Download Chapter 7 - Small-signal Amplifiers (II)

7-1
Electronics
Principles & Applications
Eighth Edition
Charles A. Schuler
Chapter 7
More About
Small-Signal Amplifiers
McGraw-Hill
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7-2
INTRODUCTION
• Amplifier Coupling
• Voltage Gain
• FET Amplifiers
• Negative Feedback
• Frequency Response
• Positive Feedback
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7-3
Capacitive coupling is convenient in cascade ac amplifiers.
VCC
These two points are at different dc voltages.
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7-4
Direct coupling is required for dc gain.
VCC
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7-5
The Darlington is a popular dc arrangement.
VCC
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7-6
VCC
Transformer coupling offers the
advantage of impedance matching.
10:1
P
S
10 W
ZRATIO =
2
TRATIO
2
= 10 = 100
ZCOLLECTOR = 100 x 10 W = 1000 W
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7-7
Transformer coupling
is used in 70.7 volt
sound systems.
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7-8
VCC
Transformer coupling can be
used in bandpass amplifiers
to achieve selectivity.
Gain
fR
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7-9
Amplifier coupling quiz
Capacitive coupling is not useful for
_________ amplifiers.
dc
Dc frequency response requires ________
coupling.
direct
Transformer coupling offers the advantage
of _________ matching.
impedance
Tuned transformer coupling provides
frequency _____________.
selectivity
A Darlington amplifier is an example of
_________ coupling.
direct
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More about solving the practical
circuit for its ac conditions:
7-10
VCC = 12 V
RB1 22 kW
RL= 2.2 kW
C
Zin = ?
B
RB2 2.7 kW
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E
RE = 220 W
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Zin is a combination of RB1, RB2,
and rin of the transistor.
VCC = 12 V
RB1 22 kW
Determine rin first:
RL= 2.2 kW rin = b (RE + rE)
C
B
RB2 2.7 kW
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7-11
E
rin = 150 (220 W + 9.03 W)
rin = 34.4 kW
RE = 220 W
Note: rin = brE
when RE is bypassed.
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7-12
RB1, RB2, and rin act in parallel
to load the input signal.
VCC = 12 V
1
Zin =
1
1
1
+
+ r
RB2
in
RB1 22 kW
RL= 2.2 kW RB1
C
B
RB2 2.7 kW
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Zin =
E
RE = 220 W
1
1
1
+
+
22 kW 2.7 kW 34.4 kW
1
Zin = 2.25 kW
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7-13
What happens when an amplifier is loaded?
VCC = 12 V
RL and the Load act in parallel.
RP = 1.1 kW
RB1 22 kW
RL= 2.2 kW
Load = 2.2 kW
RB2 2.7 kW
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RE = 220 W
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7-14
There are two saturation currents for a loaded amplifier.
VCC
= 4.96 mA
ISAT(DC) =
RL + RE
V = 12 V
CC
RB1 22 kW
VCC
ISAT(AC) =
= 9.09 mA
RP + RE
RL= 2.2 kW
RP = 1.1 kW
Load = 2.2 kW
RB2 2.7 kW
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RE = 220 W
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There are two load lines for a loaded amplifier.
7-15
The DC load line connects VCC and ISAT(DC).
100 mA
14
12
10
IC in mA 8
TEMPORARY AC
6
4 DC
2
0 2 4 6
80 mA
60 mA
40 mA
20 mA
8 10 12 14 16 18
VCE in Volts
0 mA
A temporary AC load line connects VCC and ISAT(AC).
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The quiescent VCE is projected to the DC load line to
establish the Q-point. The AC load line is drawn through
the Q-point, parallel to the temporary AC load line.
7-16
100 mA
14
12
10
IC in mA 8
AC
6
4 DC
2
80 mA
60 mA
40 mA
TEMP. AC
0 2 4 6
20 mA
8 10 12 14 16 18
VCE in Volts
0 mA
5.3 V
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7-17
The AC load line shows the limits for VCE
and if the Q-point is properly located.
14
12
10
IC in mA 8
6
4
2
100 mA
80 mA
60 mA
40 mA
20 mA
AC
0 2 4 6
8 10 12 14 16 18
VCE in Volts
0 mA
5.3 V
With loaded amplifiers, the Q-point is often closer to saturation.
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7-18
What about voltage gain for a loaded amplifier?
RP
AV =
R
+
r
E
E
VCC = 12 V
1.1 kW
= 4.8
AV =
220 W + 9.03 W
RL= 2.2 kW
RB1 22 kW
RP = 1.1 kW
Load = 2.2 kW
RB2 2.7 kW
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RE = 220 W
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7-19
When analyzing cascade amplifiers, remember:
VCC
1st
2nd
Zin of the 2nd stage loads the 1st stage.
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7-20
Amplifier ac conditions quiz
Emitter bypassing _________ an amplifier’s
input impedance.
decreases
Loading at the output of an amplifier
________ its voltage gain.
decreases
A loaded amplifier has two load lines: dc
and ___________.
ac
The clipping points of a loaded amplifier are
set by its _______ load line.
ac
In a cascade amplifier, the Zin of a stage
_______ the prior stage.
loads
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7-21
Common-source JFET amplifier.
