Download ECE 352 Electronics II

Comparison of Amplifier Configurations
Midband Characteristics*
• These are approximate expressions neglecting the effects of the biasing resistors R 1 and R2
and the source resistance RS.
J. Millman and A. Grabel, Microelectronics, 2nd Ed., McGraw Hill, NY (1987), p. 420.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
1
Characteristics of Amplifier Configurations
Current gain is large ( β)
for CE and EF, but < 1 for CB.
Voltage gain is large
for CE and CB, but < 1 for EF.
Input resistance is
• Very small (few Ωs) for CB,
• Medium (few KΩs) for CE, but
• Very large (~ 10’s of KΩs) for EF.
Output resistance is
• Very small (few Ωs) for EF,
• Very large (~ 100’s of KΩs) for
CE and CB.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
2
Numerical Comparison of Amplifier Configurations
for the Same Transistor and DC Biasing
• These are approximate expressions neglecting the effects of the biasing resistors R 1 and R2
and the source resistance RS.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
3
Comparison of CB to CE Amplifier (with same Rs = 5 Ω)
CE (with RS = 5 Ω)
 r   rx  r  RB  
Vo Vo V Vi

  g m RL RC    

Vs V Vi Vs
 rx  r   Rs  rx  r  RB 
AVo   2180.940.93  191 V / V
AVo 
Midband Gain
AVo dB   20 log(  191 )  45.6dB
ZP1  ZP 2 0
Low Frequency
Poles and Zeros
ZP 3 

 ZP1   ZP 2 0
1
1

RS  RB rx  r CC1 0.7 K 2F   714 rad / s
 PL1 
 PL 2 
1
1

 33 rad / s
RL  RC CC 2 10.2 K 3F 
 PL2 

 RE

 rx  r  Rs RB 

C E
 1


ZH 1  , ZH 2 
 PH 1 
 PH 2 
1

 1.7 x10 4 rad / s
0.005K 12F 
g m 206 mA / V

 1.6 x1011 rad / s
C
1.3 pF
1
1

 9.0 x108 rad / s
r rx  RB RS  C 0.065K 17 pF 


1

 
 
1
1 
 r rx  RB RS   C

RC RL 1   g m 
RC RL 

 
 
1

 5.0 x107 rad / s
15.4 K 1.3 pF 
ECES 352 Winter 2007




AVo dB  40.2dB
1
1

 252 rad / s
RE C E 0.33K 12 F 
1
    
RE re
  r
V V V

AVo  o  e   g m RL RC
V Ve Vs
Rs  RE re  rx  r
AVo   218 0.940.5  102.4 V / V
 PL1 
 PL3 
High Frequency
Poles and Zeroes
CB (with RS = 5Ω)


 ZP 3 
1
1

 42 rad / s
RBCB 2 K 12F 
1
RB rx  r  1  gmr RE RS CB
1


1
 83 rad / s
12F 1K 
1
 5.0 x104 rad / s
2F 0.010K 

 r  r  
CC1 Rs  RE  x   

 1  g m r  
1
1
 PL3 

 33 rad / s
RL  RC CC 2 10.2K 3F 
ZH 1  , ZH 2  
1
1

 2.5 x1010 rad / s
re RE Rs C 2.417 pF 
 PH 1 

 PH 2 
1
1

 7.1x108 rad / s
RC RL C 1.05K 1.3 pF 
Ch. 7 Frequency Response Part 5

4
Comparison of EF to CE Amplifier (For RS = 5Ω )
CE
EF
 r   rx  r  RB  
Vo Vo V Vi
RL RE   Ri  RB rx  Ri 

  g m RL RC    
 AVo  Vo V Vi Vb 




Vs V Vi Vs
r

r
R

r

r
R

x

B 
 x    s
V Vi Vb Vs r  RE RL   rx  Ri  RS  RB rx  Ri 
AVo   2180.940.93  191 V / V
AVo  660.0150.999(0.998)  0.987 V / V
AVo 
Midband Gain
AVo dB  0.1dB
AVo dB   20 log(  191 )  45.6dB
ZP1  ZP 2 0
Low Frequency
Poles and Zeros
1
1

