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
●
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 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 (  )
CB Amplifier
high
low (about 1)
Input Resistance
high
high
low
Output Resistance
high
low
high
CE BJT amplifier => CS MOS amplifier
CC BJT amplifier => CD MOS amplifier
CB BJT amplifier => CG MOS amplifier
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Common Collector ( Emitter Follower)
Amplifier
vout
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.”
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Emitter Follower Biasing
Then, choose/specified IE, and
the rest of the design follows:
iC
Vb
iB
iE
vout
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
⇒ R1 = R2
V B=
2
As with CE bias design, stable op.
pt. => RB ≪1 RE , i.e.
R1
RE
RB= R1∥ R2 = =1 ≈10 RE
2
10
R1 = R2=20 RE
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Typical Design
Choose: I E =1 mA
V CC =12 V
And the rest of the design
follows immediately:
V E 12/ 2−0.7
RE = =
=5.3 k 
−3
IE
10
iC
Vb
iB
iE
vout
Use standard sizes!
RE =5.1 k 
R1 = R2=100 k 
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Equivalent Circuits
RB= R1∥R2
Rb
vout
vout
<=>
VCC/2
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Multisim Bias Check
V Rb= I B R B=
IE
R =0.495 V
1 B
iB
-
vout
<=>
VRb
+
Rb
vout
Identical results – as expected!
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Emitter Follower Small Signal
Circuit
Mid-band equivalent circuit:
RB
50
vsig ' =
vsig =
vsig ≈ vsig
RB RS
50.05
50
RTH = RS∥ RB=
RS ≈ RS
50.05
vout
Rb
Small signal mid-band circuit - where CB has negligible reactance
(above min). Thevenin circuit consisting of RS and RB shows
effect of RB negligible, since it is much larger than RS.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Follower Small Signal Analysis - Voltage Gain
Circuit analysis:
vsig = RS  r 1 RE  i b
ib
ie
vout
Solving for ib
v sig
i b=
RS  r 1 RE
RE 1 v sig
vout =
RS  r  1 RE
vout
RE v sig
AV = =
≈1
vsig RS r 
 RE
1
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Small Signal Analysis – Voltage Gain - cont.
vout
RE
=
v sig RS  r 
 RE
1
vout
Since, typically:
RS  r 
≪ RE
1
vout RE
AV =
≈ =1
v sig RE
Note: AV is non-inverting
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Blocking Capacitor - CB - Selection
+
Rb
vbg
-
ib
Use the base current expression:
vbg = r  i b  RE i E = r  1 i b
vbg
i b=
r 1 RE
vbg
r bg = = r 1 RE ≈1 RE =101⋅5.1 k =515 k 
RS =50
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:
RRinB=50 k ≫ RS
515
Ri n = RS  RB∥r bg ≈ RB∥r bg =
50 k =45.6 k 
51550
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
CB – Selection cont.
Choose CB such that its reactance is ≤ 1/10 of Rin at min:
Ri n
1
Assume the lowest frequency
=
 C B 10
is 20 Hz:
10
CB ≥
min Ri n
 min=2 20≈125=1.25⋅102
=100
−7
CB ≥
10⋅10
≈1.73 F
1.25⋅0.46
Ri n≈46 k 
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.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Final Design
2.0 uF
vout
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Multisim Simulation
Results
20 Hz. Data
1 Khz. Data
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Of What value is a Unity Gain Amplifier?
ib
ie
vout
To answer this question,
we must examine the
output impedance of the
amplifier and its power
gain.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Emitter Follower Output Resistance
ib
0
vout
ix
vx
Rout
RB=50 k ≫ RS
Assume:
−i x
i x=−i b− i b =−1 i b ⇒ i b =
1
RS  r 
v x=−i b  Rs r  =
i
1 x
r
v x RS  r 
Rout = =
≈
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
RS =50
2550
Rout ≈
=25.5
100
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Multisim Verification of R
out
Rout
Rin
Av*vsig
AV = 1
i x=−1 i x
i sc =i x
vsig = RS i b  r  i b
<=>
Rin = RS  r 1 RE∥ RL≈1 RL
+
voc= AV v sig
-
Thevenin equivalent for the
short-circuited emitter follower.
If  was 200, as for most good
NPN transistors, Rout would be
lower - close to 12 .
AV v sig RS  r 
Rout =
=
ix
1
i sc =i x
Multisim short circuit check
(  = 100, vout = vsig):
voc AV v sig  rms 
1
Rout = =
=
=25.25
i sc
i sc  rms 
0.0396
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Equivalent Circuits with Load R
L
ib
vout
ie
Rout
v
Z in = sig
i 'e
<=>
RL
+
+
Rin
Av*vsig vload RL||Re
-
vsig  rms 
1
Rout =
=
=25.25
i sc  rms  0.0396
Rin= RS  r  1 RE∥RL≈1 RL
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 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
equivalent:
isig
Rout≈25  C =∞
vsig
Rin
iload
+
- Av*vsig
vth=G vsig
RL AV vsig 50
2
vload =
= v = v
RL Rout 75 sig 3 sig
AV vsig
v sig
i load =
=
Rout  RL 75
2 2
pload =vload i load =
vsig
225
+
vload RL≈50
AV ≈1
-
50  load is in parallel with 5.1k 
RE and dominates:
vsig
vsig
v sig
i sig =i b = ≈
≈
Rin 1 RE∥RL 101⋅50
1
v2sig
5000
pload 25000
G pwr =
=
=44.4≫1
p sig
225
p sig =v sig i sig ≈
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
The Common Base Amplifier
vout
vout
Voltage Bias Design
Current Bias Design
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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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.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Voltage Bias Common Base Design
47k Ohm 4.7 k Ohm
vout
5.1k Ohm
470 Ohm
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.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Common Base Small Signal Analysis - C
i'b
ib
Determine CIN: (let C B=∞)
4.7 k Ohm
∞
ic
i'e
I'c
ie
470 Ohm
Find a equivalent impedance for
the input circuit, RS, CIN, and RE2:
RE2∥r e
vRe2=
isig
RE2∥r e  RS 
Zin
re
IN
1
j  C IN
RE2∥r e
ideally vRe2= R ∥r  R vsig
E2
e
S
vsig r = r 
e
1
for  ≥  min
RS  r e
1
1
10
≪ RS  RE2∥r e ⇒
=
⇒ C IN =
10
 min C IN
 min C IN
min  RS  r e 
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Determine C IN cont.
A suitable value for CIN for a 20 Hz lower frequency:
10
10
 min C IN  RS  r e ≫1⇒ C IN ≥
=
F
2 min  RS  r e  2  20⋅75
10
C IN =
≈1062  F !
125.6⋅75
Not too practical!
Must choose smaller value of CIN.
1. Choose:  min C IN  RS  r e =1
or
2. Choose larger  min
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Small-signal Analysis - C
i'b
i'c
ib
RE2 >> RS
ib'
Determine CB: (let C IN =∞)
Note the reference current
reversals (due to vsig polarity)!


