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CHAPTER 3
FEEDBACK
AMPLIFIERS
Dr.Saju Subramanian
Associate Professor
Outline
1. Introduction to Feedback
2. Feedback Amplifier – Positive & Negative
3. Advantages/Disadvantages of Negative
Feedback
4. Basic Feedback Concept
5. Classification of Amplifiers
6. Series – Shunt Configuration
7. Shunt – Series Configuration
8. Series - Series Configuration
9. Shunt – Shunt Configuration
Introduction to Feedback
 Feedback is used in virtually all amplifier system.
 Invented in 1928 by Harold Black – engineer in
Western Electric Company
 methods to stabilize the gain of amplifier for use
in telephone repeaters.
 In feedback system, a signal that is proportional to
the output is fed back to the input and combined
with the input signal to produce a desired system
response.
 However, unintentional and undesired system
response may be produced.
Feedback Amplifier
 Fe e dback is a technique where a proportion of
the output of a system (amplifier) is fed back and
recombined with input
 There are 2 types of feedback amplifier:
 Positive feedback
 Negative feedback
A
β
input output
Positive Feedback
 Positive feedback is the process when the
output is adde d to the input, amplified again,
and this process continues.
 Positive feedback is used in the design of
oscillator and other application.
A
β
input output
+
Positive Feedback - Example
 In a PA system
get feedback when you put the microphone in
front of a speaker and the sound gets
uncontrollably loud (you have probably heard
this unpleasant effect).
Negative Feedback
 Negative feedback is when the output is
subtracted from the input.
 The use of negative feedback reduces the gain.
Part of the output signal is taken back to the input
with a negative sign.
A
β
input output
Negative Feedback - Example
 Speed control
If the car starts to speed up above the desired
set-point speed, negative feedback causes the
throttle to close, thereby reducing speed;
similarly, if the car slows, negative feedback
acts to open the throttle
Feedback Amplifier - Concept
Basic structure of a single - loop feedback amplifier
Advantages of Negative
Feedback
1. Gain Sensitivity – variations in gain is
reduced.
2. Bandwidth Extension – larger than that of
basic amplified.
3. Noise Sensitivity – may increase S-N ratio.
4. Reduction of Nonlinear Distortion
5. Control of Impedance Levels – input and
output impedances can be increased or
decreased.
Disadvantages of Negative
Feedback
1. Circuit Gain – overall amplifier gain is
reduced compared to that of basic amplifier.
2. Stability – possibility that feedback circuit will
become unstable and oscillate at high
frequencies.
Basic Feedback Concept
Basic configuration of a feedback amplifier
Basic Feedback Concept
 The output signal is:
where A is the amplification factor
 Feedback signal is
where ß is the feedback transfer function
 At summing node:
 Closed-loop transfer function or gain is
if
εAS=oS
oSβ=fbS
fbi SS −=εS
A
A
S
S
i
o
β+
==
1
fA
ββ
β
1
1 =≅>>
A
A
then fAA
Classification of Amplifiers
Classify amplifiers into 4 basic categories
based on their input (parameter to be
amplified; voltage or current) & output signal
relationships:
 Voltage amplifier (series-shunt)
 Current amplifier (shunt-series)
 Transconductance amplifier (series-series)
 Transresistance amplifier (shunt-shunt)
Feedback Configuration
Series:
connecting theconnecting the
feedback signalfeedback signal
in series within series with
thethe
input signalinput signal
voltage.voltage.
Shunt:
connecting
the
feedback
signal
in shunt
(parallel) with
an input
Series - Shunt Configuration
vv
v
vf
A
A
A
β+
=
1
Series - Shunt Configuration
if Lo RR <<
then the output of feedback network is an open
circuit;
Output voltage is:
εVAV vo =
feedback voltage is:
ovVVfb β=
By neglecting Rs due to ; error voltage
is:
si RR >>
fbi V−=VVε
vv
v
i
o
vf
A
A
V
V
A
β+
==∴
1
where ßv is closed-loop voltage transfer function
Series - Shunt Configuration
Or
 Input current
 Rif with feedback
Assume Vi=0 and Vx
applied to output
terminal.
Or
 Input current
 Rof with feedback
Input Resistance, Rif Output Resistance, Rof
)( εεε β VAVVV vvfbi +=+=V
)1( vv
i
A
V
V
β
ε
+
=
)1( vvi
i
i
i
AR
V
R
V
I
β
ε
+
==
)1( vvi
i
i
if AR
I
V
R β+==
0=+=+ xvfb VVVV βεε
xvVV βε −=
o
vvx
o
vx
i
R
AV
R
VAV
I
)1( βε +
=
−
=
)1( vv
o
x
x
of
A
R
I
V
R
β+
==
 Series input connection increase input resistance –
avoid loading effects on the input signal source.
 Shunt output connection decrease the output resistance
- avoid loading effects on the output signal when output
load is connected.
Equivalent circuit of shunt - series feedback circuit or
voltage amplifier
Series - Shunt Configuration
For ideal non-inverting op-
amp amplifier
Feedback transfer function;
Series - Shunt Configuration
 Non-inverting op-amp is an example of the
series-shunt configuration.






