ESE319 Introduction to Microelectronics
Common Base BJT Amplifier
Common Collector BJT Amplifier
Common Collector (Emitter Follower) Configuration
● Common Base Configuration
● Small Signal Analysis
● Design Example
● Amplifier Input and Output Impedances
●
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Basic Single BJT Amplifier Features
CE Amplifier
Voltage Gain (AV) moderate (-RC/RE)
CC Amplifier
low (about 1)
Current Gain (AI)
moderate ( 1)
moderate (  )
Input Resistance
high
high
Output Resistance
high
low
VCVS
CB Amplifier
high
low (about 1)
low
CCCS
high
CE BJT amplifier => CS MOS amplifier
CC BJT amplifier => CD MOS amplifier
CB BJT amplifier => CG MOS amplifier
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Common Collector ( Emitter Follower) Amplifier
R1
ro
CB
V CC
RE
vs
vo
R2
vs
vo
RE
ro
Voltage Bias Design
Current Bias Design
In the emitter follower, the output voltage is taken between emitter
and ground. The voltage gain of this amplifier is nearly one – the
output “follows” the input - hence the name: emitter “follower.”
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Emitter Follower Biasing
R1
CB
RS
vs
VB
Vb
iB
R2
Then, choose/specified IE, and
the rest of the design follows:
iC
iE
vout
vo
RE
V CC
V E V CC /2−0.7
R E=
=
IE
IE
For an assumed  = 100:
Split bias voltage drops about
equally across the transistor
VCE (or VCB) and VRe (or VB).
For simplicity,choose:
V CC
⇒ R1 =R 2
V B=
2
2009 Kenneth R. Laker, updated 06Oct09 KRL
As with CE bias design, stable op.
pt. => R B ≪1 R E , i.e.
R1
RE
R B =R1∥R2 = =1 ≈10 R E
2
10
R1 =R 2=20 R E
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ESE319 Introduction to Microelectronics
Typical Design
Choose: I E =1 mA
V CC =12V
And the rest of the design
follows immediately:
V E 12/2−0.7
R E= =
=5.3 k 
−3
IE
10
Use standard sizes
100 kΩ
12V
100 kΩ
5.1 kΩ
R E =5.1 k 
R1 =R 2=100 k 
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Equivalent Circuits
CB
RS
vout
<=>
vs
R B =R1∥R2
RB
Rb
VB
V CC
vout
vo
RE
VCC/2
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Multisim Bias Check
V Rb= I B R B=
IE
R =0.495V
1 B
iB
-
vout
<=>
VRb
+
Rb
vout
Identical results – as expected!
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Emitter Follower Small Signal Circuit
RS
Mid-band equivalent circuit:
RB
50
v=
v s=
v s≈v s
R B R S
50.05
'
s
50
RTH = R S∥R B =
RS ≈ RS
50.05
vs
vout
vo
R
Rb
B
RE
Small signal mid-band circuit - where CB has negligible reactance
(above fmin). Thevenin circuit consisting of RS and RB shows
effect of RB negligible, since it is much larger than RS.
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Follower Small Signal Analysis - Voltage Gain
Circuit analysis:
v s = R S r  1 R E  i b
RS
ib
vs
ie
vout
vo
RE
Solving for ib
vs
i b=
R S r 1 R E
v o =R E i e =R E 1i b
R E 1v s
vo=
R S r  1 R E
for Current Bias Design
replace RE with ro||ro = ro/2 >> RE
2009 Kenneth R. Laker, updated 06Oct09 KRL
vo
R E r o∥r o
AV = =
≈1
v s R S r 
R E r o∥r o
1
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ESE319 Introduction to Microelectronics
Small Signal Analysis – Voltage Gain - cont.
vo
RE
=
v s R S r 
 RE
1
Since, typically:
R S r 
1
≪ R E (or ro||ro = ro/2)
vo R E
AV = ≈ =1
vs RE
Note: AV is non-inverting
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
RS C B
ib
vs
Blocking Capacitor - CB - Selection
+
ib
Rb i
vbg e
v bg R
B
-
Rin RRinB=50k ≫RS
RS =50
Use the base current expression:
v bg =r  i b R E i E = r 1 i b
vo
RE
v bg
i b=
r  1 R E
v bg
r bg = =r 1 R E ≈1 R E =101⋅5.1 k =515 k 
ib
To obtain the base to ground resistance of the transistor:
This transistor input resistance is in parallel with the 50 k 
RB, forming the total amplifier input resistance:
515
Rin = R S R B∥r bg ≈ R B∥r bg =
50 k =45.6 k ≈ R B =50 k 
51550
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
CB – Selection cont.
Choose CB such that its reactance is ≤ 1/10 of Rin at fmin:
Rin
1
Assume fmin = 20 Hz
=
2 f C B 10
Rin ≈50 k 
with
10
CB ≥
2  f min Rin
10
CB ≥
≈1.59  F
3
2 ⋅20⋅50⋅10
Pick CB = 2  F (two 1 F caps in parallel), the nearest standard
value in the RCA Lab. We could be (unnecessarily) more precise
and include Rs as part of the total resistance in the loop. It is very
small compared to Rin.
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Final Design
CB
R1
2.0 uF
RS
vs
2009 Kenneth R. Laker, updated 06Oct09 KRL
V CC
R2
vo
RE
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ESE319 Introduction to Microelectronics
