Electronic Devices
10th ed., Global Edition
Chapter 5
Transistor Bias
Circuits
Copyright © 2018 Pearson Education, Ltd. All Rights Reserved.
Electronic Devices
10th ed.
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 circuit
◆ Troubleshoot faults in transistor bias circuits
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Electronic Devices
The DC Operating Point
Bias establishes the operating point (Q-point) of a transistor
amplifier; the ac signal
I (mA)
moves above and below
Load line
this point.
C
Ic
40
ICQ
Ib
A
Q
30
0
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300 µ A = IBQ
B
20
1.2
3.4
VCEQ
400 µ A
200 µ A
VCE (V)
5.6
Vce
Electronic Devices
The DC Operating Point
Bias establishes the operating point (Q-point) of a transistor
amplifier; the ac signal
I (mA)
moves above and below
Load line
this point.
C
For this example, the dc
base current is 300 µA.
When the input causes the
base current to vary between
200 µA and 400 µA, the
collector current varies
between 20 mA and 40 mA.
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Ic
40
ICQ
Ib
A
Q
30
0
300 µ A = IBQ
B
20
1.2
3.4
VCEQ
400 µ A
200 µ A
VCE (V)
5.6
Vce
Electronic Devices
The DC Operating Point
A signal that swings
outside the active
region will be clipped.
IB
Q
IC
ICQ
Q
Cutoff
0
VCC
Cutoff
Vce
VCEQ
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Input
signal
VCE
Electronic Devices
The DC Operating Point
A signal that swings
outside the active
region will be clipped.
For example, the bias
has established a low Qpoint. As a result, the
signal is will be clipped
because it is too close to
cutoff.
IB
Q
IC
ICQ
Q
Cutoff
0
VCC
Cutoff
Vce
VCEQ
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Input
signal
VCE
Electronic Devices
Voltage-Divider Bias - Example
A practical way to establish a Q-point is to form a voltagedivider from VCC.
R1 and R2 are selected to establish VB. If the
divider is stiff, IB is small compared to I2. Then,
 R2 
VB ≈ 
 VCC
+
R
R
 1
2 
+VCC
+VCC
+15
V
R1
R
1
27 kΩ
IB
I2
R2
R
2
12 kΩ
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RC
R
C
1.2 kΩ
βDC = 200
RE
R
E
680 Ω
Electronic Devices
Voltage-Divider Bias - Example
A practical way to establish a Q-point is to form a voltagedivider from VCC.
R1 and R2 are selected to establish VB. If the
divider is stiff, IB is small compared to I2. Then,
 R2 
VB ≈ 
 VCC
+
R
R
 1
2 
Determine the base voltage for the circuit.
+VCC
+VCC
+15
V
R1
R
1
27 kΩ
IB
I2
R2
R
2
12 kΩ
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RC
R
C
1.2 kΩ
βDC = 200
RE
R
E
680 Ω
Electronic Devices
Voltage-Divider Bias - Example
A practical way to establish a Q-point is to form a voltagedivider from VCC.
R1 and R2 are selected to establish VB. If the
divider is stiff, IB is small compared to I2. Then,
 R2 
VB ≈ 
 VCC
+
R
R
 1
2 
Determine the base voltage for the circuit.
 R2 
VB = 
 VCC
 R1 + R2 
12 kΩ


= 
=
 ( +15 V )
Ω
+
Ω
27
k
12
k


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+VCC
+VCC
+15
V
R1
R
1
27 kΩ
IB
I2
R2
R
2
12 kΩ
RC
R
C
1.2 kΩ
βDC = 200
RE
R
E
680 Ω
Electronic Devices
Voltage-Divider Bias - Example
A practical way to establish a Q-point is to form a voltagedivider from VCC.
R1 and R2 are selected to establish VB. If the
divider is stiff, IB is small compared to I2. Then,
 R2 
VB ≈ 
 VCC
+
R
R
 1
2 
Determine the base voltage for the circuit.