VDD = 20 V
20 V
= 4 mA
ISAT =
5 kW
RL = 5 kW
Drain
Gate
Input
signal
CC
VGS = 1.5 V
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RG
Source
Phase-inverted
output
Fixed bias
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ID in mA 2
-1.5
1
-2.0
0
8 VP-P
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5
10
20
15
VDS in Volts
VGS in Volts
N-channel JFET characteristic curves
The Q-point is set by the fixed bias.
0
Load line 4
1 VP-P
-0.5
3
-1.0
7-22
-2.5
25
AV = 8
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Determining forward transfer admittance:
0
4
ID in mA
-0.5
3
-1.0
1.6 mA
2
-1.5
1
-2.0
0
Yfs =
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DID
DVGS
5
10
15
VGS in Volts
7-23
-2.5
20
25
VDS in Volts
VDS = 1.6 mS
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7-24
When the forward transfer admittance is known,
the voltage gain can be determined using:
VDD = 20 V
RL = 5 kW
D
= 1.6 mS x 5 kW
G
CC
VGS = 1.5 V
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RG
AV = Yfs x RL
S
=8
This agrees with the
graphic solution.
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7-25
Source bias eliminates the need for a separate VGS supply.
VDD
IS = I D
RL
D
G
CC
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RG
VGS = ID x RS
S
RS
This resistor also provides
ac negative feedback which
decreases the voltage gain.
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7-26
JFET amplifier quiz
In a common-source amplifier, the input
signal goes to the _______.
gate
In a common-source amplifier, the input
to output phase relationship is ____. 180o
The voltage gain of a C-S amplifier is equal
to Yfs x _________.
load resistance
Source bias is produced by current flow
through the _______ resistor. source
An unbypassed source resistor _______ the
voltage gain of a C-S amp.
decreases
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Summing
junction
7-27
A negative feedback model
A(Vin - BVout)
Vin
Vin - BVout
A
Vout
A = open loop gain
BVout
B
Feedback
B = feedback ratio
Vin
V
Vin
AV
A AV
in
out
in)
VVout
=
A(V
BV
AB
+1
==AV
ABV
in - =
out
AB
out1AB
in
out
+1 =A
Vout VV
Vout
A +1
out
AB
in
AB +1
Vout
A simplified model
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7-28
The feedback ratio (B) for this circuit
is easy to determine since the source and
drain currents are the same.
VDD
RL = 5 kW
D
G
CC
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RG
B=
800 W
5 kW
= 0.16
S
RS = 800 W
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7-29
Vin
A
AB +1
Vout
Use the simplified model:
8
A(WITH NEG. FEEDBACK) =
= 3.51
(8)(0.16) + 1
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7-30
The source bypass capacitor will
eliminate the ac negative feedback
and restore the voltage gain.
VDD
RL
D
G
CC
RG
RS
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7-31
Amplifier Negative Feedback
• DC reduces
sensitivity to device
parameters
• DC stabilizes
operating point
• DC reduces
sensitivity to
temperature change
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• AC reduces gain
• AC increases
bandwidth
• AC reduces signal
distortion and noise
• AC may change
input and output
impedances
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7-32
The frequency response curve of an ac amplifier
A
Midband
Amax
0.707 Amax
-3dB
f
Bandwidth
The gain is maximum in the midband.
The bandwidth spans the -3 dB points
which are called the break frequencies.
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7-33
The emitter bypass capacitor in this amplifier has
a significant effect on both gain and bandwidth.
6.8 kW
1 kW
10 mF
50 W
1k W
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100 W
10 mF
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7-34
Gain and bandwidth with and without the emitter bypass
50
Gain in dB
BW1
BW2
0
10 Hz
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Frequency
100 MHz
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7-35
Amplifier frequency response
• The lower break frequency is partly
determined by coupling capacitors.
• It is also influenced by emitter bypass
capacitors.
• The upper break frequency is partly
determined by transistor internal
capacitance.
• Both break frequencies can be influenced
by negative feedback.
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7-36
Positive Feedback
• Is the opposite of negative feedback
• Increases gain and reduces bandwidth
• Can be used in some circuits to reduce the
effects of noise
• The next slide shows a circuit with a noise
problem.
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7-37
This circuit is
supposed to convert
the input signal to a
rectangular output
signal. It works, but
the output waveform
shows an extra pulse
caused by noise.
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The trip points are equal.
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7-38
This circuit has positive
feedback and two trip points.
The hysteresis is the
difference between the trip
points (UTP and LTP) and
that makes this circuit less
sensitive to noise. The
output waveform is noise
free.
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R5 provides positive feedback
back from the output amplifier
to the input amplifier.
UTP
LTP
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7-39
REVIEW
• Amplifier Coupling
• Voltage Gain
• FET Amplifier
• Negative Feedback
• Frequency Response
• Positive Feedback
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