Chapter 5. BJT AC Analysis
Outline:
 The re transistor model
CB, CE & CC
 AC analysis through re model
common-emitter fixed-bias
voltage-divider bias
emitter-bias & emitter-follower
common-base configuration
BJT AC Analysis
Transistor Modeling
A model is a combination of circuit elements,
properly chosen, that best approximates the
actual behavior of a semiconductor device
under specific operating condition.
There are three models:
 re model
 hybrid  model
 hybrid equivalent model.
BJT AC Analysis
Before discussing model, we must make
preparation to the discussion.
 The superposition theorem is applicable
and the the investigation of the dc
conditions can be totally separated from
the ac response.
 An suitable Q-point has been chosen.
Then the dc levels can be ignored in the
ac analysis network.
BJT AC Analysis
 The coupling capacitor C1, C2 and
bypass capacitor C3 are chosen to have
a very small reactance. Therefore, each
is replaced by a short circuit.
This also results in the “shorting out”
of dc biasing resistor RE .
 Those important parameters, such as Zi ,
Zo , Vi , Vo , Ii and Io should be kept
unchanged.
BJT AC Analysis
The input voltage Vi is defined between
base and ground.
The input current Ii is defined as the
base current of the transistor.
The input impedance Zi is defined from
base to ground.
The output voltage Vo is defined between
collector and ground.
BJT AC Analysis
The output current Io is defined as the
current through the load resistor RC.
The output impedance Zo is defined
from collector to ground.
 By introducing a common ground, R1
and R2 will be in parallel.
RC will appear from collector to emitter.
BJT AC Analysis
 The transistor equivalent circuit also
consists of resistors and independent
controlled source.
The circuit analysis techniques such as
superposition, Thevenin theorem can be
applied to determine the desired
quantities.
 There are important quantities indicating
amplification effects:
BJT AC Analysis
Voltage gain: Vo / Vi
Current gain: Io / Ii
 Briefly, steps to obtain ac equivalent circuit:
 dc sources to zero
 capacitors to short circuits
 Removing elements bypassed by short
circuits and rearranging network
BJT AC Analysis
Figure: Transistor circuit for ac analysis
BJT AC Analysis
The re Model for CB
As shown in the figure, it is the commonbase BJT circuit.
Now, re model is introduced.
 On the input port, there is a resistor, re .
where re = 26mV / IE
(Eq. 5.1)
The subscript e of re indicates that it is
the dc level of IE that determines the ac
level of the resistance.
BJT AC Analysis
 On the output port, there is a controlled
current source, denoted by a diamond
shape.
The Ic is controlled by the level of Ie .
 The input impedance Zi is re .
 The output impedance Zo  .
In general, for common-base configuration,
the Zi is relatively small and Zo quite high.
BJT AC Analysis
 Voltage gain:
Vo = - Io RL = - (- Ic) RL =Ie RL
Vi = Ii Zi = Ie Zi =Ie re
so that
Av = Vo / Vi = (Ie RL)/(Ie re) = ( RL)/ re  RL / re
 Current gain:
Ai = Io / Ii = (-Ic)/(Ie) = (- Ie )/(Ie)  -1
 For common-base configuration, the Vo and
Vi are in phase.
BJT AC Analysis
Figure: re model for common-base circuit
BJT AC Analysis
The re Model for CE
As shown in the figure, it is the commonemitter BJT circuit.
Now, replace the transistor with re model
 On the input side, there exists re .
Zi = Vi / Ii = Vbe /Ib = (Ie re)/ Ib = [( +1)Ib re]/ Ib
= ( +1) re   re
( if  is large enough)
Typically, Zi is at moderate level.
BJT AC Analysis
 On the output side, the controlledcurrent source is connected between
collector and emitter, Isource=  Ib.
 Ideally, the output impedance Zo  .
Or, Zo = ro , as shown in the figure. At
this time, Ic  Ib .
 Normally, voltage gain Av and current
gain Ai are high.