 252 rad / s
RE C E 0.33K 12 F 
 ZP1   ZP 2  0
1
1
 PL1 

RS  RB rx  r CC1 0.7 K 2F   714 rad / s
 PL 2
 PL1 
1
1


 33 rad / s
RL  RC CC 2 10.2 K 3F 
 PL3 
1

 RE

ZH 1 
High Frequency
Poles and Zeroes
ZP 3 
 PH 1 
 PH 2 
 rx  r  Rs RB 

C E
 1


ZH 2 
 PL2 
g m 206 mA / V

 1.6 x1011 rad / s
C
1.3 pF

1

 
 
1
1 
 r rx  RB RS   C

RC RL 1   g m 
RC RL 

 
 
1

 5.0 x107 rad / s
15.4 K 1.3 pF 
ECES 352 Winter 2007


1
RL  RE re CC 2

1
 37 rad / s
3F 9 K 
1

 1.7 x10 4 rad / s
0.005K 12F 
1
1

 9.0 x108 rad / s
r rx  RB RS  C 0.065K 17 pF 

1
1

 256 rad / s
CC1 Rs  RB rx  Ri  2F 1.95K 


 1  g m r  1
201

ZH 2  

 1.2 x1010 rad / s


r
C
0
.
97
K
17
pF


 
1
1
 PH 1 

 1.0 x1010 rad / s


0
.
386
K
0
.
26
pF


C
RxC ' 

1

g
m RE RL 

ZH 1 
 PH 2 
r 1  g


1
m
RE RL   RE RL 
r  R
x
S
RB C
1
 1.1x1010 rad / s
0.07 K 1.3 pF 
Ch. 7 Frequency Response Part 5
5
Comparison of Amplifier Configurations
Midband Gain and High and Low Frequency Performance
CE
CB
EF
Midband Voltage Gain
-191 V/V
45.6dB
+102 V/V
40.2dB
+0.987 V/V
- 0.1dB
Low 3dB Frequency
1.7x104 rad/s
5.0x104 rad/s
2.6x102 rad/s
High 3dB Frequency
5.0x107 rad/s
7.1x108 rad/s
1.0x1010 rad/s
RS= 5 Ω
• Results for all three amplifiers with the smaller (5Ω) source resistance RS.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
6
Cascade Amplifier
EF
CE
1   2  100
rx1  rx 2  0
C 1  C 2  13.9 pF
C1  C 2  2 pF
*
*
*
*
Emitter Follower + Common Emitter (EF+CE)
Voltage gain from CE stage, gain of one for EF.
Low output resistance from EF provides a low source resistance for CE amplifier so
good matching of output of EF to input of CE amplifier
High frequency response (3dB frequency) for Cascade Amplifier is improved over CE
amplifier.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
7
Cascade Amplifier - DC analysis
VTh1 
IB1
IE1
R2
100 K
VCC 
10V  5V
R1  R2
200 K
RTh1  R1 R2  100 K 100 K  50 K
IB2
1  2  100
KVL Base Q1
VTh1  VBE1  I B1 RTh1  [ 1  1]I B1  I B 2 RE1
Neglecting I B 2 as a first approximation
IRE1
I B1 
Then
I E1  1  1I B1  (101)8.9 A  899 A
Small Signal Parameters
I C1 1 I B1 100(8.9 A)
mA


 34.8
VT
VT
0.0256V
V
g m1 
r 1 
1
g m1
gm2 
r 2 
100

 2.9 K
34.8mA / V
gm2
100

 2.9 K
34.0mA / V
ECES 352 Winter 2007
Now calculate VB 2 and I B 2
VB 2  I E1 RE1  899 A(4.3K )  3.87V
VB 2  VBE 2   2  1I B 2 RE 2
I C 2  2 I B 2 100(8.7 A)
mA


 34.0
VT
VT
0.0256V
V
2
5V  0.7V
 8.9 A
50 K  100  14.3K
3.87V  0.7V
 8.7 A
(101)3.6 K
 I E1 so approximate analysis is okay.
I B2 
I B2
Ch. 7 Frequency Response Part 5
8
Cascade Amplifier - Midband Gain Analysis
Iπ1
+
Note: rx1 = rx2 = 0 so equivalent circuit is simplified.
+
Vπ1
_
+
Vπ2
_
Vi
_
Ri
AVo 
Vo Vo V 2 V 1 Vi

