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 EE 332 DEVICES AND CIRCUITS II Lecture 17 AC Coupled Multi-Stage Amplifier ACCOUPLED MULTISTAGE AMPLIFIER /INTRODUCTION Coupling capacitors decouple the circuit into 3 stages CE amplifier R 1 R 2 R 4 R 3 R E2 C 3 R L C 2 C E1 R E1 R C1 C 1 V CC Q 2 Q 1 v o CC amplifier (voltage follower) v i  ACCOUPLED MULTISTAGE AMPLIFIER /EXAMPLE AND GAIN R 1 R 2 R 4 R 3 R E2 C 3 R L C 2 C E1 R E1 R C1 C 1 V CC Q 2 Q 1 v o v i , , , 1,2 70 0.7 70 , 15 BE ON A CC V V Q V V V V  β = = = = = Analyze the given multistage amplifier Given:  1 2 1 1 3 4 160 , 43 , 4.7 , 1.6 , 91 , 120 , 2.5 , C E L R k R k R k R k  R k R k R k   1 1 2 3 10 , E C C C C F  = Ω = Ω = Ω = Ω = Ω = Ω = Ω = = = =  Lecture Goals Understanding of concepts related to:  AC-Coupled Multi-Stage Amplifier  AC-Coupled Multi-Stage Amplifier/Gain  AC-Coupled Multi-Stage Amplifier/Current Gain 73

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EE 332

DEVICES AND CIRCUITS II

Lecture 17

AC Coupled Multi-Stage Amplifier 

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/INTRODUCTION

Coupling capacitors de‐couple the circuit into 3 stages

CE amplifier

R1

R2 R4

R3

RE2

C3

RL

C2

CE1RE1

RC1

C1

VCC

Q 2Q 1

vo

CC amplifier (voltage follower)

vi

 

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/EXAMPLE AND GAIN

R1

R2R4

R3

RE2

C3

RL

C2

CE1RE1

RC1

C1

VCC

Q 2Q 1

vo

vi

, ,,

1,2

70 0.7

70 , 15

B E O N  

A CC 

V V Q 

V V V V  

β = = ⎫= ⎨ ⎬

= = ⎭

Analyze the given multi‐stage amplifier

Given:

 1 2 1 1

3 4

160 , 43 , 4.7 , 1.6 ,

91 , 120 , 2.5 ,

C E 

L

R k R k R k R k  

R k R k R k  

 1 1 2 3

10 ,E 

C C C C F  

= Ω = Ω = Ω = Ω ⎫⎪ ⎪

= Ω = Ω = Ω ⎬⎪ ⎪= = = = ⎭  

Lecture Goals

Understanding of concepts related to:

•  AC-Coupled Multi-Stage Amplifier 

•  AC-Coupled Multi-Stage Amplifier/Gain

•  AC-Coupled Multi-Stage Amplifier/Current Gain

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 AC‐COUPLED MULTI‐STAGE AMPLIFIER

/1ST STAGE DC

 ANALYSIS

R1

R2 R4

R3

RE2

C3

RL

C2

CE1RE1

RC1

C1

VCC

Q 2Q 1

vo

vi

Step 1: DC analysis DC effects on the dash box

→ Two “4‐resistor” BJT circuits

 

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/ 1ST STAGE DC ANALYSIS

and

1

1 1

16.82

,

1.18 1.19

C E 

I A

then 

I m A I m A

μ

= =

Solve RHS loop

1 1 1

1 1 1

1 1 1

5.53

1.91

7.56

RC C C  

RE E E  

CE CC RC RE  

V R I V  

V R I V  

V V V V V  

= × =

= × =

= − − =

Since VCE=7.56 >VCE,sat=0.1V

The BJT operates in Forward Active region

REQ 

VEQ 

RC

RE

Q1

VCC

vo

IC

IE

IB

 

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/ 1ST STAGE SMALL SIGNAL PARAMETERS

Step 2: Small Signal Parameters:

1

1

1

1

1

1

1

1.18

47.225

701.48

47.2

77.5665.7

1.18

A CE 

I  mA

g mS V mV 

r k g mS 

V V  V r k 

I mA

π

β

= = =

= = = Ω

+= = = Ω  

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/ 1

ST

STAGE DC

 ANALYSIS

Assuming BJT operating in forward active mode,

Theveninizing LHS circuit

2

1 2

1 2

3.18

( / / ) 33.89

EQ CC  

EQ 

R V V V 

R R 

R R R k  

= × =+

= = Ω

Solve the loop

1 1 , 1

1 1

,

1

1

( 1)

2.4816.82

( 1) 33.89 71 1.6

EQ E E BE ON B E Q  

E B 

EQ BE ON  

EQ E 

V I R V I R  

I I 

V V I A

R R k k  

β

μβ

= + +⎨

= +

−⇒ = = =

+ + + ×

REQ 

VEQ 

RC

RE

Q1

VCC

vo

IC

IE

IB

 

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 AC‐COUPLED MULTI‐STAGE AMPLIFIER

