MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)
MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)
4th Edition
ISBN: 9781266368622
Author: NEAMEN
Publisher: MCG
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Chapter 12, Problem 12.69P
To determine

To derive: The expression for the loop gain.

Expert Solution & Answer
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Answer to Problem 12.69P

The expression for the loop gain is (gm1gm2RC2(1RE3+1RF)(1rπ3+gm3))(RC1||rπ2)[[1rπ+gm3+(1RE1+1RF)]RF(1E3+1RF)[(1RC2+1rπ3)RC2(1RE3+1RF)]+(1rπ+gm3)[1RF(1+[(1RC2+1rπ3)RC2(1RE3+1RF)]+(1rπ+gm3))]] .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 12, Problem 12.69P , additional homework tip  1

Figure 1

Calculation:

The small signal equivalent diagram for the Figure is shown in Figure 2

  MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 12, Problem 12.69P , additional homework tip  2

Figure 2

Apply KCL at node Ve1 .

  Vπ1rπ1+gm1Vπ1=Ve1RE1+Ve1VORFVπ1[1rπ1+gm1]=Ve1[1RE1+1RF]VORF ....................(1)

Apply KCL at node Vr .

  gm1Vπ1+VrRC1||rπ2=0Vπ1=(gm1Vπ1)(RC1||rπ2) ....................(2)

The expression for the voltage VC2 is given by,

  VC2=Vπ3+VO

Apply KCL at node VC2 .

  gm2Vt+VC2RC2+Vπ3rπ3=0

Substitute Vπ3+VO and VC2 in the above equation.

  gm2Vt+Vπ3+VORC2+Vπ3rπ3=0gm2Vt+Vπ3(1RC2+1rπ3)+VORC2=0 ....................(3)

Apply KCL at node VO .

  Vπ3rπ3+gm3Vπ3=VORE3+VOVe1RFVπ3(1rπ3+gm3)+Ve1RE3=VO(1RE3+1RF)

Substitute Vπ1 for Ve1 in the above equation.

  Vπ3(1rπ3+gm3)Vπ1RE3=VO(1RE3+1RF)VO=Vπ3(1rπ3+gm3)Vπ1RE3(1RE3+1RF)

Substitute Vπ3(1rπ3+gm3)Vπ1RE3(1RE3+1RF) for VO and Vπ1 for Ve1 in equation (1).

  Vπ1[1rπ1+gm1]=Vπ1[1RE1+1RF](Vπ3(1rπ3+gm3)Vπ1RE3(1RE3+1RF))RFVπ1[1rπ1+gm1]+Vπ1[1RE1+1RF]=Vπ3(1rπ3+gm3)+Vπ1RE3RF(1RE3+1RF) ....................(4)

Substitute Vπ3(1rπ3+gm3)Vπ1RE3(1RE3+1RF) for VO in equation (3).

  gm2Vt+Vπ3(1RC2+1rπ3)+Vπ3(1rπ3+gm3)Vπ1RE3(1RE3+1RF)RC2=0Vπ3=Vπ1RFgm2VtRC2(1RE3+1RF)[(1RC2+1rπ3)RC2(1RE3+1RF)+gm3]

Substitute Vπ1RFgm2VtRC2(1RE3+1RF)[(1RC2+1rπ3)RC2(1RE3+1RF)+gm3] for Vπ3 in the above equation.

   V π1 [ 1 r π1 + g m1 ]+ V π1 [ 1 R E1 + 1 R F ]= ( V π1 R F g m2 V t R C2 ( 1 R E3 + 1 R F ) [ ( 1 R C2 + 1 r π3 ) R C2 ( 1 R E3 + 1 R F )+ g m3 ] )( 1 r π3 + g m3 )+ V π1 R E3 R F ( 1 R E3 + 1 R F )

   V π1 [ 1 r π1 + g m1 + 1 R E1 + 1 R F ][ R F ( 1 E 3 + 1 R F ) ]= V π1 R F + g m2 V t R C2 ( 1 R E3 + 1 R F )( 1 r π3 + g m3 ) [ ( 1 R C2 + 1 r π3 ) R C2 ( 1 R E3 + 1 R F )+ g m3 ]+( 1 r π + g m3 ) + V π1 R E3

   V π1 = V t ( g m2 R C2 ( 1 R E3 + 1 R F )( 1 r π3 + g m3 ) ) [ [ 1 r π + g m3 +( 1 R E1 + 1 R F ) ] R F ( 1 E 3 + 1 R F ) [ ( 1 R C2 + 1 r π3 ) R C2 ( 1 R E3 + 1 R F ) ]+( 1 r π + g m3 ) [ 1 R F ( [ 1+[ ( 1 R C2 + 1 r π3 ) R C2 ( 1 R E3 + 1 R F ) ]+( 1 r π + g m3 ) ] ) ] ] .

Substitute Vt(gm2RC2(1RE3+1RF)(1rπ3+gm3))[[1rπ+gm3+(1RE1+1RF)]RF(1E3+1RF)[(1RC2+1rπ3)RC2(1RE3+1RF)]+(1rπ+gm3)1RF([1+[(1RC2+1rπ3)RC2(1RE3+1RF)]+(1rπ+gm3)])] for Vπ1 in equation (2).

   V r = g m1 V t ( g m2 R C2 ( 1 R E3 + 1 R F )( 1 r π3 + g m3 ) )( R C1 || r π2 ) [ [ 1 r π + g m3 +( 1 R E1 + 1 R F ) ] R F ( 1 E 3 + 1 R F ) [ ( 1 R C2 + 1 r π3 ) R C2 ( 1 R E3 + 1 R F ) ]+( 1 r π + g m3 ) [ 1 R F ( [ 1+[ ( 1 R C2 + 1 r π3 ) R C2 ( 1 R E3 + 1 R F ) ]+( 1 r π + g m3 ) ] ) ] ]

   V r V t = ( g m1 g m2 R C2 ( 1 R E3 + 1 R F )( 1 r π3 + g m3 ) )( R C1 || r π2 ) [ [ 1 r π + g m3 +( 1 R E1 + 1 R F ) ] R F ( 1 E 3 + 1 R F ) [ ( 1 R C2 + 1 r π3 ) R C2 ( 1 R E3 + 1 R F ) ]+( 1 r π + g m3 ) [ 1 R F ( 1+[ ( 1 R C2 + 1 r π3 ) R C2 ( 1 R E3 + 1 R F ) ]+( 1 r π + g m3 ) ) ] ]

Conclusion:

Therefore, the expression for the loop gain is (gm1gm2RC2(1RE3+1RF)(1rπ3+gm3))(RC1||rπ2)[[1rπ+gm3+(1RE1+1RF)]RF(1E3+1RF)[(1RC2+1rπ3)RC2(1RE3+1RF)]+(1rπ+gm3)[1RF(1+[(1RC2+1rπ3)RC2(1RE3+1RF)]+(1rπ+gm3))]] .

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Chapter 12 Solutions

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

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