Vs V 2 V 1 Vi Vs
Ri 
Vo  g m 2V 2 RL RC 

 68
V 2
V 2
I  1r 1  1  g m1r 1 I  1 RE1 r 2 
I 1
 r 1  1  g m1r 1 RE1 r 2 

V 2 I  1  g m1V 1 RE1 r 2  1
mA

   g m1  RE1 r 2  35.1
(4.3K 2.9 K )  60.8
V 1
V 1
V
 r 1

V 1
I  1r 1
2.9 K


 0.016
Vi
I  1r 1  1  g m1r 1 I  1 ( RE1 r 2 ) 2.9 K  (101)( 4.3K 2.9 K )
(100 K 100 K ) 178K
( R1 R2 ) Ri
Vi


 0.91
Vs RS  RB ( R1 R2 ) 4 K  (100 K 100 K ) 178K
 2.9 K  (101)( 4.3K 2.9 K )  178K
AVo   68(60.8)0.0160.91  60.2 V / V
AVo dB   20 log(  60.2 )  35.6dB
Note: Voltage gain is nearly equal to
that of the CE stage, e.g. – 68 !
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
9
Cascade Amplifier - Low Frequency Poles and Zeroes
s  Z 1L s  Z 2 L s  Z 3L 
FL ( s ) 
s  P1L s  P 2 L s   P3L 
  Z 1L   Z 2 L   Z 3 L 

1 
1 
1 
s 
s 
s 


  P1L   P 2 L   P 3 L 

1 
1 
1 
s
s
s




ECES 352 Winter 2007
*
*
Ch. 7 Frequency Response Part 5
Use Gray-Searle (Short Circuit)
Technique to find the poles.
● Three low frequency poles
● Equivalent resistance may
depend on rπ for both
transistors.
Find three low frequency zeroes.
10
Cascade Amplifier - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 1 μF
RxC1 
Vx
 Rs  RB Ri
Ix
Ri 
Vi
I
rπ1
Vi  I  1r 1  I  1  g m1V 1 RE1 r 2   I  1 r 1  1  g m1r 1 RE1 r 2 
Ri 
Vi
 r 1  1  g m1r 1 RE1 r 2 
I
+
RE1
 2.9 K  (101)4.3K 2.9 K   178K
RxC1  Rs  RB Ri
rπ2
IX
Iπ1
Ri
CC1 RxC1  1F 43.0 K   4.3x10 2 sec
1
1

 23 rad / s
CC1 RxC1 4.3x10 2 sec
ECES 352 Winter 2007
Vπ2
_
 4 K  50 K 178K  43.0 K
 PL1 
Vπ1
Vi
RE1
Ch. 7 Frequency Response Part 5
rπ2
11
Cascade Amplifier - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
CC2
rX2
Vo
gm2Vπ2
Vπ2
rπ2
RE2
*
RC
RL
CE
Output coupling capacitor CC2 = 1 μF
VX
Vo
RC
RL
RC 2  RL  RC  4 K  4 K  8K
 PL 2 
ECES 352 Winter 2007
1
1

 125 rad / s
RC 2CC 2 8 K 1F 
Ch. 7 Frequency Response Part 5
12
Cascade Amplifier - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
Emitter bypass capacitor CE = 47 μF
REx
Iπ1
V
 x  RE 2 re 2
Ix
re 2 
 I  2 ( r 2  RE1 re1 ) r 2  RE1 re1
VE 2


I e2
 I  2 (1  g m 2 r 2 )
1  g m 2 r 2
re1 
 I  1 ( r 1  RS ' )
r  RS '
VE1

 1
I e1
 I  1 (1  g m1r 1 ) 1  g m1r 1
2.9 K  3.7 K

 0.065 K
101
2.9 K  4.3K 0.065 K
re 2 
 0.029 K
101
REx  RE 2 re 2  3.6 K 0.029 K  0.029 K
 PL3 
1
1

 734 rad / s
RExC E
0.029 K 47 F 
r π1
Vπ1
Ie1
RS '  RS R1 R2
 3.7 K
gm1Vπ1
VE1
re1
RE1
Iπ2
rπ2
re2
Vπ2
Ie2
VE2
gm2Vπ2
Ix
RE
VX
2
IE2
Low 3 dB Frequency
 PL   PL1   PL 2   PL3  23  125  734  882 rad / s
The pole for CE is the largest and therefore the
most important in determining the low 3 dB frequency.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
13
Cascade Amplifier - Low Frequency Zeros
*
*
*
*
s 

