1. Principles & Applications Small-Signal Amplifiers
Electronics
Principles & Applications
Seventh Edition
Charles A. Schuler
Chapter 6
Introduction to
Small-Signal Amplifiers
(student version)
McGraw-Hill
© 2008 The McGraw-Hill Companies Inc. All rights reserved.

2. INTRODUCTION Gain Common-Emitter Amplifier Stabilizing the Amplifier
Other Configurations

3. Dear Student:
This presentation is arranged in segments. Each segment
is preceded by a Concept Preview slide and is followed by a
Concept Review slide. When you reach a Concept Review
slide, you can return to the beginning of that segment by
clicking on the Repeat Segment button. This will allow you
to view that segment again, if you want to.

4. Concept Preview Voltage gain is the ratio of Vout to Vin.
Power gain is the ratio of Pout to Pin.
Common logarithms are exponents of 10.
Gain or loss in decibels is equal to 10 times the log of the power ratio or 20 times the log of the voltage ratio.
dB voltage gain equals dB power gain when the input impedance equals the output impedance.
System gain or loss is found by adding dB stage gains or losses.

5. Amplifier Out In The units cancel Out 5 V Gain = = 3.33 1.5 V In 1.5 V

6. Gain can be expressed in decibels (dB).
The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
102 = 100
103 = 1000
10-2 = 0.01
= 1
103.6 = 3981
The log of 1000 is 3
The log of 0.01 is -2
The log of 1 is 0
The log of 3981 is 3.6

7. The dB unit is based on a power ratio.
POUT
PIN
50 W
1 W
dB = 10 x log 17
1.70 50
The dB unit can be adapted to a voltage ratio.
dB = 20 x log
VOUT
VIN
This equation assumes VOUT and VIN
are measured across equal impedances.

8. dB units are convenient for evaluating systems.
Total system gain = +46 dB

9. Gain quiz
Amplifier output is equal to the input
________ by the gain.
multiplied
Common logarithms are ________ of the
number 10.
exponents
Doubling a log is the same as _________
the number it represents.
squaring
System performance is found by ________
dB stage gains and losses.
adding
Logs of numbers smaller than one are
____________.
negative

10. Concept Review Voltage gain is the ratio of Vout to Vin.
Power gain is the ratio of Pout to Pin.
Common logarithms are exponents of 10.
Gain or loss in decibels is equal to 10 times the log of the power ratio or 20 times the log of the voltage ratio.
dB voltage gain equals dB power gain when the input impedance equals the output impedance.
System gain or loss is found by adding dB stage gains or losses.
Repeat Segment

11. Concept Preview
In a common emitter amplifier (C-E), the base is the input and the collector is the output.
C-E amplifiers produce a phase inversion.
The circuit limits are called saturation and cutoff.
If a signal drives the amplifier beyond either or both limits the output will be clipped.
The operating point (Q-point) should be centered between saturation and cutoff.
Beta dependent amplifiers are not practical.

12. A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
and common to the input and
output signal circuits.
Connect a signal source CC
A coupling capacitor is often required
VCC
Add a power supply RL
Next, a load resistor RB
Then a base bias resistor C B E
Start with an NPN bipolar junction transistor

13. The output
is phase inverted. RB RL
VCC C B CC E

14. RB RL VCC C B CC E When the input signal goes positive:
The base current increases. C
The collector current increases b times. RL
So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive. RB
VCC CC E

15. RB RL VCC C B CC E When the input signal goes negative:
The base current decreases. C
The collector current decreases b times. RL
So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive. RB
VCC CC E

16. 350 kW 1 kW 14 V C B CC E 14 V IC(MAX) = 1 kW
The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
IC(MAX) =
These are the limits for this circuit.
14 V
1 kW
350 kW
1 kW C B CC E

17. The load line connects the limits.
SAT.
This end is called
saturation.
LINEAR
The linear region is between the limits.
100 mA 14 12
80 mA 10
60 mA
IC in mA 8
40 mA 6 4
20 mA 2
0 mA 2 4 6 8 10 12 14 16 18
VCE in Volts
CUTOFF
This end is called cutoff.