 R2 
VB = 
 VCC
 R1 + R2 
12 kΩ


= 
=
 ( +15 V )
Ω
+
Ω
27
k
12
k


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+VCC
+VCC
+15
V
R1
R
1
27 kΩ
IB
I2
R2
R
2
12 kΩ
4.62 V
RC
R
C
1.2 kΩ
βDC = 200
RE
R
E
680 Ω
Electronic Devices
Voltage-Divider Bias - Example
+VCC
+15 V
What is the emitter voltage, VE, and current, IE?
R1
27 kΩ
4.62 V
R2
12 kΩ
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RC
1.2 kΩ
βDC = 200
RE
680 Ω
Electronic Devices
Voltage-Divider Bias - Example
+VCC
+15 V
What is the emitter voltage, VE, and current, IE?
VE is one diode drop less than VB:
VE = 4.62 V – 0.7 V =
R1
27 kΩ
4.62 V
R2
12 kΩ
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RC
1.2 kΩ
βDC = 200
RE
680 Ω
Electronic Devices
Voltage-Divider Bias - Example
+VCC
+15 V
What is the emitter voltage, VE, and current, IE?
VE is one diode drop less than VB:
VE = 4.62 V – 0.7 V = 3.92 V
R1
27 kΩ
4.62 V
R2
12 kΩ
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RC
1.2 kΩ
βDC = 200
3.92 V
RE
680 Ω
Electronic Devices
Voltage-Divider Bias - Example
+VCC
+15 V
What is the emitter voltage, VE, and current, IE?
VE is one diode drop less than VB:
VE = 4.62 V – 0.7 V = 3.92 V
Applying Ohm’s law:
=
IE
VE 3.92 V
=
=
RE 680 Ω
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R1
27 kΩ
4.62 V
R2
12 kΩ
RC
1.2 kΩ
βDC = 200
3.92 V
RE
680 Ω
Electronic Devices
Voltage-Divider Bias - Example
+VCC
+15 V
What is the emitter voltage, VE, and current, IE?
VE is one diode drop less than VB:
VE = 4.62 V – 0.7 V = 3.92 V
Applying Ohm’s law:
=
IE
VE 3.92 V
=
=
RE 680 Ω
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5.76 mA
R1
27 kΩ
4.62 V
R2
12 kΩ
RC
1.2 kΩ
βDC = 200
3.92 V
RE
680 Ω
Electronic Devices
Voltage-Divider Bias
The unloaded voltage divider approximation for VB gives
reasonable results. A more exact solution is to Thevenize
+V
the input circuit.
+15 V
CC
VTH = VB(no load)
= 4.62 V
R1
27 kΩ
RTH = R1||R2 =
= 8.31 kΩ
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RC
1.2 kΩ
βDC = 200
R2
12 kΩ
RE
680 Ω
Electronic Devices
Voltage-Divider Bias
The unloaded voltage divider approximation for VB gives
reasonable results. A more exact solution is to Thevenize
+V
+ VCC
the input circuit.
+15 V
+15 V
VTH = VB(no load)
RC
1.2 kΩ
= 4.62 V
RTH = R1||R2 =
= 8.31 kΩ
The Thevenin input
circuit can be drawn
and used to determine
a more precise value
of IE.
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CC
+V TH
4.62 V
R TH
+
–
+
VBE –
IB
8.31 kΩ
IE
+
–
R1
27 kΩ
βDC = 200
βDC = 200
RE
680 Ω
RC
1.2 kΩ
R2
12 kΩ
RE
680 Ω
Electronic Devices
Voltage-Divider Bias
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 kΩ
+V TH
4.62 V
R TH
+
–
+
VBE –
IB
8.31 kΩ
IE
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+
–
βDC = 200
RE
680 Ω
Electronic Devices
Voltage-Divider Bias
Write KVL around the base emitter circuit and solve for IE.
VTH= I B RTH + VBE + I E RE
VTH − VBE
R
RE + TH
β DC
Substituting and solving,
4.62 V − 0.7 V
IE =
680 Ω + 8.31 kΩ
200
+ VCC
+15 V
IE =
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RC
1.2 kΩ
+V TH
4.62 V
R TH
+
–
+
VBE –
IB
8.31 kΩ
IE
+
–
βDC = 200
RE
680 Ω
Electronic Devices
Voltage-Divider Bias
Write KVL around the base emitter circuit and solve for IE.