 For common-collector configuration,
this model is also applicable.
BJT AC Analysis
Figure: re model for common-emitter circuit
BJT AC Analysis
CE Fixed Bias Circuit
As shown in the figure, it is the commonemitter fixed-bias configuration.
 The input signal Vi is applied to the base
and the output Vo is off the collector.
 The input current Ii is not the base current
and the Io is the collector current.
 For small-signal analysis, VCC is replaced
with ground.
BJT AC Analysis
 Those dc blocking capacitors C1 and C2 are
replaced with short circuits.
 So the RB is in parallel with the input port
and RC output port.
 The parameters of Zi , Zo , Ii and Io should
be in the same places as original network.
 Finally, substitute the re model for the
transistor.
BJT AC Analysis
 In practice,
The  is read from the datasheet or
measured with testing instrument.
The re is determined from dc analysis.
The ro is obtained from the datasheet.
 In the small-signal analysis, we assume
that , ro and re have been determined in
advance.
BJT AC Analysis
 Input impedance Zi :
From the figure, it is obvious that
Zi = RB|| re
For the majority of situation, RB is greater
than  re by more than a factor of 10.
So we get
Zi   re
BJT AC Analysis
 Output impedance Zo:
From the definition of Zo , we get
Zo = RC ||ro
If ro  10RC , then
Zo RC
 Voltage gain Av:
From the figure, we get
Vo =( - Ib )(RC ||ro)
BJT AC Analysis
And
Ib =Vi /( re)
So
Vo =( - Ib )(RC ||ro) =( - Vi /( re))(RC ||ro)
=( - Vi / re)(RC ||ro)
Finally,
Av = Vo / Vi
= - (RC ||ro) / re
BJT AC Analysis
 Phase relationship:
The negative sign in Av ,reveals that a
180 phase shift occurs between the input
and output signals.
This also means that a single-stage of
amplifier of this type is not enough.
The magnitude of output signal is larger
than that of input signal. But the
frequencies of them should be the same.
BJT AC Analysis
Figure: re model for CE fixed-bias circuit
BJT AC Analysis
Figure: phase shift of input & output
BJT AC Analysis
Example 5.4
As shown in the figure, it is the commonemitter fixed-bias configuration. Determine:
re , Zi , Zo, Av with ro = 50k .
Solution:
 From dc analysis, we get
12V  0.7V
VCC  VBE

IB 
 24.04 A
470k
RB
BJT AC Analysis
IE= (β+1) IB = (100+1)  24.04A
= 2.428mA
re = (26mV)/IE = (26mV)/ 2.428mA
= 10.71 
 Then, ac analysis
re = (100)(10.71  ) = 1.071 k
Zi = RB|| re = (470 k ) || (1.071 k )
= 1.069 k
BJT AC Analysis
Zo = RC || ro = (3 k ) || (50 k )
= 2.83k
Av = - (RC ||ro) / re = - 2.83k / 10.71 
= - 264
From the Av ,we can see that the output
signal has been amplified but out of
phase with the input signal.
BJT AC Analysis
Figure: Example 5.4
BJT AC Analysis
Voltage Divider Bias
As shown in the figure, it is the voltage
divider bias configuration.
Substituting re equivalent circuit, note that:
 RE is absent due to the low impedance of
the bypass capacitor CE .
 When VCC is set to zero, one end of R1
and RC are connected to ground.
 R1 and R2 remain part of the input circuit
while RC is part of output circuit.
BJT AC Analysis
Some parameter of the equivalent circuit:
 Input impedance Zi :
Zi = R1|| R2|| re
 Output impedance Zo:
Zo = RC ||ro
If ro  10RC , then
Zo RC
 Voltage gain Av:
BJT AC Analysis
Vo =( - Ib )(RC ||ro) =[- Vi /( re)] (RC ||ro)
=( - Vi / re)(RC ||ro)
So,
Av = Vo / Vi = - (RC ||ro) / re
If ro  10RC , then
Av = Vo / Vi  - RC / re
 Phase relationship:
180 phase shift occurs between the input
and output signals.
