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Bipolar Junction Transistors
Physical Structure and Modes of Operation
Emitter
(E)
Metal
contact
n-type
p-type
n-type
Emitter
region
Base
region
Collecter
region
Emitter-base
junction
(EBJ)
Mode
Base
(B)
EBJ
Cutoff
Reverse
Active
Forward
Saturation Forward
© REP 5/1/2017 ENGR224
Collecter
(C)
Collecter-base
junction
(CBJ)
CBJ
Reverse
Reverse
Forward
Page BJT 4.1-1
Bipolar Junction Transistors
Operation of the npn Transistor in the Active Mode
p
n
E
Injected
electrons
n
Diffusing
electrons
Collected
electrons
C
iC
Injected holes (iB1)
iE
iB
 v BE 
iE
iE
 VBE 
© REP 5/1/2017 ENGR224
 vCB 
iB
B
iC
iC
 VCB 
Page BJT 4.1-2
Bipolar Junction Transistors
Current Flow


Only diffusion-current components are considered
Profiles of minority-carrier concentrations in the base and in the emitter of an npn
transistor operating in the active mode; vBE > 0 and vCB  0.
EBJ
depletion
region
Carrier concentration
Emitter
(n)
Base
(p)
CBJ
depletion
region
Collector
(n)
vBE
n p (0)  n p 0 e VT
dn p x 
I n  AE qDn
dx
 n p 0 

 AE qDn  
 W 
Electron
concentration
np (ideal)
Hole
concentration
pn0
pn(0)
np(0)
np (with
recombination)
Distance (x)
Effective base
width W
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Page BJT 4.1-3
Bipolar Junction Transistors
The Collector Current



Most of the diffusing electrons will reach the boundary of the collector-base depletion
region
These successful electrons will be swept across the CBJ depletion region into the
collector
By convention, the direction of iC is opposite to that of electron flow
iC  I n and n p 0  n p 0 e vBE
iC  I S e vBE
VT
VT
saturation current
IS 
AE qDn ni2
N AW
n p 0  ni2 N A
I S  AE qDn n p 0 W
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Page BJT 4.1-4
Bipolar Junction Transistors
The Base Current
Two components of base current, iB1 and iB2.
Hole diffusivity
in the emitter
iB1 
AE qD p ni2
N D Lp
e
vBE VT
Hole diffusion length
in the emitter
iB 2 
Qn
b
minority-carrier
lifetime
Doping concentration
of the emitter
1
Qn  AE q  n p 0W
2
AE qWni2 vBE VT
Qn 
e
2N A
iB 2
1 AE qWni2 vBE

e
2 bNA
© REP 5/1/2017 ENGR224
VT
 Dp N A W 1 W 2  v
e BE
iB  I S 

D N L

 n D p 2 Dn b 
iB 
iC

I 
iB   S e vBE

VT
VT
 Dp N A W 1 W 2 

  1 


 Dn N D L p 2 Dn b 
common-emitter
current gain
Page BJT 4.1-5
Bipolar Junction Transistors
The Emitter Current
common-base
current gain
iE  iC  iB
 1
iC

 1 v
iE 
IS

iC  iE

iE 
© REP 5/1/2017 ENGR224

 1
iE  I S  e vBE
BE
VT

VT

1
Page BJT 4.1-6
Bipolar Junction Transistors
First Order Equivalent Circuit Models




The externally controlled signals for this model are
Voltage Controlled Current Source Model
the three currents shown outside the gray box.
C
The voltage VBE, exists internally as a result of the
currents and can be externally measured. We can
iC
force a current and measure a voltage.
The diode in the model is designated as DE since the
VBE
current flowing through the diode is the same as the
emitter current. The collector current is dependent on
iB
I S e VT
the base-emitter voltage VBE.
B
The model is a non-linear voltage controlled current

source
C



The externally controlled
signals for this model are two
currents and the voltage VBE
shown outside the gray box.
The current iE exists internally
as a result of the voltage VBE
and can be externally
measured.
The collector current is
dependent on the emitter
current iE.
iE
iB
B
iC
DE
vBE

iE
E

DE
vBE
iE

Current Controlled Current Source
E
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Page BJT 4.1-7
Bipolar Junction Transistors
Equivalent Circuit Models, cont’d


In this version of the model the diode conducts the BASE current which is beta times
smaller.
In one version the dependent current source is voltage controlled (vBE), in the other
version the dependent current source is current controlled ().
iB
B
iC
iC

v BE

C
DB
I S  
I S e vBE
VT
B

v BE
iB
C
DB
I S  
I B

iE
iE
E
Note connection point is now
on the opposite side of the diode
Voltage Controlled Current Source Model
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E
Current Controlled Current Source
Page BJT 4.1-8
Bipolar Junction Transistors
Two Port Model of the Common-Base Configuration
iC
C
iB
B
ISe

vBE
B
VBE
VT
iin
B
DE
iE
E
C
The base lead is common to both ports

E
E
B
Two port
Network
E
C
B
B
If we switch the leads within the network
the common base aspect is more apparent
iE
B
iC
C iout
B
Two-Port representation of a BJT Transistor
symbol in a common-base configuration
© REP 5/1/2017 ENGR224
iin  iE
iout  ic
Ai 
iout iC
 
iin iE
The common-base current gain is 
Page BJT 4.1-9
Bipolar Junction Transistors
Two Port Model of the Common-Emitter Configuration
iC
B

v BE
iB
C
DB
I S  
B
I B
E

Two port
Network
C
E
The emitter lead is common to both ports
iE
E
iC is out of phase with iB
iC
iin
B
iB
E
C iout
iin  iB
E
Two-Port representation of a BJT Transistor
symbol in a common-emitter configuration
© REP 5/1/2017 ENGR224
iout  iC
Ai 
iout  iC

 
iin
iB
The common-emitter current gain is 
Page BJT 4.1-10
Bipolar Junction Transistors
Operation of the pnp Transistor in the Active Mode
n
p
E
Injected
holes
p
Diffusing
holes
Collected
holes
C
iC
Injected electrons (iB1)
iE
iB
 vEB 
iE
iE
 VEB 
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 vBC 
iB
B
iC
iC
 VBC 
Page BJT 4.1-11
Bipolar Junction Transistors
Equivalent pnp Circuit Models
E
iB
B
iE

vEB
E
DE
I S

iE

ISe

vEB
VEB
VT

B
DB
I S

I S e vEB
VT
C
iC
iB
iC
C
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Page BJT 4.1-12
Bipolar Junction Transistors
Circuit Symbols and Conventions - npn
C
C
VCB
iC
B
B
iB
VBE
E
iE
E
npn BJT
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Voltage polarities and current flow in a
transistor biased in the active mode.
Page BJT 4.1-13
Bipolar Junction Transistors
Circuit Symbols and Conventions - pnp
E
E
VEB
iE
B
B
iB
VBC
C
iC
C
pnp BJT
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Voltage polarities and current flow in a
transistor biased in the active mode.
Page BJT 4.1-14
Bipolar Junction Transistors
Example 4.1

