Download Chap9part3

Common-Base (CB) Amplifier
*
*
*
*
Input at emitter, output at collector.
ECES 352 Winter 2007
DC biasing
● Calculate IC, IB, VCE
● Determine related small signal
equivalent circuit parameters
 Transconductance gm
 Input resistance rπ
Midband gain analysis
Low frequency analysis
● Gray-Searle (Short Circuit)
Technique
 Determine pole frequencies
ωPL1, ωPL2, ... ωPLn
● Determine zero frequencies
ωZL1, ωZL2, ... ωZLn
High frequency analysis
● Gray-Searle (Open Circuit)
Technique
 Determine pole frequencies
ωPH1, ωPH2, ... ωPHn
● Determine zero frequencies
ωZH1, ωZH2, ... ωZHn
Ch. 7 Frequency Response Part 3
1
CB Amplifier - DC Analysis (Same as CE Amplifier)
*
GIVEN: Transistor parameters:
●
Current gain β = 200
●
Base resistance rx = 65 Ω
●
Base-emitter voltage VBE,active = 0.7 V
●
Resistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K
*
Form Thevenin equivalent for base; given VCC = 12.5V
●
RTh = RB = R1||R2 = 10K||2.5K = 2K
●
VTh = VBB = VCC R2 / [R1+R2] = 2.5V
● KVL base loop
 IB = [VTh-VBE,active] / [RTh+(β +1)RE]
 IB = 26 μA
DC collector current IC = β IB
IC = 200(26 μ A) = 5.27 mA
Transconductance gm = IC / VT ; VT = kBT/q = 26 mV
gm = 5.27 mA/26 mV = 206 mA/V
Input resistance
rπ = β / gm = 200/[206 mA/V]= 0.97 K
Check on transistor region of operation
● KVL collector loop
● VCE = VCC - IC RC - (β +1) IB RE = 4.4 V
(okay since not close to zero volts).
*
*
*
R1 = 10K
R2 = 2.5K
RC = 1.2K
RE = 0.33K
ECES 352 Winter 2007
*
Ch. 7 Frequency Response Part 3
2
CB Amplifier - Midband Gain Analysis
*
*
*
High and Low Frequency AC Equivalent Circuit
Construct small signal ac equivalent circuit
(set DC supply to ground)
Substitute small signal equivalent circuit
(hybrid-pi model) for transistor
Neglect all capacitances
● Coupling and emitter bypass capacitors become shorts
at midband frequencies (~ 105 rad/s)
 Why? Impedances are negligibly small, e.g.
few ohms because CC1, CC2, CE ~ few μF (10-6F)
1
1
ZC 

 10 
C (105 rad / s)(1F )
●
Transistor capacitances become open circuits at
midband frequencies
 Why? Impedances are very large, e.g. ~ 10’s M Ω
because Cπ , Cμ ~ pF (10-12 F)
ZC 
*
ECES 352 Winter 2007
1
1

 107 
5
C (10 rad / s)(1 pF )
Calculate small signal voltage gain
Ch. 7 Frequency Response Part 3
AVo = Vo /Vs
3
CB Amplifier - Midband Gain Analysis
V
 Ve
I   
r rx  r
V
 r

Ve rx  r
RL  9 K
RC  1.2 K
Iπ
re
+
Ve
_
RS  5K
Equivalent resistance re
Vo  g mV RL RC 

  g m RL RC   206mA / V 1.2 K 9 K   218
V
V
V
r I 
 0.97 K


 0.94
Ve  I  (rx  r ) 0.97 K  0.065K
RE re   0.33K 0.0051K   0.0050K  0.001
Ve

Vs Rs  RE re  5K  0.33K 0.0051K  5.0050 K
AVo   218 0.940.001  0.20 V / V
ECES 352 Winter 2007
R1  10K
R2  2.5K
V V V V
AVo  o  o  e
Vs V Ve Vs
AVo dB   20 log( 0.20)  14dB
RE  0.33K
βIπ
Voltage gain is
less than one !
re 
Ve
Ie
KCL at node E
I e  g mV 
V
0
r

1
I e  V  g m 
r


 1  g m r 
  V 

r




 r  r  r
V
V
r
re  e   e
   x  
Ie
V 1  g m r
r  1  g m r


Ch. 7 Frequency Response Part 3
rx  r
r r
0.065 K  0.97 K
 x  
 0.0051 K
1  g m r
1 
1  200
4
What Happened to the CB Amplifier’s Midband Gain?
+
Ve
_
re
AVo  2180.940.001  0.20 V / V




