Transcript lecture2_singlestageamplifiers

Lecture 2
Single-Transistor Amplifiers
Dr. Ahmed Nader
Adapted from presentation by Richard C. Jaeger
Travis N. Blalock
4/12/2017
Faculty of Engineering
Cairo University
Chap 14 - 1
Chapter Goals
• Detailed study of three broad classes of amplifiers
– Inverting amplifiers- that provide high voltage gain with a 1800
phase shift (common-emitter and common-source configurations)
– Followers- that provide nearly unity gain similar to op amp voltage
follower (common-collector and common-drain configurations)
– Noninverting amplifiers- that provide high voltage gain with no
phase shift (common-base and common-gate configurations).
• Detailed design
– Voltage gain & Current gain
– Input voltage range
– Input and output resistances
– Coupling and bypass capacitor design and lower cutoff frequency
for each type of amplifier.
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Chap 14 - 2
Signal Injection and Extraction: BJT
• In forward-active region,






v

BE
i  I exp

C
S
V 
T 
v

I


i  S exp BE 
E 
 V

T


F
I





v 
i  S exp BE 
B b
V 
T 
FO
• To cause change in current, vBE = vB - vE must be changed.
Base or emitter terminals are used to inject signal because
even if Early voltage is considered, collector voltage has
negligible effect on terminal currents.
• Substantial changes in collector or emitter currents can
create large voltage drops across collector and emitter
resistors and collector or emitter can be used to extract
output. Since iB is a factor of bF smaller than iC or iE
currents, base terminal is not used to extract output.
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Chap 14 - 3
Signal Injection and Extraction: FET
• In pinch-off region,
i i 
S D
Kn 
2


v V 
GS TN 
2
• To cause change in current, vGS = vG - vS must be changed.
Gate or source terminals are used to inject signal because
even with channel-length modulation, drain voltage has
negligible effect on terminal currents.
• Substantial changes in drain or source currents can create
large voltage drops across drain and source resistors and
drain or source can be used to extract output. Since iG is
always zero, gate terminal is not used to extract output.
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Chap 14 - 4
Amplifier Families
• Constraints for signal injection and extraction yield three families of
amplifiers
– Common-Emitter(C-E) / Common-Source(C-S)
– Common-Collector(C-C) / Common-Drain(C-D)
– Common-Base(C-B) / Common-Gate(C-G)
• All circuit examples here use the four-resistor bias circuits to establish
Q-point of the various amplifiers
• Coupling and bypass capacitors are used to change the ac equivalent
circuits.
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Chap 14 - 5
Inverting Amplifiers: Common-Emitter (C-E)
and Common-Source (C-S) Circuits
AC equivalent for C-E Amplifier
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AC equivalent for C-S Amplifier
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Chap 14 - 6
Followers: Common-Collector (C-C) and
Common-Drain (C-D) Circuits
Some Redundant Components!
AC equivalent for C-C Amplifier
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AC equivalent for C-D Amplifier
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Chap 14 - 7
Inverting Amplifiers: Common-Base (C-B)
and Common-Gate (C-G) Circuits
AC equivalent for C-B Amplifier
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AC equivalent for C-G Amplifier
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Chap 14 - 8
Inverting Amplifiers: C-E and C-S
Amplifier Review
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Chap 14 - 9
Inverting Amplifiers: Summary
• C-E and C-S amplifiers have similar voltage gains.
• C-S amplifier provides extremely high input resistance but that of C-E is also
substantial due to the mf RE term.
• Output resistance of C-E amplifier is much higher than that of C-S amplifier as
mf is much larger for BJT than for FET.
• Input signal range of C-S amplifier is higher than that of C-E amplifier.
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Chap 14 - 10
Follower Circuits: Common-Collector
and Common-Drain Amplifiers
AC equivalent for C-C Amplifier
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AC equivalent for C-D Amplifier
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Chap 14 - 11
Follower Circuits: Terminal Voltage
Gain
Neglecting ro,
For C-S Amplifier, take limit of voltage
gain of C-E amplifier as r  
and bo  gmr
gm R
CD
L , What about body effect?
 Avt  
1 g m R
L
In most C-C and C-D amplifiers, gmR 1
L
CC
CD
 Avt  Avt  1
( bo 1)R
vo
L
Avt  
v r  ( bo 1)R
L
b
gm R
CC 
L Assuming b 1
A
o
vt
1 g m R
L
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Output voltage follows input voltage,
hence theses circuits are called
followers. BJT gain is closer to unity
than FET. Mostly, 0.75  Avt  1
ro can be neglected as gain<< mf
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Chap 14 - 12
Follower Circuits: Input Signal Range
For small-signal operation, magnitude of vbe developed across r in smallsignal model must be less than 5 mV.
v


