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ElectronicsElectronics
Principles & ApplicationsPrinciples & ApplicationsSixth EditionSixth Edition
Chapter 6Introduction to
Small-Signal Amplifiers(student version)
©2003 Glencoe/McGraw-Hill
Charles A. Schuler
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations
INTRODUCTION
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 youto view that segment again, if you want to.
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.
Amplifier Out
InGain =
In
Out= 3.33
1.5 V 5 V
1.5 V
5 VThe units cancel
Gain can be expressed in decibels (dB).
The dB is a logarithmic unit.
Common logarithms are exponents of the number 10.
102 = 100103 = 100010-2 = 0.01100 = 1103.6 = 3981
The log of 100 is 2
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
The dB unit is based on a power ratio.
dB = 10 x log POUT
PIN
50 W
1 W501.7017
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.
+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
Gain Quiz
Amplifier output is equal to the input________ by the gain. multiplied
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
Common logarithms are ________ of the number 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
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.
A small-signal amplifier can also be called a voltage amplifier.
Common-emitter amplifiers are one type.
C
BE
Start with an NPN bipolar junction transistor
VCC
Add a power supply
RL
Next, a load resistor
RB
Then a base bias resistor
CC
A coupling capacitor is often requiredConnect a signal sourceThe emitter terminal is grounded
and common to the input andoutput signal circuits.
RB RL
VCC
CC
C
BE
The outputis phase inverted.
RB
VCC
CC E
When the input signal goes positive:
B
The base current increases.
C
The collector current increases times.
RL
So, RL drops more voltage and VCE must decrease.
The collector terminal is now less positive.
RB
VCC
CC E
When the input signal goes negative:
B
The base current decreases.
C
The collector current decreases times.
RL
So, RL drops less voltage and VCE must increase.
The collector terminal is now more positive.
350 k
CC EB
C
1 k14 V
The maximum value of VCE for this circuit is 14 V.
The maximum value of IC is 14 mA.
IC(MAX) =14 V
1 k
These are the limits for this circuit.
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
The load line connects the limits.
SAT.
This end is called saturation.
CUTOFFThis end is called cutoff.
LINEAR
The linear region is between the limits.
350 k
CC EB
C
1 k14 V
IB =14 V
350 k
Use Ohm’s Law to determine the base current:
= 40 A
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
An amplifier can be operated at any point along the load line.
The base current in this case is 40 A.
Q
Q = the quiescent point
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
The input signal varies the base current above and below the Q point.
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
Overdriving the amplifier causes clipping.
The output is non-linear.
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
What’s wrong with this Q point?
How about this one?
350 k
CC EB
C
1 k14 V
IB =14 V
350 k
= 150
IC = x IB
= 40 A
= 150 x 40 A = 6 mA
VRL = IC x RL = 6 mA x 1 k = 6 V
This is a good Q point for linear amplification.
VCE = VCC - VRL = 14 V - 6 V = 8 V
350 k
CC EB
C
1 k14 V
IB =14 V
350 k
= 350
IC = x IB
= 40 A (IB is not affected)
= 350 x 40 A = 14 mA (IC is higher)
VRL = IC x RL = 14 mA x 1 k = 14 V (VRL
is higher)
This is not a good Q point for linear amplification.VCE = VCC - VRL
= 14 V - 14 V = 0 V (VCE is lower)
is higher
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
The output is non-linear.
The higher causessaturation.
RB
CC EB
C
RL
VCC
It’s dependent!
This common-emitter amplifier is not practical.
It’s also temperature dependent.
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.
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
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.
RB1
CC
EB
C
RL
VCC
RB2 RE
This common-emitter amplifier is practical.
It uses voltage divider bias and
emitter feedback to reduce sensitivity.
+VCC
RL
RE
RB1
RB2{RB1 and RB2 form
a voltage divider
Voltage divider bias
+VCC
RB1
RB2
+VB
Voltage dividerbias analysis:
VB =RB2
RB1 + RB2
VCC
The base current is normallymuch smaller than the dividercurrent so it can be ignored.
RB1
EB
C
RL
VCC
RB2 RE = 220
= 12 V
2.7 k
22 k = 2.2 k
Solving the practical circuit for its dc conditions:
VB = RB2
RB1 + RB2
x VCC
VB = 2.7 k
22 k2.7 k +x 12 V
VB = 1.31 V
RB1
EB
C
RL
VCC
RB2 RE = 220
= 12 V
2.7 k
22 k = 2.2 k
Solving the practical circuit for its dc conditions:
VE = VB - VBE
VE = 1.31 V - 0.7 V = 0.61 V
RB1
EB
C
RL
VCC
RB2 RE = 220
= 12 V
2.7 k
22 k = 2.2 k
Solving the practical circuit for its dc conditions:
IE = RE
VE
IE = 0.61 V
220 = 2.77 mA
IC IE
RB1
EB
C
RL
VCC
RB2 RE = 220
= 12 V
2.7 k
22 k = 2.2 k
Solving the practical circuit for its dc conditions:
VRL = IC x RL
VRL = 2.77 mA x 2.2 k
VRL = 6.09 V
VCE = VCC - VRL - VE
VCE = 12 V - 6.09 V - 0.61 V
VCE = 5.3 V
A linear Q point!
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.
RB1
EB
C
RL
VCC
RB2 RE = 220
= 12 V
2.7 k
22 k = 2.2 k
Solving the practical circuit for its ac conditions:
The ac emitter resistance is rE:
rE = 25 mV
IE
rE =25 mV
2.77 mA= 9.03
RB1
EB
C
RL
VCC
RB2 RE = 220
= 12 V
2.7 k
22 k = 2.2 k
Solving the practical circuit for its ac conditions:
The voltage gain from base to collector:
AV =RL
RE + rE
AV =2.2 k
220 9.03= 9.61
RB1
EB
C
RL
VCC
RB2 RE
= 12 V
2.7 k
22 k = 2.2 k
Solving the practical circuit for its ac conditions:
AV =RL
rE
AV =2.2 k
9.03= 244
An emitter bypass capacitorcan be used to increase AV:
CE
Practical C-E Amplifier Quiz
-dependency is reduced with emitterfeedback 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
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
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.
RB1
EB
C
RL
VCC
RB2 RE CE
The common-emitter configuration is used most often.
It has the best power gain.
Medium output Z
Medium input Z
RB1
EB
C
RC
VCC
RB2 RL
The common-collector configuration is shown below.
Its input impedance and current gain are both high.
It’s often called an emitter-follower.
In-phaseoutput
Low output Z
RB1
EB
C
RL
VCC
RB2 RE
The common-base configuration is shown below.
Its voltage gain is high. It’s used mostat RF.
In-phaseoutputLow input Z
PNP C-E Amplifier
47
1 k
1.5 k
22 k
10 k
+12 V
VB = - 3.75 V
VE = - 3.05 V
IE = 2.913 mA
VRL = 4.37 V
VCE = - 4.58 V
VC = - 7.63 V
rE = 8.58
AV = 27
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
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
REVIEW
• Gain
• Common-Emitter Amplifier
• Stabilizing the Amplifier
• Other Configurations