+==
1
2
1
R
R
V
V
A
i
o
vf






+
=
1
2
1
1
R
R
β
Series - Shunt Configuration
Equivalent circuit )1(
/
1
11
1
221
1
21
1
21
1
vi
i
i
i
i
if
v
oi
v
v
v
v
i
o
vf
ofb
fbi
AR
RV
V
I
V
R
R
R
VA
VV
RR
R
VV
A
A
RR
R
A
A
V
V
A
V
RR
R
V
VV
β
β
ε
ε
εε
ε
ε
+===






+
+=





+
+=
+
=






+
+
==






+
≅
−=
=
V
VAV vo
Series - Shunt Configuration
Example:
Calculate the feedback amplifier gain of the
circuit below for op-amp gain, A=100,000;
R1=200 Ω and R2=1.8 kΩ.
Solution: Avf = 9.999 or 10
Series - Shunt Configuration
 Basic emitter-follower and source-follower circuit
are examples of discrete-circuit series-shunt
feedback topologies.
• vi is the input signal
• error signal is base-
emitter/gate-source
voltage.
• feedback voltage =
output voltage 
feedback transfer
function, ß v = 1
Series - Shunt Configuration
 Small-signal voltage gain:
 Open-loop voltage gain:
 Closed-loop input resistance:
 Output resistance:
e
E
e
E
Em
Em
i
o
vf
r
R
r
R
Rg
r
Rg
r
V
V
A
+
=






++






+
==
11
1
1
π
π
e
E
Emv
r
R
Rg
r
A =





+=
π
1












++=++= EmEmif Rg
r
rRrgrR
π
πππ
1
1)1(
Em
E
m
Eof
Rg
r
R
rg
r
RR






++
=
+
=
π
π
π
1
1
)1(
Shunt – Series Configuration
ii
i
if
A
A
A
β+
=
1
Shunt – Series Configuration
 Basic current amplifier with input resistance, Ri
and an open-loop current gain, Ai.
 Current IE is the difference between input signal
current and feedback current.
 Feedback circuit samples the output current –
provide feedback signal in shunt with signal
current.
 Increase in output current – increase feedback
current – decrease error current.
 Smaller error current – small output current –
stabilize output signal.
Shunt – Series Configuration
if si RR <<
then the output is a short circuit; output current is:
εIAI io =
feedback current is:
oi II fb β=
Input signal current:
fbi II += εI
ii
i
i
o
if
A
A
I
I
A
β+
==∴
1
then εIIi ≈
where ßi is closed-loop current transfer function
Shunt – Series Configuration
Or
 Input current
 Rif with feedback
Input Resistance, Rif
)( εεε β IAIII iifbi +=+=I
)1( ii
i
A
I
I
β
ε
+
=
)1( ii
ii
ii
A
RI
RIV
β
ε
+
==
)1( ii
i
i
i
if
A
R
I
V
R
β+
==
Assume Ii=0 and Ix applied
to output terminal.
 Rof with feedback
Output Resistance, Rof
[ ]
oiixx
oxiixx
oixx
xi
xifb
RAIV
RIAIV
RIAIV
II
IIII
)1(
)(
)(
0
β
β
β
β
ε
ε
εε
+=
−−=
−=
−=
=+=+
( )iio
x
x
of AR
I
V
R β+== 1
Shunt - Series Configuration
 Shunt input connection decrease input resistance –
avoid loading effects on the input signal current source.
 Series output connection increase the output resistance
- avoid loading effects on the output signal due to load
connected to the amplifier output.
Equivalent circuit of shunt - series feedback circuit or
voltage amplifier
Shunt - Series Configuration
 Op-amp current amplifier – shunt-series
configuration.
 Ii’ from equivalent source of Ii and Rs.• Iε is negligible and
Rs>>Ri;
• assume V1 virtually
ground;
• Current I1:
• Output current:
• Ideal current gain:
fbii II == 'I
FiFfbo RIRI −=−=V
1/ RVo−=1I






