Lecture 9
OUTLINE
• BJT Amplifiers (cont’d)
– Common-base topology
– CB core
– CB stage with source resistance
– Impact of base resistance
Reading: Chapter 5.3.2
EE105 Fall 2007
Lecture 9, Slide 1
Prof. Liu, UC Berkeley
Common Base (CB) Amplifier
• The base terminal is biased at a fixed voltage; the input
signal is applied to the emitter, and the output signal
sensed at the collector.
EE105 Fall 2007
Lecture 9, Slide 2
Prof. Liu, UC Berkeley
Small-Signal Analysis of CB Core
• The voltage gain of a CB stage is gmRC, which is
identical to that of a CE stage in magnitude and
opposite in phase.
Av  g m RC
EE105 Fall 2007
Lecture 9, Slide 3
Prof. Liu, UC Berkeley
Tradeoff between Gain and Headroom
• To ensure that the BJT operates in active mode, the
voltage drop across RC cannot exceed VCC-VBE.
IC
VCC  VBE
Av 
RC 
VT
VT
EE105 Fall 2007
Lecture 9, Slide 4
Prof. Liu, UC Berkeley
Simple CB Stage Example
VCC = 1.8V
IC = 0.2mA
IS = 5x10-17 A
b = 100
Vb  1.354V 
R2
VCC if I1  I B
R1  R2
Choose I1  10 I B  20A 
VCC
R1  R2
1
Av  g m RC 
 2230  17.2
 R1  22.3k, R2  67.7k
130
EE105 Fall 2007
Lecture 9, Slide 5
Prof. Liu, UC Berkeley
Input Impedance of a CB Stage
• The input impedance of a CB stage is much smaller
than that of a CE stage.
1
Rin 
if VA  
gm
EE105 Fall 2007
Lecture 9, Slide 6
Prof. Liu, UC Berkeley
CB Stage with Source Resistance
• With the inclusion of a source resistance, the input
signal is attenuated before it reaches the emitter of
the amplifier; therefore, the voltage gain is lowered.
– This effect is similar to CE stage emitter degeneration.
Av 
EE105 Fall 2007
Lecture 9, Slide 7
RC
1
 RS
gm
Prof. Liu, UC Berkeley
Practical Example of a CB Stage
• An antenna usually has low output impedance;
therefore, a correspondingly low input impedance is
required for the following stage.
EE105 Fall 2007
Lecture 9, Slide 8
Prof. Liu, UC Berkeley
Output Impedance of a CB Stage
• The output impedance of a CB stage is equal to RC in
parallel with the impedance looking into the collector.
Rout1  1  g m ( RE || r )rO  RE || r 
Rout 2  RC || Rout1
EE105 Fall 2007
Lecture 9, Slide 9
Prof. Liu, UC Berkeley
Output Impedance: CE vs. CB Stages
• The output impedances of emitter-degenerated CE and
CB stages are the same. This is because the circuits for
small-signal analysis are the same when the input port
is grounded.
EE105 Fall 2007
Lecture 9, Slide 10
Prof. Liu, UC Berkeley
Av of CB Stage with Base Resistance
(VA = ∞)
• With base resistance, the voltage gain degrades.
vout
vout   g m v RC  v  
g m RC
v
vout
r  RB 
vP   r  RB  
r
r g m RC
vout
r  RB 
vP 
bRC
vout
r  RB   vin




v
v v
v
bR
1
KCL at node P :   g m v  P in    g m   out   C
r
RE
RE
 r
 g m RC 
EE105 Fall 2007
vout
bRC
RC


1
RB
vin r  b  1RE  RB
 RE 
gm
b 1
Lecture 9, Slide 11
Prof. Liu, UC Berkeley
Voltage Gain: CE vs. CB Stages
• The magnitude of the voltage gain of a CB stage with
source and base resistances is the same as that of a CE
stage with base resistance and emitter degeneration.
EE105 Fall 2007
Lecture 9, Slide 12
Prof. Liu, UC Berkeley
Rin of CB Stage with Base Resistance
(VA = ∞)
• The input impedance of a CB stage with base
resistance is equal to 1/gm plus RB divided by (b+1).
This is in contrast to a degenerated CE stage, in which
the resistance in series with the emitter is multiplied
by (b+1) when seen from the base.
v
KCL   g m v  ix
r
r
v  
vx
r  RB
EE105 Fall 2007
Lecture 9, Slide 13
1


r
  g m  
vx   ix
 r
 r  RB 
vx r  RB 1
RB
Rin  


ix
b 1 gm b 1
Prof. Liu, UC Berkeley
Input Impedance Seen at Emitter vs. Base
Common Base Stage
EE105 Fall 2007
Common Emitter Stage
Lecture 9, Slide 14
Prof. Liu, UC Berkeley
Input Impedance Example
• To find RX, we have to first find Req, treat it as the
base resistance of Q2 and divide it by (b+1).
1
RB
Req 

g m1 b  1
EE105 Fall 2007
1
1  1
RB 


Rx 


g m 2 b  1  g m1 b  1 
Lecture 9, Slide 15
Prof. Liu, UC Berkeley