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EEE 4670 -Electronics Engineering III
Lecture 4: Biasing of Bipolar Junction Transistors
Dr. Brilliant Habeenzu
Email: brilliant.habeenzu@unza.zm
hakhabeen@gmail.com
CC (Emitter-Follower) configuration
The output can be taken off the emitter terminal
in the CC configuration (emitter follower configuration).
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DC analysis of CC configuration
If (β+1)RE is much larger than the R2, the current IB will be much smaller than I2. If we
accept the approximation that IB is essentially 0A compared to I1 or I2, then I1=I2 and R1
and R2 can be considered series elements.
The voltage across R2 is actually the base voltage, can be determined using the voltagedivider rule.
R2
VB 
VCC
R1  R2
If (  1) RE  RE  10R2
the approximate approach can be applied with
, a high degree of accuracy.
VE  VB  VBE
VE VB  VBE
IE 

RE
RE
VE  I E RE
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Biasing of Common Base configuration
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Common-Base configuration
The applied signal is connected to the emitter terminal and the base is ground in the
common base configuration.
It has a very low input impedance, high output impedance, and good voltage gain. A
typical common-base configuration appears in the Fig. below. The base is the common
terminal between the input emitter terminal and output collector terminal.
Common-Base configuration
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Common-Base configuration
Applying KVL will result in
 VEE  I E RE  VBE  0
VEE  VBE
IE 
RE
Applying KVL to the entire outside perimeter of the
network will result in
 VEE  I E R E VCE  I C RC  VCC  0
Because
I E  IC
Input DC
equivalent
VCE  VEE  VCC  I E ( RC  RE )
The voltage VCB can be found by applying KVL to the
output loop to obtain:
VCB  I C RC  VCC  0 Using I C  I E
VCB  VCC  I C RC
Determining
VCE and VCB
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Example:
Determine the current IE and IB and the voltage VCE and VCB for the common-base
configuration of Fig.
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Solution:
VEE  VBE 4V  0.7V
 2.75mA

IE 
1.2k
RE
2.75mA
IE
 45.08A

IB 
  1 60  1
VCE  VEE  VCC  I E ( RC  RE )
 4V  10V  (2.75mA)( 2.4k  1.2k)  4.1V
VCB  VCC  I C RC  VCC   I B RC
 10V  (60)( 45.08A)( 2.4k)  3.51V
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Working principle of transistor amplification
Input signal V1=0.02 sinωt is applied to the base. Base current varies depending on the input
signal. Collector current varies depending on the base current. Collector current is much larger
than the base current by the amplifying effect.
Base current
Base emitter
voltage
Collector
current
Base current
Collector emitter
voltage
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10
Input
voltage(V)
Working principle of
transistor amplification
The variation of base current is 5 μA
The variation of collector current is 0.65mA
Base
current
The variation of input voltage is 0.02V
The current gain is (0.65mA)/(5μA)=130
The voltage gain is (-2V)/(0.02V)=-100
The power gain is 130×100=13000
Collector voltage(V)
Collector
current
The variation of collector voltage is 2.0V.
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Transistor Switching Networks
The application of transistors is not limited solely to the amplification of signals. Transistors
can be used as switches for computer and control applications. The network shown below
can be employed as an inverter in computer logic circuitry. The operating point switch from
cutoff to saturation along the load line. When Vi=5V, the transistor will be “on” and the
design must ensure that the network is heavily saturated by a level of IB greater than that
associated with the IB curve appearing near the saturation level.
The saturation collector current
is defined by
VCC
I C sa t 
RC
For the saturation level, the following condition
must be satisfied.
IB 
I Csat
 dc
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For the network of Fig. when Vi=5V, the resulting level of IB is
Vi  0.7V 5V  0.7V
V
5V
I B

 63A and I C  CC 
 6.1mA
sat
RB
68k
RC 0.82k
Therefore,
Which is satisfied.
I Csat
6.1mA
I B  63A 

 48.8A
 dc
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For Vi=0V, IB=0A, IC=0mA,
voltage drop across RC is 0V, resulting
in VC=+5V.
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At saturation, the current IC is quite high and voltage VCE is very low.
resistance level between the two terminals determined by
VCEsat
Rsat 
I Csat
The result is a
Using a typical average value of VCEsat such as 0.15V gives
VCEsat 0.15V
Rsat 

 24.6
I Csat 6.1mA
Which is a relatively low and can be considered as approximately 0Ω
when placed in
series with resistors in the kΩ range. For Vi=0V, the cutoff condition results in a resistance
level of the following magnitude:
VCC
5V
Rcutoff 

 
I CEO 0mA
resulting in the open-circuit equivalence
For a typical value of ICEO=10μA, the magnitude of the cutoff resistance is
VCC
5V
Rcutoff 

 500k
I CEO 10 A
Saturation conditions and resulting terminal resistance
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Cutoff conditions and resulting terminal resistance
Example:
Determine RB and RC for the transistor inverter of Fig. below,
if Icsat =10mA
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Solution:
VCC
10V
10V
and 10mA 
so that RC 
 1k
RC
RC
10mA
I Csat 10mA
At saturation , I B 

 40 A Choosing I B  60 A
 dc
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At saturation , I C 
Vi  0.7V
to ensure saturation and using I B 
RB
Vi  0.7V 10V  0.7V
We obtain RB 

 155k
IB
60A
Choose RB  150k, which is a standard value. Then
Vi  0.7V 10V  0.7V
IB 

 62A
RB
150k
and
I B  62A 
I Csat
 DC
 40 A
Therefore, use RB  150k and RC  1k
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The impact on the speed of response of the collector output is defined by the collector
current response of Fig. below.
The total time required for
the transistor to switch from the “off” to the “on” state is designated as ton and is defined by
ton  t r  t d
with td the delay time between the changing state of the input and the beginning of a response at
the output. tr is the rise time from 10% to 90% of the final value.
The total time required for a transistor to
switch from the “on” to the “off” state is
referred to as toff and is defined by
toff  t s  t f
Where ts is the storage time and tf is
the fall time from 90% to 10% of the
final value.
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