Chapter 5
Transistor Bias
Circuits
Objectives
 Discuss the concept of dc biasing of a transistor for
linear operation
 Analyze voltage-divider bias, base bias, and
collector-feedback bias circuits.
 Basic troubleshooting for transistor bias circuits
Lecture’s outline
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•
•
•
•
Objectives
Introduction
DC operating point
Voltage-divider bias
Other bias methods
– Base bias
– Emitter bias
– Collector-feedback bias
• Troubleshooting
• Summary
Introduction
The term biasing is used for application of dc
voltages to establish a fixed level of current and
voltage.
Transistor must be properly biased with dc
voltage to operate as a linear amplifier.
If amplifier is not biased with correct dc
voltages on input and output, it can go into
saturation or cutoff when the input signal
applied.
There are several methods to establish DC
operating point.
We will discuss some of the methods used for
biasing transistors.
DC OPERATING POINT
The DC Operating Point
• The goal of amplification in most cases is to increase the
amplitude of an ac signal without altering it.
• Improper biasing can cause distortion in the output signal.
The DC Operating Point
The purpose of biasing a circuit is to establish a proper stable
dc operating point (Q-point). The dc operating point between
saturation and cutoff is called the Q-point. The goal is to set
the Q-point such that that it does not go into saturation or
cutoff when an ac signal is applied.
• Q-point of a circuit: dc operating point of
amplifier specified by voltage and current values
(VCE and IC). These values are called the coordinates
of Q-point.
• Refer to figure a, given IB = 200μA and βDC=100.
IC=βDCIB so IC=20mA and
VCE  VCC  I C RC  10V  (20mA)(220)  10  4.4  5.6V
• Figure b, VBB is increased to produce IB of 300μA
and IC of 30mA.
VCE  VCC  I C RC  10V  (30mA)(220)  10  6.6  3.4V
• Figure c, VBB is increased to produce IB of 400μA
and IC=40mA. So, VCE is:
• VCE  VCC  IC RC  10V  (40mA)(220)  10  8.8  1.2V
DC Operating Point-DC load line
•Recall that the collector characteristic curves graphically show the
relationship of collector current and VCE for different base currents.
• When IB increases, IC increases and VCE decreases or vice-versa. Each
separate Q-point is connected through dc load line. At any point along line,
values of IB, IC and VCE can be picked off the graph.
•Dc load line intersect VCE axis at 10V, where VCE=VCC. This is cutoff point
because IB and IC zero. Dc load line also intersect IC axis at 45.5mA ideally.
This is saturation point because IC is max and VCE=0.
DC Operating Point-Linear operation
•Region between saturation and cutoff is linear region of
transistor’s operation. The output voltage is ideally linear
reproduction of input if transistor is operated in linear region.
•Let’s look at the effect a superimposed ac voltage has on the
circuit. IB vary sinusoidally 100μA above and below Q-point of
300μA. IC vary up and down 10mA of its Q-point(30mA). VCE
varies 2.2V above and below its Q-point of 3.4V.
•However, as you might already know, applying too much ac
voltage to the base would result in driving the collector current
into saturation or cutoff resulting in a distorted or clipped
waveform.
•When +ve peak is limited, transistor is in cutoff. When –ve peak
is limited, transistor is in saturation.
Variations in IC and VCE as a result of variation in IB.
Graphical load line illustration of transistor being driven into
saturation or cutoff
Graphical load line for transistor in saturation and cutoff
Example 1
• Determine Q-point in figure below and find the maximum
peak value of base current for linear operation. Assume
βDC=200.
Solution
• Q-point is defined by values of IC and VCE.
VBB  VBE 10  0.7
IB 

 198A
RB
47k
I C   DC I B  200(198 )  39.6mA
VCE  VCC  I C RC  20V  (39.6mA)(330)  6.93V
• Q-point is at IC=39.6mA and VCE=6.93V. Since
IC(cutoff)=0, we need to know IC(sat) to determine variation
in IC can occur and still in linear operation.
I C ( sat )
VCC
20