The transistor in the circuit below has  = 100 and exhibits a vBE of 0.7V at iC = 1 mA.
Design the circuit so that a current of 2 mA flows through the collector and a voltage of +5V
appears at the collector.
RC
15V  5V 10V

 5k
2mA
2mA
since VBE  0.7V at iC  1mA,
 15V
 15V
I C  2mA
RC 
RC
VC  5V
 2
VBE  0.7  VT ln    0.717V
1
and VE  VBE  0.717V
VE  VBE
RE
 15V
RE
 15V
I E  IC  I B
for   100,   100 101  0.99
I
2
thus I E  C 
 2.02mA
 0.99
RE 

© REP 5/1/2017 ENGR224
VE   15
IE
 .0717  15
 7.07k
2.02
Page BJT 4.1-15
Bipolar Junction Transistors
Graphical Representation of Transistor Characteristics

Similar to diodes, except we talk about the voltage across one junction VBE and the
current through the other terminal iC.
For most of the conditions we will encounter in working with BJTs the ideality factor, n
will be considered to be 1.

iC
iC  I S e vBE VT
T1 T2 T3
iC
I
0
0.5
0.7
iC-vBE characteristics
0
0.5 0.7
vBE V 
vBE
I
iB  S

I
iE  S

vBE V 
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0
T1>T2>T3
0.5 0.7
Effect of temperature on iC-vBE characteristic.
At a constant current, vBE changes by –2mV/oC.
vBE V 
Page BJT 4.1-16
Bipolar Junction Transistors
iC versus vCB Characteristics

npn transistor in active mode
saturation
vCB
iC
iC mA  iE
Current controlled
current source
iE  4 mA
4
 3
3 mA
2
 1
2 mA
1 mA
iE
-Vnp = forward bias
saturation
vCB V 
+Vnp = reverse bias
0 1 2 3
iC mA
iB 4
iC vs vCE
iB 3 See next page
iB 2
i B1
0 1 2 3
© REP 5/1/2017 ENGR224
vCE V 
Page BJT 4.1-17
Bipolar Junction Transistors
iC=vCE Characteristics


The Early Voltage (typically 50 -100 Volts), also known as the Base-Width Modulation
parameter.
As the base-collector junction reverse bias is
increased the depletion layer expands and
consumes some of the base narrowing it and
causing an increase in the collector current.
 v
iC  I S evBE VT 1  CE
 VA
iC

vBE



Saturation
region
iC
 i
rO   C
 vCE


vBE constant 

rO 
1
VA
IC
Active
region
vBE  . . .
vBE  . . .
vBE  . . .

vCE
vBE  . . .


 VA
© REP 5/1/2017 ENGR224
0
vCE
Page BJT 4.1-18
Bipolar Junction Transistors
Example 4.2

We wish to analyze this circuit to determine all node voltages and branch currents. We
will assume that  is specified to be 100.
 10V
 4V
RC  4.7k
RC  4.7k
10 V
4V
RE  3.3k
© REP 5/1/2017 ENGR224
RE  3.3k
Page BJT 4.1-19
Bipolar Junction Transistors
Example 4.2, cont’d


We don’t know whether the transistor is in the active mode or not.
A simple approach would be to assume that the device is in the active mode, and then
check our results at the end
1
VE  4  VBE  4  0.7  3.3 V
 10V
3 IC
 4V
4.7 k
VC 4
5 IB
3.3 k
VE 1
IE 2
V  0 3.3
 1 mA

IE  E
3.3
RE

2
100
 0.99
  1 1013
I C  0.99  1  0.99 mA
I C  I E ,  

4
VC  10  I C RC  10  0.99  4.7  5.3 V
1
I
 0.01 mA
IB  E 
  1 101
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5
Page BJT 4.1-20
Bipolar Junction Transistors
Example 4.3

We wish to analyze the circuit shown below to determine the voltages at all nodes and
the currents through all branches. Note that this circuit is identical to the previous
circuit except that the voltage at the base is now +6 V.
 10V
 10V
Assuming active-mode:
VE  5.6  VBE  6  0.7  5.3 V
3 IC
 6V
RC  4.7k
 4V
4.7 k
VC
IB
RE  3.3k
3.3 k
VE 1
IE
4
IE 
5.3
 1.6 mA
3.3
I C  I C  1.6 mA
2
VC  10  4.7  I C  10  7.52  2.48 V
Collector voltage > base voltage
saturation mode, not active mode
© REP 5/1/2017 ENGR224
Page BJT 4.1-21
Bipolar Junction Transistors
Example 4.4

We wish to analyze the circuit below to determine the voltages at all nodes and the
currents through all branches. This circuit is identical to that considered in the previous
two examples except that now the base voltage is zero.
 10V
 10V
RC  4.7k
RE  3.3k
© REP 5/1/2017 ENGR224
I C  0 mA 3
5
I B  0 mA
3.3 k
4.7 k
VC  10 V
4
VE  0V cutoff
I E  0 mA
2
1
Page BJT 4.1-22
Bipolar Junction Transistors
Example 4.6

We will analyze the following circuit to determine the voltages at all nodes and currents
through all branches. Assume =100.
 10V
 10V
IC
 5V
VB  VBE  0.7 V
I C  4.3 mA
RC  2 k
IB
RB  100 k
 5V
100 k
RC  2 k
VC  1.4 V
IB 
5  VBE 5  0.7

 0.043 mA
RB
100
I C  I B  100  0.043  4.3 mA
2
3
4
I B  0.043 mA
2
© REP 5/1/2017 ENGR224
3
1
I E  4.343 mA
5
VC  10  I C RC  10  4.3  2  1.4 V 4
I E    1I B  101 0.043  4.3 mA
Page BJT 4.1-23
5
Bipolar Junction Transistors
Example 4.7