VTH= I B RTH + VBE + I E RE
VTH − VBE
R
RE + TH
β DC
Substituting and solving,
4.62 V − 0.7 V
IE =
680 Ω + 8.31 kΩ
200
+ VCC
+15 V
IE =
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RC
1.2 kΩ
5.43 mA
+V TH
4.62 V
R TH
+
–
+
VBE –
IB
8.31 kΩ
IE
+
–
βDC = 200
RE
680 Ω
Electronic Devices
Voltage-Divider Bias
Write KVL around the base emitter circuit and solve for IE.
VTH= I B RTH + VBE + I E RE
VTH − VBE
R
RE + TH
β DC
Substituting and solving,
4.62 V − 0.7 V
I E = 5.43 mA
680 Ω + 8.31 kΩ
200
and VE = IERE = (5.43 mA)(0.68 kΩ)
+ VCC
+15 V
IE =
=
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RC
1.2 kΩ
+V TH
4.62 V
R TH
+
–
+
VBE –
IB
8.31 kΩ
IE
+
–
βDC = 200
RE
680 Ω
Electronic Devices
Voltage-Divider Bias
Write KVL around the base emitter circuit and solve for IE.
VTH= I B RTH + VBE + I E RE
VTH − VBE
R
RE + TH
β DC
Substituting and solving,
4.62 V − 0.7 V
I E = 5.43 mA
680 Ω + 8.31 kΩ
200
and VE = IERE = (5.43 mA)(0.68 kΩ)
+ VCC
+15 V
IE =
= 3.69 V
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RC
1.2 kΩ
+V TH
4.62 V
R TH
+
–
+
VBE –
IB
8.31 kΩ
IE
+
–
βDC = 200
RE
680 Ω
Electronic Devices
Multisim Check
Multisim allows you to do a
quick check of your result.
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Electronic Devices
Voltage-Divider Biasing of pnp Transistors
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.
− VEE
+ VEE
R1
RC
R1
RC
R2
RE
R2
RE
(a)
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(b)
+ VEE
R2
RE
R1
RC
(c)
Electronic Devices
Voltage-Divider Biasing of pnp Transistors - Example
Determine IE for the pnp circuit. Assume a stiff
voltage divider (no loading effect).
+VEE
+15 V
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R2
12 kΩ
RE
680 Ω
R1
27 kΩ
RC
1.2 kΩ
Electronic Devices
Voltage-Divider Biasing of pnp Transistors - Example
Determine IE for the pnp circuit. Assume a stiff
voltage divider (no loading effect).
 R1 
VB = 
 VEE
+
R
R
 1
2 
27 kΩ


= 
15.0 V )
 ( +=
 27 kΩ + 12 kΩ 
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+VEE
+15 V
10.4 V
R2
12 kΩ
RE
680 Ω
R1
27 kΩ
RC
1.2 kΩ
10.4 V
Electronic Devices
Voltage-Divider Biasing of pnp Transistors - Example
Determine IE for the pnp circuit. Assume a stiff
voltage divider (no loading effect).
 R1 
VB = 
 VEE
+
R
R
 1
2 
27 kΩ


= 
15.0 V )
 ( +=
 27 kΩ + 12 kΩ 
+VEE
+15 V
10.4 V
VE =VB + VBE = 10.4 V + 0.7 V = 11.1 V
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R2
12 kΩ
RE
680 Ω
10.4 V
11.1 V
R1
27 kΩ
RC
1.2 kΩ
Electronic Devices
Voltage-Divider Biasing of pnp Transistors - Example
Determine IE for the pnp circuit. Assume a stiff
voltage divider (no loading effect).
 R1 
VB = 
 VEE
+
R
R
 1
2 
27 kΩ


= 
15.0 V )
 ( +=
 27 kΩ + 12 kΩ 
+VEE
+15 V
10.4 V
VE =VB + VBE = 10.4 V + 0.7 V = 11.1 V
=
IE
VEE − VE 15.0 V − 11.1 V
=
=
RE
680 Ω
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R2
12 kΩ
RE
680 Ω
10.4 V
11.1 V
R1
27 kΩ
RC
1.2 kΩ
Electronic Devices
Voltage-Divider Biasing of pnp Transistors - Example
Determine IE for the pnp circuit. Assume a stiff
voltage divider (no loading effect).