 23  125  734 
1  1 
1 

s 
s 
s 

 5.9 
1 

s 

FL ( s ) 
 23  125  734 
1  1 
1 

s 
s 
s 

ECES 352 Winter 2007
Z E'  RE 2 Z CE
1
1
1
1
1  sRE 2CE




sC

E
Z E'
RE 2 Z CE RE 2
RE
  ZL1   ZL 2   ZL 3 

1 
1 
1 

s   ZL1 s  ZL 2 s  ZL 3  
s 
s 
s 
FL ( s ) 

s  PL1 s  PL2 s  PL3  1  PL1 1  PL2 1  PL3 




s 
s 
s 

1  01  01  ZL 3 
What are the zeros for the Cascade
amplifier?
For CC1 and CC2 , we get zeros at ω = 0
since ZC = 1 / jωC and these capacitors
are in the signal line, i.e. ZC  at ω =
0 so Vo  0.
Consider RE in parallel with CE
Impedance given by
Z E' 
*

*
RE 2
1  sRE 2CE
When Z’E  , Iπ  0, so gmVπ  0,
so Vo  0
Z’E   when s = - 1 / RE2CE so pole
for CE is at
ZL 3 
Ch. 7 Frequency Response Part 5
1
1

 5.9 rad / s
RE 2C E 3.6 K 47 F 
14
Cascade Amplifier - High Frequency Poles and Zeroes
*
*
Use Gray-Searle (Open Circuit)
Technique to find the poles.
● Four high frequency poles
● Equivalent resistance may
depend on rπ for both
transistors.
Find four high frequency zeroes.

s
1 
 Z 1H
FH ( s)  

s
1 
 P1H

s
1 
 Z 2 H

s
1 
 P 2 H

s
1 
 Z 3 H

s
1 
 P 3 H

s
1 
 Z 4 H

s
1 
 P 4 H
High Frequency Equivalent Circuit
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
15






Cascade Amplifier - High Frequency Poles
Pole for Capacitor Cπ1 = 13.9 pF
RS '  R1 R2 RS  3.7 K
Ix- Iπ1
KVL around base loop
Ix
Iπ1
+
VX
_
 VX  ( I X  I  1 ) RS ' I e1 ( RE1 r 2 )  0
But
I 1 
VX
and I e1  I  1  g m1V 1  I X  I  1 1  g m1r 1   I X
r 1
So

V

V 
 VX   I X  X  RS '  X 1  g m1r 1   I X  ( RE1 r 2 )  0
r 1 

 r 1

Rearrangin g and solving for VX /I X
Ie1
PH1 
1
 8.0 x108 rad / s
0.09 K 13.9 pF 
ECES 352 Winter 2007
 R ' R r

VX 1  S  E1  2 1  g m1r 1   I X RS ' RE1 r 2 
r 1
r 1


 R ' R r

VX
 RS ' RE1 r 2  / 1  S  E1  2 1  g m1r 1 
IX
r 1
r 1


 3.7 K 3.6 K 2.9 K

 3.7 K  3.6 K 2.9 K  / 1 

(101) 
2.9 K
 2.9 K

5.3K

 0.09 K
58.2
Ch. 7 Frequency Response Part 5
16
Cascade Amplifier - High Frequency Poles
Pole for Capacitor Cμ1 = 2 pF
RS '  R1 R2 RS
I e1  I  1  g m1V 1  I  1 1  g m1r 1 
 3.7 K
+
Ix- Iπ1
VX
_
So
Ix
VX  I  1r 1  1  g m1r 1 I  1 ( RE1 r 2 )
Iπ1
 I  1 r 1  1  g m1r 1 ( RE1 r 2 )
Also
VX  I X  I  1 RS ' or I  1  I X 
Ie1
VX
RS '
Substituti ng for I  1 in above and
Rearrangin g and solving for VX /I X
PH 2 
1
 1.4 x108 rad / s
3.6 K 2 pF 

V 
VX   I X  X r 1  1  g m1r 1 ( RE1 r 2 )
RS ' 