18. 350 kW 1 kW 14 V C B CC E Use Ohm’s Law to determine the base current:
IB =
= 40 mA
350 kW
350 kW
1 kW
14 V C B CC E

19. An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.
100 mA 14 Q 12
80 mA 10
60 mA
IC in mA 8
40 mA 6 4
20 mA 2
0 mA 2 4 6 8 10 12 14 16 18
VCE in Volts
Q = the quiescent point

20. The input signal varies the base current above and below the Q point.
100 mA 14 12
80 mA 10
60 mA
IC in mA 8
40 mA 6 4
20 mA 2
0 mA 2 4 6 8 10 12 14 16 18
VCE in Volts

21. Overdriving the amplifier causes clipping.
100 mA 14 12
80 mA 10
60 mA
IC in mA 8
40 mA 6 4
20 mA 2
0 mA 2 4 6 8 10 12 14 16 18
VCE in Volts
The output is non-linear.

22. What’s wrong with this Q point?
100 mA 14 12
80 mA 10
60 mA
IC in mA 8
40 mA 6 4
20 mA 2
0 mA 2 4 6 8 10 12 14 16 18
VCE in Volts
How about this one?

23. This is a good Q point for linear amplification.
14 V
IB =
= 40 mA
350 kW
IC = b x IB
= 150 x 40 mA
= 6 mA
VRL = IC x RL = 6 mA x 1 kW
= 6 V
VCE = VCC - VRL = 14 V - 6 V = 8 V
This is a good Q point for linear amplification.
350 kW
1 kW
14 V C B CC E
b = 150

24. This is not a good Q point for linear amplification.
14 V
IB =
= 40 mA (IB is not affected)
350 kW
IC = b x IB
= 350 x 40 mA
= 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 kW
= 14 V (VRL is higher)
VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)
This is not a good Q point for linear amplification.
350 kW
1 kW
14 V C B CC E
b = 350
b is higher

25. The higher b causes
saturation.
100 mA 14 12
80 mA 10
60 mA
IC in mA 8
40 mA 6 4
20 mA 2
0 mA 2 4 6 8 10 12 14 16 18
VCE in Volts
The output is non-linear.

26. It’s b dependent! This common-emitter amplifier is not practical.
It’s also temperature dependent. RB RL
VCC C B CC E

27. Basic C-E amplifier quiz
The input and output signals in C-E are
phase ______________.
inverted
The limits of an amplifier’s load line are
saturation and _________.
cutoff
Linear amplifiers are normally operated near
the _________ of the load line.
center
The operating point of an amplifier is also
called the ________ point.
quiescent
Single resistor base bias is not practical since
it’s _________ dependent. 

28. Concept Review
In a common emitter amplifier (C-E), the base is the input and the collector is the output.
C-E amplifiers produce a phase inversion.
The circuit limits are called saturation and cutoff.
If a signal drives the amplifier beyond either or both limits the output will be clipped.
The operating point (Q-point) should be centered between saturation and cutoff.
Beta dependent amplifiers are not practical.
Repeat Segment

29. Concept Preview
C-E amplifiers can be stabilized by using voltage divider base bias and emitter feedback.
The base current can be ignored when analyzing the divider for the base voltage (VB).
Subtract VBE to find VE.
Use Ohm’s Law to find IE and VRL.
Use KVL to find VCE.
IE determines the ac emitter resistance (rE).
RL, RE and rE determine the voltage gain.
Emitter bypassing increases the voltage gain.

30. This common-emitter amplifier is practical.
RB1 RL
VCC C CC B E
RB2 RE
It uses voltage divider bias and
emitter feedback to reduce b sensitivity.

31. { Voltage divider bias +VCC RL RB1 RB1 and RB2 form a voltage dividerRE

32. Voltage divider bias analysis:
+VCC
Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
+VB
RB1 + RB2
RB2
The base current is normally
much smaller than the divider
current so it can be ignored.

33. Solving the practical circuit for its dc conditions:
VCC
RB2
= 12 V
VB =
x VCC
RB1 + RB2
RB1
22 kW RL
= 2.2 kW
2.7 kW C
VB =
x 12 V
2.7 kW +
22 kW B E
VB = 1.31 V
RB2
2.7 kW RE
= 220 W

34. Solving the practical circuit for its dc conditions:
VCC
= 12 V
RB1
22 kW RL
= 2.2 kW
VE = VB - VBE C
VE = 1.31 V V = 0.61 V B E
RB2
2.7 kW RE
= 220 W

35. Solving the practical circuit for its dc conditions:
VCC
= 12 V VE
IE =
RB1 RL RE
22 kW
= 2.2 kW C
0.61 V
IE =
= 2.77 mA
220 W B E
RB2
2.7 kW IE RE
= 220 W

36. VCC RB1 RL C B E RB2 RE A linear Q point!
Solving the practical circuit for its dc conditions:
VCC
VRL = IC x RL
= 12 V
VRL = 2.77 mA x 2.2 kW
RB1
22 kW RL
= 2.2 kW
VRL = 6.09 V C
VCE = VCC - VRL - VE B E
VCE = 12 V V V
RB2
2.7 kW RE
= 220 W
VCE = 5.3 V
A linear Q point!

