Electronics II
Lecture 3(c): Transistor Bias Circuits
A/Lectr. Khalid Shakir
Dept. Of Electrical Engineering
College of Engineering
Maysan University
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Copyright @2013 by Dept. of Electrical Engineering,
College of Engineering, Maysan University
Page 1-28
Electronics II- Lecture 3(c)/1st Semester 013/014
Voltage-Divider Bias
- A method of biasing a transistor for linear operation using a
single-source resistive voltage divider.
- Earlier, a separate dc source, VBB, was used to bias the base-emitter
- junction because it could be varied independently of VCC,
& it helped to illustrate transistor operation.
- More practical bias method: use VCC as the single bias source.
- A dc bias voltage at the base of the transistor can be developed by
a resistive voltage divider that consists of R1 & R2. VCC: the dc
collector supply voltage
- 2 current paths are between point A & ground;
(1) through R2, and (2) through the base-emitter junction of the
transistor & RE
- Generally, voltage-divider bias circuits are designed so that the base current is much
smaller than the current (I2) through R2.
For this case, the voltage-divider circuit is very straightforward to analyze because the
loading effect of the base current can be ignored.
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College of Engineering, Maysan University
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Electronics II- Lecture 3(c)/1st Semester 013/014
Voltage-Divider Bias
Stiff voltage divider: A voltage divider in which the base current is small
compared to the current in R2 (the base voltage is relatively independent of
different transistors & temperature effects).
Ideally, a voltage-divider circuit is stiff; where transistor does not appear as
a significant load.
To determine if the divider is stiff or not, then need to examine the dc input
resistance looking in at the base.
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Electronics II- Lecture 3(c)/1st Semester 013/014
Loading-Effect of Voltage-Divider Bias
(1) DC Input Resistance at the
Transistor Base
(2) Stability of VoltageDivider Bias
(3) Voltage-Divider Biased
PNP Transistor
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Electronics II- Lecture 3(c)/1st Semester 013/014
DC Input Resistance at the Transistor Base
 The dc input resistance of the transistor is proportional to
βDC, thus it will change for different transistors.
 When a transistor is operating in its linear region, the
emitter current is βDC IB.
 When the emitter resistor is viewed from the base circuit,
the resistor appears to be larger than its actual value by a
factor of is βDC because of the current gain in the transistor.
Thus,
b V
RIN (BASE ) =
DC B
IE
 This is the effective load on the voltage divider illustrated
in earlier Figure.
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Electronics II- Lecture 3(c)/1st Semester 013/014
DC Input Resistance at the Transistor Base
 We can estimate the loading effect by comparing RIN(BASE)
to the resistor R2 in the voltage divider.
 As long as RIN(BASE) is at least 10x larger than R2, the
loading effect will be 10% or less & the voltage divider is
stiff.
 If RIN(BASE) is less than 10x R2, it should be combined in
parallel with R2.
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Electronics II- Lecture 3(c)/1st Semester 013/014
Stability of Voltage-Divider Bias
 Thevenin’s theorem is applied when
analyze a voltage-divider biased transistor
circuit for base current loading effects.
 1st, get an equivalent base-emitter
circuit for this circuit using Thevenin’s
theorem.
 Looking out from the base terminal, the bias
circuit can be redrawn as shown in this figure.
 Apply Thevenin’s theorem to the circuit left of
point A, with VCC replaced by a short to ground &
the transistor disconnected from the circuit.
 The voltage at point A with respect to
ground is,
 and the resistance is,
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College of Engineering, Maysan University
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Electronics II- Lecture 3(c)/1st Semester 013/014
Stability of Voltage-Divider Bias
 The Thevenin equivalent of the bias circuit, connected to the
transistor base, is as shown in the grey box in the figure.
 Applying Kirchhoff’s voltage law around the
equivalent base-emitter loop gives,
 Substituting, using Ohm’s law, & solving for
VTH,
 Substituting IE / βDC for IB,
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College of Engineering, Maysan University
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Electronics II- Lecture 3(c)/1st Semester 013/014
Stability of Voltage-Divider Bias
 Solving for IE,
 If RTH / βDC is small compared to RE, the result is the
same as for unloaded voltage divider.
 Reason of why a voltage-divider bias is widely used;
 Reasonably good bias stability is achieved with a
single supply voltage.
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Electronics II- Lecture 3(c)/1st Semester 013/014
Stability of Voltage-Divider Bias
The unloaded voltage divider approximation for VB gives reasonable results. A
more exact solution is to Thevenize the input circuit.
+ VCC
+15 V
VTH = VB(no load)
= 4.62 V
RTH = R1||R2 =
RC
1.2 kW
+V TH
4.62 V
= 8.31 kW
The Thevenin input
circuit can be drawn
Copyright @2013 by Dept. of Electrical Engineering,
College of Engineering, Maysan University
R TH
+
–
IB
8.31 kW
+
+VCC
+15 V
R1
27 kW
RC
1.2 kW
βDC = 200
βDC = 200
VBE –
IE
+
–
Page 10-28
RE
680 W
R2
12 kW
RE
680 W
Electronics II- Lecture 3(c)/1st Semester 013/014
Stability of Voltage-Divider Bias
Now write KVL around the base emitter circuit and solve for IE.
VTH  I B RTH  VBE  I E RE
IE 
+ VCC
+15 V
VTH  VBE
R
RE  TH
β DC
RC
1.2 kW
Substituting and solving,
4.62 V  0.7 V
IE 
 5.43 mA
680  + 8.31 k
200
and
+V TH
4.62 V
R TH
+
–
IB
8.31 kW
+
βDC = 200
VBE –
IE
+
–
RE
680 W
VE = IERE = (5.43 mA)(0.68 kW )
= 3.69 V
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College of Engineering, Maysan University
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Electronics II- Lecture 3(c)/1st Semester 013/014
Voltage-Divider Biased PNP Transistor
 pnp transistor requires bias polarities opposite to the npn.
 This can be accomplished by using a negative collector supply
voltage: (a), or with a positive emitter supply voltage (b). 3rd figure is
same as (b), only that it is drawn upside down.
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Electronics II- Lecture 3(c)/1st Semester 013/014
Voltage-Divider Biased PNP Transistor
 Analysis procedure is the same as for an npn transistor circuit.
 For a stiff voltage divider (ignoring loading
effects),
 By Ohm’s law,
 Thus,
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Electronics II- Lecture 3(c)/1st Semester 013/014
Voltage-Divider Biased PNP Transistor
Determine IE for the pnp circuit. Assume a
stiff voltage divider (no loading effect).
+VEE
+15 V
 R1 
VB  
 VEE
R