+=+=
1
1 1
R
R
III F
ifboI






+==
1
1
R
R
I
A F
i
i
oI
Shunt - Series Configuration
 Ai is open-loop current
gain
and
 Assume V1 is virtually
ground:
 I1 current:
 Output current
fbifbi IIII −≅−= 'εI
)( fbii IIA −== εIAI io
Ffb RI−=oV
Closed-loop current gain:






=−=
11
1
R
R
I
R
V F
fb
o
I






+=+=
1
1
R
R
IIII F
fbfbfboI






+
+
==
1
1
1
R
R
A
A
I
I
F
i
i
i
o
ifA
Shunt - Series Configuration
 Common-base circuit is example of discrete
shunt-series configuration.
 Amplifier gain: Closed-loop current gain:
RLIo
Ii
Iε
RLIo
Ii
Iε
Ifb
βε == iAI/oI
i
i
i
o
if
A
A
I
I
A
+
=
+
==
11 β
β
Shunt - Series Configuration
 Common-base circuit with RE and RC
RCIoRE
Ii
V-
V+
RCIoRE
Ii
i
E
i
m
E
m
i
o
if
A
R
r
A
rg
R
r
rg
I
I
A
+





+
=
+





+
==
π
π
π
π
11
Series – Series Configuration
gg
g
gf
A
A
A
β+
=
1
Series – Series Configuration
 The feedback samples a portion of the output
current and converts it to a voltage – voltage-
to-current amplifier.
 The circuit consist of a basic amplifier that
converts the error voltage to an output current
with a gain factor, Ag and that has an input
resistance, Ri.
 The feedback circuit samples the output
current and produces a feedback voltage, Vfb,
which is in series with the input voltage, Vi.
Series – Series Configuration
Assume the output is a short circuit, the output
current:
εVAI go =
feedback voltage is:
oz IVfb β=
Input signal voltage (neglect Rs=∞ ):
fbi VV += εV
gz
g
i
o
gf
A
A
V
I
A
β+
==∴
1
where ßz is a resistance feedback transfer function
Series – Series Configuration
Assume Ii=0 and Ix applied
to output terminal.
 Rof with feedback
Output Resistance, Rof
[ ]
ogzxx
oxzgxx
ogxx
xz
xzfb
RAIV
RIAIV
RIAIV
II
IIII
)1(
)(
)(
0
β
β
β
β
ε
ε
εε
+=
−−=
−=
−=
=+=+
( )gzo
x
x
of AR
I
V
R β+== 1
Or
 Input current
 Rif with feedback
Input Resistance, Rif
)( εεε β VAVVV gzfbi +=+=V
)1( gz
i
A
V
V
β
ε
+
=
)1( gzi
i
i
i
AR
V
R
V
I
β
ε
+
==
)1( gzi
i
i
if AR
I
V
R β+==
Series – Series Configuration
 Series input connection increase input
resistance
 Series output connection increase the output
resistance
Equivalent circuit of series - series feedback
Series – Series Configuration
 The series output
connection samples the
output current 
feedback voltage is a
function of output current.
 Assume ideal op-amp
circuit and neglect
transistor base-current:
Ei
o
gf
Eofbi
RV
I
A
RIV
1
==
==V
Series – Series Configuration
 Assume IE≅IC and Ri≈∞
( )
( ) Egm
gm
i
o
gf
Eoigmo
Eoifbi
gmbm
E
fb
o
RArg
Arg
V
I
A
RIVArgI
RIVVVV
VArgIrg
R
V
I
π
π
π
ε
εππ
+
==
−=
−=−=
===
1
Series – Series Configuration
Series – Series Configuration
Em
LC
C
m
i
o
gf
Emfbi
Emfb
LC
C
mo
Rg
r
RR
R
g
V
I
A
Rg
r
VVVV
RVg
r
V
V
RR
R
VgI






