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 05Oct12 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 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Common Collector ( Emitter Follower) Amplifier
R1
ro
C in
V CC
RS
vs
vs
vO
R2
RE
Voltage Bias Design
Q1
Qamp
Q2
vO
ro
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 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Equivalent Circuits
C in
RE
RS
vout
vO
<=>
vs
R B =R1∥R2
RB
Rb
VB
V CC
vout
vO
RE
VCC/2
2009 Kenneth R. Laker, updated 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Multisim Bias Check
V Rb =I B R B =
Cin
IE
R =0.495V
1 B
Cin
iB
-
vout
<=>
VRb
+
Rb
vout
Identical results – as expected!
2009 Kenneth R. Laker, updated 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Follower Small Signal Analysis - Voltage Gain
Circuit analysis:
v s =R S r  ib R E i e = R S r   1 R E i b
RS
ib
vs
ie
vout
v o=v e
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
v o=
R S r  1 R E
for Current Bias Design
replace RE with ro||ro = ro/2 >> RE
2009 Kenneth R. Laker, updated 05Oct12 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 
R E
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 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Quick Review
CE Amplifier
CC Amplifier
CB Amplifier
Voltage Gain (AV)
Current Gain (AI)
Input Resistance
Output Resistance
ANSWERS: Low, Moderate or High
2009 Kenneth R. Laker, updated 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Quick Review cont.
CE Amplifier
Voltage Gain (AV) moderate (-RC/RE)
Current Gain (AI)
moderate (β)
Input Resistance
high (RB||βRE)
Output Resistance
high (RC||ro)
2009 Kenneth R. Laker, updated 05Oct12 KRL
CC Amplifier
CB Amplifier
low (about 1)
high (RC/RS)
moderate (β + 1)
low (about 1)
high (RB||βRE)
low (re)
low (re)
VCVS
high (RC||ro)
CCCS
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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 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Emitter Follower Output Resistance
−i x
i x =−i e =−1i b ⇒ i b=
1
RS
ib
vs
0
ie
ix
vx
Rout
Assume:
R S r 
v x =−i b  R S r  =
i x where r  ≫ R S
1
r
v x R S r 
1 VT
R out = =
≈
=r e= =
ix
gm IC
1
1
VT
VT
I C =1 mA ⇒ r = =
=2500 
IB
IC
=100 R S =50 
2009 Kenneth R. Laker, updated 05Oct12 KRL
R out ≈r e =
2550
=25.5 
100
Recall Rin =r bg =r  1 R E
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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 bg =R S r 1 R E ≈1 R E
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 05Oct12 KRL
v oc
R out =
i sc
vvsigs =
=R
RSSiibbr
ri bi b
vs
+
v ocv=ocA=v
V v ss
-
Rout
i sc =i x
vAs v RS r
R S r 
V=sig
RRoutout=
=i
=
1
sc i x 1
=100
Multisim short circuit check
(  = 100, vo = vs):
v oc Av s rms
1 V1
 