RE re
0.33K 0.0051K
Ve
0.0050K



 0.001
Vs Rs  RE re
5 K  0.33K 0.0051K
5.0050K




New signal source with low resistance
For Rs  5




RE re
0.33K 0.0051K
Ve
0.0050K



Vs Rs  RE re
0.005K  0.33K 0.0051K
0.005K  0.005K



and
AVo   218 0.940.5  102.5 V / V
AVo (dB)  20 log(102.5)  40.2dB
ECES 352 Winter 2007

* Source resistance Rs = 5K is
killing the gain.
● Why? Rs >> re = 0.0051 K
so Ve/Vs<<1
* Need to use a different signal
source with a very low source
resistance Rs , i.e. ~ few ohms
* Why is re so low?
● Vs drives formation of Ve
● Ve creates Vπ across rπ
● Vπ turns on dependent
current source
 0.5 ● Get large Ie for small Ve so
re =Ve/Ie is very small.
Voltage gain is
now much bigger than one !
Ch. 7 Frequency Response Part 3
5
Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
* Draw low frequency AC circuit
●
●
●
Substitute AC equivalent circuit for transistor
(hybrid-pi for bipolar transistor)
Include coupling and base capacitors CC1, CC2, CB
Ignore (remove) all transistor capacitances Cπ , Cμ
* Turn off signal source, i.e. set Vs= 0
●
Keep source resistance RS in circuit (do not remove)
* Consider the circuit one capacitor Cx at a time
●
●
●
●
Replace all other capacitors with short circuits
Solve remaining circuit for equivalent resistance Rx seen
by the selected capacitor
1
Calculate pole frequency using  Px  R C
x x
Repeat process for each capacitor finding equivalent
resistance seen and the corresponding pole frequency
* Determine the dominant (largest) pole frequency
* Calculate the final low pole frequency using
 LP 
ECES 352 Winter 2007

 Px   P1   P 2 ... Pn 
Ch. 7 Frequency Response Part 3

1
RxC x
6
Common Base - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
Low Frequency AC Equivalent Circuit
RxC B  RB rx  Ri 
Ri 
*
Vo
Vπ
Vi
 2 K 0.065K  0.97 K  2010.33K 0.005K 
 2 K 2.03K  1.0 K
C B RxC B  12 F 1.0 K   1.2 x10  2 sec
 PL1 
RxCB
V
r
RxC B  RB rx  r  1  g m r RE RS 
Iπ
+
sin ce I  
 1


Vi  V  I   g mV RE RS   V 1    g m RE RS 

  r

 1


V 1    g m RE RS 
 r

  r  1  g r R R 
Ri  r 

m 
E
S
V
Base capacitor CB = 12 μF
Vx Ix
Vi
Vi
V

 r i
I  V / r
V
1
RxC B C B

1
 83 rad / s
1.2 x10  2
_
Ri
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 3
7
Common Base - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 2 μF
V
RxCC1  x  Rs  RE re
Ix
V
re  e
Ie
I e   I  g mV  1  g m r I
Ve   I rx  r 
r  r 
V
 I rx  r 
re  e 
 x 
I e  1  g m r I 1  g m r 
 r r 
RxCC1  Rs  RE  x  
 1  g m r 
Iπ
Vo
 0.065K  0.97K 
 0.005K  0.33K 

201


Vπ
Ve
 0.005K  0.0051K  0.010K
CC1RxCC1  2 F 0.010K   2.0 x105 sec
1
1
 PL2 

 5.0 x104 rad / s

5
CC1RxCC1 2.0 x10 sec
ECES 352 Winter 2007
Vx
Ie
re
+
Ve
_
Ch. 7 Frequency Response Part 3
Ix
Rs
8
Common Base - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
* Output coupling capacitor CC2 = 3 μF
VX
 PL   PL1   PL2   PL3
Vo
RC
* Low 3dB frequency
 83  50,000  33  50,116 rad / s
RL
Dominant low frequency
pole is due to CC1 !
RC 2  RL  RC  9 K  1.2 K  10.2 K
1
1
 PL3 

 33 rad / s
RC 2CC 2 10.2 K 3F 
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 3
9
Common Base - Low Frequency Zeros
Iπ
*
*
*
*
What are the zeros for the CB
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 RB in parallel with CB
Impedance given by
  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 