1
v sig = R i  r 
i 'b
j CB
'
S e
i'e
ie
Zin
Determine Z in =
ic
B
v sig
i
'
e


'
i
1
e
v sig = RS i'e  r 
j  C B 1
1
'
e
i=
1 RS  r 
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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j CB
vsig
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ESE319 Introduction to Microelectronics
Determine – CB
i'b
i'c
'
e
i=
ib'
i'e
1
1
1 RS  r 
j CB
v sig
'
e
i=
RE2 >> RS
RS 
Zin
Z in=
v sig
i 'e
vsig

1
1
r 
1
j C B
= RS 

1
1
r 
1
j CB


r
1
ideally Z in≈ RS 
⇒
≪1 RS  r  for  ≥  min
1  C B
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Determine - C B cont.
i'b
i'c
For  ≥  min
vout
ib'
i'e
RE2 >> RS
r
1
Z in≈ RS 
⇒
≪1 RS  r 
1  C B
Choose:
C B≥
10
F
min 1 RS  r  
i.e.
C B≥
10
=10.5  F
2  20  1001502500 
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Small-signal Analysis –
∞
ib'
RE2 >> RS
Assume: C B=C IN =∞
'
e
i≈
Voltage Gain
1
r
RS 
1
1
v sig =
v sig
 RS  r e 
RC

vout = R i = R i =
vsig
1 RS  r e
'
C c
'
C e
vout
RC

100 5100
AV =
=
=
≈67
vsig 1 RS  r e 101 5025
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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ESE319 Introduction to Microelectronics
Multisim Simulation
1062 uF
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ESE319 Introduction to Microelectronics
Multisim Frequency Response
20 Hz. response
1 KHZ. Response
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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