++






+
−
==














++=+=






+=






+
−=
π
π
ππ
π
π
π
π
1
1
1
1
)(
Shunt – Shunt Configuration
zz
z
zf
A
A
A
β+
=
1
Shunt – Shunt Configuration
 The feedback samples a portion of the output
voltage and converts it to a current – current-
to-voltage amplifier.
 The circuit consist of a basic amplifier that
converts the error current to an output voltage
with a gain factor, Az and that has an input
resistance, Ri.
 The feedback circuit samples the output
voltage and produces a feedback current, Ifb,
which is in shunt with the input current, Ii.
Shunt – Shunt Configuration
Assume the output is a open circuit, the output
voltage:
εIAV zo =
feedback voltage is:
ogVI fb β=
Input signal voltage (neglect Rs=∞ ):
fbi II += εI
zg
z
i
o
zf
A
A
I
V
A
β+
==∴
1
where ßg is a conductance feedback transfer function
Shunt – Shunt Configuration
Or
 Input current
 Rif with feedback
Input Resistance, Rif
)( εεε β IAIII zgfbi +=+=I
)1( zg
i
A
I
I
β
ε
+
=
)1( zg
ii
ii
A
RI
RIV
β
ε
+
==
)1( zg
i
i
i
if
A
R
I
V
R
β+
==
Assume Vi=0 and Vx
applied to output
terminal.
Or
 Input current
 Rof with feedback
Output Resistance, Rof
0=+=+ xgfb VVVV βεε
xgVV βε −=
o
zgx
o
zx
i
R
AV
R
VAV
I
)1( βε
+
=
−
=
)1( zg
o
x
x
of
A
R
I
V
R
β+
==
Shunt – Shunt Configuration
Equivalent circuit of shunt - shunt feedback circuit
or
voltage amplifier
Shunt – Shunt Configuration
 Basic inverting op-amp circuit is an example of
shunt-shunt configuration.
 Input current splits between feedback current
and error current.
 Shunt output connection samples the output
voltage  feedback current is function of output
voltage.
2
2
R
I
V
A
IIwhere
RIV
i
o
zf
ifb
fbo
−==
=
−=
Shunt – Shunt Configuration
 Az is open-loop
transresistance gain
factor (-ve value)
( )
2
2
1
/
R
A
A
I
V
A
RVIwhere
IIAIAV
z
z
i
o
zf
ofb
fbizzo
+
−
==
−=
−−== ε
Shunt – Shunt Configuration
Shunt – Shunt Configuration






−





+





+





+






−−
==
=





+





−+





+





+
−
+=
=
−
++
F
m
FFFC
F
m
i
o
zf
F
o
i
F
m
FFC
o
F
o
i
F
o
m
C
o
R
g
RRrRR
R
g
I
V
A
R
V
I
R
g
RrRR
V
R
VV
r
V
I
R
VV
Vg
R
V
111111
1
0
11111
0
π
π
π
π
π
π
π
( )
( )





 −
+
≅=






+





−





+





+






++
==
−=












−
=
F
z
z
i
o
zf
F
C
z
FFF
C
F
C
z
i
o
zf
Cm
C
m
z
R
A
A
I
V
A
R
Rr
A
RR
r
R
R
R
Rr
A
I
V
A
Rrg
rR
g
A
1
1
1
11
11
ππ
π
π
π
Shunt – Shunt Configuration
 Open-loop transresistance gain factor Az is
found by setting RF=∞
 Multiply by (rπRC)
 Assume RC <<RF
& rπ<< RF
Feedback Amplifier
Input and output Impedances
 Summary
1. For a series connection at input or output,
the resistance is increased by (1+βA).
2. For a shunt connection at input or output,
the resistance is lowered by (1+βA).
Feedback Amplifier

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How to Optimize Your Website for Search Engines

