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
●
Kenneth R. Laker, updated 27Sep13 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)
CB Amplifier
High (RC/(Rs + re))
Current Gain (AI)
High (β)
High (β + 1)
low (about 1)
Input Resistance
High (RB||βRE)
high (RB||βRE)
low (re)
low (re)
High (RC||ro)
Output Resistance
High (RC||ro)
VCVS
CE BJT amplifier => CS MOS amplifier
CC BJT amplifier => CD MOS amplifier
CB BJT amplifier => CG MOS amplifier
Kenneth R. Laker, updated 27Sep13 KRL
CCCS
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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.”
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
DC Bias
C in
RE
RS
vout
vO
<=>
vs
R B =R1∥R2
RB
Rb
VB
iC
V CC
vout
vO
RE
VCC/2
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
Follower Small Signal Analysis - Voltage Gain
Circuit analysis:
v s = R s r  i b 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
vo=
R s r  1 R E
for Current Bias Design
replace RE with ro||ro = ro/2 >> RE
Kenneth R. Laker, updated 27Sep13 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.
Rs
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
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
Of What value is a Unity Gain Amplifier?
Rs
ib
vs
ie
vo
RE
Or Buffer
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
Emitter Follower Small Sig. 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 I C
1
1
VT
VT
I C =1 mA ⇒ r = =
=2500 
IB
IC
=100 R s =50 
Kenneth R. Laker, updated 27Sep13 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.
Kenneth R. Laker, updated 27Sep13 KRL
v oc
R out =
i sc
vvsigs=RSs i bbr
ri bi b
v s rms
vs
+
v ocv=ocA=v
V v ss
-
Rout
iiscsc=i
x
rms
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
Kenneth R. Laker, updated 27Sep13 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
Quick Review
Rs
?
ib
vs
ie
vo
RE
?
?
?
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
Quick Review
Rs
ib
vs
ie
vo
RE
Or Buffer
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
Base Voltage 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
Kenneth R. Laker, updated 27Sep13 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 
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
R s =50
R S C in
ib
Blocking Capacitor - Cin - Selection
ib
+
Rb ie
vbg
vs
v bg R
B
-
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
Rin
This transistor input resistance rbg is in parallel with RB = 50 k ;
vbg:
R B∥r bg v s
v bg =
1
R B∥r bg R s 
2  f C in
Ideally we want vbg = vs for f ≥ fmin
Kenneth R. Laker, updated 27Sep13 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.
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
Final Design
C in
R1
3.3 uF
RS
vs
Kenneth R. Laker, updated 27Sep13 KRL
V CC
R2
vo
RE
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ESE319 Introduction to Microelectronics
Multisim Simulation Results
20.74 Hz Data Av = 0.989
1 kHz Data Av = 0.995
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
The Common Base Amplifier
Voltage Bias Design
Kenneth R. Laker, updated 27Sep13 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.
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 RE feedback)
holds the emitter at signal ground, while the common collector
circuit does the same for the collector.
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
Voltage Bias Common Base Design
CE Amp
We keep the same bias (1/3, 1/3, 1/3) 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.
Kenneth R. Laker, updated 27Sep13 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
Kenneth R. Laker, updated 27Sep13 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
4.7 k Ohm
ic
∞
Cb R
B
re
25 Ohm
v Re
NOTE:
Cb is short ckt
ie
RS
C in
RE
470 Ohm
Determine Cin: (let C b =∞ )
v Re =
R E∥r e
R E∥r e R s 
vs
1
j2  f C in
vs
R E∥r e
re
ideally v Re = R ∥r R v s≈ r R v s
E
e
s
e
s
for f ≥ f min
R s R E
1
10
≪ R s r e ⇒
=
⇒ C in=
10
2 f min C in
2  f min C in
2  f min  R s r e 
1
Kenneth R. Laker, updated 27Sep13 KRL
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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 =
≈1060 F
125.6⋅75
Too Large to be Practical!
Choose Cin = 220 µF (largest standard value in Detkin Lab)
1
X Cin =
≈36 
2  f min C in
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
ignore RB
Small-signal Analysis - Cb
Determine Cb: (let C in =∞ )
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
Kenneth R. Laker, updated 27Sep13 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
Kenneth R. Laker, updated 27Sep13 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
153 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
Kenneth R. Laker, updated 27Sep13 KRL
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ESE319 Introduction to Microelectronics
Multisim Frequency Response
19.3 Hz response Av sim=61.8
vertical axis is
a linear scale
1 %
1 kHz Response
Kenneth R. Laker, updated 27Sep13 KRL
Av sim=63.3A v theory =62.7
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