 R1 
VB = 
 VEE
+
R
R
 1
2 
27 kΩ


= 
15.0 V )
 ( +=
 27 kΩ + 12 kΩ 
+VEE
+15 V
10.4 V
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 Ω
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R2
12 kΩ
RE
680 Ω
10.4 V
11.1 V
R1
27 kΩ
RC
1.2 kΩ
Electronic Devices
Emitter Bias
Emitter bias has excellent stability but requires both a
positive and a negative DC source.
V
CC
+15 V
RC
3.9 kΩ
RB
68 kΩ
RE
7.5 kΩ
VEE
−15 V
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Electronic Devices
Emitter Bias
Emitter bias has excellent stability but requires both a
positive and a negative DC source.
V
CC
For troubleshooting analysis, assume that VE
for an npn transistor is about −1 V.
+15 V
RC
3.9 kΩ
RB
68 kΩ
−1 V
RE
7.5 kΩ
VEE
−15 V
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Electronic Devices
Emitter Bias
Emitter bias has excellent stability but requires both a
positive and a negative DC source.
V
CC
For troubleshooting analysis, assume that VE
for an npn transistor is about −1 V.
Assuming that VE is −1 V, what is IE?
+15 V
RC
3.9 kΩ
RB
68 kΩ
−1 V
RE
7.5 kΩ
VEE
−15 V
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Electronic Devices
Emitter Bias
Emitter bias has excellent stability but requires both a
positive and a negative DCsource.
V
CC
For troubleshooting analysis, assume that VE
for an npn transistor is about −1 V.
Assuming that VE is −1 V, what is IE?
−VEE − 1 V −15 V − (−1 V)
IE =
=
=
RE
7.5 kΩ
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+15 V
RC
3.9 kΩ
RB
68 kΩ
−1 V
RE
7.5 kΩ
VEE
−15 V
Electronic Devices
Emitter Bias
Emitter bias has excellent stability but requires both a
positive and a negative DC source.
V
CC
For troubleshooting analysis, assume that VE
for an npn transistor is about −1 V.
Assuming that VE is −1 V, what is IE?
−VEE − 1 V −15 V − (−1 V)
IE =
=
= −1.87 mA
RE
7.5 kΩ
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+15 V
RC
3.9 kΩ
RB
68 kΩ
−1 V
RE
7.5 kΩ
VEE
−15 V
Electronic Devices
Emitter Bias – Multisim
A check with Multisim shows that
the assumption for troubleshooting
purposes is reasonable.
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Electronic Devices
Emitter Bias – Multisim
A check with Multisim shows that
the assumption for troubleshooting
purposes is reasonable.
For detailed analysis
work, you can include
the effect of βDC. In
this case,
−VEE − 1 V
IE =
R
RE + B
β DC
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Electronic Devices
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 β dependent.
Base current is derived from the collector supply
through a large base resistor.
+VCC
+V
+15CCV
RB
560 kΩ
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RC
1.8 kΩ
Electronic Devices
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 β dependent.
Base current is derived from the collector supply
through a large base resistor.
What is IB?
+VCC
+V
+15CCV
RB
560 kΩ
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RC
1.8 kΩ
Electronic Devices
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 β dependent.
Base current is derived from the collector supply
through a large base resistor.
What is IB?
VCC − 0.7 V 15 V − 0.7 V
=
IB =
=
RB
560 kΩ
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+VCC
+V
+15CCV
RB
560 kΩ
RC
1.8 kΩ
Electronic Devices
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 β dependent.
Base current is derived from the collector supply
through a large base resistor.
What is IB?
VCC − 0.7 V 15 V − 0.7 V
=
IB =
= 25.5 µA
RB
560 kΩ
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+VCC
+V
+15CCV
RB
560 kΩ
RC
1.8 kΩ
Electronic Devices
Base Bias
Compare VCE for the case where β = 100 and β = 200.
