $A υ = υ o / υ i$ and the output resistance $R o$ seen looking back into the circuit. (Ans. $A υ = 0.817$ , $R o = 0.915 k Ω$ )

Expert Solution & Answer

To determine

The values of $Av,Ro$ .

Answer to Problem 4.9EP

$Av=0.77Ro=0.914 kΩ$

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 1

$VDD=5 V, RS=5 kΩR1=70.7 kΩ, R2=9.3 kΩRsi=500 Ω, λ=0VTP=−0.8 V, KP=0.4 mAV2$

Calculation:

The coupling capacitor acts like open circuit for DC value calculation.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 2

The value of gate voltage is:

$VG=R2R1+R2×VDDVG=9.370.7+9.3×5VG=9.3×580VG=0.58 V$

From the circuit:

$Vs=VDD−IDRSVs=5−5×103ID$

The value of $VSG$ is:

$VSG=VS−VGVSG=5−5×103ID−0.58VSG=4.42−5×103ID$

Assuming the transistor operates in saturation region:

$ID=KP( V SG+ V TP)2ID=0.4×10−3(4.42−5× 10 3 I D−0.8)2ID=0.4×10−3(3.62−5× 10 3 I D)2104ID2−15.48ID+5.24×10−3=0ID=1.048 mA,0.5 mA$

The value of $VSG$ is:

$VSG=4.42−5×103IDID=1.048 mA,VSG=4.42−5×103×1.048×10−3VSG=−0.82VSG<|VTP|$

Hence, the transistor would be in cutoff mode for $ID=1.048 mA$ .

Plugging $ID=0.5 mA,$

$VSG=4.42−5×103×0.5×10−3VSG=1.92 VVSG>|VTP|$

The value of $VSD$ is:

$VSD=VS−VDVSD=5−5×103ID−0VSD=5−5×103×0.5×10−3VSD=2.5 V$

$VSD>VSG+VTP$

Hence, the assumption is correct and transistor operates in saturation region.

The DC voltage source and coupling capacitor are short-circuited for small-signal equivalent circuit. It is common drain amplifier.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 3

The value of $gm,ro$ is:

$gm=2KPIDgm=20.4× 10 −3×0.5× 10 −3gm=0.8944 mAVro=1λIDro=10×IDro=∞$

The value of output voltage is:

$Vo=(−gmV sg)(Rs||ro)Vo=(−gmV sg)Rs ...(1)$

Applying Kirchhoff’s voltage law from input to output:

$Vin=−Vsg+VoVin=−Vsg−(gmV sg)(Rs)Vsg=−V in1+gm( R s )$

Applying voltage division rule in input:

$Vin=(R1||R2)Rsi+(R1||R2)Vi$

Hence, the value of $Vsg$ is:

$Vsg=−(R1||R2)(1+gmRs)(R si+( R 1 || R 2 ))Vi$

From equation (1):

$Vo=−gm(Rs)−( R 1 || R 2 )( 1+ g m R s )( R si +( R 1 || R 2 ))ViVoVi=gmRs( 1+ g m R s )( R i R si + R i )(Ri=R1||R2)Av=0.8944×51+0.8944×5( 8.22 0.5+8.22)Av=0.77$

The value of output resistance is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 4

Applying Kirchhoff’s current law at output node:

$Ix=gmVsg+VxRs+Vxro$

The value of current in input circuit is zero.

$Vsg=Vx$

Plugging the value:

$Ix=gmVx+VxRs+VxroIx=Vx(gm+1 R s +1 r o )IxVx=gm+1Rs+1ro1Ro=gm+1Rs+1roRo=1gm||Rs||ro$

Plugging the values:

$Ro=1gm||Rs||roRo=10.8944||5 (ro=∞)Ro=0.914 kΩ$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The values of $Av,Ro$ .

Answer to Problem 4.9EP

$Av=0.77Ro=0.914 kΩ$

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 1

$VDD=5 V, RS=5 kΩR1=70.7 kΩ, R2=9.3 kΩRsi=500 Ω, λ=0VTP=−0.8 V, KP=0.4 mAV2$

Calculation:

The coupling capacitor acts like open circuit for DC value calculation.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 2

The value of gate voltage is:

$VG=R2R1+R2×VDDVG=9.370.7+9.3×5VG=9.3×580VG=0.58 V$

From the circuit:

$Vs=VDD−IDRSVs=5−5×103ID$

The value of $VSG$ is:

$VSG=VS−VGVSG=5−5×103ID−0.58VSG=4.42−5×103ID$

Assuming the transistor operates in saturation region:

$ID=KP( V SG+ V TP)2ID=0.4×10−3(4.42−5× 10 3 I D−0.8)2ID=0.4×10−3(3.62−5× 10 3 I D)2104ID2−15.48ID+5.24×10−3=0ID=1.048 mA,0.5 mA$

The value of $VSG$ is:

$VSG=4.42−5×103IDID=1.048 mA,VSG=4.42−5×103×1.048×10−3VSG=−0.82VSG<|VTP|$

Hence, the transistor would be in cutoff mode for $ID=1.048 mA$ .

