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Chapter 5
Electronics 2
EE 317
Electrical Engineering
Majmaah University
1st Semester 1433/1434 H
2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.Electronic Devices, 9th edition
Thomas L. Floyd1
Transistor Bias Circuits
Chapter Outline
51 The DC Operating Point
52 Voltage-Divider Bias
53 Other Bias Methods
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Thomas L. Floyd2
Chapter Objectives
Discuss and determine the dc operating point of a linear
amplifier.
Analyze a voltage-divider biased circuit.
Analyze an emitter bias circuit, a base bias circuit, an
emitter-feedback bias circuit, and a collector-feedback bias
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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emitter-feedback bias circuit, and a collector-feedback bias
circuit.
Introduction
As you learned in Chapter 4, a transistor must be properly biased in
order to operate as an amplifier.
DC biasing is used to establish fixed dc values for the transistor currents
and voltages called the dc operating point or quiescent point (Q-point).
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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In this chapter, several types of bias circuits are discussed.
This material lays the groundwork for the study of amplifiers, and other
circuits that require proper biasing.
The DC Operating Point
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Thomas L. Floyd5
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Thomas L. Floyd6
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1.2 3.4 5.6
The DC Operating Point
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Linear Operation
Bias establishes the operating point (Q-point) of a transistor
amplifier; the ac signal
moves above and below
IC (mA)
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Thomas L. Floyd9
moves above and below
this point.
For this example, the dc
base current is 300 A. When the input causes the
base current to vary between
200 A and 400 A, the collector current varies
between 20 mA and 40 mA.
0VCE (V)
400 A
300 A = IBQ
200 A
A
B
Q
1.2 3.4 5.6VCEQ
ICQ
Vce
IbIc
20
30
40
Load line
Waveform Distortion
A signal that swings
outside the active
region will be clipped.
I BQ
IC
Input
signal
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Thomas L. Floyd10
region will be clipped.
For example, the bias
has established a low Q-
point.
As a result, the signal
will be clipped because
it is too close to cutoff.
VCCVCE
Cutoff
QICQ
Cutoff 0
Vce
VCEQ
Waveform Distortion
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Thomas L. Floyd11
High Q- point. The signal will be
clipped because it is too close to
saturation.
Input signal too large. The signal
will be clipped from both sides.
QuizQuizQuiz
Q. A signal that swings outside the active area will be
a. clamped
b. clipped
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Thomas L. Floyd12
c. unstable
d. all of the above
Voltage-Divider Bias
A practical way to establish a Q-point is to form a voltage-
divider from VCC.
R1 and R2 are selected to establish VB. If the
divider is stiff, I is small compared to I . Then,
+VCC
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divider is stiff, IB is small compared to I2. Then,
RCR1
RER2
2B CC
1 2
RV V
R R
+
I2
IB
QuizQuizQuiz
Q. A stiff voltage divider is one in which
a. there is no load current
b. divider current is small compared to load current
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Thomas L. Floyd14
c. the load is connected directly to the source voltage
d. loading effects can be ignored
Voltage-Divider Bias
+VCC+15 V
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Thomas L. Floyd15
RCR1
RER2
DC = 200
27 k
12 k 680
1.2 k
Determine the base voltage for the circuit.
( )
2B CC
1 2
12 k15 V
27 k 12 k
RV V
R R
= +
= + = + 4.62 V
Voltage-Divider Bias
+VCC+15 V
What is the emitter voltage, VE, and current, IE?
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RCR1
RER2
DC = 200
27 k
12 k 680
1.2 k
4.62 V
VE is one diode drop less than VB:
VE = 4.62 V 0.7 V = 3.92 V
3.92 VApplying Ohms law:
EE
E
3.92 V
680
VI
R= = =
5.76 mA
QuizQuizQuiz
Q. Assuming a stiff voltage-divider for the circuit shown,
the emitter voltage is
a. 4.3 V+VCC+15 V
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Thomas L. Floyd17
b. 5.7 V
c. 6.8 V
d. 9.3 V
RCR1
RER2
DC = 200
20 k
10 k 1.2 k
1.8 k
VB = 10 / (10 + 20) * 15 = 5 V
VE = 5 0.7 = 4.3 V
QuizQuizQuiz
Q. For the circuit shown, the dc load line will intersect the
y-axis at (neglecting IB)
a. 5.0 mA+VCC+15 V
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Thomas L. Floyd18
b. 10.0 mA
c. 15.0 mA
d. none of the above
RCR1
RER2
DC = 200
20 k
10 k 1.2 k
1.8 k
@ VCE=0:
IC = (15 0) / (1.8k + 1.2k) = 15 / 3k = 5 mA
Voltage-Divider Bias
The unloaded voltage divider approximation for VB gives
reasonable results.