+VCC
+15 V
RB
560 kΩ
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RC
1.8 kΩ
Electronic Devices
Base Bias
Compare VCE for the case where β = 100 and β = 200.
I C β=
IB
For β = 100: =
(100 )( 25.5 μA=)
+VCC
+15 V
2.55 mA
V=
VCC − I C RC
CE
= 15 V − ( 2.55 mA )(1.8 kΩ
=
)
RB
560 kΩ
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RC
1.8 kΩ
Electronic Devices
Base Bias
Compare VCE for the case where β = 100 and β = 200.
I C β=
IB
For β = 100: =
(100 )( 25.5 μA=)
+VCC
+15 V
2.55 mA
V=
VCC − I C RC
CE
= 15 V − ( 2.55 mA )(1.8 kΩ
=
) 10.4 V
RB
560 kΩ
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RC
1.8 kΩ
Electronic Devices
Base Bias
Compare VCE for the case where β = 100 and β = 300.
I C β=
IB
For β = 100: =
(100 )( 25.5 μA=)
+VCC
+15 V
2.55 mA
V=
VCC − I C RC
CE
= 15 V − ( 2.55 mA )(1.8 kΩ
=
) 10.4 V
For β = 300: =
I C β=
IB
V=
VCC − I C RC
CE
( 300 )( 25.5 μA=)
= 15 V − ( 7.65 mA )(1.8 kΩ
=
)
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7.65 mA
RB
560 kΩ
RC
1.8 kΩ
Electronic Devices
Base Bias
Compare VCE for the case where β = 100 and β = 300.
I C β=
IB
For β = 100: =
(100 )( 25.5 μA=)
+VCC
+15 V
2.55 mA
V=
VCC − I C RC
CE
= 15 V − ( 2.55 mA )(1.8 kΩ
=
) 10.4 V
For β = 300: =
I C β=
IB
V=
VCC − I C RC
CE
( 300 )( 25.5 μA=)
= 15 V − ( 7.65 mA )(1.8 kΩ
=
) 1.23 V
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7.65 mA
RB
560 kΩ
RC
1.8 kΩ
Electronic Devices
Base Bias
Compare VCE for the case where β = 100 and β = 200.
I C β=
IB
For β = 100: =
(100 )( 25.5 μA=)
+VCC
+15 V
2.55 mA
V=
VCC − I C RC
CE
= 15 V − ( 2.55 mA )(1.8 kΩ
=
) 10.4 V
For β = 300: =
I C β=
IB
V=
VCC − I C RC
CE
( 300 )( 25.5 μA=)
= 15 V − ( 7.65 mA )(1.8 kΩ
=
) 1.23 V
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7.65 mA
RB
560 kΩ
RC
1.8 kΩ
Electronic Devices
Emitter-Feedback Bias
An emitter resistor changes base bias into emitter-feedback
bias, which is more predictable. The emitter resistor is a
form of negative feedback.
+VCC
RC
RB
RE
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Electronic Devices
Emitter-Feedback Bias
An emitter resistor changes base bias into emitter-feedback
bias, which is more predictable. The emitter resistor is a
form of negative feedback.
The equation for emitter current is found
by writing KVL around the base circuit.
The result is:
VCC − VBE
IE =
R
RE + E
β DC
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+VCC
RC
RB
RE
Electronic Devices
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.
+VCC
RB
Copyright © 2018 Pearson Education, Ltd. All Rights Reserved.
RC
Electronic Devices
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 equation for collector current is found
by writing KVL around the base circuit.
The result is
VCC − VBE
IC =
R
RC + B
β DC
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+VCC
RB
RC
Electronic Devices
Collector-Feedback Bias - Example
Compare IC for the case when β = 100 with the case when β = 300.
+VCC
+ 15 V
RB
330 kΩ
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RC
1.8 kΩ
Electronic Devices
Collector-Feedback Bias - Example
Compare IC for the case when β = 100 with the case when β = 300.
