(a) Using the small-signal equivalent circuit in Figure 12.25 for the circuit in Figure 12.24 ( a ) , derive the expression for the small-signal current gain A i f = I o / I s . (b) Using the circuit parameters given in Figure 12.24 ( a ) and assuming transistor parameters h F E = 100 and V A = ∞ calculate the value of A i f . Compare this answer with the results of Example 12.9.

Question

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Answer 1

Textbook Question

Chapter 12, Problem 12.50P

(a) Using the small-signal equivalent circuit in Figure 12.25 for the circuit in Figure $12.24 ( a ) ,$ derive the expression for the small-signal current gain $A i f = I o / I s .$ (b) Using the circuit parameters given in Figure $12.24 ( a )$ and assuming transistor parameters $h F E = 100$ and $V A = ∞$ calculate the value of $A i f$ . Compare this answer with the results of Example 12.9.

(a)

Expert Solution

To determine

To derive: The expression for the small signal current gain of the circuit.

Answer to Problem 12.50P

The value of the small signal current gain is ${ R F( 1 R B || R B2 ) g m1+( R F R s || R B1 || r π1 || R F )( 1 R B || R B2 )[( R F R s || R B1 || r π1 || R F )B−1 R F ]−RFB}−1g m2[( R C2 + R L R C2 )]{( 1 r π2 + g m2){( 1 R 2 + g m2 )+D A( 1 R B || R B2 )+ g m1 }[( AB)−1 R F ]}$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

Microelectronics: Circuit Analysis and Design, Chapter 12, Problem 12.50P , additional homework tip 1

Figure 1

Calculation:

Mark the nodes and redraw the circuit.

The given diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 12, Problem 12.50P , additional homework tip 2

Figure 2

By KCL the expression for the current $Ii$ is given by,

$Ii=V π1Rs||R B1||r π1+V π1−V e2RFVe2=Vπ1( R F R s || R B1 || r π1 || R F )−IiRF$

The expression for the node voltage is given by,

$VC1=Vπ2+Ve2$

Apply KCL at node $VC1$ .

$gm1Vπ1+V C1R C1||R B2+V π2r π2=0gm1Vπ1+V π2+V e2R C1||R B2+V π2r π2=0gm1Vπ1+Vπ2(1 R C1 || R B2 || r π2 )+V e2R C12||R B2=0$

Substitute $Vπ1(RFRs||R B1||r π1||RF)−IiRF$ for $Ve2$ in the above equation.

$gm1Vπ1+Vπ2(1 R C1 || R B2 || r π2 )+1R C12||R B2[Vπ1( R F R s || R B1 || r π1 || R F )−IiRF]=0Vπ1=IiRF( 1 R C12 || R B2 )−V π2( 1 R C || R B2 || r π2 )g m1+[( R F R s || R B1 || r π1 || R F )( 1 R C12 || R B2 )]$

The expression for the output current is given by,

$IO=−(g m2V π2)( R C2 R C2 + R L )Vπ2=−IOg m2( R C2 + R L R C2 )$

Apply KCL at node $Ve2$ .

$V π2r π2+gm2Vπ2=V e2R E2+V e2−V π1RF(1 r π2 +g m2)Vπ2+V π1RF=(1 R E2 +1 R F )Ve2$

Substitute $Vπ1(RFRs||R B1||r π1||RF)−IiRF$ for $Ve2$ in the above equation.

$(1 r π2 +g m2)Vπ2+V π1RF=(1 R E2 +1 R F )(V π1( R F R s || R B1 || r π1 || R F )−IiRF)(1 r π2 +g m2)Vπ2+IiRF(1 R E2 +1 R F )=Vπ1{[( R F R s || R B1 || r π1 || R F )( 1 R E2 + 1 R F )]−1 R F}$

Substitute $IiRF(1 R C12 || R B2 )−Vπ2(1 R C || R B2 || r π2 )gm1+[( R F R s || R B1 || r π1 || R F )( 1 R C12 || R B2 )]$ for $Vπ1$ in the above equation.

$[( 1 r π2 + g m2)Vπ2+IiRF( 1 R E2 + 1 R F )]=[ I i R F( 1 R C1 || R B2 )− V π2( 1 R C || R B2 || r π2 )][gm1+[ ( R F R s || R B1 || r π1 || R F ) ( 1 R C12 || R B2 )]]{[( R F R s || R B1 || r π1 || R F )( 1 R E2 + 1 R F )]−1RF}$

Consider $A=RFRs||RB1||rπ1||RF$ , $B=1RE2+1RF$ , $C=1RC1||RB2$ and $D=1RC1||RB2||rπ2$ in the above equation. So the equation is,

$[ ( 1 r π2 + g m2 ) V π2 + I i R F B ]= [ I i R F C− V π2 D ] [ g m1 +AC ] { AB− 1 R F }$

$V π2 { ( 1 r π2 + g m2 )+ D g m1 +AC [ AB− 1 R F ] }= I i { R F C g m1 +AC [ AB− 1 R F ]− R F B }$

$[ − I O g m2 ( R C2 + R L R C2 ) ]{ ( 1 r π2 + g m2 )+ D g m1 +AC [ AB− 1 R F ] }= I i { R F C g m1 +AC [ AB− 1 R F ]− R FB B }$

$I O I i = { R F C g m1 +AC [ AB− 1 R F ]− R F B } −1 g m2 [ ( R C2 + R L R C2 ) ]{ { ( 1 r π2 + g m2 )+ D AC+ g m1 }[ ( AB )− 1 R F ] }$

Thus, the expression for the small signal current gain is,

$Ai={ R FC g m1+AC[AB−1 R F ]−RFB}−1g m2[( R C2 + R L R C2 )]{{( 1 r π2 + g m2 )+D AC+ g m1 }[( AB)−1 R F ]}$ …….(1)

Conclusion:

Therefore, the value of the small signal current gain is ${ R F( 1 R B || R B2 ) g m1+( R F R s || R B1 || r π1 || R F )( 1 R B || R B2 )[( R F R s || R B1 || r π1 || R F )B−1 R F ]−RFB}−1g m2[( R C2 + R L R C2 )]{( 1 r π2 + g m2){( 1 R 2 + g m2 )+D A( 1 R B || R B2 )+ g m1 }[( AB)−1 R F ]}$ .