R
 1
2 
27 k



  15.0 V   10.4 V
 27 k  12 k 
VE  VB  VBE  10.4 V  0.7 V = 11.1 V
IE 
VEE  VE 15.0 V  11.1 V

 5.74 mA
RE
680 
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College of Engineering, Maysan University
Page 14-28
R2
12 kW
RE
680 W
10.4 V
11.1 V
R1
27 kW
RC
1.2 kW
Electronics II- Lecture 3(c)/1st Semester 013/014
EMITTER BIAS
Provides excellent bias stability in spite of changes in β or
temperature.
Use both a positive & a negative supply voltage.
In an npn circuit, the small
base current causes the base
voltage to be slightly below
ground.
The emitter voltage is one
diode drop less than this.
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Electronics II- Lecture 3(c)/1st Semester 013/014
EMITTER BIAS
The combination of this small drop across RB & VBE forces the emitter to be
at approximately -1 V.
Thus, emitter current is,
Applying the approximation that
to calculate the collector voltage, thus
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Electronics II- Lecture 3(c)/1st Semester 013/014
EMITTER BIAS
• Approximation that
& the neglect of βDC may not be accurate
enough for design work or detailed analysis.
• Kirchhoff’s voltage law (KVL) can be applied to develop a more detailed
formula for IE.
• In (a): KVL applied around the baseemitter circuit
• In (b): (a) is redrawn for analysis
• Thus,
• Substitute
transposing VEE, &
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Electronics II- Lecture 3(c)/1st Semester 013/014
EMITTER BIAS
• Thus,
• Voltages with respect to ground are indicated by a single subscript.
• Thus, the emitter voltage with respect to the ground is,
• The base voltage with respect to ground is,
• The collector with respect to the ground is,
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Electronics II- Lecture 3(c)/1st Semester 013/014
BASE BIAS
 This method of biasing is common in switching circuits.
 Analysis of this circuit for the linear region shows that it is
directly dependent on βDC.
 Starts with KVL around the base circuit,
 Substitutes IBRB for VRB,
Thus,
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Electronics II- Lecture 3(c)/1st Semester 013/014
BASE BIAS
 When KVL is applied around the collector circuit, we get,
 Thus,
 Substitute IC = βDCIB, we get,
Base bias is used in switching circuits because of its simplicity, but
not widely used in linear applications because the Q-point is b
dependent.
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Electronics II- Lecture 3(c)/1st Semester 013/014
BASE BIAS
Base current is derived from the collector supply through
a large base resistor.
What is IB?
+VCC
+15 V
RB
V  0.7 V 15 V  0.7 V
I B  CC

