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Electronic Devices Ninth Edition Floyd thired e
- 1. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Electronic Devices
Ninth Edition
Floyd
Chapter 5
- 2. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Biasing
Summary
Biasing: The DC voltages applied to a transistor in order to turn it on so
that it can amplify the AC signal.
The term biasing is an all-inclusive term for the application of dc voltages
to establish a fixed level of current and voltage.
For transistor amplifiers the resulting dc current and voltage establish an
operating point, Since the operating point is a fixed point on the
characteristics, it is also called the quiescent point (abbreviated Q-point).
DC Biasing – used to set the initial values of IB, IC, and VCE
for ac operation of the transistor
- 3. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
•If no bias were used, the device would
initially be completely off, resulting in a
Q-point at A. Since it is necessary to bias a
device so that it can respond to the entire
range of an input signal, point A would not
be suitable.
•For point B, if a signal is applied to the
circuit, the device will vary in current and
voltage from operating point, but not
enough to drive the device into cutoff or
saturation.
•Point C would allow some positive and
negative variation of the output signal, but
the peak-to-peak value would be limited by
the proximity of VCE=0
•Point D sets the device operating point near
the maximum voltage and power
level. The output voltage swing in the
positive direction is thus limited if the
maximum voltage is not to be exceeded.
Point B therefore seems the best operating
point in terms of linear gain and largest
possible voltage and current swing.
- 4. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
The DC Operating Point
Summary
The end points of the load line are:
ICsat
IC = VCC / RC
VCE = 0 V
VCEcutoff
VCE = VCC
IC = 0 mA
This line called the load
line, since it is defined by
the load resistance Rc
- 5. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
The DC Operating Point
Summary
Effect of lower values of VCC
on the load line and Q-point.
- 6. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
The DC Operating Point
Summary
Effect of increasing levels of RC on the
load line and Q-point.
- 7. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
The DC Operating Point
Summary
If the level of IB is changed by varying
the value of RB the Q-point moves up or
down the load line
- 8. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
The DC Operating Point
Bias establishes the operating point (Q-point) of a transistor
amplifier; the ac signal
moves above and below
this point.
Summary
For this example, the dc
base current is 300 mA.
When the input causes the
base current to vary between
200 mA and 400 mA, the
collector current varies
between 20 mA and 40 mA.
0
VCE (V)
400 A
IC (mA)
300 A= IBQ
200 A
A
B
Q
1.2 3.4 5.6
VCEQ
ICQ
Vce
Ib
Ic
20
30
40
µ
µ
µ
Load line
- 9. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
The DC Operating Point
A signal that swings
outside the active
region will be clipped.
Summary
For example, the bias
has established a low Q-
point. As a result, the
signal is will be clipped
because it is too close to
cutoff.
VCC
VCE
Cutoff
Q
I
B
Q
IC
ICQ
Cutoff 0
Vce
VCEQ
Input
signal
- 10. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
Voltage-Divider Bias
A practical way to establish a Q-point is to form a voltage-
divider from VCC.
Summary
R1 and R2 are selected to establish VB. If the
divider is stiff, IB is small compared to I2. Then,
+VCC
RC
R1
RE
R2
2
B CC
1 2
R
V V
R R
+VCC
RC
R1
RE
R2
βDC =200
27 kW
12 kW
+15 V
680 W
1.2 kW
Determine the base voltage for the circuit.
2
B CC
1 2
12 k
15 V
27 k 12 k
R
V V
R R
W
W W
4.62 V
I2
IB
- 11. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
Voltage-Divider Bias
Summary
+VCC
RC
R1
RE
R2
βDC =200
27 kW
12 kW
+15 V
680 W
1.2 kW
4.62 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
3.92 V
Applying Ohm’s law:
E
E
E
3.92 V
680
V
I
R
W
5.76 mA
- 12. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Voltage-Divider Bias
The unloaded voltage divider approximation for VB gives
reasonable results. A more exact solution is to Thevenize
the input circuit. +VCC
RC
R1
RE
R2
βDC =200
27 kW
12 kW
+15 V
680 W
1.2 kW
VTH = VB(no load)
= 4.62 V
RTH = R1||R2 =
= 8.31 kW
The Thevenin input
circuit can be drawn
R C
R TH
+VCC
R E
+VTH
+ –
IB
+
+
–
–
IE
VBE
8.31 kW
680 W
1.2 kW
4.62 V
+15 V
βDC = 200
- 13. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Voltage-Divider Bias
Now write KVL around the base emitter circuit and solve
for IE.