(b)

Expert Solution

To determine

The value of the gain $Aif$ for the given specifications.

To compare: The obtained value with the given value of gain. and compare it to the value of 9.58.

Answer to Problem 12.50P

The value of the current gain is $9.58$ and it is approximately equal to the given value of gain.

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 12, Problem 12.50P , additional homework tip 3

The given value of gain is 9.58.

Also, the values are:

$hFE=100VA=∞$

Calculation:

The expression to determine the value of the resistance $RB1$ is given by,

$RB1=( 80 kΩ)( 20 kΩ)( 80 kΩ)+( 20 kΩ)=16 kΩ$

The expression for the value of the voltage $VTH1$ is given by,

$VTH1=VCC( R 2 R 2 + R 1 )=10 V( 20 kΩ 100 kΩ)=2 V$

The expression to determine the value of the current $IB1$ is given by,

$IBQ1=VTH1−VBE(on)RB1+(h fe+1)RE1$

Substitute $2 V$ for $VTH1$ , $0.7 V$ for $VBE$ , $16 kΩ$ for $RB1$ and $1 kΩ$ for $RE1$ in the above equation.

$IBQ1=2 V−0.7 V16 kΩ+( 100+1)1 kΩ=0.0111 mA$ \

The value of the current $ICQ1$ is calculated as,

$ICQ1=hFEIBQ1$

Substitute $100$ for $hFE$ and $0.0111 mA$ for $IBQ1$ in the above equation.

$ICQ1=(100)(0.0111 mA)=1.11 mA$

The expression to determine the value of the resistance $RB2$ is given by,

$RB2=R3||R4=( 85 kΩ)( 15 kΩ)( 85 kΩ)+( 15 kΩ)=12.75 kΩ$

The value of the Thevenin voltage is given by,

$VTH2=VCC( R 4 R 3 + R 4 )=(10 V)( 15 kΩ 15 kΩ+85 kΩ)=1.5 V$

The expression to determine the value of the current $IB2$ is given by,

$IBQ2=VTH2−VBE(on)RB2+(h fe+1)RE2$

Substitute $1.5 V$ for $VTH2$ , $0.7 V$ for $VBE$ , $12.75 kΩ$ for $RB2$ and $0.5 kΩ$ for $RE2$ in the above equation.

$IBQ2=1.5 V−0.7 V12.75 kΩ+( 100+1)0.5 kΩ=0.01265 mA$ \

The value of the current $ICQ2$ is calculated as,

$ICQ2=hFEIBQ2$

Substitute $100$ for $hFE$ and $0.01265 mA$ for $IBQ2$ in the above equation.

$ICQ2=(100)(0.01265 mA)=1.265 mA$

The expression to determine the transconductance of first transistor is calculated as,

$gm1=1.11 mA0.026 V=42.69 mA/V$

The expression to determine the transconductance of second transistor is calculated as,

$gm1=1.265 mA0.026 V=48.65 mA/V$

The value of the small signal input resistance is calculated as,

$rπ1=h FEg m1=10042.69 mA/V=2.34 kΩ$

The value of the small signal input resistance is calculated as,

$rπ2=h FEg m2=10048.65 mA/V=2.06 kΩ$

The value of A is calculated as,

$A=RFRs||R B1||r π1||RF≅RFR B1||r π1||RF=10 kΩ16 kΩ||2.34 kΩ||10 kΩ=5.896$

The value of B is calculated as,

$B=1R E2+1RF=10.5 kΩ+110 kΩ=2.1 m$

The value of C is calculated as,

$C=1RB||RB2$

The value of D is calculated as,

$D=1RC1||RB2||rπ2$

Substitute $1.7289 kΩ$ for $RC1||RB2$ and $2.06 kΩ$ for $rπ$ in the above equation.

$D=11.7289 kΩ||2.06 kΩ=1.0627 m$ , $4 kΩ$

Substitute $10 kΩ$ for $RF$ , $42.69 mA/V$ for $gm1$ , $48.65 mA/V$ for $gm2$ , $4 kΩ$ for $RC2$ , $4 kΩ$ for $RE$ , $2.06 kΩ$ for $rπ2$ , $5.896$ for $A$ , $2.1 m$ for $B$ , $0.5784 m$ for $C$ and $1.0672 m$ for $D$ in equation (1).

$Ai={ ( 10 kΩ )( 0.5784 m ) ( 42.69 mA V )+[ ( 5.896 ) ( 0.5784 m ) ][ ( 5.896 )( 2.1 m )− 1 ( 10 kΩ ) ]−[ ( 10 kΩ ) ( 2.1 m ) ]} −1 48.65 mA V [( ( 4 kΩ )+( 4 kΩ ) ( 4 kΩ ) )]{ { ( 1 2.06 kΩ +48.65 mA V ) + ( 1.0627 m ) ( 5.896 )( 0.5784 m )+42.69 mA V } [ ( ( 5.896 ) ( 2.1 m ) )− 1 ( 10 kΩ ) ] }=−19.46( −0.0411)( 49.4174)=9.58122$

The value of the current gain is $9.58122$ and it is approximately equal to the given value of gain.

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