R ' r  1  g m1r 1 ( RE1 r 2 )
VX
 S 1
IX
RS 'r 1  1  g m1r 1 ( RE1 r 2 )
 RS ' r 1  1  g m1r 1 ( RE1 r 2 ) 
 3.7 K 2.9 K  (101)4.3K 2.9 K 
 3.6 K
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
17
Cascade Amplifier - High Frequency Poles and Zeroes
RS '  R1 R2 RS  3.7 K
Simplified Equivalent Circuit
Using Miller’s Theorem,
replace Cμ2 by two capacitors.
K
V2 Vo  g m 2V 2 RL RC 


  g m 2 RL RC 
V1 V 2
V 2
C21  C 2 (1  K )  C 2 1  g m 2 RL RC   C 2


1
1

C22  C 2 1    C 2 1 
  C 2


g
R
R
 K
m
2
L
C


ECES 352 Winter 2007
CT  C 2  C21  C 2  C 2 1  g m 2 RL '
 13.9 pF  2 pF{1  (34mA / V )2 K }  152 pF
C22  C 2
Ch. 7 Frequency Response Part 5
18
Cascade Amplifier - High Frequency Poles
RS '  R1 R2 RS  3.7 K
+
Vπ1
_
Iπ1
Pole for Capacitor CT = 152 pF
re1 
Ve1
Ie1
Ix
re1
VX
Ve1  I  1 r 1  RS ' r 1  RS '


I e1  I  1 1  g m1r 1  1  g m1r 1
2.9 K  3.7 K
 0.065K
101
RX  re1 RE1 r 2 0.065K 4.3K 2.9 K  0.063K

PH 3 
1
 1.0 x108 rad / s
0.063K 152 pF 
Pole for Output Capacitor Cμ2 = 2 pF
RxC  RL '  RC RL
gm2Vπ2 +
_
VX
RxC  4 K 4 K  2 K
 PH 4 
ECES 352 Winter 2007
1
1

 2.5 x108 rad / s
RxC C 2 2 K 2 pF 
Ch. 7 Frequency Response Part 5
19
Cascade Amplifier - High Frequency Zeroes
Ie1
*
*
*
*
When does Vo = 0?
When ω → ∞, ZCμ1→ 0, so signal shorted to ground.
ωZH1= ∞.
When ω → ∞, ZCπ2→ 0, so rπ2 shorted, so Vπ2 = 0. ωZH2= ∞.
For Cπ1 , we get a zero when Ie1 = 0.
I e1  I  1  g m1V 1  sC 1V 1  0
I e1  I  1 1  g m1r 1   sC 1r 1   0
or
s
Ie1
 1  g m1r 1 
C 1r 1
so
 ZH 3 
ECES 352 Winter 2007
1  g m1r 1  
Ch. 7 Frequency Response Part 5
C 1r 1
101
 2.5 x109 rad / s
13.9 pF (2.9 K )
20
Cascade Amplifier - High Frequency Zeroes
I Cμ2
*
*
IRL’= 0
When does Cμ2 produce a zero, i.e. make Vo = 0?
For Cμ2 , we get a zero when IRL’ = 0, or ICμ2 = gm2Vπ2 , i.e. the output load
resistance RL’ is starved of any current.
I C 2  sC  2 V 2  Vo   sC  2V 2  g m 2V 2
or
s
gm2
C 2
so
ZH 4 
ECES 352 Winter 2007
g m 2 34 mA / V

 1.7 x1010 rad / s
C 2
2 pF
Zero for Output Capacitor Cμ2 = 2 pF
Ch. 7 Frequency Response Part 5
21
Cascade Amplifier - High Frequency Poles and Zeroes

s 
s 
s 
s 
1 
1 
1 
1 





Z 1H 
Z 2 H 
Z 3 H 
Z 4H 
FH ( s )  

s 
s 
s 
s 
1 
1 
1 
1 





P1H 
P 2 H 
P 3 H 
P4H 

s 
s 
s
s



1
1  1  1 
9 
10 
     2.5 x10  1.7 x10 

s 
s 
s
s 


1
1
1
1 
8 
8 
8 
8 
 8 x10  1x10  2.5 x10  1.4 x10 
s
s



1
1 
9 
10 
 2.5 x10  1.7 x10 
FH ( s ) 
s 
s 
s
s 


1
1
1
1 
8 
8 
8 
8 
 8 x10  1x10  2.5 x10  1.4 x10 
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
22
Comparison of Cascade to CE Amplifier
CE*
AVo 
Midband Gain
Cascade (EF+CE)
Vo Vo V 2