Plugging $ID=0.5 mA,$

$VSG=4.42−5×103×0.5×10−3VSG=1.92 VVSG>|VTP|$

The value of $VSD$ is:

$VSD=VS−VDVSD=5−5×103ID−0VSD=5−5×103×0.5×10−3VSD=2.5 V$

$VSD>VSG+VTP$

Hence, the assumption is correct and transistor operates in saturation region.

The DC voltage source and coupling capacitor are short-circuited for small-signal equivalent circuit. It is common drain amplifier.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 3

The value of $gm,ro$ is:

$gm=2KPIDgm=20.4× 10 −3×0.5× 10 −3gm=0.8944 mAVro=1λIDro=10×IDro=∞$

The value of output voltage is:

$Vo=(−gmV sg)(Rs||ro)Vo=(−gmV sg)Rs ...(1)$

Applying Kirchhoff’s voltage law from input to output:

$Vin=−Vsg+VoVin=−Vsg−(gmV sg)(Rs)Vsg=−V in1+gm( R s )$

Applying voltage division rule in input:

$Vin=(R1||R2)Rsi+(R1||R2)Vi$

Hence, the value of $Vsg$ is:

$Vsg=−(R1||R2)(1+gmRs)(R si+( R 1 || R 2 ))Vi$

From equation (1):

$Vo=−gm(Rs)−( R 1 || R 2 )( 1+ g m R s )( R si +( R 1 || R 2 ))ViVoVi=gmRs( 1+ g m R s )( R i R si + R i )(Ri=R1||R2)Av=0.8944×51+0.8944×5( 8.22 0.5+8.22)Av=0.77$

The value of output resistance is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 4

Applying Kirchhoff’s current law at output node:

$Ix=gmVsg+VxRs+Vxro$

The value of current in input circuit is zero.

$Vsg=Vx$

Plugging the value:

$Ix=gmVx+VxRs+VxroIx=Vx(gm+1 R s +1 r o )IxVx=gm+1Rs+1ro1Ro=gm+1Rs+1roRo=1gm||Rs||ro$

Plugging the values:

$Ro=1gm||Rs||roRo=10.8944||5 (ro=∞)Ro=0.914 kΩ$

Answer 8

Expert Solution & Answer

To determine

The values of $Av,Ro$ .

Answer to Problem 4.9EP

$Av=0.77Ro=0.914 kΩ$

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 1

$VDD=5 V, RS=5 kΩR1=70.7 kΩ, R2=9.3 kΩRsi=500 Ω, λ=0VTP=−0.8 V, KP=0.4 mAV2$

Calculation:

The coupling capacitor acts like open circuit for DC value calculation.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 2

The value of gate voltage is:

$VG=R2R1+R2×VDDVG=9.370.7+9.3×5VG=9.3×580VG=0.58 V$

From the circuit:

$Vs=VDD−IDRSVs=5−5×103ID$

The value of $VSG$ is:

$VSG=VS−VGVSG=5−5×103ID−0.58VSG=4.42−5×103ID$

Assuming the transistor operates in saturation region:

$ID=KP( V SG+ V TP)2ID=0.4×10−3(4.42−5× 10 3 I D−0.8)2ID=0.4×10−3(3.62−5× 10 3 I D)2104ID2−15.48ID+5.24×10−3=0ID=1.048 mA,0.5 mA$

The value of $VSG$ is:

$VSG=4.42−5×103IDID=1.048 mA,VSG=4.42−5×103×1.048×10−3VSG=−0.82VSG<|VTP|$

Hence, the transistor would be in cutoff mode for $ID=1.048 mA$ .

Plugging $ID=0.5 mA,$

$VSG=4.42−5×103×0.5×10−3VSG=1.92 VVSG>|VTP|$

The value of $VSD$ is:

$VSD=VS−VDVSD=5−5×103ID−0VSD=5−5×103×0.5×10−3VSD=2.5 V$

$VSD>VSG+VTP$

Hence, the assumption is correct and transistor operates in saturation region.