b
R  


v  ir 
be
R
v  0.0051 gm  R  L    0.005(1 gm R )V
L
L b 
b
1 g m R  L


o



L r

If gmR 1 , vb can be increased beyond 5 mV limit.Since only small
L
portion of input signal appears across base-emitter or gate-source terminals,
followers can be used with relatively large input signals without violating
small-signal limits.
In case of FET, magnitude of vgs must be less than 0.2(VGS - VTN).
vg
vg  0.2(V V )(1 gmR )
v gs 
 0.2(V V )
L
GS TN
GS TN
1 g m R
L
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Chap 14 - 13
Follower Circuits: Input Resistance and
Overall Voltage Gain
Overall voltage gain is
Input resistance looking into the base
terminal is given by
vo
CC
Av  
v
i
For C-S Amplifier, r  

RCD
iG







v






v






For C-S Amplifier,







R

CD
CD
G

Av  Avt

R  R 
I G 

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vo






b  ACC b
vt v
v v
i
b i


CC


R R


B iB
 ACC




vt 
CC
R
R  
R
iB 
 I  B
v
CC
RiB  b  r (bo 1)RL
i
b








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Chap 14 - 14
Follower Circuits: Voltage Gain
Calculations (Example)
• Problem: Find overall voltage gain.
• Given data: Q-point values and values for RI, R1, R2, R4, R7 ,for both
BJT and FET.
• Assumptions: Small-signal operating conditions.
• Analysis: For C-C Amplifier,
R  R R 104k
B 1 2
R  R R 11.5k
L 4 7
 r (1 gmR )10.2k[19.8mS(11.5k)]1.16M
RCC
iB 
L
CC

A
ACC
vt
vt
gm R
L  9.80mS(11.5k)  0.991

1 gm RL 1 + 9.80mS(11.5k)
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
R RCC

B iB
CC
ACC
v  Avt 
R  R RCC

 I
B iB
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



 0.956



Chap 14 - 15
Follower Circuits: Voltage Gain
Calculations (Example cont.)
• Analysis: For C-D Amplifier,
R  R R 892k
G 1 2
R  R R 10.7k
L 6 3
ACD
vt 

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g R
m L  (0.491mS)(10.7k)  0.840

1 gm R 1(0.491mS)(10.7k)
L







R

CD
CD
G
  0.838
Av  Avt
R  R 
I
G 
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Chap 14 - 16
Follower Circuits: Output Resistance
Current is injected
into emitter of BJT.
oi Rthi
ve 

g m bo  1


v
x
x

i  i  b i 
 b 
x
o R r
o  R  r 
 th  
th 
R r
r
R
 RiE  th     th
b 1
b 1 b 1
o
o
o
v


R
R
1
o
th


 th
RiE 
g
b 1 g
b 1
m
o
m
o
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Current oi coming out of collector must
be supported by veb = oi/gm, given by
first term. ib =-i/bo+1creates voltage
drop in Rth given by second term
In case of FET,
RiS 
1
g
m
Thus equivalent resistance looking into
emitter or source of a transistor is
 1/ g .
approximately
m
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Chap 14 - 17
Follower Circuits: Output Resistance
(Example)
• Problem: Find output resistance.
• Given data: Q-point values and values for RI, R1, R2, R4, R7 ,for both
BJT and FET.
• Assumptions: Small-signal operating conditions. Small -signal values
are known.
• Analysis: For C-C Amplifier,

R 
 0.990 1.96k 
1
CC
 th  13k || 

 121
Rout  R 6 || RiE  R6 || 
g



b

1
9.80mS
101
 m
o 
For C-D Amplifier,

1
1
CD

R
||
R

R
||
12k
||
1.74k
Rout
6
iS
6
g
4.91mS
m
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
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Chap 14 - 18
Follower Circuits: Current Gain
• Terminal current gain is the ratio of the current delivered to the load
resistor to the current being supplied from the Thevenin source.
i
CC
Ait  1  bo 1
i