R C
R TH
+VCC
R E
+VTH
+ –
IB
+
+
–
–
IE
VBE
8.31 kW
680 W
1.2 kW
4.62 V
+15 V
βDC = 200
TH B TH BE E E
V I R V I R
TH BE
E
TH
E
DC
β
V V
I
R
R
Substituting and solving,
E
4.62 V 0.7 V
8.31 k
680 +
200
I
W
W
5.43 mA
and VE = IERE = (5.43 mA)(0.68 kW
= 3.69 V
- 14. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Voltage-Divider Bias
Multisim allows you to do a
quick check of your result.
- 15. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
Summary
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
V
EE
EE
EE
R
R
R
2
2
2
1
1
1
R
R
R
R
R
R
C
C
C
R
R
R
E
E
E
(a) (b) (c)
- 16. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Voltage-Divider Bias
Determine IE for the pnp circuit. Assume a stiff
voltage divider (no loading effect).
+VEE
R2
1
R RC
1.2 kW
RE
680 W
27 kW
12 kW
+15 V
1
B EE
1 2
27 k
15.0 V 10.4 V
27 k 12 k
R
V V
R R
W
W W
E B BE 10.4 V 0.7 V = 11.1 V
V V V
EE E
E
E
15.0 V 11.1 V
680
V V
I
R
W
5.74 mA
10.4 V 11.1 V
- 17. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Voltage-divider bias circuits are designed
so that the IB is much smaller than the
current (I2) through R2
In this case, the voltage-divider circuit
is very straightforward to analyze
because the loading effect of the base
current can be ignored.
This called stiff voltage divider because
the base voltage is relatively
independent of different transistors and
temperature effects.
+VCC
RC
R1
RE
R2
βDC =200
27 kW
12 kW
+15 V
680 W
1.2 kW
Voltage-Divider Bias
Summary
- 18. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Improved Biased Stability
V
V
CC
EE
RC
RE
RB
68 kW
+15 V
15 V
7.5 kW
3.9 kW
1 V
Stability refers to a circuit condition in which the
currents and voltages will remain fairly constant
over a wide range of temperatures and transistor
Beta () values.
Adding RE to the emitter improves the stability of a
transistor.
- 19. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Emitter Bias
Emitter bias has excellent stability but requires both a
positive and a negative source.
Assuming that VE is 1 V, what is IE?
V
V
CC
EE
RC
RE
RB
68 kW
+15 V
15 V
7.5 kW
3.9 kW
The combination of the small drop across RB and VBE
forces the emitter for npn to be approximately 1 V.
1 V
EE
E
E
1 V 15 V ( 1 V)
7.5 k
V
I
R
W
1.87 mA
- 20. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
The approximation that and the neglect of may not be
accurate enough for design work or detailed analysis.
In this case, Kirchhoff’s voltage law can be applied as
follows to develop a more detailed formula for IE
Emitter Bias
- 21. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
- 22. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Emitter Bias
A check with Multisim shows that
the assumption for troubleshooting
purposes is reasonable.
For detailed analysis
work, you can include
the effect of DC. In
this case,
EE
E
B
E
DC
1 V
β
V
I
R
R
- 23. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Base Bias
RC
RB
+VCC
Base bias is used in switching circuits because of its
simplicity, but not widely used in linear applications
because the Q-point is dependent.
Base current is derived from the collector supply
through a large base resistor.
What is IB?
CC
B
B
0.7 V 15 V 0.7 V
560 k
V
I
R
W
25.5 mA
RC
RB
+VCC
560 kW
+15 V
1.8 kW
- 24. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Base Bias
Compare VCE for the case where = 100 and = 200.
RC
RB
+VCC
560 kW
+15 V
1.8 kW
C B
β 100 25.5 μA 2.55 mA
I I
10.4 V
For = 100:
CE CC C C
15 V 2.55 mA 1.8 k
V V I R
W
For = 300:
C B
β 300 25.5 μA 7.65 mA
I I
CE CC C C
15 V 7.65 mA 1.8 k
V V I R
W 1.23 V
- 25. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
- 26. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Emitter-Feedback Bias
R
R
C
E
RB
+VCC
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:
CC BE
E
E
E
DC
β
V V
I
R
R
- 27. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
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.
The equation for collector current is found
by writing KVL around the base circuit.
The result is
CC BE
C
B
C
DC
β
V V
I
R
R
+VCC
RC
RB
- 28. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Collector-Feedback Bias
When = 100,
CC BE
C
B
C
DC
15 V 0.7 V
330 k
1.8 k
100
β
V V
I
R
R
W
W
+VCC
RC
RB
330 kW
1.8 kW
+ 15 V
Compare IC for the case when = 100 with the case when = 300.
2.80 mA
When = 300,
CC BE
C
B
C
DC
15 V 0.7 V
330 k
1.8 k
300
β
V V
I
R
R
W
W
4.93 mA
- 29. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Key Terms
Q-point
DC load line
Linear region
Stiff voltage
divider
Feedback
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
between saturation and cutoff.
A voltage divider for which loading effects
can be ignored.
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.