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TRANSCRIPT
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.McGraw-Hill
7-1
ElectronicsElectronics
Principles & ApplicationsPrinciples & ApplicationsEighth EditionEighth Edition
Chapter 7More About
Small-Signal Amplifiers
Charles A. Schuler
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7-2
• Amplifier Coupling• Voltage Gain• FET Amplifiers• Negative Feedback• Frequency Response• Positive Feedback
INTRODUCTION
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VCC
These two points are at different dc voltages.
Capacitive coupling is convenient in cascade ac amplifiers.
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VCC
Direct coupling is required for dc gain.
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VCC
The Darlington is a popular dc arrangement.
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VCC
P S
10:1
10 ZRATIO = TRATIO
2
= 102 = 100
ZCOLLECTOR = 100 x 10 = 1000
Transformer coupling offers the advantage of impedance matching.
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Transformer couplingis used in 70.7 volt
sound systems.
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VCC
Transformer coupling can beused in bandpass amplifiers
to achieve selectivity.
fR
Gain
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Amplifier coupling quiz
Capacitive coupling is not useful for_________ amplifiers. dc
Dc frequency response requires ________coupling. direct
Transformer coupling offers the advantageof _________ matching. impedance
Tuned transformer coupling providesfrequency _____________. selectivity
A Darlington amplifier is an example of_________ coupling. direct
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7-10
RB1
EB
C
RL
VCC
RB2 RE
= 12 V
2.7 k
22 k = 2.2 k
More about solving the practical circuit for its ac conditions:
= 220
Zin = ?
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RB1
EB
C
RL
VCC
RB2 RE
= 12 V
2.7 k
22 k = 2.2 k
Zin is a combination of RB1, RB2, and rin of the transistor.
= 220
rin = (RE + rE)
rin = (220 + 9.03 )
rin = 34.4 k
Note: rin = rE
when RE is bypassed.
Determine rin first:
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7-12
RB1
EB
C
RL
VCC
RB2 RE
= 12 V
2.7 k
22 k = 2.2 k
= 220
Zin =1
RB2
1rin
1+
RB1
1+
++Zin =
1
2.7 k1
34.4 k1
22 k1
Zin = 2.25 k
RB1, RB2, and rin act in parallelto load the input signal.
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7-13
RB1
VCC
RB2 RE
= 12 V
2.7 k
22 k RL= 2.2 k
= 220
Load = 2.2 k
What happens when an amplifier is loaded?
RL and the Load act in parallel.
RP = 1.1 k
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7-14
RB1
RB2 RE
VCC = 12 V
2.7 k
22 k RL= 2.2 k
= 220
Load = 2.2 k
There are two saturation currents for a loaded amplifier.
RP = 1.1 k
ISAT(DC) = VCC
RL + RE
= 4.96 mA
ISAT(AC) = VCC
RP + RE
= 9.09 mA
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7-15
0 2 4 6 8 10 12 14 16 18
2468
101214
VCE in Volts
IC in mA
20 A
0 A
100 A
80 A
60 A
40 A
There are two load lines for a loaded amplifier.
DC
TEMPORARY AC
The DC load line connects VCC and ISAT(DC).
A temporary AC load line connects VCC and ISAT(AC).
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0 2 4 6 8 10 12 14 16 18
2468
101214
VCE in Volts
IC in mA
20 A
0 A
100 A
80 A
60 A
40 A
5.3 V
DC
AC
TEMP. AC
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.
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0 2 4 6 8 10 12 14 16 18
2468
101214
VCE in Volts
IC in mA
20 A
0 A
100 A
80 A
60 A
40 A
5.3 V
AC
The AC load line shows the limits for VCE and if the Q-point is properly located.
With loaded amplifiers, the Q-point is often closer to saturation.
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7-18
RB1
RB2 RE
VCC = 12 V
2.7 k
22 k RL= 2.2 k
= 220
Load = 2.2 k
What about voltage gain for a loaded amplifier?
RP = 1.1 k
AV =RP
RE + rE
AV =1.1 k
220 9.03= 4.8
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VCC
Zin of the 2nd stage loads the 1st stage.