ACD
it
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Chap 14 - 19
Follower Circuits: Summary
• Both C-C and C-D amplifiers have voltage gains approaching unity.
• C-D amplifier provides extremely high input resistance because of
infinite resistance looking into gate terminal of FET as compared to CC amplifier.
• Output resistance of C-C amplifier is much lower than the C-D
amplifier due to higher transconductance of BJT than an FET for given
operating current.
• Both C-C and C-D amplifiers can handle relatively large input signal
levels..
• Current gains of FET is inherently infinite, whereas that of BJT is
limited by its finite value of bo.
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Chap 14 - 20
Follower Circuits: Summary (cont.)
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Chap 14 - 21
Noninverting Amplifiers: Common-Base
and Common-Gate Circuits
AC equivalent for C-E Amplifier
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AC equivalent for C-S Amplifier
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Chap 14 - 22
C-B and C-G Amplifiers: Terminal
Voltage Gain and Input Resistance
v
r
1
e
RiE 
  
i b 1 g
o
m
Polarities of vbe and dependent

current source gmvbe are both
reversed, signal source is
transformed to its Norton
equivalent ro is neglected.
vo
CB
Avt    gm RL
ve
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For C-S Amplifier, take limit of voltage
gain of C-E amplifier as r  
ACG
vt   gmRL
RiS 
1
g
m
What if ro is not neglected?
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Chap 14 - 23
C-B and C-G Amplifiers: Overall Voltage
Gain
Overall voltage gain is 





R RiE 
v
v v 

o
o  e  CB
6
ACB
v      Avt 



v ve  v 
R


R

R


iE

 i 
i

 I  6



gm R
R


L

6


R  R // R


th 6 I
1 gm (Rth ) R  R 
 I
6 
For C-S Amplifier,




gm R
R


CG
L

6

Av 


1 gm (R ) R  R 
th  I 6 
For R6 >> RI, CB,CG gm RL

Av
1 g m R
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For gmR 1 (very low RI ),
I
ACB
vt   gmRL
ACG
vt   gmRL
This is the upper bound on gain.
For gm R  1 ,
th
R
CB
CG
Avt  Avt   L
R
I
ro can be neglected as gain<< mf
I
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Chap 14 - 24
C-B and C-G Amplifiers: Input Signal Range









R

6
i
v 
eb 1 g (R R ) R  R 
m I 6
I
6 
v
v  v (1 gmR )
I
i eb
…for R6 >> RI.
For small-signal operation,
v  0.005(1 gmR )V
I
b
In case of FET,
v  vsg (1 gmR )
I
i
v  0.2(V V )(1 gmR )
i
I
GS TN
Relative size of gm and RI determine signal-handling limits.
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Chap 14 - 25
C-B and C-G Amplifiers: Voltage Gain
Calculations (Example)
• Problem: Find overall voltage gain.
• Given data: Q-point values and values for R1, R2, R3, R7 ,for both BJT
and FET, RI =2 k, R4 =12 k.
• Assumptions: Small-signal operating conditions.
• Analysis: For C-E Amplifier,
CB   g R 176
R 1/gm 102
A
m L
vt
iE


CB
R


A vt
R  R R 18k
CB

6
8.48
L 3 C
Av 


R

R


1 gm (R R ) 

I
6
I 6
For C-S Amplifier,

  gmR  8.84
ACS
R  R R 18k
vt
L
L 3 D



CG


R
R 1/gm  2.04k
Avt

CG
4   4.11
iS
Av 


1 gm ( R R )  R  R 
4
I 4  I

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Chap 14 - 26
C-B and C-G Amplifiers: Output
Resistance
Desired resistance is that looking into collector
with base grounded and resistor Rth in emitter.
The redrawn equivalent circuit is same as that
for C-E amplifier except resistance in base is
zero and resistance in emitter is relabeled as Rth.





