Download Chapter 6 - Small-signal Amplifiers

6-1
Electronics
Principles & Applications
Eighth Edition
Charles A. Schuler
Chapter 6
Introduction to
Small-Signal Amplifiers
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6-2
INTRODUCTION
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations
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6-3
1.5 V
In
5V
Amplifier
Out
The units cancel
Gain =
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5V
Out
InV
1.5
= 3.33
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Gain can be expressed in decibels (dB).
6-4
The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
The log of 100 is 2
2
10 = 100
3
10 = 1000
-2
10 = 0.01
0
10 = 1
3.6
10 = 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
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6-5
The dB unit is based on a power ratio.
P
50OUT
W
1.7050
dB = 17
10 x log
1PIN
W
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.
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6-6
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
dB units are convenient for evaluating systems.
+10 dB
-6 dB
+30 dB
-8 dB
+20 dB
Total system gain = +46 dB
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6-7
Acoustical sound levels
are often measured
using dBA units. This
instrument also has a
dBC scale, which has a
different frequency
response curve.
Exposure to loud
sounds is a concern for
employers and
employees, and
citizens in general.
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6-8
Gain quiz
Amplifier output is equal to the input
________ by the gain.
multiplied
Common logarithms are ________ of the
exponents
number 10.
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
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6-9
A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
The emitter terminal is grounded
Next,
load
resistor
a base
bias
resistor
A coupling
capacitor
issupply
often
required
Add
aapower
Start
with
an
NPN
junction
transistor
Connect
abipolar
signal
sourceand
andThen
common
to
the
input
output signal circuits.
RB
RL
C
CC
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B
VCC
E
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6-10
The output
is phase inverted.
RB
RL
C
CC
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B
VCC
E
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6-11
When the input signal goes positive:
The base current increases.
The collector current increases b times.
So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.
RB
RL
C
CC
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B
VCC
E
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6-12
When the input signal goes negative:
The base current decreases.
The collector current decreases b times.
So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.
RB
RL
C
CC
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B
VCC
E
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6-13
The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
These are the limits for this circuit.
14 V
IC(MAX) =
1 kW
350 kW
1 kW
C
CC
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B
14 V
E
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6-14
The load line connects the limits.
This end is called
saturation.
The linear region is between the limits.
100 mA
14
12
10
IC in mA 8
6
4
2
SAT.
LINEAR
80 mA
60 mA
40 mA
20 mA
0 2 4 6
8 10 12 14 16 18
0 mA
VCE in Volts
This end is called cutoff.
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CUTOFF
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6-15
Use Ohm’s Law to determine the base current:
14 V
IB =
350 kW
350 kW
1 kW
C
CC
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= 40 mA
B
14 V
E
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6-16
An amplifier can be operated at any point along the load line.
The base current in this case is 40 mA.
100 mA
14
12
10
IC in mA 8
6
4
2
Q
80 mA
60 mA
40 mA
20 mA
0 2 4 6
8 10 12 14 16 18
0 mA
VCE in Volts
Q = the quiescent point
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6-17
The input signal varies the base
current above and below the Q point.
100 mA
14
12
10
IC in mA 8
6
4
2
80 mA
60 mA
40 mA
20 mA
0 2 4 6
8 10 12 14 16 18
0 mA
VCE in Volts
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Overdriving the amplifier causes clipping.
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100 mA
14
12
10
IC in mA 8
6
4
2
80 mA
60 mA
40 mA
20 mA
0 2 4 6
8 10 12 14 16 18
0 mA
VCE in Volts
The output is non-linear.
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6-19
What’s wrong with this Q point?
100 mA
14
12
10
IC in mA 8
6
4
2
80 mA
60 mA
40 mA
20 mA
0 2 4 6
8 10 12 14 16 18
0 mA
VCE in Volts
How about this one?
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14 V
= 40 mA
IB =
350 kW
IC = b x IB = 150 x 40 mA = 6 mA
VRL = IC x RL = 6 mA x 1 kW = 6 V
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VCE = VCC - VRL = 14 V - 6 V = 8 V
This is a good Q point for linear amplification.
350 kW
1 kW
C
CC
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B
E
14 V
b = 150
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14 V
= 40 mA (IB is not affected)
IB =
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 (VR is higher)
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L
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
C
CC
B
E
14 V
b = 350
b is higher
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The higher b causes
saturation.