Vs V 2 Vs
AVo 
AVo dB   20 log   60.2   35.6dB
AVo dB   20 log(  30 )  29.5dB
Low Frequency
Poles and Zeros
ZP 3 
1

ZP1  ZP 2  0
1
 157 rad / s
4.4 K 1F 
 PL1 

 PL 2 
1
1

 125 rad / s
RL  RC CC 2 8K 1F 
 PL3 
1
1

 734 rad / s
( RE 2 re 2 )C E 47 F 0.03K 
R
 PL 2 
1
1

 125 rad / s
RL  RC CC 2 8K 1F 
 PL3 
1

 RE 2

 r 2  Rs RB 

C E
   1 
ZH 1  , ZH 2 
High Frequency
Poles and Zeroes
1
1

 5.9 rad / s
RE 2C E 3.6 K 47 F 
 PL1 
S  RB r 2 CC1
 PH 1 
r
2 X improvement
in voltage gain !
AVo   6860.80.0160.91  60.2 V / V
AVo   81.20.37   30 V / V
ZP1  ZP 2  0
Vo V 2 V 1 Vi
V 2 V 1 Vi VS

1
 591 rad / s
0.036 K 47 F 
g m 2 40.6 mA / V

 2.0 x1010 rad / s
C 2
2 pF
1
1

 4.8 x107 rad / s
RB RS C 2 1.5K 13.9 pF 

ZP 3 
1
1

 5.9 rad / s
RE 2C E 3.6 K 47 F 
1
1

 23 rad / s
RS  RB r 1  1  g m1r 1 RE 2 r 2  CC1 1F 43K 

ZH 1  , ZH 2  ,
ZH 3  2.5 x109 rad / s,  ZH 4  1.7 x1010 rad / s
1
 8.0 x108 rad / s,
0.09 K 13.9 pF 
25 X improvement
1
1
8
 PH 2 
 PH 2 
 1.4 x10 rad / s, in bandwidth !

 
 
1
1 
3.6 K 2 pF 






R
R
1

g


r
R
R
C
 C L   m2
2
B
S   2
RC RL 

1
 
 
 PH 3 
 1.0 x108 rad / s,
0.063K 152 pF 
1

 4.0 x106 rad / s
125K 2 pF 
1
 PH 4 
 2.5 x108 rad / s
2 K 2 pF 
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
23
2

 PH 1 

* CE stage with same transistor, biasing resistors, source resistance and load as cascade.
Comparison of Cascade to CE Amplifier
*
Why the better voltage gain for the cascade?
●
●
Ri1
Emitter follower gives no voltage gain!
Cascade has better matching with source than CE.
 Cascade amplifier has an input resistance that is
higher due to EF first stage.
Ri1  r 1    1RE1 r 2
 2.9 K  (101)4.3K 2.9 K   178K
Ri2
*
Pole for Capacitor CT = 152 pF
 Versus Ri2 = rπ2 = 2.5 K for CE
 So less loss in voltage divider term (Vi / Vs )
with the source resistance.
* 0.91 for cascade vs 0.37 for CE.
Why better bandwidth?
●
Low output resistance re1 of EF stage gives smaller
effective source resistance for CE stage and higher
frequency for dominant pole due to CT (including
Cμ2)
V
 I r  R ' r  R '
re1 
re1
ECES 352 Winter 2007

1 1
S
S
 1
 I  1 1  g m1r 1  1  g m1r 1
2.9 K  3.7 K
 0.065K
101
RX  re1 RE1 r 2 0.065K 4.3K 2.9 K  0.063K