We want to analyze the circuit shown below to determine the voltages at all nodes and
currents through all branches. Assume =100.
 15 V
RB1 
100 k
RB 2 
50 k
 15 V
RC  5 k
VBB  5 V
RC  5 k
RBB 
RE  3 k
33.3 k
VBB  15
RE  3 k
RB 2
50
 15
 5V
RB1  RB 2
100  50
RBB  RB1 // RB 2   100 // 50   33.3 k
VBB  I B RBB  VBE  I E RE
IB 
© REP 5/1/2017 ENGR224
IE
 1
Page BJT 4.1-24
Bipolar Junction Transistors
Example 4.7, cont’d
 15 V
1.28 mA
5V
0.013 mA
1.29
 0.0128 mA
101
VB  VBE  I E RE
8.6 V
33.3 k
3.87 V
0.013 mA
1.29 mA
IE 
100 k
5 k
50 k
VBB  VBE
RE  RBB   1
0.103 mA
0.09 mA
IB 
 0.7  1.29  3  4.57 V
3 k
assuming active - mode operation,
I C  I E  0.99 1.29  1.28 mA
VC  15  I C RC  15  1.28  5  8.6 V
© REP 5/1/2017 ENGR224
Page BJT 4.1-25
Bipolar Junction Transistors
The BJT as an Amplifier

Objectives
1. Biasing
2. DC equations
3. Transconductance
4. Input resistance looking into the base
5. Input resistance looking into the emitter
6. Voltage gain
7. Gummel plots

Lesson
1. Biasing
1) For our amplifiers, the BJT must be biased in the FORWARD-ACTIVE
2) However, it’s a difficult challenge to establish a CONSTANT DC CURRENT
3) Our goal: A Q point insensitive to TEMPERATURE , ß , VBE .
© REP 5/1/2017 ENGR224
Page BJT 4.1-26
Bipolar Junction Transistors
The BJT as an Amplifier
2. DC Equations (learn ‘em now)
1)
I C  I S eVBE / VT
2)
I E  IC / 
3)
I B  IC / 
4)
VC  VCC  I C RC


 0.99
 1


 100
 1
3. Transconductance (remember the small-signal approximation from before?)
- Valid only for vBE< 10 mV
- Defined as the incremental change in output current for an incremental change in input
voltage at a DC operating point…..
iC
gm 
vBE
© REP 5/1/2017 ENGR224
iC  I C
Page BJT 4.1-27
Bipolar Junction Transistors
The BJT as an Amplifier
iC  I S eVBE /VT  I S e(VBE vbe ) /VT  I S eVBE /VT evbe /VT  I C evbe /VT
ic
If vbe<< VT
IC
Vbe
vbe
IC
iC  I C (1  )  I C  I C  iC  vbe  g m vbe
VT
VT
VT
Q
VBE
x 2 x3
(e  1  x 

 .......)
2! 3!
x
vBE
 Note that iC = IC at vBE = VBE, so…...

v BE / VT
gm 
ISe
vBE
© REP 5/1/2017 ENGR224
iC  I C
IC

VT
Page BJT 4.1-28
Bipolar Junction Transistors
The BJT as an Amplifier
Input Resistance “ looking into “ the Base ( highlight this in your text & on this page!)
 Defined as the incremental change input voltage for an incremental change in base
current at a DC operating point…
vBE
r 
 iB
iC  I C
vBE

 iB
iC  I C
VT

IC
 Other important relationships ( be prepared to use any of these!)
r 
© REP 5/1/2017 ENGR224

gm
VT
r 
IB
Page BJT 4.1-29
Bipolar Junction Transistors
The BJT as an Amplifier
Input Resistance “looking into “ the Emitter (hightlight this in your text & on this page)
 Define as the incremental change in input voltage for an incremental change in
emitter current at DC operating point…..
vBE
re 
iE

iC  I C
1

 1 i
C
 IC
  1
VT
 
 
IC
  
 Other important relationship ( be prepared to use either of them!)
VT
re 
IE
© REP 5/1/2017 ENGR224

1
re 

gm gm
Page BJT 4.1-30
Bipolar Junction Transistors
The BJT as an Amplifier
Relationship between r and re
- The same input resistance . . . just “ viewed from two different places ! “
VT
r 
IB
VT
re 
IE
© REP 5/1/2017 ENGR224
I E  (   1) I B
r  (   1)re
r
re 
 1
Page BJT 4.1-31
Bipolar Junction Transistors
The BJT as an Amplifier
Lets look at Voltage Gain again
 A BJT senses vbe
 This is a
and causes a proportional current
gm vbe
VOLTAGE - CONTROLED CURRENT SOURCE
 So . . . How do we obtain an output voltage so that we get a voltage gain?
Out of phase with
the input
© REP 5/1/2017 ENGR224
Page BJT 4.1-32
Bipolar Junction Transistors
Voltage Gain
vC  VCC  iC RC
 VCC  I C  iC RC
 VCC  I C RC   iC RC
 VC  iC RC
Signal voltage:
vC  iC RC   g m vbe RC
  g m RC vbe
© REP 5/1/2017 ENGR224
Voltage gain:
vc
Voltage gain 
  g m RC
vbe
Page BJT 4.1-33
Bipolar Junction Transistors
Small-signal equivalent circuit models



Every current and voltage in the amplifier circuit is composed of two components: a dc
component and a signal component.
The dc components are determined from the dc circuit below on the left.
By eliminating the dc voltages, we are left with the signal components (on the right). The
resulting circuit is equivalent to the transistor as far as small-signal operation is
concerned.
IC
IB
vbe
VBE
+
-
RC
RC

v CE
VCC

IE
Amplifier circuit with dc sources
© REP 5/1/2017 ENGR224
ib  vbe r 
vbe
+
-

vbe

ic  g m vbe

vce

v
ie  be
re
Amplifier circuit with dc sources eliminated
Page BJT 4.1-34
Bipolar Junction Transistors
The Hybrid- Model
voltage-controlled
current source
current-controlled
current source
ic  g m vbe and ib  vbe r
g m vbe  g m ib r 
v
v
ie  be  g m vbe  be 1  g m r 
r
r

vbe
1     vbe
r
 g m r ib
g m vbe  ib
 r 


1




ie  vbe re
ib
B
ic

v BE
ic
C
r
B

v BE
g m vbe

C
ib
I B
r

g m  I C VT
ie
E
© REP 5/1/2017 ENGR224
r   g m
ie
E
Page BJT 4.1-35
Bipolar Junction Transistors
The T Model