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100 mA
14
12
10
IC in mA 8
6
4
2
80 mA
60 mA
40 mA
20 mA
0 2 4 6
8 10 12 14 16 18
VCE in Volts
0 mA
The output is non-linear.
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6-23
This common-emitter amplifier is not practical.
It’s b dependent!
It’s also temperature dependent.
RB
RL
C
CC
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B
VCC
E
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6-24
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
b
it’s _________ dependent.
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6-25
This common-emitter amplifier is practical.
RB1
RL
C
CC
B
RB2
VCC
E
RE
It uses voltage divider bias and
emitter feedback to reduce b sensitivity.
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6-26
Voltage divider bias
+VCC
RB1
RL
RB1 and RB2 form
a voltage divider
RB2
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RE
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6-27
+VCC
Voltage divider
bias analysis:
RB1
RB2
VB =
VCC
RB1 + RB2
The base current is normally
much smaller than the divider
current so it can be ignored.
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+VB
RB2
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6-28
Solving the practical circuit for its dc conditions:
VCC = 12 V
RB1 22 kW
B
RB2 2.7 kW
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VB =
RB2
RB1 + RB2
x VCC
RL= 2.2 kW
C
2.7 kW
x 12 V
VB =
2.7 kW + 22 kW
E
VB = 1.31 V
RE = 220 W
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6-29
Solving the practical circuit for its dc conditions:
VCC = 12 V
RB1 22 kW
RL= 2.2 kW
C
B
RB2 2.7 kW
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VE = VB - VBE
VE = 1.31 V - 0.7 V = 0.61 V
E
RE = 220 W
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6-30
Solving the practical circuit for its dc conditions:
VCC = 12 V
RB1 22 kW
RL= 2.2 kW
C
B
RB2 2.7 kW
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IE =
IE =
VE
0.61 V
220 W
RE
= 2.77 mA
E
RE = 220 W
IC @ IE
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6-31
Solving the practical circuit for its dc conditions:
VRL = IC x RL
VCC = 12 V
VRL = 2.77 mA x 2.2 kW
RB1 22 kW
RL= 2.2 kW
C
B
RB2 2.7 kW
E
VRL = 6.09 V
VCE = VCC - VRL - VE
VCE = 12 V - 6.09 V - 0.61 V
RE = 220 W
VCE = 5.3 V
A linear Q point!
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6-32
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.
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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.
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6-33
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
B
RB2 2.7 kW
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E
RE = 220 W
25 mV
rE =
IE
25 mV
= 9.03 W
rE =
2.77 mA
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6-34
Solving the practical circuit for its ac conditions:
VCC = 12 V
The voltage gain from base to collector:
RB1 22 kW
RL= 2.2 kW
C
B
RB2 2.7 kW
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E
AV =
AV =
RL
RE + rE
2.2 kW
220 W + 9.03 W
= 9.61
RE = 220 W
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6-35
Solving the practical circuit for its ac conditions:
VCC = 12 V
RB1 22 kW
An emitter bypass capacitor
can be used to increase AV:
RL= 2.2 kW
RL
AV =
rE
C
B
RB2 2.7 kW
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E
RE
AV =
2.2 kW
9.03 W
= 244
CE
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6-36
Practical C-E amplifier quiz
b-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
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6-37
The common-emitter configuration is used most often.
It has the best power gain.
VCC
RB1
RL
C
B
RB2
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E
RE
CE
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6-38
The common-collector configuration is shown below.
Its input impedance and current gain are both high.
VCC
RB1
It’s often called an
emitter-follower.
RC
C
B
RB2
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E
RL
In-phase
output
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6-39
The common-base configuration is shown below.
It’s used most
at RF.
Its voltage gain is high.
VCC
RB1
RL
C
B
RB2
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E
RE
In-phase
output
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6-40
PNP C-E amplifier
VB = - 3.75 V
1.5 kW
22 kW
10 kW
47 W
12 V
+
VE = - 3.05 V
IE = 2.913 mA
rE = 8.58 W
VRL = 4.37 V
AV = 27
VCE = - 4.58 V
1 kW
VC = - 7.63 V
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6-41
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
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6-42
REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations
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