Z B'  RB Z CB
1  01  01  ZL 3 
s 


 83  33  50,000 
1  1  1 

s 
s 
s 

 42 
1  
s 

FL ( s ) 
 83  33  50,000 
1  1  1 

s 
s 
s 

Z B' 
ECES 352 Winter 2007
1
1
1
1
1  sRB C B



 sC B 
'
Z B RB Z CB RB
RB
*
*
RB
1  sRB C B
When Z’B  , Iπ  0, so
gmVπ  0, so Vo  0
Z’B   when s = - 1 / RBCB
so pole for CB is at
ZL 3 
Ch. 7 Frequency Response Part 3
1
1

 42 rad / s
RB CB 2 K 12F 
10
Common Base - Low Frequency Poles and Zeros
Magnitude Bode Plot
ZL1  0, ZL 2  0, ZL 3  42
42 
 42 

1  
1  j 

PL1  83

s 



A( )  102.5
 102.5

83 
33 
50,000 
PL 2  33
 83  33  50,000 


1  1  1 

1  j 1  j 1  j


s 
s 
s 
 
 
 


PL3  50,000
AMo (dB)  20 log 102.5  40.2dB
  42  2 
  83  2 
  33  2 
  50,000  2 
A(dB)  40.2dB  10 log 1      10 log 1      10 log 1      10 log 1  
 
    
    
    
    
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 3
11
Common Base - Low Frequency Poles and Zeros
Phase Shift Bode Plot
42 
 42 

1  
1  j 
s 



A( )  102.5
 102.5
83 
33 
50,000 
 83  33  50,000 

1  1  1 

1  j 1  j 1  j

s 
s 
s 
 
 
 


ZL1  0, ZL 2  0, ZL 3  42

 PL1  83


 PL2  33


 PL3  50,000
 42 
1  83 
1  33 
1  50,000 
  tan    tan    tan 


 
 
  
 ( )   tan 1 
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 3
12
Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
*
*
*
Draw high frequency AC equivalent circuit
● Substitute AC equivalent circuit for transistor
(hybrid-pi model for transistor with Cπ, Cμ)
● Consider coupling and emitter bypass
capacitors CC1, CC2, CB as shorts
● Turn off signal source, i.e. set Vs = 0
● Keep source resistance RS in circuit
● Neglect transistor’s output resistance ro
Consider the circuit one capacitor Cx at a time
● Replace all other transistor capacitors with
open circuits
● Solve remaining circuit for equivalent
resistance Rx seen by the selected capacitor
1
● Calculate pole frequency using
 PHx 
RxC x
● Repeat process for each capacitor
Calculate the final high frequency pole using


 PH  

ECES 352 Winter 2007
1 

 Px 
1
 1
1
1 


... 

 PHn 
  PH1  PH 2
Ch. 7 Frequency Response Part 3
1

1
R C
x x
13
Common Base - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
High frequency AC equivalent circuit
NOTE: We neglect rx here
since the base is grounded.
This simplifies our analysis,
but doesn’t change the
results appreciably.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 3
14
Common Base - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
Ze 
Ve
Ie
Ve  V
KCL at node E gives
1

Ve
Ve

 g mV  Ve   sC  g m 
r  1

 r

 sC 
 

1  g m r

 Ve 
 sC 
 r

Ie 
Ze
+
Ve
_
Replace this
with this.
* Equivalent circuit for Ze
Ve

Ie
Ve
1

1  g m r
 1  g m r

Ve 
 sC  
 sC 
 r
  r

 r   1
1



  sC 

 1  g m r  



1
1



  r
  1

  sC  
 
 

Parallel combination
  1  g m r 

Ze 
So Z e  re Z C
of a resistor and
capacitor.
where
Ze
ECES 352 Winter 2007
re 
r
r
 
1  gmr 1  
Ch. 7 Frequency Response Part 3
15
Common Base - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
Turn off signal source
when finding resistance
seen by capacitor.
*
Pole frequency for Cπ =17pF
RxC  re RE Rs
re 
r
0.97K

 0.0048K  4.8
1  g m r 1  200
RxC  0.0048K 0.33K 0.005K  0.0024K  2.4
 PH1 
ECES 352 Winter 2007
1
RxC C

1
1

 2.5 x1010 rad / s
2.417 pF  4.1x1011 s
Ch. 7 Frequency Response Part 3
16
Common Base - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
*
Pole frequency for Cμ =1.3pF
RxC   RC RL
RxC   1.2 K 9 K  1.05K
 PH 2 
* Equivalent circuit for Capacitor Cμ = 1.3 pF
=0
1
RxC  C
1
1