Multisim Simulation Results
20 Hz Data
1 kHz Data
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ESE319 Introduction to Microelectronics
Of What value is a Unity Gain Amplifier?
RS
ib
vs
ie
vo
To answer this question,
we must examine the small-signal
output impedance of the amplifier
and its power gain.
RE
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Emitter Follower Output Resistance
RS
ib
vs
0
ix
vx
Rout
R B =50 k ≫ R S
Assume:
−i x
i x =−i b−i b =−1i b ⇒i b =
1
R S r 
v x =−i b  R S r  =
i
1 x
r
v x R S r 
R out = =
≈
=r e
ix
1
1
Rout is the Thevenin resistance looking
into the open-circuit output.
VT
VT
I C =1 mA ⇒ r  = = =2500 
IB
IC
=100
R S =50
2009 Kenneth R. Laker, updated 06Oct09 KRL
2550
R out ≈
=25.5
100
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ESE319 Introduction to Microelectronics
Multisim Verification of Rout
RS
Rout
vs
Rin
Av*vsig
AV = 1
i x=−1i b
i sc =i x
<=>
Rin =R S r 1 R E ≈1 R E
Thevenin equivalent for the
short-circuited emitter follower.
If β = 200, as for most good
NPN transistors, Rout would
be lower - close to 12 Ω.
2009 Kenneth R. Laker, updated 06Oct09 KRL
v sig = RS i b r  i b
vs
AV v sig R S r 
Rout =
=
ix
1
+
v oc= AV v s
-
i sc =i x
=100
Multisim short circuit check
(  = 100, vo = vs):
v oc AV v srms
1
R out = =
=
=25.25
i sc
i sc rms
0.0396
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ESE319 Introduction to Microelectronics
Equivalent Circuits with Load RL
RS
ib
vo
vs
ie
RE
RL
RL
Z in =
Rout
vs
+
'
ie
<=>
vs
Rin +
-
Av vs
vo
R L∥R E
-
v s rms
1
R out =
=
=25.25 
i sc rms  0.0396
R in=R S r  1 R E∥R L≈1 R L
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Emitter Follower Power Gain
Consider the case where a RL = 50  load is connected through an infinite
capacitor to the emitter of the follower we designed. Using its Thevenin
is
equivalent:
Rout ≈25 
C=∞
vs
Rin
+
-
io
Avth=G
v v s vsig
R L AV v s 50
2
vo=
= v= v
R L Rout 75 s 3 s
AV v s
vs
i o=
=
R out R L 75
2 2
po =v o i o =
vs
225
2009 Kenneth R. Laker, updated 06Oct09 KRL
+
vo
R L≈50
AV ≈1
-
50
 load is in parallel with 5.1k 
R E∥R L =5.1 k ∥50≈50
RE and dominates:
vs
vs
vs
vs
i s=ib = ≈
≈
≈
R in 1 R E∥R L 101⋅50 5000
1
2
p s =v s i s≈
vs
5000
p o 25000
A pwr = =
=44.4≫1
ps
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ESE319 Introduction to Microelectronics
The Common Base Amplifier
Voltage Bias Design
2009 Kenneth R. Laker, updated 06Oct09 KRL
Current Bias Design
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ESE319 Introduction to Microelectronics
Common Base Configuration
Both voltage and current biasing follow the same rules as
those applied to the common emitter amplifier.
As before, insert a blocking capacitor in the input signal path
to avoid disturbing the dc bias.
The common base amplifier uses a bypass capacitor – or a
direct connection from base to ground to hold the base at
ground for the signal only!
The common emitter amplifier (except for intentional RE
feedback) holds the emitter at signal ground, while the common
collector circuit does the same for the collector.
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Voltage Bias Common Base Design
We keep the same bias that we established for the gain
of 10 common emitter amplifier.
All that we need to do is pick the capacitor values and
calculate the circuit gain.
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Common Base Small Signal Analysis - CIN
ib
Determine CIN: (let C B =∞)
4.7 k Ohm
∞
ic
Find a equivalent impedance for
the input circuit, RS, Cin, and RE2:
ie
v Re2
NOTE:
RB is shorted
by CB = ∞
470 Ohm
vs
v Re2 =
R E2∥r e R S 
ideally
1
2 f min C in
re
≪ R S R E2∥r e ⇒
2009 Kenneth R. Laker, updated 06Oct09 KRL
R E2∥r e
1
j2  f C in
R E2∥r e
v Re2 =
vs
R E2∥r e R S
for
vs
r
re =
1
f ≥ f min
R S r e
10
=
⇒C in =
10
2 f min C in
2  f min  R S r e 
1
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ESE319 Introduction to Microelectronics
Determine CIN cont.
A suitable value for Cin for a 20 Hz lower frequency:
10
10
2 f min C in  R S r e ≫1⇒ C in ≥
=
F
2  f min  R S r e  2 20⋅75
10
C in =
≈1062 F !
125.6⋅75
Not too Practical!
Must choose smaller value of Cin.
1. Choose: 2 f min C in  R S r e =1
or
2. Choose larger f min
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Small-signal Analysis - CB
i'b
i'c
ib
ib'
ic
Determine CB: (let C in =∞ )
Note the ac reference current
reversals (due to vs polarity)!