The DC voltage source and coupling capacitor are short-circuited for small-signal equivalent circuit. It is common drain amplifier.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 3

The value of $gm,ro$ is:

$gm=2KPIDgm=20.4× 10 −3×0.5× 10 −3gm=0.8944 mAVro=1λIDro=10×IDro=∞$

The value of output voltage is:

$Vo=(−gmV sg)(Rs||ro)Vo=(−gmV sg)Rs ...(1)$

Applying Kirchhoff’s voltage law from input to output:

$Vin=−Vsg+VoVin=−Vsg−(gmV sg)(Rs)Vsg=−V in1+gm( R s )$

Applying voltage division rule in input:

$Vin=(R1||R2)Rsi+(R1||R2)Vi$

Hence, the value of $Vsg$ is:

$Vsg=−(R1||R2)(1+gmRs)(R si+( R 1 || R 2 ))Vi$

From equation (1):

$Vo=−gm(Rs)−( R 1 || R 2 )( 1+ g m R s )( R si +( R 1 || R 2 ))ViVoVi=gmRs( 1+ g m R s )( R i R si + R i )(Ri=R1||R2)Av=0.8944×51+0.8944×5( 8.22 0.5+8.22)Av=0.77$

The value of output resistance is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 4

Applying Kirchhoff’s current law at output node:

$Ix=gmVsg+VxRs+Vxro$

The value of current in input circuit is zero.

$Vsg=Vx$

Plugging the value:

$Ix=gmVx+VxRs+VxroIx=Vx(gm+1 R s +1 r o )IxVx=gm+1Rs+1ro1Ro=gm+1Rs+1roRo=1gm||Rs||ro$

Plugging the values:

$Ro=1gm||Rs||roRo=10.8944||5 (ro=∞)Ro=0.914 kΩ$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The values of $Av,Ro$ .

Answer 12

Answer to Problem 4.9EP

$Av=0.77Ro=0.914 kΩ$

Answer 13

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 1

$VDD=5 V, RS=5 kΩR1=70.7 kΩ, R2=9.3 kΩRsi=500 Ω, λ=0VTP=−0.8 V, KP=0.4 mAV2$

Calculation:

The coupling capacitor acts like open circuit for DC value calculation.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 2

The value of gate voltage is:

$VG=R2R1+R2×VDDVG=9.370.7+9.3×5VG=9.3×580VG=0.58 V$

From the circuit:

$Vs=VDD−IDRSVs=5−5×103ID$

The value of $VSG$ is:

$VSG=VS−VGVSG=5−5×103ID−0.58VSG=4.42−5×103ID$

Assuming the transistor operates in saturation region:

$ID=KP( V SG+ V TP)2ID=0.4×10−3(4.42−5× 10 3 I D−0.8)2ID=0.4×10−3(3.62−5× 10 3 I D)2104ID2−15.48ID+5.24×10−3=0ID=1.048 mA,0.5 mA$

The value of $VSG$ is:

$VSG=4.42−5×103IDID=1.048 mA,VSG=4.42−5×103×1.048×10−3VSG=−0.82VSG<|VTP|$

Hence, the transistor would be in cutoff mode for $ID=1.048 mA$ .

Plugging $ID=0.5 mA,$

$VSG=4.42−5×103×0.5×10−3VSG=1.92 VVSG>|VTP|$

The value of $VSD$ is:

$VSD=VS−VDVSD=5−5×103ID−0VSD=5−5×103×0.5×10−3VSD=2.5 V$

$VSD>VSG+VTP$

Hence, the assumption is correct and transistor operates in saturation region.

The DC voltage source and coupling capacitor are short-circuited for small-signal equivalent circuit. It is common drain amplifier.

The modified circuit is:

Microelectronics: Circuit Analysis and Design, Chapter 4, Problem 4.9EP , additional homework tip 3

The value of $gm,ro$ is:

$gm=2KPIDgm=20.4× 10 −3×0.5× 10 −3gm=0.8944 mAVro=1λIDro=10×IDro=∞$

The value of output voltage is:

$Vo=(−gmV sg)(Rs||ro)Vo=(−gmV sg)Rs ...(1)$

Applying Kirchhoff’s voltage law from input to output:

$Vin=−Vsg+VoVin=−Vsg−(gmV sg)(Rs)Vsg=−V in1+gm( R s )$

Applying voltage division rule in input:

$Vin=(R1||R2)Rsi+(R1||R2)Vi$

Hence, the value of $Vsg$ is:

$Vsg=−(R1||R2)(1+gmRs)(R si+( R 1 || R 2 ))Vi$

From equation (1):

$Vo=−gm(Rs)−( R 1 || R 2 )( 1+ g m R s )( R si +( R 1 || R 2 ))ViVoVi=gmRs( 1+ g m R s )( R i R si + R i )(Ri=R1||R2)Av=0.8944×51+0.8944×5( 8.22 0.5+8.22)Av=0.77$