A more exact solution is to Thevenize the input circuit.
Looking from the base to the left:+VCC+15 V
+VCC+15 V
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Thomas L. Floyd19
RCR1
RER2
DC = 200
27 k
12 k
+15 V
680
1.2 k
VTH = VB(no load)
= 4.62 V
RTH = R1||R2 =
= 8.31 k
The Thevenin input
circuit can be drawn
R C
R TH
R E
+V TH +
IB
+
+
I E
VBE
8.31 k
680
1.2 k
4.62 V
+15 V
DC = 200
Thevenin Resistance, RTH
To find RTH, replace voltage sources by a short circuit.
Which means VCC will be grounded.
RTH = R1||R2
+VCC+15 V
1 2 3
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RCR1
RER2
DC = 200
27 k
12 k 680
1.2 k
QuizQuizQuiz
Q. If you Thevenize the input voltage divider, the Thevenin
resistance is
a. 5.0 k
+VCC+15 V
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Thomas L. Floyd21
b. 6.67 k
c. 10 k
d. 30 k
RCR1
RER2
DC = 200
20 k
10 k 1.2 k
1.8 k
RTH = R1 R2 / (R1 + R2)
= 20 * 10 / (20 + 10) = 200 / 30 = 6.67 k
Voltage-Divider Bias
Now write KVL around the base emitter circuit and solve
for IE.
+VCC+15 V
TH B TH BE E EV I R V I R= + +
IE = IB + IC = IB + DC IB = (DC + 1) IB DC IBVTH VBE = IE RE + IB RTH IE RE + IE RTH / DC= IE ( RE + RTH / DC )
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Thomas L. Floyd22
R C
R TH
R E
+V TH +
IB
+
+
I E
VBE
8.31 k
680
1.2 k
4.62 V
+15 V
DC = 200
TH BEE
THE
DC
V VI
RR
=
+
Substituting and solving,
E
4.62 V 0.7 V
8.31 k680 +200
I
= =
5.43 mA
and VE = IERE = (5.43 mA)(0.68 k)
= 3.69 V
Voltage-Divider Bias
A pnp transistor can be biased from either a positive or negative supply.
Notice that (b) and (c) are the same circuit; both with a positive supply.
+ V V
EE
EE
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Thomas L. Floyd23
+
+
V
V
EE
EE
R
RR
2
221
11
R
RR
R
RR
C
CCR
RR
E
EE
(a) (b) (c)
Voltage-Divider Bias
Determine IE for the pnp circuit. Assume a stiff
voltage divider (no loading effect).
+VEE
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Thomas L. Floyd24
+VEE
R2
1RRC1.2 k
R E680
27 k
12 k
+15 V
( )
1B EE
1 2
27 k15.0 V 10.4 V
27 k 12 k
RV V
R R
= +
= + = +
E B BE 10.4 V 0.7 V = 11.1 VV V V= + = +
EE EE
E
15.0 V 11.1 V
680
V VI
R
= = =
5.74 mA
10.4 V 11.1 V
QuizQuizQuiz
Q. For the circuit shown, the emitter voltage is
a. less than the base voltage
b. less than the collector voltage
+VEE
R R
+15 V
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Thomas L. Floyd25
c. both of the above
d. none of the above
R2
1RRC1.2 k
R E680
27 k
12 k
pnp circuit
(higher than the base voltage by 0.7 V)
Emitter Bias
Emitter bias has excellent stability but requires both a
positive and a negative source.
VCC+15 V
For troubleshooting analysis, assume that VEfor an npn transistor is about 1 V.
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Assuming that VE is 1 V, what is IE?
VEE
RC
RE
RB68 k
+15 V
15 V
7.5 k
3.9 k
1 VEEE
E
1 V 15 V ( 1 V)
7.5 k
VI
R
= = =
1.87 mA
The minus sign means the
current is directed downwards
(opposite to initial assumption).
Emitter Bias
The approximation that VE 1 V and the neglect of DC may not be accurate
enough for design work or detailed analysis.
In this case, KVL can be applied to develop a more detailed formula for IE.