When analyzing cascade amplifiers, remember:
2nd1st
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7-20
Amplifier ac conditions quiz
Emitter bypassing _________ an amplifier’sinput impedance. decreases
Loading at the output of an amplifier________ its voltage gain. decreases
A loaded amplifier has two load lines: dcand ___________. ac
The clipping points of a loaded amplifier areset by its _______ load line. ac
In a cascade amplifier, the Zin of a stage_______ the prior stage. loads
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7-21
Drain
Source
Gate
VDD = 20 V
VGS = 1.5 V
RGCC
RL = 5 k
Inputsignal
Common-source JFET amplifier.
Fixed bias
ISAT = 20 V
5 k= 4 mA
Phase-invertedoutput
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7-22
0
2
4
1
VDS in Volts
ID in mA
5 10 15 20 25
3
-2.5
-2.0
-1.5
-1.0
-0.5
0
N-channel JFET characteristic curves
VG
S in
Vol
ts
Load line
The Q-point is set by the fixed bias.
8 VP-P
1 VP-P
AV = 8
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0
2
4
1
VDS in Volts
ID in mA
5 10 15 20 25
3
-2.5
-2.0
-1.5
-1.0
-0.5
0Determining forward transfer admittance:
Yfs = ID
VGS
VG
S in
Vol
ts
VDS
1.6 mA
= 1.6 mS
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7-24
D
S
G
VDD = 20 V
VGS = 1.5 V
RGCC
RL = 5 k
When the forward transfer admittance is known,the voltage gain can be determined using:
AV = Yfs x RL
= 1.6 mS x 5 k
= 8
This agrees with the graphic solution.
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7-25
D
S
G
VDD
VGS = ID x RS
RGCC
RL
RS
Source bias eliminates the need for a separate VGS supply.
IS = ID
This resistor also providesac negative feedback whichdecreases the voltage gain.
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JFET amplifier quiz
In a common-source amplifier, the inputsignal goes to the _______. gate
In a common-source amplifier, the inputto output phase relationship is ____. 180o
The voltage gain of a C-S amplifier is equalto Yfs x _________. load resistance
Source bias is produced by current flowthrough the _______ resistor. source
An unbypassed source resistor _______ thevoltage gain of a C-S amp. decreases
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7-27
Vin - BVout
A(Vin - BVout)
BVout
A = open loop gain
Summingjunction
Vin VoutA
B Feedback
A negative feedback model
B = feedback ratio
Vout = A(Vin - BVout)Vout = AVin - ABVout
AVin
Vout
1 = - ABAVin
Vout
AB +1 =Vin
Vout
AB +1A
=Vin
Vout
AB +1
A=
AB +1AVin Vout
A simplified model
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7-28
D
S
G
VDD
RGCC
RL
RS
= 5 k
= 800
The feedback ratio (B) for this circuitis easy to determine since the source and
drain currents are the same.
B = 800 5 k
= 0.16
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7-29
AB +1AVin Vout
Use the simplified model:
A(WITH NEG. FEEDBACK) =8
(8)(0.16) + 1= 3.51
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7-30
CS
DG
VDD
RGCC
RL
RS
The source bypass capacitor will eliminate the ac negative feedback
and restore the voltage gain.
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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
• AC reduces gain• AC increases
bandwidth• AC reduces signal
distortion and noise• AC may change
input and output impedances
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0.707 Amax
A
f
The frequency response curve of an ac amplifier
Bandwidth
The gain is maximum in the midband.
Amax
Midband
The bandwidth spans the -3 dB points which are called the break frequencies.
-3dB
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50
10 F
10 F
1 k
100 1k
6.8 k
The emitter bypass capacitor in this amplifier hasa significant effect on both gain and bandwidth.
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7-34G
ain
in d
B
0
50
Frequency10 Hz 100 MHz
BW1
BW2
Gain and bandwidth with and without the emitter bypass
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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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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.
The trip points are equal.
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7-38
UTP LTP
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.
R5 provides positive feedback back from the output amplifier to the input amplifier.
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7-39
REVIEW
• Amplifier Coupling• Voltage Gain• FET Amplifier• Negative Feedback• Frequency Response• Positive Feedback