  • 2. Outline 1. Introduction to Feedback 2. Feedback Amplifier – Positive & Negative 3. Advantages/Disadvantages of Negative Feedback 4. Basic Feedback Concept 5. Classification of Amplifiers 6. Series – Shunt Configuration 7. Shunt – Series Configuration 8. Series - Series Configuration 9. Shunt – Shunt Configuration
  • 3. Introduction to Feedback  Feedback is used in virtually all amplifier system.  Invented in 1928 by Harold Black – engineer in Western Electric Company  methods to stabilize the gain of amplifier for use in telephone repeaters.  In feedback system, a signal that is proportional to the output is fed back to the input and combined with the input signal to produce a desired system response.  However, unintentional and undesired system response may be produced.
  • 4. Feedback Amplifier  Fe e dback is a technique where a proportion of the output of a system (amplifier) is fed back and recombined with input  There are 2 types of feedback amplifier:  Positive feedback  Negative feedback A β input output
  • 5. Positive Feedback  Positive feedback is the process when the output is adde d to the input, amplified again, and this process continues.  Positive feedback is used in the design of oscillator and other application. A β input output +
  • 6. Positive Feedback - Example  In a PA system get feedback when you put the microphone in front of a speaker and the sound gets uncontrollably loud (you have probably heard this unpleasant effect).
  • 7. Negative Feedback  Negative feedback is when the output is subtracted from the input.  The use of negative feedback reduces the gain. Part of the output signal is taken back to the input with a negative sign. A β input output
  • 8. Negative Feedback - Example  Speed control If the car starts to speed up above the desired set-point speed, negative feedback causes the throttle to close, thereby reducing speed; similarly, if the car slows, negative feedback acts to open the throttle
  • 9. Feedback Amplifier - Concept Basic structure of a single - loop feedback amplifier
  • 10. Advantages of Negative Feedback 1. Gain Sensitivity – variations in gain is reduced. 2. Bandwidth Extension – larger than that of basic amplified. 3. Noise Sensitivity – may increase S-N ratio. 4. Reduction of Nonlinear Distortion 5. Control of Impedance Levels – input and output impedances can be increased or decreased.
  • 11. Disadvantages of Negative Feedback 1. Circuit Gain – overall amplifier gain is reduced compared to that of basic amplifier. 2. Stability – possibility that feedback circuit will become unstable and oscillate at high frequencies.
  • 12. Basic Feedback Concept Basic configuration of a feedback amplifier
  • 13. Basic Feedback Concept  The output signal is: where A is the amplification factor  Feedback signal is where ß is the feedback transfer function  At summing node:  Closed-loop transfer function or gain is if εAS=oS oSβ=fbS fbi SS −=εS A A S S i o β+ == 1 fA ββ β 1 1 =≅>> A A then fAA
  • 14. Classification of Amplifiers Classify amplifiers into 4 basic categories based on their input (parameter to be amplified; voltage or current) & output signal relationships:  Voltage amplifier (series-shunt)  Current amplifier (shunt-series)  Transconductance amplifier (series-series)  Transresistance amplifier (shunt-shunt)
  • 15. Feedback Configuration Series: connecting theconnecting the feedback signalfeedback signal in series within series with thethe input signalinput signal voltage.voltage. Shunt: connecting the feedback signal in shunt (parallel) with an input
  • 16. Series - Shunt Configuration vv v vf A A A β+ = 1
  • 17. Series - Shunt Configuration if Lo RR << then the output of feedback network is an open circuit; Output voltage is: εVAV vo = feedback voltage is: ovVVfb β= By neglecting Rs due to ; error voltage is: si RR >> fbi V−=VVε vv v i o vf A A V V A β+ ==∴ 1 where ßv is closed-loop voltage transfer function
  • 18. Series - Shunt Configuration Or  Input current  Rif with feedback Assume Vi=0 and Vx applied to output terminal. Or  Input current  Rof with feedback Input Resistance, Rif Output Resistance, Rof )( εεε β VAVVV vvfbi +=+=V )1( vv i A V V β ε + = )1( vvi i i i AR V R V I β ε + == )1( vvi i i if AR I V R β+== 0=+=+ xvfb VVVV βεε xvVV βε −= o vvx o vx i R AV R VAV I )1( βε + = − = )1( vv o x x of A R I V R β+ ==