RE2 >> RS
ignore RB
i'e
ie

1
v s= R i  r  
i 'b
j C B
'
S e
vs
Zin
Determine Z in =
vs
i
'
e
2009 Kenneth R. Laker, updated 06Oct09 KRL


'
ie
1
'
v s= R S i e  r  
j  C B 1
'
e
i=
1
1 R S r 
1
j C B
vs
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ESE319 Introduction to Microelectronics
Determine – CB
i'b
i'c
1 R S r 
ib'
ideally Z in≈
i'e
RE2 >> RS
1
i'e =
vs
or
ignore RB
Z in=
vs
'
ie
1
j 2 f C B
1 R S r 
1
=R S 
vs
re
r
1
1
≪1 R S r 
2 f C B
f ≥ f min
f ≥ f min
Choose (conservatively):
10
C B≥
F
2  f min  1 R S r  
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Determine - CB cont.
i'b
i'c
vout
ib'
i'e
RE2 >> RS
vs
ignore RB
Choosing (conservatively):
10
C B≥
F
2  f min  1 R S r  
for fmin = 20 Hz
10
C B≥
=10.6 F
2 20  100502500 
i.e.
Choose (less conservatively):
1
C B≥
=1.06 F
2  20  100502500 
2009 Kenneth R. Laker, updated 06Oct09 KRL
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ESE319 Introduction to Microelectronics
Small-signal Analysis – Voltage Gain
∞
RE2 >> RS
v out
ib'
vs
ignore RB
Assume: C B =C in =∞
2009 Kenneth R. Laker, updated 06Oct09 KRL
'
e
i≈
1
r
RS 
1
1
v s=
vs
 R S r e 
RC

v out =R i = R i =
vs
1 R S r e
'
C c
'
C e
v out
RC

100 5100
AV =
=
=
≈67
v s 1 R S r e 101 5025
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ESE319 Introduction to Microelectronics
Multisim Simulation
10.6 uF
1060 uF
vs
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ESE319 Introduction to Microelectronics
Multisim Frequency Response
20 Hz response
1 kHz Response
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