BJT AC Analysis
Figure: Voltage divider bias & its equivalent circuit
BJT AC Analysis
Example 5.5
As shown in the figure, it is the voltage
divider bias configuration. Determine:
re , Zi , Zo, Av with ro = 50k .
Solution:
 dc analysis, testing RE >10 R2 ,
901.5k > 108.2k
135k > 82k (satisfied)
BJT AC Analysis
Using the approximate approach (Sec. 4.5,
p155), we obtain:
R2
8.2k
VB 
VCC 
 22V  2.81V
R1  R2
56k  8.2k
VE = VB - VBE = 2.81V-0.7V = 2.11V
IE = VE / RE = 2.11V/1.5k = 1.41mA
re = 26mV / IE = 26mV / 1.41mA
= 18.44 
BJT AC Analysis
 ac analysis,
Zi = R1|| R2|| re
= 56k || 8.2k || (90)(18.44 )
= 1.347k
Zo = RC ||ro
= 6.8k || 50k = 5.99k
Av= - (RC ||ro) / re = - 5.99k /18.44 
= - 324
BJT AC Analysis
Figure: Example 5.5
BJT AC Analysis
CE Emitter Bias
As shown in the figure, it is the emitter bias
configuration, without CE .
Substituting re equivalent circuit, note that:
 The resistance ro is ignored for simplicity.
 First, let us obtain Zb .
Vi =Ibre+ Ie RE =Ibre+(+1) Ib RE
So, Zb = Vi / Ib = re+(+1) RE
BJT AC Analysis
 At input port, Zi = RB ||Zb
 At output port,
Zo = RC
Vo = -Io RC = - Ib RC = - (Vi / Zb) RC
So, Av = Vo / Vi = - RC / Zb
 Phase relationship:
180 phase shift occurs between the
output and input signals.
BJT AC Analysis
Figure: Emitter bias & its equivalent circuit
BJT AC Analysis
Example 5.6
As shown in the figure, it is the emitter bias
configuration. Determine:
re , Zi , Zo, Av.
Solution:
 dc analysis,
VCC  VBE
IB 
RB  (   1) RE
20V  0.7V

 35.89 A
470k  (120  1)  0.56k
BJT AC Analysis
IE= (β+1) IB = (120+1)  35.89A
=4.34mA
re = (26mV)/IE = (26mV)/ 4.34mA
= 5.99 
 Then, ac analysis
Zb = re+(+1) RE
=1205.99 + 121560
= 68.48 k
BJT AC Analysis
Zi = RB|| Zb = (470 k ) || (68.48k )
= 59.77 k
Zo = RC = 2.2 k
Av = - RC / Zb = -120  2.2k / 68.48k
= -3.86
For emitter bias with CE , see Ex. 5.7.
For emitter bias + voltage divider, see Ex. 5.8
& 5.9.
BJT AC Analysis
Figure: Example 5.6
BJT AC Analysis
Emitter-Follower
As shown in the figure, it is the emitterfollower configuration.
Actually, it is a common-collector network.
The output is always slightly less than the
input, but this is good for practical use.
Also the Vo is in phase with Vi and this
accounts for the name of “emitter-follower”.
BJT AC Analysis
The emitter-follower configuration has a high
input impedance and a low output impedance.
This is the reason why it is used for impedance
matching purpose.
This can give a weak load to the previous
stage and a strong output ability to the next
stage.
BJT AC Analysis
Substituting re equivalent circuit, note that:
 The resistance ro is ignored because for
most applications a good approximation
for the actual results can still be obtained.
 First, the same as before, Zb is obtained:
Zb = re+(+1) RE
Then,
Zi = RB ||Zb
BJT AC Analysis
 At output port,
Ie = (+1) Ib = (+1) (Vi /Zb )
So,
(   1)Vi
Ie 
re  (   1) RE
If  is sufficiently large, we get
Vi
Ie 
re  RE
This means that is Ie generated by Vi .