boR





th
R  ro 1
iC
R r
th
 ro[1 gm(R ||r )] for bo  gmr
th

4/12/2017
And for the FET C-G amp
RiD  ro (1 gm Rth )
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Chap 14 - 27
Noninverting Amplifiers: Current Gain
• Terminal current gain is the ratio of the current delivered to the load
resistor to the current being supplied to the base terminal.
i
CB
Ait  1 o  1
ie
 1
ACG
it
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Chap 14 - 28
C-B and C-G Amplifiers: Summary
• C-B and C-G amplifiers have similar voltage and current gains.
Numerical differences occur due to difference in parameter values of
BJT and FET at similar operating points.
• C-B amplifier can achieve high output resistance due to higher
amplification factor of BJT.
• C-B amplifier can more easily reach low levels of output resistance due
to higher transconductance of BJT for a given operating current.
• Input signal range of C-G amplifier is inherently larger than that of C-B
amplifier.
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Chap 14 - 29
C-B and C-G Amplifiers: Summary
(cont.)
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Chap 14 - 30
Simplified Characteristics of BJT
Single-Stage Amplifiers
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Chap 14 - 31
Simplified Characteristics of BJT
Single-Stage Amplifiers
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Chap 14 - 32
Simplified Characteristics of FET
Single-Stage Amplifiers
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Chap 14 - 33
Simplified Characteristics of FET
Single-Stage Amplifiers
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Chap 14 - 34
Selecting Amplifier Configuration
• A single-transistor amplifier with a gain of 80 dB and input resistance of 100
k.
– Av = 1080/20 = 10,000. For even best BJTs, gain< mf = 40VA = 40(150) =
6000 and FET typically has much lower intrinsic gain. Hence such large
gain can’t be achieved by single-transistor amplifier.
• A single-transistor amplifier with gain of 52 dB, input resistance of 250 k.
– Av = 1052/20 = 400. Since we need large gain and relatively large input
resistance, we can use C-E amplifier. Av = 20VCC , so, VCC =20 V.
boV
T  250kΩ  I  100(0.025V)  10mA which is small but acceptable.
r 
C 2.5105
I
C
For FET, even with small gate overdrive, VDD =100 V which is too large
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Chap 14 - 35
Coupling and Bypass Capacitor Design
•
Since impedance of a capacitor increases with decreasing frequency,
coupling and bypass capacitors reduce amplifier gain at low
frequencies.
• To choose capacitor values, short-circuit time constant method is used:
each capacitor is considered separately with all other capacitors
replaced by short circuits.
• To neglect a capacitor, the magnitude of capacitive impedance must be
much smaller than the equivalent resistance appearing at its terminals.
4/12/2017
Faculty of Engineering
Cairo University
Chap 14 - 36
Coupling and Bypass Capacitor Design:
C-E and C-S Amplifiers
For C-E amplifier,
CE
R  R RCE
R

R
R
out 3 out
in B in
For C-S amplifier,
R  R RCS
R  R RCS
out 3 out
in G in
For coupling capacitor C1,
1
C 
1 w  R  R 

in 
 I
For coupling capacitor C3,
1
C 
3 w  R  R 

out 
 7
w is chosen to be lowest frequency
for which midband operation is
needed in given application.
4/12/2017
Faculty of Engineering
Cairo University
Chap 14 - 37
Coupling and Bypass Capacitor Design:
C-E and C-S Amplifiers (contd.)
In this case, we can neglect impedances
of capacitors C1 and C3 , the find the
equivalent resistance looking up into
emitter or source of amplifier.
1
C 

2

 


1  


wR R 

 6  E
g
 
m



C 
2
4/12/2017
1






1
R 
6 S g
m
w R
Faculty of Engineering
Cairo University






 



 

Chap 14 - 38
Coupling and Bypass Capacitor Design:
C-E and C-S Amplifiers (Example)
• Problem: Choose values of coupling and bypass capacitors.
• Given data: f = 1000Hz, values of all resistors and input and output
resistances for both C-E and C-S amplifiers.
• Analysis:
For C-S amplifier:
For C-E amplifier:
R  R  892k
R  R RCE  78.1k
in G
in B in
1
1
C

178pF C 1800pF
C 
1.99nF C  0.02mF


1
1
1 w  R  R 
1
 w  R  R 


in 
 I
in 
 I
1
1
C

 55.3nF
C 
 67.2nF


2




2
w  R  R  (1/ gm )  
w  R  R  (1/ gm )  