These models explicitly show the emitter resistance re rather than the base resistance rp
featured in the hybrid- model.
ib 
vbe
v
 g m vbe  be 1  g m re 
re
re
vbe
1     vbe 1   
re
re    1 
vbe
v
ib 
 be
  1re r

C
ic
g m vbe
ib
B

v BE
re

ie
E
voltage-controlled
current source
© REP 5/1/2017 ENGR224
g m vbe  g m ie re 
 g m re ie
g m vbe  ie
C
ic
B
ib
i e

v BE ie

re
E
current-controlled
current source
Page BJT 4.1-36
Bipolar Junction Transistors
Application of the Small-Signal Equivalent Circuits

The availability of the small-signal BJT circuit models makes the analysis of transistor
amplifier circuits a systematic process consisting of the following steps:
 Determine the dc operating point of the BJT and in particular the dc collector current IC.
 Calculate the values of the small-signal model parameters:
gm 
IC
V

1
, r 
, and re  T 
VT
gm
I E gm
 Eliminate the dc sources by replacing each dc voltage source with a short circuit and
each dc current source with an open circuit.
 Replace the BJT with one of its small-signal equivalent circuit models. Although any one
of the models can be used, one might be more convenient than the others for the
particular circuit being analyzed. This point will be made clearer later in this chapter.
 Analyze the resulting circuit to determine the required quantities (e.g., voltage gain, input
resistance).
© REP 5/1/2017 ENGR224
Page BJT 4.1-37
Bipolar Junction Transistors
Example 4.9

We wish to analyze the transistor amplifier shown below to determine its voltage gain.
Assume  = 100.
VCC  10V
2
10V
2.3 mA
RC  3 k
RBB  100 k
3V
3 k
100 k
3.1 V
3
vi
0.023 mA
+
-
1
VBB  3 V
Assume vi  0 to find the dc base current
1
IB 
VBB  VBE
RBB
3  0.7
IB 
 0.023 mA
100
© REP 5/1/2017 ENGR224
2
0.7 V
2.323 mA
The dc collector current wi ll be
I C  I B  100  0.023  2.3 mA
The dc voltage at the collector will be
3 VC  VCC  I C RC
VC  10  2.3  3  3.1V
Page BJT 4.1-38
Bipolar Junction Transistors
Example 4.9, cont’d

Having determined the operating point, we now proceed to determine the small-signal
model parameters
RBB  100 k
vi
+
-
B
r

v BE
C
g m vbe

RC  3 k
E
re 
VT
25 mV

 10.8 
I E 2.3 0.99 mA
gm 
r 
I C 2.3 mA

 92 mA / V
VT 25 mV

gm

100
 1.09 k
92
© REP 5/1/2017 ENGR224
vbe  vi
 vi
r
r  RBB
1.09
 0.011 vi
101.09
Voltage gain :
vo
 3.04 V / V
vi
vo   g m vbe RC
vo  92  0.011 vi  3  3.04vi
Page BJT 4.1-39
Bipolar Junction Transistors
A note about Output Signal Swing






The collector voltage (and vo) can have a maximum value of zero volts before the
transistor goes from forward active mode to saturation mode since the base is
grounded.
When no input (ac) voltage is applied the output (collector) was found to be at a DC level
of -5.4V.
If we desire a symmetric output signal (about the -5.4V DC level) the signal would have
to go to -5.4 - 5.4 or -10.8 Volts (this is a large output signal swing).
This causes a problem, since our lower voltage supply is only -10V.
In order to avoid possibly producing a distorted output signal the input signal range
must be limited so that the output is not clipped as shown below. Limiting the input
signal to smaller values to limit clipping is not the same as using a small signal to
invoke the linear approximation as indicated in the next bullet item.
Another important point to be made about the output signal is that it is shown to be
linear in the figure below but in fact the iC-vbe characteristic is not linear for a large
output signal swing.
vC V 
t
0
-5.4
-10
clipping
© REP 5/1/2017 ENGR224
Page BJT 4.1-40
Bipolar Junction Transistors
Modifying the Hybrid- Model to Include the Early Effect



The Early effect causes the collector current to depend on vBE as well as vCE.
The dependence on vCE is modeled by assigning a finite output resistance to the
controlled current-source.
By including ro in the equivalent circuit shown below, the gain will be somewhat
reduced.
voltage-controlled
current source
B

v
current-controlled
current source
C
r
g m v

B
ib
C
r
ro
ib
1
E
 i 
V
ro   C 
 A
 vCE  vbe 0 I C
E
vo   g mvbe RC // ro 
© REP 5/1/2017 ENGR224
Page BJT 4.1-41
ro
Bipolar Junction Transistors
Summary of the BJT Small-Signal Model Parameters


Keep these at your fingertips (I.e. formula sheet for an exam or homework or in lab)
 See Table 4.3
Model parameters in terms of DC bias currents
V 
I
V
gm  C
re  T    T 
VT
IE
 IC 
r 

V 
VT
   T 
IB
 IC 

r 
gm

gm
Model parameters in terms of re
gm 

VA
IC
Model parameters in terms of the transconductance, gm
re 

ro 

re
r    1re
gm 
1 1

r re
Relationships between the common-emitter current gain and the common-base gain


1

© REP 5/1/2017 ENGR224

 1
 1 
1
1
Page BJT 4.1-42
Bipolar Junction Transistors
Graphical Analysis
iC
vi
iB

vBE

© REP 5/1/2017 ENGR224

vCE

Page BJT 4.1-43
Bipolar Junction Transistors
Graphical Analysis (cont.)
vCE  VCC  iC RC
VCC
1
iC 

vCE
RC
RC
© REP 5/1/2017 ENGR224
Page BJT 4.1-44
Bipolar Junction Transistors
Graphical Analysis (cont.)
© REP 5/1/2017 ENGR224
Page BJT 4.1-45
Bipolar Junction Transistors
Graphical Analysis (cont.)
© REP 5/1/2017 ENGR224
Page BJT 4.1-46
Bipolar Junction Transistors
Single Power Supply