CT  C 2  C 2 1  g m 2 RL '  13.9 pF  2 pF{1  (34mA / V )2K }  152 pF
RL '  RL RC  4 K 4 K  2 K
e1
I e1
 PH 3 
1
 1.0 x108 rad / s for the cascade
0.063K 152 pF 
versus 4.0 x106 rad / s for the CE amplifier.
Ch. 7 Frequency Response Part 5
24
Another Useful Amplifier – Cascode (CE+CB) Amplifier
*
*
*
*
For CE amplifier
Common Emitter + Common Base
(CE + CB) configuration
Voltage gain from both stages
Low input resistance from second CB stage
provides first stage CE with low load
resistance so Miller Effect multiplication of
Cμ1 is much smaller.
High frequency response dramatically
improved (3 dB frequency increased).
●
Bandwidth is much improved (~130 X).
the high frequency performanc e is limited
by
Cin1  C 1  C 1 1  g m1 RL RC   17 pF  1.3 pF 1  206mA / V 9 K 1.2 K   17 pF  1.3 pF 1  218  302 pF
1
 5.6 x10 6 rad / s
8
Large Miller Effect
1.8 x10 sec
For cascode amplifier , RL for first stage CE amplifier becomes Ri of CB amplifier where Ri is given by
 PHin 
Ri 
rx  r
 5
1  g m r
Small Miller Effect
so Cin1  C 1  C 1 1  g m RL RC   17 pF  1.3 pF 1  206mA / V 0.005 K 1.2 K   17 pF  1.3 pF (1  1)  19.6 pF
 PHin 
1
 7.3 x108 rad / s
1.2 K 9 K 2.6 pF 
ECES 352 Winter 2007
Bandwidth is improved by a factor of 130X
over that for the CE amplifier !
Ch. 7 Frequency Response Part 5
25
Example of Cascode (CE +CB) Amplifier
http://www.freescale.com ECTW Conf. Proceedings 2003.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
26
Other Examples of Multistage Amplifiers
CE
CE
EF
EF
Darlington Pair
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
27
Other Examples of Multistage Amplifiers
Push – Pull Amplifier
Amplifier with Npn and Pnp Transistors
Amplifier with FETs and
Bipolar Transistors
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
28
Differential Amplifier
Similar to CE amplifier, but two CE’s
operated in parallel
* Signal applied between two equivalent inputs
instead of between one input and ground
* Common emitter resistor or current source
used
* Current shared or switched between two
transistors (they compete)
* Analyze using equivalent half-circuit
● 1/2 of signal at input
● 1/2 of signal at output
● 1/2 of source resistance
* Gain and frequency response similar to CE
amplifier for high frequencies
* Advantage:
● Rejects common noise pickup on input
● No coupling capacitors so can operate
down to zero frequency.
*
+V
o
ECES 352 Winter 2007
_
Ch. 7 Frequency Response Part 5
29
Differential Amplifier Analysis
Midband Gain
 Vo
Vo  2
2
AVo 


Vs
Vs  V
2



   g mV RC 
r

  

V
 
 r  rx  RS
2


0.97K


  206m A/ V 9 K 
  509
 0.97K  0.065K  2.5K 
Vo
Vo
Vo /2

 V
 V
 s
 2





AVo (dB)  20 log 509  54.1dB
Low Frequency Poles and Zeros
* Direct coupled so no coupling capacitors
and no emitter bypass capacitor
* No low frequency poles and zeros
* Flat (frequency independent) gain
down to zero frequency
Vo /2
High Frequency Poles and Zeros
Dominant pole using Miller’s Thoerem
 PH 


1
r rx  Rs / 2C  1  gm RC C 
1
1

0.97K 0.065K  5K / 2 17 pF  2011.3 pF  0.70K 278pF 


1
1.95x107 sec
 5.1x106 rad / s
High frequency performance is very similar to CE amplifier.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
30
Summary
*
In this chapter we have shown how to analyze the high and low frequency
dependence of the gain for an amplifier.
● Analyzed the effects of the coupling capacitors on the low frequency response
 Found the expressions for the corresponding poles and zeros.
 Demonstrated Bode plots of magnitude and phase.
● Analyzed the effects of the capacitances within the transistor on the high
frequency response.
 Found the expressions for the corresponding poles and zeros.
 Demonstrated Bode plots of the magnitude and phase.
*
Analyzed the high and low frequency performance of the three bipolar transistor
amplifiers: common emitter, common base and emitter follower.
● Found the expressions for the corresponding poles and zeros.
● Demonstrated Bode plots of the magnitude and phase.
* Demonstrated how to find the expressions for the gain and the high and low
frequency poles and zeros for multistage amplifiers.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 5
31