 25.5 m A
RB
560 k
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College of Engineering, Maysan University
Page 21-28
RC
1.8 kW
560 kW
Electronics II- Lecture 3(c)/1st Semester 013/014
Q-Point Stability of Base Bias
 Equation
shows that IC is dependent on βDC. Thus, a variation in βDC causes
IC and, as a result, VCE to change, thus changing the Q-point of the
transistor. Therefore, the base bias circuit is extremely betadependent & unpredictable.
 βDC varies with temperature & collector current.
 There is a large spread of βDC values from one transistor to
another of the same type due to manufacturing variations.
 Therefore, base bias is rarely used in linear circuits.
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Electronics II- Lecture 3(c)/1st Semester 013/014
Emitter-Feedback Bias
 If an emitter resistor is added to the base-circuit, we get an
emitter-feedback bias.
 Help make base bias more predictable with
negative feedback, which negates any attempted
change in collector current with an opposing change
in base voltage.
 If the collector current tries to increase, the
emitter voltage increases, causing an increase in
base voltage due to VB = VE + VBE
 This increase in base voltage reduces the voltage
across RB, thus reducing the base current & keeping
the collector current from increasing.
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Electronics II- Lecture 3(c)/1st Semester 013/014
Emitter-Feedback Bias
The same scenario happens if the collector
current tries to decrease.
 This circuit is better for linear circuits than base
bias, but still dependent on βDC & is not predictable
as voltage-divider bias.
 To calculate IE, write KVL around the base
circuit.
Substitute IE/βDC for IB, can see that IE is still
dependent on βDC.
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Electronics II- Lecture 3(c)/1st Semester 013/014
Collector-Feedback Bias
 The collector voltage provides the bias for the base-emitter junction.
 The negative feedback creates an “offsetting” effect that tends to keep
the Q-point stable.
 If IC tries to increase, it drops more voltage across RC, and thus
causing VC to decrease.
 When VC decreases, there is a decrease in voltage across RB, which
decreases IB.
 The decrease in IB produces less IC which, in turn, drops less voltage
across RC & thus offsets the decrease in VC.
 By Ohm’s law,
 Assume that IC >> IB, thus the collector voltage,
 Also,
Thus,
 Thus,
 Since the emitter is ground, VCE = VC, then
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Electronics II- Lecture 3(c)/1st Semester 013/014
Q-Point Stability Over Temperature
 Equation
shows that IC is dependent to some extent on βDC and VBE .
 The dependency can be minimized by making RC >> RB/βDC &
VCE >>VBE.
 The collector-feedback bias is essentially eliminates the βDC &
VBE dependency even if the stated conditions are met.
 βDC varies directly with temperature, & VBE varies inversely with
temperature. As the temperature goes up in a collector-feedback
circuit, βDC goes up & VBE goes down.
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Electronics II- Lecture 3(c)/1st Semester 013/014
Q-Point Stability Over Temperature
 The increase in βDC acts to increase IC. The decrease in
VBE acts to increase IB which, in turn also acts to increase
IC. As IC tries to increase, the voltage drop across RC also
tries to increase. This tends to reduce the collector voltage
& therefore the voltage across RB, thus reducing IB &
offsetting the attempted increase in IC & the attempted
decrease in VC.
 The result is that the collector-feedback circuit
maintains a relatively stable Q-point. The reverse action
occurs when the temperature decreases.
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Electronics II- Lecture 3(c)/1st Semester 013/014
Q-Point Stability Over Temperature
Compare IC for the case when b = 100 with the case when b =
300.
When b = 100,
VCC  VBE
15 V  0.7 V
IC 

 2.80 mA
RB
330
k

1.8 k 
RC 
100
β DC
RB
+VCC
+ 15 V
RC
1.8 kW
330 kW
When b = 300,
IC 
VCC  VBE
15 V  0.7 V

 4.93 mA
RB
330
k

1.8 k 
RC 
300
β DC
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Electronics II- Lecture 3(c)/1st Semester 013/014