v
v
oc
V
s
rms
= =
=
==
=22.25

RRoutout=
=25.25
39.61
mA
iiscsc i iscscrms
0.0396
rms 
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ESE319 Introduction to Microelectronics
Emitter Follower Power Gain
R S C in
ib
+
ib
Rb ie
v bg vbg
RB
vs
-
Rin
vo
RE
Consider the case where a RL=50 Ω load is connected
through an infinite capacitor to the emitter of the follower.
Using its Thevenin equivalent:
RL
r bg
Rin
@ midband where Av = 1
R L AV v s
50
2
vo=
= v= v
R L∥R E R out 75 s 3 s
AV v s
vs
i o=
=
R out R L∥R E 75
2 2
p o =v o i o =
vs
225
2009 Kenneth R. Laker, updated 05Oct12 KRL
is
vs
R out≈ 25
i
Rin
o
+
- A
vth=G
v v s vsig
C=∞
+
vo
R E∥R L=50 
-
50  load is in parallel with 5.1k 
vs
vs
vs
vs
REb =
and ≈
dominates: ≈
i s =i
≈
R in
1 R L∥R E 101⋅50
R E∥R L =5.1 k ∥50 ≈50 
1 2
p s =v s i s ≈
vs
5000
p o 25000
A pwr = =
=44.4≫1
ps
225
5000
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ESE319 Introduction to Microelectronics
Emitter Follower Biasing – Typical Design
R1
C in
RS
vs
VB
Vb
iB
R2
Then, choose/specified IE, and
the rest of the design follows:
iC
iE
vO =v E
vout
RE
V CC
V E V CC /2−0.7
RE= =
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
⇒ R 1=R 2
V B=
2
2009 Kenneth R. Laker, updated 05Oct12 KRL
As with CE bias design, stable op. pt.
=>
R B ≪1 R E , i.e.
R 1 1
R B =R 1∥R 2= =
R E ≈10 R E
2
10
R 1=R 2 =20 R E
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ESE319 Introduction to Microelectronics
Typical Design - Cont.
Given: R out =r e =25 
V CC =12 V
And the rest of the design follows:
VT
I E ≈ I C = =1 mA
re
V E 12/ 2−0.7
RE= =
=5.3 k 
−3
IE
10
Use standard sizes
100 kΩ
C in
v O =v E
12V
100 kΩ
5.1 kΩ
R E =5.1 k 
R1 =R 2=100 k 
2009 Kenneth R. Laker, updated 05Oct12 KRL
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ESE319 Introduction to Microelectronics
R S =50 
R S C in
ib
Blocking Capacitor - Cin - Selection
Rb ie
vbg
vs
v bg R
B
-
Rin
ib
+
Rin
Use the base current expression:
v bg =r  i b R E i E =r  1 R E  ib
vo
RE
r bg
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 RB = 50 k ,
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 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Cin – Selection cont.
Choose Cin such that its reactance is ≤ 1/10 of RB at fmin:
RB
1
Assume fmin = 20 Hz
=
2 f C in 10
with R B =50 k 
10
C in ≥
2  f min R B
10
C in ≥
=1.59 F
3
2 ⋅20⋅50⋅10
Pick Cin = 3.3 µ F ), the nearest standard value in the Detkin Lab.
We could be (unnecessarily) more precise and include rbg and Rs
as part of the total resistance in the loop.
2009 Kenneth R. Laker, updated 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Final Design
C in
R1
3.3 uF
RS
vs
2009 Kenneth R. Laker, updated 05Oct12 KRL
V CC
R2
vo
RE
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ESE319 Introduction to Microelectronics
Multisim Simulation Results
20 Hz Data
Av = 0.995
1 kHz Data
2009 Kenneth R. Laker, updated 05Oct12 KRL
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ESE319 Introduction to Microelectronics
The Common Base Amplifier
Voltage Bias Design
2009 Kenneth R. Laker, updated 05Oct12 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!
RECALL: 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 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Voltage Bias Common Base Design
CE Amp
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 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Mid-band Small Signal Analysis
Input Impedance
RC
ro
is
Zin
C in =∞
Note: is = - ie
RE >>re
2009 Kenneth R. Laker, updated 05Oct12 KRL
v Re =r e∥R E i s
v Re
VT
Z in = =r e∥R E ≈r e =
is
IC
vo
Z out = =RC∥r o
ic
Current Gain
ic
Ai = =≈1
ie
Voltage Gain
v s =−i e R S −r e∥R E i e
RC
1
v o =−RC i c =− RC i e =
vs
 R S r e∥R E
v o 1 RC
Av = ≈
v s  R S r e
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ESE319 Introduction to Microelectronics
Common Base Small Signal Analysis - Cin
ib
vo
RC
Determine Cin: (let C b =∞ )
4.7 k Ohm
ic
∞
Cb R
B
re
25 Ohm
NOTE:
RB is shorted
v Re
ie
RS
C in
RE
470 Ohm
v Re =
R E∥r e R S 
vs
by Cb = ∞
1
2 f min C in
R E∥r e
re
≪ R S R E∥r e ⇒
2009 Kenneth R. Laker, updated 05Oct12 KRL
1
j2  f C in
R E∥r e
ideally v Re = R ∥r R v s
E
e
S
vs
r
re=
1
for 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 fmin with re = 25 Ω and RS = 50 Ω:
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 Practical!
Must choose smaller value of Cin.
Choose: 2 f min C in R S r e =1
1
C in =
≈106.2  F
125.6⋅75
2009 Kenneth R. Laker, updated 05Oct12 KRL
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ESE319 Introduction to Microelectronics
ignore RB
Small-signal Analysis - Cb
Determine Cb: (let C in =∞ )
Note the ac reference current
reversals (due to vs polarity)!
1

j 2 f C b 1
v Re =
vs
1
R E∥r e 
R S
j 2 f C b 1
R E∥r e 
R E∥r e
ideally v Re = R ∥r R v s for f ≥ f min
E
e
S
re
1
1
10
≪ re ⇒
= ⇒ C b=
2 f min C b 1
2  f min C b 1 10
2  f min r e 1
2009 Kenneth R. Laker, updated 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Determine - CB cont.
i'b
i'c
vout
vo
R E∥r e≈r e
is
R
4.7 Ck Ohm
ib'
i'e
vs
ignore RB
2009 Kenneth R. Laker, updated 05Oct12 KRL
Choose (conservatively):
10
C b=
F
2  f min  1r e 
for fmin = 20 Hz
i.e.
10
C b=
=31.8  F
2  20  10025 
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ESE319 Introduction to Microelectronics
Multisim Simulation
4.7 k Ohm
vO
31.8 uF
1060 uF
470 Ohm
vs
v o 1 RC
RC
4700
AV = =
≈
=
=62.7
v s  R S r e R S r e 5025
2009 Kenneth R. Laker, updated 05Oct12 KRL
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ESE319 Introduction to Microelectronics
Multisim Frequency Response
20 Hz response
vertical axis is
a linear scale
1 %
1 kHz Response
2009 Kenneth R. Laker, updated 05Oct12 KRL
Av sim=63.3A v theory =62.7
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