BJT AC Analysis
Also, Vo is the potential drop across RE .
So we construct a network from the
viewpoint of output port.
 So, by setting the Vi to zero, we get
Zo = RE ||re
 Furthermore, from this network it is
obvious that
RE
Vo 
Vi
RE  re
BJT AC Analysis
This leads to
Vo
RE
Av  
Vi RE  re
Since, RE is usually much greater than re ,
Av  1
 Phase relationship:
Vo and Vi are in phase.
BJT AC Analysis
Figure: Emitter-follower & its equivalent circuit
BJT AC Analysis
Example 5.10
As shown in the figure, it is the emitterfollower configuration. Determine:
re , Zi , Zo, Av.
Solution:
 dc analysis,
VCC  VBE
IB 
RB  (   1) RE
20V  0.7V

 20.43A
220k  (100  1)  3.3k
BJT AC Analysis
IE= (β+1) IB = (100+1)  20.43A
=2.063mA
re = (26mV)/IE = (26mV)/ 2.063mA
= 12.6 
 Then, ac analysis
Zb = re+(+1) RE
=10012.6 + 1013.3k
= 334.56 k
BJT AC Analysis
Zi = RB|| Zb = (220 k ) || (334.56k )
= 132.72 k
Zo = RE ||re = 3.3k ||12.6  = 12.55 
Vo
RE
3.3k
Av  

 0.9962
Vi RE  re 3.3k  12.6
For some variations of emitter follower
configuration, see Fig. 5.54 & 5.55
BJT AC Analysis
Figure: Example 5.10
BJT AC Analysis
Common-base Configuration
As shown in the figure, it is the commonbase configuration.
It has a relatively low Zi and high Zo and a
current gain less than 1.
However, the Av can be quite large.
Substituting re equivalent circuit into the
network, note that:
BJT AC Analysis
 The resistance ro is typically in the M
and can be ignored in parallel with RC.
 First, Zi = RE || re
 Then, Zo= RC
And Vo = - Io RC = Ic RC =  Ie RC
With Vi = Ie re ,
So that Av = Vo / Vi =  RC / re
BJT AC Analysis
 At last, assuming RE >> re , we get
Ii = Ie
And
Io = - Ie = - Ii
So that Ai = Io / Ii = -  -1
 Phase relationship:
Vo and Vi are in phase in commonbase configuration.
BJT AC Analysis
Figure: Common-base & its equivalent circuit
BJT AC Analysis
Example 5.11
As shown in the figure, it is the commonbase configuration. Determine:
re , Zi , Zo, Av and Ai.
Solution:
 dc analysis,
VEE  VBE 2V  0.7V
 1.3mA
IE 

RE
1k
re = (26mV)/IE = (26mV)/ 1.3mA
= 20 
BJT AC Analysis
 Then, ac analysis
Zi = RE || re = 1 k || 20 = 19.61
Zo= RC = 5 k
Av= RC / re =0.985 k / 20 = 245
Ai = - = -0.98  -1
BJT AC Analysis
Figure: Example 5.11
BJT AC Analysis
Content not discussed but still
important
• The hybrid equivalent mode (Sec. 5.5)
• The hybrid  model (Sec. 5.6)
• Collector Feedback (Sec. 5.13)
• Cascaded Systems (Sec. 5.19)
• Darlington Connection (Sec. 5.20)
• Current Source (Sec. 5.22, 5.23)
BJT AC Analysis
Summary of Chapter 5
The re model is applied to the analysis of
transistor configurations:
 Common-emitter:
Fixed bias; Voltage divider; Emitter bias
 Common-collector: emitter-follower
 Common-base.
The calculation of those parameters, like Vi ,
Vo , Ii , Io, Zi , Zo, Av and Ai ,in the circuits.
BJT AC Analysis