Biasing the BJT involves establishing a constant dc current in the emitter which is
calculable, predictable, and insensitive to temperature variations and to  (for transistors
of the same type).
The bias point should allow for maximum output signal swing.
To design for a stable IE, the design constraints (shown below) must be satisfied.
VCC
R1
VCC
VCC
VBB
RC
RC
RBB
IC
IE
R2
RE
 R2 

VBB  VCC 
R

R
2 
 1
RR
RBB  1 2
R1  R2
IE 
VBB  VBE
RE  RB   1
RE
Design constraints:
Circuit topology for
biasing a BJT amplifier
© REP 5/1/2017 ENGR224
VBB  VBE
and
RE 
Page BJT 4.1-47
RB
 1
Bipolar Junction Transistors
Single Power Supply, cont’d

From the previous page, our design constraints are as follows:
VBB  VBE
and
RE 
RB
 1

When biasing the BJT, we must make sure of the following:
 VBB must not be too large, or it will lower the sum of voltages across RC and VCB
 RC should be large enough to obtain high voltage gain and large signal swing
 VCB (or VCE) should be large enough to provide a large signal swing

Rules of thumb:
1
VBB  VCC
3

1
VCB (or VCE )  VCC
3
1
I C RC  VCC
3
We’d like for RB to be small, which is achieved by low values of R1 and R2. This could
result in higher current drain from the power supply, hence lower input resistance (if the
input signal is coupled to the base). This means that we want to make the base voltage
independent of  and solely determined by the voltage divider. In order to achieve this,
another rule of thumb is practiced: select R1 and R2 such that their current is in the
range of I E to 0.1I E
© REP 5/1/2017 ENGR224
Page BJT 4.1-48
Bipolar Junction Transistors
Example 4.12

We wish to design the bias network of the amplifier shown below to establish a current IE =
1mA using a power supply VCC = +12 V.
VCC
VCC
neglecting base current:
for voltage divider current = 0.1IE
R1
RC
R1  R2 
12
 120 k
0.1I E
R2
VCC  4 V
R1  R2
Thus R2  40 k and
R2
R1  80 k
RE
for nonzero base current:
VB  4 V
VE  4  VBE  3.3 V
I
RE  E  3.3 k
VE
© REP 5/1/2017 ENGR224
IE 
VBB  VBE
3.3

 0.93 mA
RE  RB   1 3.3  0.267
for voltage divider current = IE
R1  8k and
IE 
R2  4k
3.3
 1 mA
3.3  0.0266
Page BJT 4.1-49
Bipolar Junction Transistors
Example 4.12, cont’d

Depending on what the emitter current is, we can have two designs:
 Design 1: the voltage divider current = 0.1IE, and
 Design 2: the voltage divider current = IE
Design 1
R1  80 k and
R2  40 k
Design 2
R1  8k and
R2  4k
IE 
VBB  VBE
3.3

 0.93 mA
RE  RB   1 3.3  0.267
IE 
RC 
12  VC
IC
RC 
12  VC
IC
RC 
12  8
 4.34k
0.99  0.93
RC 
12  8
 4.04k
0.99  1
© REP 5/1/2017 ENGR224
3.3
 1 mA
3.3  0.0266
Page BJT 4.1-50
Bipolar Junction Transistors
Biasing Using Two Power Supplies



A somewhat simpler bias arrangement is possible if two power supplies are available.
If the transistor is to be used with the base grounded, then RB can be eliminated
altogether.
If the input signal is to be coupled to the base, then RB is needed.
VCC
IE 
RC
I
IB  E
 1
VEE  VBE
RE  RB   1
Design constraints:
VBB  VBE
and
RE 
RB
 1
IE
RB
RE
 VEE
© REP 5/1/2017 ENGR224
Page BJT 4.1-51
Bipolar Junction Transistors
An Alternative Biasing Arrangement
VCC  I E RC  I B RB  VBE
VCC
VCC  I E RC 
RC
VC  VBE  I B RB
IB
RB
VBE
IE
R B  V BE
 1
VCC  V BE
RC  R B   1
IC
IE 
IE
select RB   1  RC
VCB  I B R B  I E
© REP 5/1/2017 ENGR224
RB
 1
Page BJT 4.1-52
Bipolar Junction Transistors
Biasing Using a Current Source



The BJT can be biased using a current source
The advantage is that the emitter current is independent of the values  and RB
Current-source biasing leads to significant design simplification
VCC
VCC
I
I REF
R
RC
IC
RB
I
V
IC
Q1

V BE

Q2
I REF 
VCC   VEE   VBE
R
 VEE
VCC  VEE  VBE
R
since Q1 and Q2 have the same VBE
I  I REF 
© REP 5/1/2017 ENGR224
Page BJT 4.1-53
Bipolar Junction Transistors
Common-Emitter Amplifier
Circuit
© REP 5/1/2017 ENGR224
AC Hybrid -based model
Page BJT 4.1-54
Bipolar Junction Transistors
Common-Emitter Amplifier (cont.)
Input Resistance:
Ri  r
Voltage Gain:
v
r

vs Rs  r
vo   g m v Rc ro 
vo
  g m Rc ro 
v
 Rc ro 
vo
Av   
vs
Rs  r
© REP 5/1/2017 ENGR224
Current Gain:
Ai 
io

ib
Ai   
 g m v ro
v
ro  RC 
r
ro
ro  RC
Output Resistance:
Ro  RC ro
Page BJT 4.1-55
Bipolar Junction Transistors
Exercise 4.31
For a CE amplifier, let I=1 mA, RC=5k, =100, VA=100V, and Rs=5k. Find Ri, Av, Ai, and Ro:
I C  I  1mA
gm 
IC
r  
ro  VA
VT
gm
IC
 1mA
 100
.25mV
 40mA / V
40mA / V
 100V
1mA
 2.5k
 100k
 Rc ro   1005k 100k 
vo
Av   