 60.6mA
RC 330
• Before saturation is reached, IC can increase an amount equal
to: IC(sat) – ICQ = 60.6mA – 39.6mA = 21mA.
Solution cont..
• However, IC can decrease by 39.6mA before cutoff (IC=0) is
reached. Since the gap of Q-point with saturation point is less
than gap between Q-point and cutoff, so 21mA is the max
peak variation of IC.
• The max peak variation of IB is:
I b( peak ) 
I c ( peak )
 DC
21m

 105A
200
VOLTAGE-DIVIDER BIAS
Voltage-Divider Bias
• Voltage-divider bias is the most
widely used type of bias circuit.
Only one power supply is needed
and voltage-divider bias is more
stable( independent) than other
bias types. For this reason it will be
the primary focus for study.
• dc bias voltage at base of
transistor is developed by a resistive
voltage-divider consists of R1 and
R2.
• Vcc is dc collector supply voltage.
2 current path between point A and
ground: one through R2 and the
other through BE junction and RE.
Voltage divider bias
• If IB is much smaller than I2, bias
circuit is viewed as voltage divider
of R1 and R2 as shown in Figure a.
• If IB is not small enough to be
neglected, dc input resistance
RIN(base) must be considered.
RIN(base) is in parallel with R2 as
shown in figure b.
Input resistance at transistor base
• VIN is between base and ground
and IIN is the current into base.
•By Ohm’s Law,
RIN(base) = VIN / IIN
• Apply KVL, VIN=VBE+IERE
• Assume VBE<<IERE, so VIN≈IERE
•Since IE≈IC=βDCIB,
VIN≈ βDCIBRE
•IN=IB, so
RIN(base)= βDCIBRE / IB
RIN(base) = DCRE
Analysis of Voltage-Divider Bias Circuit
Analysis of voltage divider bias circuit
 Total resistance from base to ground is:
R2 R IN ( base )
R2  DC R E
 A voltage divider is formed by R1 and resistance from
base to ground in parallel with R2.
 R2  DC R E 
VCC
VB  
 R1  R  DC RE 
2


 If DCRE >>R2, (at least ten times greater), then the
formula simplifies to
 R2 
VCC
VB  
 R1  R 2 
Analysis of Voltage-Divider Bias Circuit
• Now, determine emitter voltage VE.
VE=VB – VBE
• Using Ohm’s Law, find emitter current IE.
IE = VE / RE
• All the other circuit values
IC ≈ I E
VC = VCC – ICRC
• To find VCE, apply KVL:
VCC – ICRC – IERE – VCE =0
• Since IC ≈ IE,
VCE ≈ VCC – IC (RC + RE)
Example 2
• Determine VCE and IC in voltage-divider biased transistor
circuit below if βDC=100.
Solution
1.
Determine dc input resistance at base to see if it can be
neglected.
RIN ( base )   DC RE  100(560)  56k
2.
RIN(base)=10R2, so neglect RIN(base). Then, find base voltage
 R2
VB  
 R1  R 2
3.

 5.6k 
VCC  
10V  3.59V
 15.6k 

So, emitter voltage
VE  VB  VBE  3.59  0.7  2.89V
4.
And emitter current
5.
6.
Thus, I C  5.16mA
And VCE is VCE  VCC  I C ( RC  RE )  10  5.16m(1.56k )  1.95V
IE 
V E 2.89

 5.16mA
RE
560
Voltage-Divider Bias for PNP Transistor
Pnp transistor has opposite polarities from npn. To obtain
pnp, required negative collector supply voltage or with a
positive emitter supply voltage. The analysis of pnp is
basically the same as npn.
Analysis of voltage bias for pnp transistor

R1

VB 
R R  R
2
DC E
 1
• Emitter voltage
• Base voltage
VE  VB  VBE
• By Ohm’s Law,
V EE  V E
IE 
RE
• And,
VC  I C RC
VEC  VE  VC

V EE


OTHER BIAS METHODS
BASE BIAS
EMITTER BIAS
COLLECTOR-FEEDBACK BIAS
Other bias methods - Base Bias
• KVL apply on base circuit.
VCC – VRB – VBE = 0 or VCC – IBRB – VBE =0
• Solving for IB,
VCC  VBE
IB 
RB
• Then, apply KVL around collector
circuit.
VCC – ICRC – VCE = 0
• We know that IC = βDCIB,
 VCC  VBE
I C   DC 
 RB