KVL applied around the base-emitter circuit in Figure 517(a), which has
been redrawn in part (b) for analysis, gives the following equation:
2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Emitter Bias
VRB+ VBE + VRE
VEE = 0
IB RB + VBE + IE RE VEE = 0
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IE ( RB / DC) + IE RE = VEE VBE
IE (RE + RB / DC) = VEE VBE
QuizQuizQuiz
Q. Emitter bias
a. is not good for linear circuits
b. uses a voltage-divider on the input
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Thomas L. Floyd29
c. requires dual power supplies
d. all of the above
QuizQuizQuiz
Q. With the emitter bias shown, a reasonable assumption
for troubleshooting work is that the
a. base voltage = +1 V VCC+15 V
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Thomas L. Floyd30
b. emitter voltage = +5 V
c. emitter voltage = 1 V
d. collector voltage = VCC
VEE
RC
RE
RB68 k
15 V
7.5 k
3.9 k
Base Bias
Base bias is used in switching circuits because of its
simplicity, but not widely used in linear applications
because the Q-point is dependent.
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Thomas L. Floyd31
RC
R B
+VCC
Base current is derived from the collector supply
through a large base resistor.
What is IB?
CCB
B
0.7 V 15 V 0.7 V
560 k
VI
R
= = =
25.5 A
RC
R B
+VCC
560 k
+15 V
1.8 k
IB
+
VBE
Base Bias
Compare VCE for the case where = 100 and = 300.
( )( )C B 100 25.5 A 2.55 mAI I= = =For = 100:V V I R=
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Thomas L. Floyd32
RC
R B
+VCC
560 k
+15 V
1.8 k
10.4 V( )( )CE CC C C
15 V 2.55 mA 1.8 k
V V I R=
= =
For = 300: ( ) ( )C B 300 25.5 A 7.65 mAI I= = =
( )( )CE CC C C
15 V 7.65 mA 1.8 k
V V I R=
= = 1.23 V
IC
+
VCE
The Q-point is dependent.
QuizQuizQuiz
Q. The circuit shown is an example of
a. base bias
b. collector-feedback bias+VCC
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Thomas L. Floyd33
c. emitter bias
d. voltage-divider bias
RC
R B
Emitter-Feedback Bias
An emitter resistor (RE) changes base bias into emitter-
feedback bias, which is more predictable.
The emitter resistor (RE) is a form of negative feedback.
If IC tries to increase, VE increases, causing an
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Thomas L. Floyd34
R
R
C
E
R B
+VCCIf IC tries to increase, VE increases, causing an
increase in VB because VB = VE + VBE.
This increase in VB reduces the voltage across RB(VRB
= VCC VB), thus reducing IB and keeping ICfrom increasing.
A similar action occurs if IC tries to decrease.
Emitter-Feedback Bias
The equation for emitter current (IE) is found by writing KVL
around the base circuit.
IE RE + IB RB = VCC VBE
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Thomas L. Floyd35
R
R
C
E
R B
+VCC
IB
IE+
+
+
IE RE + IB RB = VCC VBE
IE RE + (IE / DC) RB = VCC VBE
IE (RE + RB / DC) = VCC VBE
Collector-Feedback Bias
Collector feedback bias uses another form of negative
feedback to increase stability. (more on next slide)
Instead of returning the base resistor (RB) to VCC, it is
returned to the collector.
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The equation for collector current (IC) is found
by writing KVL around the base circuit.
(details on page 247 in the book)
The result isCC BE
CB
CDC
V VI
RR
=
+
+VCC
RCRB
Collector-Feedback Bias
Collector feedback bias uses another form of negative
feedback to increase stability. Instead of returning the base
resistor to VCC, it is returned to the collector.
The negative feedback creates an offsetting effect
that tends to keep the Q-point stable.
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Thomas L. Floyd37
+VCC
RCRB
If IC tries to increase, it drops more voltage across
RC, thereby 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 and thus offsets the
decrease in VC.
Collector-Feedback Bias
Compare IC for the case when = 100 with the case when = 300.
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Thomas L. Floyd38
When = 100,
CC BEC
BC
DC
15 V 0.7 V
330 k1.8 k100
V VI
RR
= = =
++
+VCC
RCR B
330 k
1.8 k
+ 15 V
2.80 mA
When = 300,
CC BEC
BC
DC
15 V 0.7 V
330 k1.8 k300
V VI
RR
= = =
++4.93 mA
QuizQuizQuiz
Q. The circuit shown is an example of
a. base bias
b. collector-feedback bias
+VCC
RC
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Thomas L. Floyd39
c. emitter bias
d. voltage-divider bias
RCRB
Key TermsKey TermsKey Terms
Q-point
DC load line
Linear region
The dc operating (bias) point of an amplifier
specified by voltage and current values.
A straight line plot of IC and VCE for a
transistor circuit.
The region of operation along the load line
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Linear region
Stiff voltage
divider
Feedback
The region of operation along the load line
between saturation and cutoff.
A voltage divider for which loading effects
can be ignored.
The process of returning a portion of a
circuits output back to the input in such a way
as to oppose or aid a change in the output.