vs
Rs  r
5k  2.5k
 63.5V / V
Ai   
ro
100k
 100
ro  RC
100k  5k
 95.2 A / A
Ro  RC ro  5k 100k  4.76k
Ri  r  2.5k
What is Av if a 5k load resistor is added to the circuit:
Av load  Av
noload
© REP 5/1/2017 ENGR224
RL
5k
 63.5
 32.5V / V
RL  Ro
5k  4.76k
Page BJT 4.1-56
Bipolar Junction Transistors
Common-Emitter Amplifier with an Emitter Resistor
© REP 5/1/2017 ENGR224
Page BJT 4.1-57
Bipolar Junction Transistors
Common-Emitter Amplifier with an Emitter Resistor (cont.)
Input Resistance:
Voltage Gain:
vb  ie re  Re 
vo  ie Rc
vo  Rc

vb re  Re
ie
ib  1   ie 
 1
vb
Ri     1re  Re 
ib
vo
 Rc

vb re  Re
Ri (with R e included)   1re  Re 

  1re
Ri (without R e )
Re
 1
 1  g m Re
re
© REP 5/1/2017 ENGR224
since   1
vb
Ri

vs Ri  Rs
Av 
vo
Rc

vs Rs    1re  Re 
Page BJT 4.1-58
Bipolar Junction Transistors
Common-Emitter Amplifier with an Emitter Resistor (cont.)
Characteristics of CE amplifier with resistance Re:
v
re
1


vb re  Re 1  g m Re
Output Resistance:
Ro  Rc





Input resistance is increased by the factor of
(1+gmRe)
An input signal of (1+gmRe) times larger can
be applied to the input without inducing
nonlinear distortion
The voltage gain is reduced
The voltage gain is less dependent on the
value of 
The high frequency response is significantly
improved
Current Gain:
io
Ai    
ib
© REP 5/1/2017 ENGR224
Page BJT 4.1-59
Bipolar Junction Transistors
Exercise 4.32
For a CE amplifier with Re, let I=1 mA, RC=5k, =100, and Rs=5k. Find Re such that
the amplifier has an input resistence of 4 times that of the source. Find Av, Ai, and Ro:
re 
VT 25mV

 25
IE
1mA
Ro  Rc  5k
Ri  20k    1re  Re 
 100  125  Re 
Re  173
Av  

Rc
Rs    1re  Re 

100(5k)
5k  100  125  173 
500k
 20V / V
25k
Ai     100 A / A
© REP 5/1/2017 ENGR224
Page BJT 4.1-60
Bipolar Junction Transistors
Exercise 4.32
For the same CE amplifier, find the maximum vs without Re and with Re if v is to be
limited to 5mV:

100


 .9901
  1 101
v (max) with Re   v (max) without Re 1  g m Re 

.9901
gm 

 39.604 X 103
re
25
r 

gm

100
 2525
3
39.604 X 10
v (max)  5mV  vs (max)
r
Rs  r

 5mV 1  39.064 X 10 3 *173
 40mV
R  Ri
vs (max)  v (max) s
Ri
5k  20k
 40mV
20k
 50mV
2525
5000  2525
 14.9mV
 vs (max)
vs (max)
© REP 5/1/2017 ENGR224
Page BJT 4.1-61

Bipolar Junction Transistors
Common-Base Amplifier
© REP 5/1/2017 ENGR224
Page BJT 4.1-62
Bipolar Junction Transistors
Common-Base Amplifier (cont.)
Input Resistance:
Ri  re
Voltage Gain:
Current Gain:
io  ie
Ai  

ib
 ie
vo  ie Rc
vs
ie  
Rs  re
Av 
Output Resistance:
Ro  Rc
vo
Rc

vs Rs  re
© REP 5/1/2017 ENGR224
Page BJT 4.1-63
Bipolar Junction Transistors
Exercise 4.33
For a CB amplifier with Re, let I=1 mA, RC=5k, =100, and Rs=5k. Find RiAv, Ai, and
Ro:
Ri  re 

VT 25mV

 25
IE
1mA

100
 .9901
  1 101

Rc
0.9901(5k)
Av 

Rs  re
5k  25
Ai    0.99 A / A
Ro  Rc  5k
 0.985V / V
Note that RiAv, Ai are much lower than the CE amplifier using the same components
although the voltage gain of the CB amplifier can be almost equivalent if Rs is low.
© REP 5/1/2017 ENGR224
Page BJT 4.1-64
Bipolar Junction Transistors
Common-Collector Amplifier - Emitter Follower
© REP 5/1/2017 ENGR224
Page BJT 4.1-65
Bipolar Junction Transistors
Common-Collector Amplifier - Emitter Follower (cont.)
© REP 5/1/2017 ENGR224
Page BJT 4.1-66
Bipolar Junction Transistors
Common-Collector Amplifier - Emitter Follower (cont.)
Input Resistance:
Ri    1re  ro RL 
if re  RL  ro
Ri    1RL
Current Gain:
io
ro
Ai     1
ib
ro  RL
© REP 5/1/2017 ENGR224
Voltage Gain:
  1re  ro RL 
vb

vs Rs    1re  ro RL 

ro RL 
vo

vb re  ro RL 
  1RL ro 
vo
Av  
vs Rs    1re  RL ro 
v
Av  o 
vs
R r 
 r  R r 
L
Rs
  1
o
e
Page BJT 4.1-67
L
o
Bipolar Junction Transistors
Common-Collector Amplifier - Emitter Follower (cont.)
Output Resistance:
vx  ie re  1   ie Rs
vx
ie  
re  1   Rs
vx
ix   ie
ro
vx
vx
ix  
ro re  1   Rs
© REP 5/1/2017 ENGR224
since Ro 
vx
ix
ix 1
1
1
  
Ro v x ro re  1   Rs
Thus Ro is the parallel equivalent
of ro and re  1   Rs 

Rs 
Ro  ro re 



1


Page BJT 4.1-68
Bipolar Junction Transistors
Common-Collector Amplifier - Emitter Follower (cont.)
Output Resistance (cont.):
when ro is large
Ro  re 
Rs
 1
Voltage Gain revisited:
Open-circuit voltage gain:
Av
RL  
Av  Av
© REP 5/1/2017 ENGR224

ro
Rs
 re  ro
  1
RL  
RL
RL  Ro
Page BJT 4.1-69
Bipolar Junction Transistors
Exercise 4.34
For an emitter follower with a load resistence, RL=1k, let I=5 mA, =100, VA=100V, and
v v v
Rs=10k. Find Ri , b , o , o , Ro , Av R  , and Ai :
L
vs vb vs
re 
VT 25mV