Base bias
• From the equation of IC, note that IC is
dependent on DC. When DC vary, VCE also
vary, thus changing Q-point of transistor.
• This type of circuit is beta-dependent and very
unstable. Recall that DC changes with
temperature and collector current. Base biasing
circuits are mainly limited to switching
applications.
Emitter Bias
Npn transistor with emitter bias
Emitter base
•
This type of circuit is independent of DC making it as
stable as the voltage-divider type. The drawback is that
it requires two power supplies.
• Apply KVL and Ohm’s Law,
IBRB + IERE + VBE = -VEE
• Since IC≈IE and IC= DC IB,
IE
IB 
 DC
• Solve for IE or IC,
IC 
 V EE  V BE
RE  RB /  DC
• Voltage equations for emitter base circuit.
VE = VEE + IERE
VB = VE + VBE
VC = VCC – ICRC
Collector-Feedback Bias
Collector-feedback bias is kept stable
with negative feedback, although it
is not as stable as voltage-divider or
emitter. With increases of IC, VC
decrease and causing decrease in
voltage across RB, thus IB also
decrease. With less IB ,IC go down as
well.
Analysis of collector-feedback circuit
• By Ohm’s Law,
IB 
VC  VBE
RB
• Collector voltage with assumption IC>>IB.
VC = VCC – ICRC
• And IB = IC / DC
• So, collector current equation
IC 
VCC  VBE
RC  RB /  DC
• Since emitter is ground, VCE = VC.
VCE = VCC - ICRC
TROUBLESHOOTING
Troubleshooting
Figure below show a typical voltage divider circuit with
correct voltage readings. Knowing these voltages is a
requirement before logical troubleshooting can be
applied. We will discuss some of the faults and
symptoms.
All indicated faults
Troubleshooting
Fault 1: R1 Open
Fault 2: Resistor RE Open
With no bias the
transistor is in cutoff.
Transistor is in cutoff.
Base voltage goes
down to 0 V.
Collector voltage
goes up to10 V(VCC).
Emitter voltage goes
down to 0 V.
Base reading voltage will
stay approximately the
same.
Since IC=0, collector
voltage goes up to 10
V(VCC).
Emitter voltage will be
approximately the base
voltage - 0.7 V.
Troubleshooting
Fault 3: Base lead
internally open
Transistor is nonconducting
(cutoff), IC=0A .
Base voltage stays
approximately the same,
3.2V.
Collector voltage goes up to
10 V(VCC).
Emitter voltage goes down
to 0 V because no emitter
current through RE.
Fault 4: BE junction open
Transistor is in cutoff.
Base voltage stays
approximately the same,3.2V.
Collector voltage goes up to
10 V(VCC)
Emitter voltage goes down to
0 V since no emitter current
through RE.
Troubleshooting
Fault 5: BC junction open
Base voltage goes down to 1.11 V
because of more base current flow
through emitter.
Collector voltage goes up to 10
V(VCC).
Emitter voltage will drop to 0.41 V
because of small current flow from
forward-biased base-emitter junction.
Troubleshooting
Fault 6: RC open
Base voltage goes down to
1.11 V because of more
current flow through the
emitter.
Collector voltage will drop
to 0.41 V because of
current flow from forwardbiased collector-base
junction.
Emitter voltage will drop to
0.41 V because of small
current flow from forwardbiased base-emitter
junction.
Troubleshooting
Fault 7: R2 open
Transistor pushed close to
or into saturation.
Base voltage goes up
slightly to 3.83V because
of increased bias.
Emitter voltage goes up to
3.13V because of
increased current.
Collector voltage goes
down because of
increased conduction of
transistor.
SUMMARY
Summary
 The purpose of biasing is to establish a stable operating
point (Q-point).
 The Q-point is the best point for operation of a transistor
for a given collector current.
 The dc load line helps to establish the Q-point for a
given collector current.
 The linear region of a transistor is the region of
operation within saturation and cutoff.
Summary
 Voltage-divider bias is most widely used because it is
stable and uses only one voltage supply.
 Base bias is very unstable because it is  dependent.
 Emitter bias is stable but require two voltage supplies.
 Collector-back is relatively stable when compared to base
bias, but not as stable as voltage-divider bias.