/2ND STAGE DC

 ANALYSIS

R1

R2 R4

R3

RE2

C3

RL

C2

CE1RE1

RC1

C1

VCC

Q 2Q 1

vo

vi

Step 1: DC analysis DC effects on the dash box

→ Two “4‐resistor” BJT circuits

 

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/ 2ND STAGE SMALL SIGNAL PARAMETERS

Step 2: Small Signal Parameters:

2

2

2

2

2

2

2

2.63105

25

70665

105

81.8831.13

2.63

A CE 

I  mAg mS 

V mV 

r g mS 

V V  V r k 

I mA

π

β

= = =

= = = Ω

+= = = Ω  

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/ SMALL SIGNAL EQUIVALENT CIRCUIT

Step 3: Small Signal Equivalent Circuit

R1//R2

RE2

C2

RL

C1

E1

rπ1

RC1

vo

vi gmvbe1

vbe1

B1

ro1R3//R4

B2

E2

rπ2

gmvbe2

vbe2

ro2

Stage 1 :CE amplifier

Stage 2 :CC amplifier  

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/ 2

ND

STAGE DC

 ANALYSIS

42

3 4

2 3 4

8.53

( / / ) 51.75

EQ CC  

EQ 

R V V V 

R R 

R R R k  

= × =+

= = Ω

2 ,

2

2

7.8337.6

( 1) 51.75 71 2.2

EQ BE ON  

EQ E 

V V I A

R R k k  μ

β

−= = =

+ + + ×

and2 2

2.63 2.67C E 

I mA I mA=

Solve RHS

 loop

2 2 211.88

CE CC RC RE  V V V V V  − − =

Since VCE2=11.88V >VCE,sat=0.1V

The BJT operates in Forward Active region

REQ 

VEQ 

RC

RE

Q1

VCC

vo

IC

IE

IB

 

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 AC‐COUPLED MULTI‐STAGE AMPLIFIER

/ VOLTAGE

 GAIN

R1//R2

RE2

C2

RL

C1

E1

rπ1

RC1

vo

vi gmvbe1

vbe1

B1

ro1R3//R4

B2

E2

rπ2

gmvbe2

vbe2

ro2

1 1 2

1 1 2

o be c b o  

v i i be c b  

v v v v v  

A v v v v v  = = × × ×

1 1be 

v =

1 1 1 1 1 3 4 2

2 2 2

1

1

[ / / / / / / / / ]

( 1)[ / / ] 84.55

47.2 3.85 181.72

c m be C o b  

b E L

be 

v g v R r R R r  

r r R R k  

πβ

= − ×

= + + = Ω

⇒ = − × = −

Voltage gain

Calculate one term at a time:

 

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/ CURRENT GAIN

R1//R2

RE2

C2

RL

C1

E1

rπ1

RC1

vo

vi gmvbe1

vbe1

B1

ro1R3//R4

B2

E2

rπ2

gmvbe2

vbe2

ro2

i v 

L

R A A

R = ×

1 1 2 1 1 2 1

1 3 4 2

/ / / / / / / / 1.42

/ / / /

91 / /120 / /84.55 31.19

1.42

i b 

L b 

1181.72 8.27

31.19i 

R R R r R R r k  

R R R r  

k k k k  

π

Ak 

= = = Ω

=

= Ω Ω Ω = Ω

= − × = −

1st stage:

11 1

1

i v 

L

R A A

R = ×

1 1 11

1

181.72c be c  

i i be  

v v v A

v v v = = × = −  

AC‐COUPLED MULTI‐STAGE AMPLIFIER

/ CURRENT GAIN

R1//R2

RE2

C2

RL

C1

E1

rπ1

RC1

vo

vi gmvbe1

vbe1

B1

ro1R3//R4

B2

E2

rπ2

gmvbe2

vbe2

ro2

2 3 4 2

2

/ / / / 31.19

2.5

31.190.960 12.35

2.5

i b 

L

R R R r k  

R k 

k A

1 28.27 12.35 102.13

i i i A A A

= = Ω

= Ω

= × =

2nd stage:

= × = − × = −

2

2 2

2

i v 

L

R A A

R = ×

2

2

1 1 2

0.99o b o 

c c b 

v v v A

v v b = = × =

i v 

L

R A A

R = ×

 

AC‐COUPLED MULTI‐STAGE AMPLIFIER

VOLTAGE 

GAIN

R1//R2

RE2

C2

RL

C1

E1

rπ1

RC1

vo

vi gmvbe1

vbe1

B1

ro1R3//R4

B2

E2

rπ2

gmvbe2

vbe2

ro2

Voltage gain

2

2 2 2

( 1)[ / / ]

( 1)[ / / ]

( 1) 11700.99

665 ( 1) 1170

1 ( 181.72) 1 0.99 179.9

o E L

b E L

v R R 

v r R R  

A

π

β

β

β

β

+=+ +

+ ×= =

+ + ×

= × − × × = −

Calculate one term at a time:

1 1 2

1 1 2

o be c b o  

v i i be c b  

v v v v v  A

v v v v v  = = × × ×

2

1

1b 

v =

 

76