 5
IE
5mA


100
 .9901
  1 101
r0 
VA VA
100V

 20.2 K
I C I E .99015mA

Ri    1re  ro RL 
vb
Ri
96.7k


 0.906V / V
vs Rs  Ri 10k  96.7k
ro RL   20k 1k  0.995V / V
vo

vb re  ro RL  5  20k 1k 
vo vo vb
   0.995  0.906  0.901V / V
vv vb vs
 100  15  20.2k 1k 
 96.74k
© REP 5/1/2017 ENGR224
Page BJT 4.1-70
Bipolar Junction Transistors
Exercise 4.34 (cont.)
For an emitter follower with a load resistence, RL=1k, let I=5 mA, =100, VA=100V, and
v v v
Rs=10k. Find Ri , b , o , o , Ro , Av R  , and Ai :
L
vs vb vs

R 
Ro  ro re  s   20k
  1

 103.5
Av
RL  
ro

Rs
 re  ro
 1
 0.995V / V
Ai    1

10k 

5



100  1
20k
10k
 5  20k
101
ro
20k
 101
r0  RL
20k  1k
 96.2 A / A
© REP 5/1/2017 ENGR224
Page BJT 4.1-71
Bipolar Junction Transistors
The BJT as a switch-cutoff and saturation

The BJT has 4 modes of operation:
 Cutoff
 Forward Active
 Saturation
 Inverse Active

So far, we have studied the forward active mode in great detail. Now we will look at the
BJT in cutoff mode and at the BJT in saturation mode. These two extreme modes of
operation are very useful if the transistor is used as a switch, such as in digital logic
circuits.
Mode
EBJ
Cutoff
Reverse
Forward Active Forward
Saturation
Forward
Inverse Active Reverse
© REP 5/1/2017 ENGR224
CBJ
Reverse
Reverse
Forward
Forward
Page BJT 4.1-72
Bipolar Junction Transistors
Cutoff Region


Consider the circuit shown below. If voltage source vI is goes lesss than about 0.5V, the
Emitter-Base Junction will conduct negligible current (reverse-biased). The CBJ is also
reverse-biased since VCC is positive.
The device will be in the cutoff mode. It follows that:
iE  0
iC  0
iB  0
vC  VCC
VCC
iC
RC
RB
vI
© REP 5/1/2017 ENGR224
+
-
vC
iB
Page BJT 4.1-73
Bipolar Junction Transistors
Active Region

To turn the transistor on, vI must be increased to above 0.7V. This gives base current:
iB 


v I  VBE v I  0.7

RB
RB
The collector current is given by iC  i B which applies only if the device is in active mode. At
this point, we don’t know for sure, therefore we assume active mode and calculate the collector
current from vC  VCC  RC iC
VCC
Next, we check whether vCB  0.7 or not.
In our case, just check whether vC  0
or not . If so, then our original assumption
is true. If not, the device is in saturation.
iC
RC
RB
vI
© REP 5/1/2017 ENGR224
+
-
vC
iB
Page BJT 4.1-74
Bipolar Junction Transistors
Saturation Region

Saturation occurs when we attempt to force a current in the collector higher than the
collector circuit can support while maintaining active-mode operation.
By setting vCB  0, we calculate
V  V B VCC  0.7
IˆC  CC

RC
RC
IˆB 
IC

VCC
I Csat
Maximum collector current
RC
MAXIMUM BASE current in forward active
RB
ˆ
 Increasing i B above I B , the collector
current will increase and the collector
voltage will fall below that of the base.
This will continue until the CBJ becomes
forward-biased.
I Csat 
VCC  VCE
RC
I B ( EOS ) 
I Csat

© REP 5/1/2017 ENGR224
Constant current
vI
+
-

IB V
BE

VCEsat
vC


IB is usually higher than IB(EOS) by a
factor of 2 to 10 -- overdrive factor.
EOS=edge of saturation
 forced 
I Csat
IB
This value can be set “at will.”
Page BJT 4.1-75
Bipolar Junction Transistors
Model for the Saturated BJT

A simple model for the npn and pnp transistors in saturation mode is shown on the left.

For quick approximate calculations one may consider VBE and VCEsat to be zero and use the
three-terminal short circuit shown on the right to model a saturated transistor.
C
B
C
VCEsat  0.2 V
VBE  0.7 V
npn
B
E
E
E
approximate model
VEB  0.7 V
VECsat  0.2 V
C
B
pnp
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Page BJT 4.1-76
Bipolar Junction Transistors
Example 4.13

We wish to analyze the circuit to determine the voltages at all nodes and currents in all
branches. Assume the transistor  is specified to be at least 50.
Assuming saturation:
 10V
 10V
VE  6  VBE  6  0.7  5.3 V
4 I
C
 6V
RC  4.7k
 6V
IE 
VC
5
RE  3.3k
4.7 k
IB
3.3 k
VE 1
IE
2
3
5.3
 1.6 mA
3.3
1
2
VC  VE  VCEsat  5.3  0.2  5.5 V
IC 
10  5.5
 0.96 mA
4.7
3
4
I B  I E  I C  1.6  0.96  0.64 mA
 forced 
I C 0.96

 1.5
I B 0.64
 forced   min ,  transistor is saturated
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Page BJT 4.1-77
5
Bipolar Junction Transistors
Example 4.14

The transistor shown below is specified to have  in the range 50 to 150. Find the value
of RB that results in saturation with an overdrive factor of at least 10.
 10V
VC  VCEsat  0.2 V
I Csat
1 k
 5V
RB
IB

VCEsat

VC
I Csat 
10  0.2
 9.8 mA
1
To saturate the transisto r with the lowest  ,
I
9.8
I B ( EOS )  Csat 
 0.196 mA
 min 50
For an overdrive factor of 10,
I B  10  0.196  1.96 mA
5  0.7
4.3
 1.96  RB 
 2.2 k
RB
1.94
© REP 5/1/2017 ENGR224
Page BJT 4.1-78
Bipolar Junction Transistors
Example 4.15

We want to analyze the circuit below to determine the voltages at all nodes and the
currents through all branches. The minimum value of  is specified to be 30.
 5V
IE
10 k
1 k