 7.1x108 rad / s
1.05K 1.3 pF  1.4 x109 s
* High 3 dB frequency
 PH 
1
1

RxC C  RxC C 4.1x10 11  1.4 x10 9
 PH 
1
8

6
.
9
x
10
rad / s
1.44 x10 9 s
Rs || RE || r π
ECES 352 Winter 2007

Dominant high frequency
pole is due to Cμ !
Ch. 7 Frequency Response Part 3
17
Common Base - High Frequency Zeros
*
*
What are the high frequency zeros
for the CB amplifier?
Voltage gain can be written as
V
V V
AV  o  o 
Vs V Vs
*
*
*
*
ECES 352 Winter 2007
When Vo/Vπ  0, we have found a
zero.
For Cμ , we get Vo  0 when ω  
since the output will be shorted to
ground thru Cμ .
Similarly,we get a zero from Cπ when
when ω   since ZC π = 1/sCπ  0,
so the voltage Vπ  0.
Both Cπ and Cμ give high
frequency zeros at ω  !
Ch. 7 Frequency Response Part 3
18
Common Base - High Frequency Poles and Zeros
Magnitude
s 
s

1  1  
    
A( s )  102.5
s
s



1
1 
10 
8 
2.5 x10  7.1x10 

 A( )  102.5
1

 


1 j
1  j

10 
2.5 x10 
7.1x108 

 ZH 1  



  ZH 2
10
 PH 1  2.5 x10
8

  PH 2  7.1x10
AMo (dB )  20 log 102.5  40.2dB
2
 
   2 

 
A(dB )  40.2dB  10 log 1  
 10 log 1  
10  
8  
  2.5 x10  
  7.1x10  
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 3
19
Common Base - High Frequency Poles and Zeros
Phase Shift
s 
s

1  1  
    
A( s)  102.5
s
s 


1
1 
10 
8 
 2.5 x10  7.1x10 
 A( )  102.5
1
 
 

1 j
1  j

10 
2.5 x10 
7.1x108 

ZH 1  



  ZH 2
10
 PH 1  2.5 x10
  PH 2  7.1x108
 

  
 tan 1 
10 
8 
 2.5 x10 
 7.1x10 
 ( )   tan 1 
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 3
20
Comparison of CB to CE Amplifier
CE (with RS = 5K)
Midband Gain

CB (with RS = 5Ω)

 r   rx  r  RB
V
V V V
AVo  o  o  i   g m RL RC    
Vs V Vi Vs
 rx  r   Rs  rx  r  RB
AVo   2180.940.12  24.6 V / V








AVo dB  20 log(  24.6 )  27.7dB
 ZP1   ZP 2 0
Low Frequency
Poles and Zeros
1
1
 PL1 

 88 rad / s
RS  RB rx  r  CC1 5.7 K 2F 
 PL2 
 PL3 


 RE


 rx  r  Rs RB 

C E


 1


ZH 1  , ZH 2 
High Frequency
Poles and Zeroes
 PH 1 
 PH 2 

1
 5,342 rad / s
0.016K 12F 
g m 206 mA / V

 1.6 x1011 rad / s
C
1.3 pF
1
1

 1.0 x108 rad / s
r rx  RB RS  C 0.59 K 17 pF 


1

 
 
1
1 
 r rx  RB RS   C

RC RL 1   g m 
RC RL 

 
 
1

 5.9 x106 rad / s
130 K 1.3 pF 
ECES 352 Winter 2007

 ZP1   ZP 2 0
 PL1 

 ZP 3 




1
1

 42 rad / s
RBCB 2 K 12F 
1
RB rx  r  1  gmr RE RS CB
1
 PL2 
1
1

 33 rad / s
RL  RC CC 2 10.2K 3F 
1

AVo dB  40.2dB
1
1
 ZP 3 

 252 rad / s
RE C E 0.33K 12F 

    
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


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 3

Note: CB amplifier has much better
high frequency performance!
21
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 3

Note: CB amplifier has much better
high frequency performance!
22
Conclusions
* Voltage gain
● Can get good voltage gain from both CE and CB amplifiers.
● Low frequency performance similar for both amplifiers.
● CB amplifier gives better high frequency performance !
CE amplifier has dominant pole at 5.0x107 rad/s.
CB amplifier has dominant pole at 7.1x108 rad/s.
* Bandwidth approximately 14 X larger!
* Miller Effect multiplication of C by the gain is avoided in
CB configuration.
* Current gain
● For CE amplifier, current gain is high AI = Ic / Ib
● For CB amplifier, current gain is low AI = Ic / Ie (close to one)!
● Frequency dependence of current gain similar to voltage gain.
* Input and output impedances are different for the two amplifiers!
● CB amplifier has especially low input resistance.
ECES 352 Winter 2007
Ch. 7 Frequency Response Part 3
23