ESE319 Introduction to Microelectronics
Common Emitter BJT Amplifier
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Adding a signal source to the
single power supply bias amplifier
Desired effect – addition
of bias and signal sources
Starting point single dc source
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Wrong Way to Combine Sources
<=>
=100
Source with low output
impedance upsets bias.
=100
Thevenin equivalent at base:
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Wrong Way - Continued
V th =
Vth
To analyze,
1. Isolate base circuit,
2 .Use superposition,
3. Apply Thevenin
equivalent again for the
base bias source, VB:
50
5
V B≈
VB
2000050
2000
Rth= R'B= RB∥RS =
50⋅20000
≈50 
5020000
We effectively “short out” the
bias source!
The signal source essentially
is unaffected. (Why? What is its
Thevenin equivalent?)
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
The Right Way – Use a “Blocking” Capacitor
iCb
vCb
+ Cb
=100
For convenience lets
continue to use the “base
bias Thevenin equiv”.
A capacitor is an OPEN at
dc and RS does not affect the
bias! The capacitor charges
to the dc bias source, VB, to
satisfy Kirchoff's voltage law
and the dc bias is in series
with the signal source.
DESIGN GOAL: for f ≥ fmin, set the
value of Cb so that its voltage drop vCb
is negligible at the lowest operating
frequency fmin.
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Blocking Capacitor Selection
ic
ib
apply
ie
=100
evaluate
<=>
ib
ib
vbg
r bg =
ib
r bg
Use the small signal equivalent
circuit and superposition to
estimate the input resistance of
the transistor.
ie
ic
i e =i b  i b
vbg = i b r  i e RE =i b r   i b 1 RE
vbg
r bg = = r  1 RE ≈1 RE
ib
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Capacitor Selection- continued
The signal source “sees” the 20 k  bias source resistance
in parallel with r bg. So the signal source equivalent circuit is:
iCb
vCb
+ Cb
vsig
=100
r bg ≈100 RE =400 k 
Therefore:
400
RB∣∣r BG=
20≈20 k = RB
420
The capacitor voltage drop is:
1
vCb=
i Cb
j  Cb
Where the capacitor current is:
1
v sig = i Cb  RS  RB∥r bg 

j Cb
v sig
vsig
iCb≈
=> vCb≈
1
j  C b RB1
RB
j Cb
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Capacitor Selection - continued
vsig
1

=
vCb≈
Cb
C b RB
j  C b RB1 or
iCb
DESIGN GOAL: Choose Cb for negligible voltage
vCb
+ -
drop at vCb for f ≥ fmin, i.e.
Cb
OR
vsig
v sig
for f ≥ fmin
vCb≈
≤
j  C b RB1 10
vsig
10
vCb ≤
⇒ min C b RB ≥10⇒ C b ≥
10
2  f min RB
DESIGN GOAL: Choose Cb so that Cb=0.1 min ,
i.e.
1
10
Cb=
=0.1 min ⇒ C b =
C b RB
2 f min RB
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Capacitor Selection - continued
Select the LOWEST frequency of
interest. This sets the lower bound
on Cb. Using fmin = 20 Hz frequency for
our example circuit:
2 f min=2 20≈6.28⋅20=125.6 sec−1
Cb ≥
10
2 f min RB
10
C b −min=
2 f min RB
10
10−6
C b −min=
=
=4  F
3
125.6⋅20⋅10 0.25
ANY capacitor larger than 4  F will
do the job! This is an inequality, not an
equality! Choose Cb = 5  F
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Common Emitter Unity Gain Amplifier
Cb
5 uF
Cb
5 uF
Equivalent circuits
How can we achieve reasonable gain with this circuit?
Solution: Split RE and use capacitor bypassing.
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Bypass for Gain
Cb
5 uF
+
ie
vRE1
+
vZE2
-
=100
Z E2 = RE2∥
1
j  C byp
Procedure:
1. Split the emitter resistor in
two. Later, we will show that
the voltage gain will be close
to -RC/RE1.
2. Bypass RE2 with a capacitor
Cbyp that looks like a near
“short circuit” at the lowest
frequency of interest (fmin).
i.e. |vZE2| << |vRE1| for f ≥ fmin)
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
v RE1=i e RE1 & v ZE2=i e Z E2
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ESE319 Introduction to Microelectronics
Bypass for Gain - continued
Rb
20 k Ohm
+
Rb
vRE1
+
20 k Ohm
vZE2
-
=100
=100
Small signal circuit
Desired circuit for f ≥ fmin
i.e. DESIGN GOAL: Choose Cbyp
s.t. |vZE2| << |vRE1| for f ≥ fmin
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Rs||Rb
c
b
ib
ic
e
ie
where
+
vRE1
+
vZE2
(Thevenin resistance) R
Need to develop a design equation for Cbyp
s.t. DESIGN GOAL: |vRE1| >> |vZE2|
or DESIGN GOAL: | 1/ j  C byp | << RCbyp
RE2
Z E2 =
j  C byp RE2 1
v ZE2
Z E2
RE2 / RE1
1
|
|=|
|=|
|
≤
vRE1
RE1
j  C byp RE2 1
10
Cbyp
i e =1 i b
=100
r e ≪ RE1
vRE1=i e RE1 & v ZE2=i e Z E2
RE2
1
10
C byp ≥10
=
RE1 2 f min RE2 2 f min RE1
10
C byp− min=
2  f min RE1
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
DESIGN GOAL: |1/ j  C byp| << RCbyp
set v-sig = 0
RS∥RB
RCbyp= RE2∥ RE1 r e 
≈ RE2∥ RE1
1
1
1
≪ RE2∥RE1= RE2∥ RE1
10
 C byp
(Thevenin resistance)
10
10
C byp ≥
=
 min RE2∥ RE1 2  f min RE2∥RE1
=100
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
C byp− min=
10
10
=
F =79.6  F
2  f min RE1 2  x 20 x 1000
Choose Cbyp = 100  F
Final Design
Cb
5 uF
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Gain Calculation in Passband
Simple gain calculation:
+
vout -
v sig
v sig
i b=
≈
Rs r 1 RE1 1 RE1
vout =− RC i c =− RC  i b
− RC
vout =
v sig
1 RE1
=100
Passband model
v out
RC
G=
≈−
=−4
vsig
RE1
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
Multisim Simulation
20 Hz Gain
I kHz Gain
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
What if R E is Fully Bypassed?
vout
RC
G=
≈−
=∞ ?
vsig
RE1
0
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
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ESE319 Introduction to Microelectronics
What if R E is Fully Bypassed?
v sig
vsig
i b=
=
Rs r 1 RE1 RS  r 
Cb
5 uF
0
vout
vout =− RC i c =− RC  i b
− RC
− RC
vout =
v sig ≈
vsig
RS  r 
r
I C =1 mA
g m=0.04 S
r  =/ g m=2.5 k 
=100
2008 Kenneth R. Laker (based on P.V. Lopresti) updated 24Sep08 KRL
− RC
.=
v =− g m RC vsig
 / g m sig
vout
G=
≈− g m RC =−160
vsig
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