+VCC
+ 15 V
When β = 100,
=
IC
VCC − VBE
15 V − 0.7 V
=
=
330
kΩ
RB
1.8 kΩ +
RC +
100
β DC
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RB
330 kΩ
RC
1.8 kΩ
Electronic Devices
Collector-Feedback Bias - Example
Compare IC for the case when β = 100 with the case when β = 300.
+VCC
+ 15 V
When β = 100,
=
IC
VCC − VBE
15 V − 0.7 V
=
=
330
kΩ
RB
1.8 kΩ +
RC +
100
β DC
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2.80 mA
RB
330 kΩ
RC
1.8 kΩ
Electronic Devices
Collector-Feedback Bias - Example
Compare IC for the case when β = 100 with the case when β = 300.
+VCC
+ 15 V
When β = 100,
VCC − VBE
15 V − 0.7 V
=
=
330
kΩ
RB
1.8 kΩ +
RC +
100
β DC
When β = 300,
=
IC
IC
=
VCC − VBE
15 V − 0.7 V
=
=
R
1.8 kΩ + 330 kΩ
RC + B
300
β DC
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2.80 mA
RB
330 kΩ
RC
1.8 kΩ
Electronic Devices
Collector-Feedback Bias - Example
Compare IC for the case when β = 100 with the case when β = 300.
+VCC
+ 15 V
When β = 100,
VCC − VBE
15 V − 0.7 V
=
=
330
kΩ
RB
1.8 kΩ +
RC +
100
β DC
When β = 300,
=
IC
IC
=
2.80 mA
VCC − VBE
15 V − 0.7 V
=
= 4.93 mA
R
1.8 kΩ + 330 kΩ
RC + B
300
β DC
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RB
330 kΩ
RC
1.8 kΩ
Electronic Devices
Troubleshooting Bias Problems
Troubleshooting requires a basic understanding of circuit operation and
logical thinking. The first step is to analyze the problem and understanding
the conditions that caused the failure?
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Electronic Devices
Troubleshooting Bias Problems
Troubleshooting requires a basic understanding of circuit operation and
logical thinking. The first step is to analyze the problem and understanding
the conditions that caused the failure?
Assuming you have determined that a specific transistor is not functioning
correctly, plan steps that isolate the problem. For example, a good starting
point is to check that power is connected and a good ground is present.
Copyright © 2018 Pearson Education, Ltd. All Rights Reserved.
Electronic Devices
Troubleshooting Bias Problems
Troubleshooting requires a basic understanding of circuit operation and
logical thinking. The first step is to analyze the problem and understanding
the conditions that caused the failure?
Assuming you have determined that a specific transistor is not functioning
correctly, plan steps that isolate the problem. For example, a good starting
point is to check that power is connected and a good ground is present.
Make measurements following a logical sequence to isolate the specific
problem.
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Electronic Devices
Troubleshooting - Example
Assume you have analyzed a failure of a basic amplifier using a pnp
transistor with voltage divider bias. You have checked the power
supply and ground and found they are okay.
+VEE
+15 V
What would you plan to check next?
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R2
12 kΩ
RE
680 Ω
R1
27 kΩ
RC
1.2 kΩ
Electronic Devices
Troubleshooting - Example
Assume you have analyzed a failure of a basic amplifier using a pnp
transistor with voltage divider bias. You have checked the power
supply and ground and found they are okay.
+VEE
+15 V
What would you plan to check next?
Answers can vary. A good next step is to check
the base voltage at the base itself. This will
check for open or shorted bias resistors and a
good connection to the transistor. Note that a bad
reading here is NOT proof the fault is in the bias
circuit but requires further checking.
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R2
12 kΩ
RE
680 Ω
R1
27 kΩ
RC
1.2 kΩ
Electronic Devices
Selected Key Terms
Q-point
DC load line
Linear region
Stiff voltage
divider
Feedback
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Electronic Devices
Selected Key Terms
Q-point The dc operating (bias) point of an amplifier
specified by voltage and current values.
DC load line
Linear region
Stiff voltage
divider
Feedback
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Electronic Devices
Selected Key Terms
Q-point The dc operating (bias) point of an amplifier
specified by voltage and current values.
DC load line A straight line plot of IC and VCE for a
transistor circuit.