VE
V ECsat

IB
IC
VC
10 k
 5V
VE  VB  VBE  VB  0.7
VC  VE  VECsat  VB  0.7  0.2  VB  0.5
© REP 5/1/2017 ENGR224
IE 
5  VE 5  VB  0.7

 4.3  V B mA
1
1
IB 
VB
 0.1VB mA
10
IC 
VC   5 V B  0.5  5

 0.1V B  0.55 mA
10
10
using I E  I B  I C
4.3  VB  0.1VB  0.1VB  0.55
VB 
3.75
 3.13 V
1.2
Page BJT 4.1-79
Bipolar Junction Transistors
Example 4.15, cont’d

Substituting in the equations on the previous page, we obtain the following:
V E  3.83 V
VC  3.63 V
I E  1.17 mA
I C  0.86 mA
 forced 
0.86
 2.8
0.31
 forced   min  the transist or is clearly saturated
I B  0.31 mA
© REP 5/1/2017 ENGR224
Page BJT 4.1-80
Bipolar Junction Transistors
Introduction - Equation forms for use in SPICE

Consider the equation for the emitter current in an ideal pnp bipolar junction transistor
 DE n E 0 DB p B 0  qVEB
  DB p B 0  qVkTCB
 
kT
I E  qA

 1  
e

1


 e
W 
  W 
 
 LE

We can simplify the equations by collecting the terms into only a few constants, giving
the coefficients different names, for example, half of the equation given above becomes;
 D n
 
 qVEB

D p  qVEB
qA E E 0  B B 0  e kT  1   I F 0  e kT  1  I F
W 
 


 LE
Therefore:

 DE
D 
I F 0  qAni2 
 B 
 LE N E WN B 
The other half of the equation is similarly reduced
qV

 qVkTCB

 aADB p B 0  kTCB
e

1


I
 1   R I R


R R0  e
 W






Therefore:

 D 
I R 0  qAni2  B 
WN B 
This allows the emitter current to be written in a much more compact form:
IE  IF  RIR
© REP 5/1/2017 ENGR224
Page BJT 4.1-81
Bipolar Junction Transistors
Equation forms for use in SPICE (continued)

Now consider the equation for the collector current in an ideal pnp bipolar junction
transistor
 DB p B 0  qVEB
  DC nC 0 DB p B 0  qVkTCB
 
kT
I C  qA
e

1


e

1







W 
  LC
 
 W 

The right half of the equation reduces to:
 DC nC 0 DB p B 0  qVCB
 
 qVkTCB

kT
qA

 1   I R 0  e
 1  I R
 e
W 
 


 LC
Therefore:

 D
D 
I R 0  qAni2  C  B 
 LC N C WN B 
The left half of the equation is similarly reduced
qV

 qVkTEB

 aADB p B 0  kTEB
e

1


I
 1   F I F


F F0 e
 W







This allows the collector current to be written in a much more compact form:
IC   F I F  I R

If we know two of the terminal currents we can find the current in the third terminal
I B  I E  I C  1   F  I F  1   R  I R
© REP 5/1/2017 ENGR224
Page BJT 4.1-82
Bipolar Junction Transistors
The Ebers-Moll equations for an ideal PNP BJT

The key results are
IC   F I F  I R
IE  IF  RIR
I B  I E  I C  1   F  I F  1   R  I R

The equivalent circuit (just another way to state the equations)
C
IC
IC
FIF
B
C
P
IR
IB
N
RIR
IF
B
IB
IE
IE
© REP 5/1/2017 ENGR224
E
P
E
Page BJT 4.1-83
Bipolar Junction Transistors
Use in SPICE

The current sources illustrate the interaction of the two junctions due to the narrow base
region
 F IF   R IR  IS


IS is one of the three required SPICE parameters for a BJT in SPICE2
The reduced equations can be manipulated to show that:
F 


F
1  F
and
R 
R
1  R
Only three numbers, F , R and IS are needed for the Ebers-Moll equations to be
completely specified.
 All other parameters can be calculated from these three
 The equations can be applied to all regions of operation
Can be extended to the nonideal case by defining coefficients in front of the exponential
terms
 For NPN transistors the diodes, currents and voltage polarities
are reversed
© REP 5/1/2017 ENGR224
Page BJT 4.1-84
Bipolar Junction Transistors
An Alternative form of the Ebers-Moll Model,
the Transport Model


In this model the diodes DBE and DBC have saturation currents IS/F and IS/F respectively
The base current, ib can be written as
v BE
v BC




I S  VT
I
V
S
 e T  1
iB 
e  1 
 R 

 F 




iT is the current component of iC which arises from the
minority carrier diffusion (transport) across the base,
hence the name of this model
 vVBE

 vVBC

T
T
iT  I S  e  1  I S  e  1









iC
DBC
B
The transport model is exactly equivalent to the EbersMoll model but it highlights different aspects of BJT
behavior. It uses one less circuit element and one less
parameter in SPICE
© REP 5/1/2017 ENGR224
C
IS/R
iB
iT
DBE
IS/F
iE
Page BJT 4.1-85
E
Bipolar Junction Transistors
Common-Base Characteristics
 First-Order iC-vCB Characteristics
 Second-Order iC-vCB Characteristics
© REP 5/1/2017 ENGR224
Page BJT 4.1-86
Bipolar Junction Transistors
Common-Emitter Characteristics
 Second-Order iC-vCE Characteristics (refer to lesson 17)
© REP 5/1/2017 ENGR224
Page BJT 4.1-87
Bipolar Junction Transistors
DC and ac 

“BETA DC” (spice)
 More accurate than F, because its determined at Q
 Common-emitter current gain at DC
 “BETAAC” (spice)
 Input ac signal results in DiB  DiC.
 Thus changes, because
 Since vCE is constant, ac=short-circuit current gain.
 Typically, ac=dc. Use ac in small-signal model.
© REP 5/1/2017 ENGR224
Page BJT 4.1-88
Bipolar Junction Transistors
“Complete” BJT Models
 Low-frequency model
 rx=base resistance of bulk Si
1
 r=  i B



VC 
© REP 5/1/2017 ENGR224
Page BJT 4.1-89
Bipolar Junction Transistors
“Complete” BJT Models
 High-frequency model
 V 
Cμ  C JC  C JC 1  CB 
 VJC 
 MJC
 VBE 
Cπ  C D  C JE  g m TF  C JE 1 

V
JE 

f 3dB
 MJE
1
gm

f 
2ππ rπ Cμ  Cπ  T 2π Cμ  Cπ 
© REP 5/1/2017 ENGR224
Page BJT 4.1-90