Basic Electronics Lecture Electronic Devices Chapter 5.ppt

This is a lecture excerpt from a reference material

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.
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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.
SummarySummary
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.
0
V
CE
(V)
400 A
I
C (mA)
300 A = I
BQ
200 A
A
B
Q
1.2 3.4 5.6
V
CEQ
I
CQ
V
ce
I
bI
c
20
30
40
µ
µ
µ
Load line

3. © 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.
SummarySummary
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.
V
CC
V
CE
Cutoff
Q
I
C
I
CQ
Cutoff0
V
ce
V
CEQ
Input
signal

4. © 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 V
CC.
SummarySummary
R
1 and R
2 are selected to establish V
B. If the
divider is stiff, I
B
is small compared to I
2
. Then,
+V
CC
R
C
R
1
R
ER
2
2
B CC
1 2
R
V V
R R
 
 

 
+V
CC
R
CR
1
R
ER
2
β
DC= 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
 
 

 
 
  
 


 
4.62 V
I
2
I
B

5. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Voltage-Divider Bias
SummarySummary
+V
CC
R
CR
1
R
ER
2
β
DC= 200
27 kW
12 kW
+15 V
680 W
1.2 kW
4.62 V
What is the emitter voltage, V
E, and current, I
E?
V
E is one diode drop less than V
B:
V
E = 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
  

5.76 mA

6. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Voltage-Divider Bias
The unloaded voltage divider approximation for V
B gives
reasonable results. A more exact solution is to Thevenize
the input circuit.
+V
CC
R
CR
1
R
ER
2
β
DC= 200
27 kW
12 kW
+15 V
680 W
1.2 kW
V
TH = V
B(no load)
= 4.62 V
R
TH = R
1||R
2 =
= 8.31 k
The Thevenin input
circuit can be drawn
RC
R
TH
+VCC
R
E
+V
TH
+–
IB
+
+
–
–
I
E
V
BE
8.31 kW
680 W
1.2 kW
4.62 V
+15 V
β
DC= 200

7. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Voltage-Divider Bias
Now write KVL around the base emitter circuit and solve
for I
E.
RC
R
TH
+VCC
R
E
+V
TH
+–
IB
+
+
–
–
I
E
V
BE
8.31 kW
680 W
1.2 kW
4.62 V
+15 V
β
DC= 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

 


5.43 mA
and V
E = I
ER
E = (5.43 mA)(0.68 k


=3.69 V

8. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Voltage-Divider Bias
Multisim allows you to do a
quick check of your result.

9. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
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
RR
2
22
1
11
R
RR
R
RR
C
CC
R
RR
E
EE
-
(a) (b) (c)

10. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Voltage-Divider Bias
Determine I
E
for the pnp circuit. Assume a stiff
voltage divider (no loading effect).
+V
EE
R
2
1
R R
C
1.2 kW
R
E
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
 
 

 
 
  
 


 
E B BE
10.4 V 0.7 V = 11.1 VV V V   
EE E
E
E
15.0 V 11.1 V
680
V V
I
R
 
  

5.74 mA
10.4 V
11.1 V

11. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Emitter Bias
Emitter bias has excellent stability but requires both a
positive and a negative source.
Assuming that V
E
is 1 V, what is I
E
?
V
V
CC
EE
R
C
R
E
R
B
68 kW
+15 V
-15 V
7.5 kW
3.9 kW
For troubleshooting analysis, assume that V
E
for an npn transistor is about 1 V.
1 V
EE
E
E
1 V15 V ( 1 V)
7.5 k
V
I
R
    
  

1.87 mA

12. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
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
 



13. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Base Bias
RC
R
B
+V
CC
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 I
B?
CC
B
B
0.7 V15 V 0.7 V
560 k
V
I
R
 
  

25.5 A
R
C
R
B
+V
CC
560 kW
+15 V
1.8 kW

14. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Base Bias
Compare V
CE
for the case where  = 100 and  = 200.
R
C
R
B
+V
CC
560 kW
+15 V
1.8 kW
 
C B
β 100 25.5 μA 2.55 mAI I  
10.4 V
For  = 100:
  
CE CC C C
15 V 2.55 mA 1.8 k
V V I R 
 
For  = 300:  
C B
β 300 25.5 μA 7.65 mAI I  
  
CE CC C C
15 V 7.65 mA 1.8 k
V V I R 
 1.23 V

15. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Emitter-Feedback Bias
R
R
C
E
R
B
+V
CC
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




16. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Collector-Feedback Bias
Collector feedback bias uses another form of negative
feedback to increase stability. Instead of returning the base
resistor to V
CC
, it is returned to the collector.
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



+V
CC
R
C
R
B

17. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
SummarySummary
Collector-Feedback Bias
When  = 100,
CC BE
C
B
C
DC
15 V 0.7 V
330 k
1.8 k
100β
V V
I
R
R
 
  


+V
CC
R
C
R
B
330 kW
1.8 kW
+ 15 V
Compare I
C
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
300β
V V
I
R
R
 
  




4.93 mA

18. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
Key TermsKey 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 I
C
and V
CE
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.

19. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
1. A signal that swings outside the active area will be
a. clamped
b. clipped
c. unstable
d. all of the above

20. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
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

21. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
3. Assuming a stiff voltage-divider for the circuit shown,
the emitter voltage is
a. 4.3 V
b. 5.7 V
c. 6.8 V
d. 9.3 V
+V
CC
R
CR
1
R
ER
2
β
DC= 200
20 kW
10 kW
+15 V
1.2 kW
1.8 kW

22. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
4. For the circuit shown, the dc load line will intersect the
y-axis at
a. 5.0 mA
b. 10.0 mA
c. 15.0 mA
d. none of the above
+V
CC
R
CR
1
R
ER
2
β
DC= 200
20 kW
10 kW
+15 V
1.2 kW
1.8 kW

23. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
5. If you Thevenize the input voltage divider, the Thevenin
resistance is
a. 5.0 k
b. 6.67 k
c. 10 k
d. 30 k
+V
CC
R
CR
1
R
ER
2
β
DC= 200
20 kW
10 kW
+15 V
1.2 kW
1.8 kW

24. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
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
+V
EE
R
2
1
R R
C
1.2 kW
R
E
680 W
27 kW
12 kW
+15 V

25. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
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

26. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
8. With the emitter bias shown, a reasonable assumption
for troubleshooting work is that the
a. base voltage = +1 V
b. emitter voltage = +5 V
c. emitter voltage = 1 V
d. collector voltage = V
CC
V
V
CC
EE
RC
R
E
R
B
68 kW
+15 V
-15 V
7.5 kW
3.9 kW

27. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
9. The circuit shown is an example of
a. base bias
b. collector-feedback bias
c. emitter bias
d. voltage-divider bias
RC
R
B
+V
CC

28. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
10. The circuit shown is an example of
a. base bias
b. collector-feedback bias
c. emitter bias
d. voltage-divider bias
+V
CC
R
C
R
B

29. © 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
QuizQuiz
Answers:
1. b
2. d
3. a
4. a
5. b
6. d
7. c
8. c
9. a
10. b