 6 S

 6 E
 C  0.56mF
 C  0.68mF
2
2
1
1
C 
1.31nF  C  0.015mF C 
1.31nF  C  0.015mF
3 w  R  R 
3
3 w  R  R 
3


out 
out 
 7
 7
4/12/2017
Faculty of Engineering
Cairo University
Chap 14 - 39
Coupling and Bypass Capacitor Design:
C-C and C-D Amplifiers
For C-E amplifier,
R  R RCC
R  R RCC
in
in B
out 4 out
For C-S amplifier,
CD
R  R RCS
R

R
R
out 4 out
in G in
For coupling capacitor C1,
1
C 
1 w  R  R 

in 
 I
For coupling capacitor C3,
1
C 
3 w  R  R 

out 
 7
4/12/2017
Faculty of Engineering
Cairo University
Chap 14 - 40
Coupling and Bypass Capacitor Design:
C-C and C-D Amplifiers (Example)

• Problem: Choose values of coupling and bypass capacitors.
• Given data: f = 1000Hz, values of all resistors and input and output
resistances for both C-E and C-S amplifiers.
• Analysis:
For C-D amplifier:
For C-C amplifier:
R  R  892k
in G
R  R RCC  95.5k
in
in B
R  R RCD
 1.74k
out
out
CC
4
R  R Rout  120
1
out 4
 C 
 89pF
1


1
C 
 816pF
w  R  R 
1 w  R  R 
in 
 I


in 
 I
 C  1000pF
1
 C  8200pF
1
1
1
C

 782pF
C 
 795pF


3
3 w  R  R 
w  R  R 
out 

 7
out 
 7
 C  8200pF
 C  8200pF
3
3
4/12/2017
Faculty of Engineering
Cairo University
Chap 14 - 41
Coupling and Bypass Capacitor Design:
C-B and C-G Amplifiers
For C-E amplifier,
CB
R  R RCB
R

R
R
in
in 4
out 3 out
For C-S amplifier,
R  R RCG
R  R RCG
out 3 out
in 4 in
For coupling capacitor C1,
1
C 
1 w  R  R 

in 
 I
For coupling capacitor C3,
1
C 
3 w  R  R 

out 
 7
4/12/2017
Faculty of Engineering
Cairo University
Chap 14 - 42
Coupling and Bypass Capacitor Design:
C-B and C-G Amplifiers (contd.)
In this case, we can neglect impedances
of capacitors C1 and C3 , the find the
equivalent resistance looking up into
emitter or source of amplifier.


CB

R
R
r

(
b

1
)(
R
R
)


Req
o
1 2  
4 I 
RCG
eq  R1 R2
1
C 
2
,CG
w RCB
eq
4/12/2017
Faculty of Engineering
Cairo University
Chap 14 - 43
Coupling and Bypass Capacitor Design:
C-B and C-G Amplifiers (Example)
• Problem: Choose values of coupling and bypass capacitors.
• Given data: f = 1000Hz, values of all resistors and input and output
resistances for both C-E and C-S amplifiers.
• Analysis: For C-B amplifier:
R  R RCB 13k102k100
in 4 in
R  R RCB
 22k 3.93M21.9k
out 3 out

1
C 
 75.8nF C  0.82mF


1
1
w  R  R 

in1
 I
C 
 2.38nF  0.027mF


2
w  R R r  ( bo 1)( R R ) 
2
4 I 
 1
1
C 
1.31nF  C  0.015mF
3 w  R  R 
3


out 
 7

4/12/2017

Faculty of Engineering
Cairo University
Chap 14 - 44
Coupling and Bypass Capacitor Design: CB and C-G Amplifiers (Example contd.)
For C-G amplifier:


4/12/2017
R  R RCG 12k 2.04k1.74k
in 4 in
R  R RCG
 22k 410k20.9k
out 4 out
1
C 
 42.6nF C  0.42mF
1 w  R  R 
1


in 
 I
1
C 
 178pF  1800pF


2
w  R R 
2
 1
1
C 
1.31nF  C  0.015mF
3 w  R  R 
3

out 
 7
Faculty of Engineering
Cairo University
Chap 14 - 45