37. Review of the analysis thus far:
1. Calculate the base voltage using the voltage divider equation.
2. Subtract 0.7 V to get the emitter voltage.
3. Divide by emitter resistance to get the emitter current.
4. Determine the drop across the collector resistor.
5. Calculate the collector to emitter voltage using KVL.
6. Decide if the Q-point is linear.
7. Go to ac analysis.

38. Solving the practical circuit for its ac conditions:
VCC
= 12 V
The ac emitter resistance is rE:
RB1
22 kW RL
= 2.2 kW C
rE =
25 mV IE B E
rE =
25 mV
2.77 mA
= 9.03 W
RB2
2.7 kW RE
= 220 W

39. Solving the practical circuit for its ac conditions:
VCC
= 12 V
The voltage gain from base to collector:
RB1
22 kW RL
AV = RL
RE + rE
= 2.2 kW C
AV =
2.2 kW
220 W W
= 9.61 B E
RB2
2.7 kW RE
= 220 W

40. An emitter bypass capacitor can be used to increase AV:
Solving the practical circuit for its ac conditions:
VCC
= 12 V
An emitter bypass capacitor
can be used to increase AV:
RB1 RL
AV = RL rE
22 kW
= 2.2 kW C
AV =
2.2 kW
9.03 W
= 244 B E
RB2
2.7 kW RE CE

41. Practical C-E amplifier quiz
-dependency is reduced with emitter feedback
and voltage _________ bias.
divider
To find the emitter voltage, VBE is subtracted
from ____________. VB
To find VCE, VRL and VE are subtracted
from _________.
VCC
Voltage gain is equal to the collector resistance
_______ by the emitter resistance.
divided
Voltage gain can be increased by ________
the emitter resistor.
bypassing

42. Concept Review
C-E amplifiers can be stabilized by using voltage divider base bias and emitter feedback.
The base current can be ignored when analyzing the divider for the base voltage (VB).
Subtract VBE to find VE.
Use Ohm’s Law to find IE and VRL.
Use KVL to find VCE.
IE determines the ac emitter resistance (rE).
RL, RE and rE determine the voltage gain.
Emitter bypassing increases the voltage gain.
Repeat Segment

43. Concept Preview C-E amplifiers are the most widely applied.
C-E amplifiers are the only ones that provide a phase inversion.
C-C amplifiers are also called emitter followers.
C-C amplifiers are noted for their low output impedance.
C-B amplifiers are noted for their low input impedance.
C-B amplifiers are used mostly in RF applications.
The analysis procedure for PNP amplifiers is the same as for NPN.

44. The common-emitter configuration is used most often.
It has the best power gain.
VCC
Medium output Z
RB1 RL C B E
Medium input Z
RB2 RE CE

45. The common-collector configuration is shown below.
Its input impedance and current gain are both high.
VCC
It’s often called an
emitter-follower.
RB1 RC C
In-phase
output B E
RB2 RL
Low output Z

46. VCC RB1 RL C B E RB2 RE The common-base configuration is shown below.
Its voltage gain is high.
It’s used most
at RF.
VCC
RB1 RL C B E
In-phase
output
RB2 RE
Low input Z

47. PNP C-E amplifier + VB = - 3.75 V VE = - 3.05 V IE = 2.913 mA
47 W
1 kW
1.5 kW
22 kW
10 kW +
12 V
VE = V
IE = mA
rE = W
VRL = V
AV = 27
VCE = V
VC = V

48. Amplifier configuration quiz
In a C-E amplifier, the base is the input and
the __________ is the output.
collector
In an emitter-follower, the base is the input
and the ______ is the output.
emitter
The only configuration that phase-inverts is
the ________.
C-E
The configuration with the best power gain
is the ________.
C-E
In the common-base configuration, the
________ is the input terminal.
emitter

49. Concept Review C-E amplifiers are the most widely applied.
C-E amplifiers are the only ones that provide a phase inversion.
C-C amplifiers are also called emitter followers.
C-C amplifiers are noted for their low output impedance.
C-B amplifiers are noted for their low input impedance.
C-B amplifiers are used mostly in RF applications.
The analysis procedure for PNP amplifiers is the same as for NPN.
Repeat Segment

50. REVIEW Gain Common-Emitter Amplifier Stabilizing the Amplifier
Other Configurations