Linear region
Stiff voltage
divider
Feedback
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Electronic Devices
Selected Key Terms
Q-point The dc operating (bias) point of an amplifier
specified by voltage and current values.
DC load line A straight line plot of IC and VCE for a
transistor circuit.
Linear region The region of operation along the load line
between saturation and cutoff.
Stiff voltage
divider
Feedback
Copyright © 2018 Pearson Education, Ltd. All Rights Reserved.
Electronic Devices
Selected Key Terms
Q-point The dc operating (bias) point of an amplifier
specified by voltage and current values.
DC load line A straight line plot of IC and VCE for a
transistor circuit.
Linear region The region of operation along the load line
between saturation and cutoff.
Stiff voltage A voltage divider for which loading effects
divider can be ignored.
Feedback
Copyright © 2018 Pearson Education, Ltd. All Rights Reserved.
Electronic Devices
Selected Key Terms
Q-point The dc operating (bias) point of an amplifier
specified by voltage and current values.
DC load line A straight line plot of IC and VCE for a
transistor circuit.
Linear region The region of operation along the load line
between saturation and cutoff.
Stiff voltage A voltage divider for which loading effects
divider can be ignored.
Feedback The process of returning a portion of a
circuit’s output back to the input in such a way
as to oppose or aid a change in the output.
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Electronic Devices
Quiz Q1
1. A signal that swings outside the active area will be
a. clamped
b. clipped
c. unstable
d. all of the above
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Electronic Devices
Quiz Q2
2. A stiff voltage divider is one in which
a. there is no load current
b. divider current is small compared to load current
c. the load is connected directly to the source voltage
d. loading effects can be ignored
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Electronic Devices
Quiz Q3
3. Assuming a stiff voltage-divider for the circuit shown,
the emitter voltage is
+VCC
+15 V
a. 4.3 V
b. 5.7 V
R1
20 kΩ
c. 6.8 V
d. 9.3 V
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RC
1.8 kΩ
βDC = 200
R2
10 kΩ
RE
1.2 kΩ
Electronic Devices
Quiz Q4
4. For the circuit shown, the dc load line will intersect the
y-axis at
+VCC
+15 V
a. 5.0 mA
b. 10.0 mA
R1
20 kΩ
c. 15.0 mA
d. none of the above
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RC
1.8 kΩ
βDC = 200
R2
10 kΩ
RE
1.2 kΩ
Electronic Devices
Quiz Q5
5. If you Thevenize the input voltage divider, the Thevenin
resistance is
+VCC
+15 V
a. 5.0 kΩ
b. 6.67 kΩ
R1
20 kΩ
c. 10 kΩ
d. 30 kΩ
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RC
1.8 kΩ
βDC = 200
R2
10 kΩ
RE
1.2 kΩ
Electronic Devices
Quiz Q6
6. For the circuit shown, the emitter voltage is
a. less than the base voltage
b. less than the collector voltage
c. both of the above
d. none of the above
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+VEE
+15 V
R2
12 kΩ
RE
680 Ω
R1
27 kΩ
RC
1.2 kΩ
Electronic Devices
Quiz Q7
7. Emitter bias
a. is not good for linear circuits
b. uses a voltage-divider on the input
c. requires dual power supplies
d. all of the above
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Electronic Devices
Quiz Q8
8. With the emitter bias shown, a reasonable assumption for
troubleshooting work is that the
a. base voltage = +1 V
VCC
+15 V
b. emitter voltage = +5 V
RC
3.9 kΩ
c. emitter voltage = −1 V
d. collector voltage = VCC
RB
68 kΩ
RE
7.5 kΩ
VEE
−15 V
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Electronic Devices
Quiz Q9
9. The circuit shown is an example of
a. base bias
+VCC
b. collector-feedback bias
c. emitter bias
d. voltage-divider bias
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RC
RB
Electronic Devices
Quiz Q10
10. The circuit shown is an example of
a. base bias
b. collector-feedback bias
c. emitter bias
d. voltage-divider bias
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+VCC
RB
RC
Electronic Devices
Answers
Answers:
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1. b
6. d
2. d
7. c
3. a
8. c
4. a
9. a
5. b
10. b