  • 19.  Series input connection increase input resistance – avoid loading effects on the input signal source.  Shunt output connection decrease the output resistance - avoid loading effects on the output signal when output load is connected. Equivalent circuit of shunt - series feedback circuit or voltage amplifier Series - Shunt Configuration
  • 20. For ideal non-inverting op- amp amplifier Feedback transfer function; Series - Shunt Configuration  Non-inverting op-amp is an example of the series-shunt configuration.       +== 1 2 1 R R V V A i o vf       + = 1 2 1 1 R R β
  • 21. Series - Shunt Configuration Equivalent circuit )1( / 1 11 1 221 1 21 1 21 1 vi i i i i if v oi v v v v i o vf ofb fbi AR RV V I V R R R VA VV RR R VV A A RR R A A V V A V RR R V VV β β ε ε εε ε ε +===       + +=      + += + =       + + ==       + ≅ −= = V VAV vo
  • 22. Series - Shunt Configuration Example: Calculate the feedback amplifier gain of the circuit below for op-amp gain, A=100,000; R1=200 Ω and R2=1.8 kΩ. Solution: Avf = 9.999 or 10
  • 23. Series - Shunt Configuration  Basic emitter-follower and source-follower circuit are examples of discrete-circuit series-shunt feedback topologies. • vi is the input signal • error signal is base- emitter/gate-source voltage. • feedback voltage = output voltage  feedback transfer function, ß v = 1
  • 24. Series - Shunt Configuration  Small-signal voltage gain:  Open-loop voltage gain:  Closed-loop input resistance:  Output resistance: e E e E Em Em i o vf r R r R Rg r Rg r V V A + =       ++       + == 11 1 1 π π e E Emv r R Rg r A =      += π 1             ++=++= EmEmif Rg r rRrgrR π πππ 1 1)1( Em E m Eof Rg r R rg r RR       ++ = + = π π π 1 1 )1(
  • 25. Shunt – Series Configuration ii i if A A A β+ = 1
  • 26. Shunt – Series Configuration  Basic current amplifier with input resistance, Ri and an open-loop current gain, Ai.  Current IE is the difference between input signal current and feedback current.  Feedback circuit samples the output current – provide feedback signal in shunt with signal current.  Increase in output current – increase feedback current – decrease error current.  Smaller error current – small output current – stabilize output signal.
  • 27. Shunt – Series Configuration if si RR << then the output is a short circuit; output current is: εIAI io = feedback current is: oi II fb β= Input signal current: fbi II += εI ii i i o if A A I I A β+ ==∴ 1 then εIIi ≈ where ßi is closed-loop current transfer function
  • 28. Shunt – Series Configuration Or  Input current  Rif with feedback Input Resistance, Rif )( εεε β IAIII iifbi +=+=I )1( ii i A I I β ε + = )1( ii ii ii A RI RIV β ε + == )1( ii i i i if A R I V R β+ == Assume Ii=0 and Ix applied to output terminal.  Rof with feedback Output Resistance, Rof [ ] oiixx oxiixx oixx xi xifb RAIV RIAIV RIAIV II IIII )1( )( )( 0 β β β β ε ε εε += −−= −= −= =+=+ ( )iio x x of AR I V R β+== 1
  • 29. Shunt - Series Configuration  Shunt input connection decrease input resistance – avoid loading effects on the input signal current source.  Series output connection increase the output resistance - avoid loading effects on the output signal due to load connected to the amplifier output. Equivalent circuit of shunt - series feedback circuit or voltage amplifier
  • 30. Shunt - Series Configuration  Op-amp current amplifier – shunt-series configuration.  Ii’ from equivalent source of Ii and Rs.• Iε is negligible and Rs>>Ri; • assume V1 virtually ground; • Current I1: • Output current: • Ideal current gain: fbii II == 'I FiFfbo RIRI −=−=V 1/ RVo−=1I       +=+= 1 1 1 R R III F ifboI       +== 1 1 R R I A F i i oI
  • 31. Shunt - Series Configuration  Ai is open-loop current gain and  Assume V1 is virtually ground:  I1 current:  Output current fbifbi IIII −≅−= 'εI )( fbii IIA −== εIAI io Ffb RI−=oV Closed-loop current gain:       =−= 11 1 R R I R V F fb o I       +=+= 1 1 R R IIII F fbfbfboI       + + == 1 1 1 R R A A I I F i i i o ifA
  • 32. Shunt - Series Configuration  Common-base circuit is example of discrete shunt-series configuration.  Amplifier gain: Closed-loop current gain: RLIo Ii Iε RLIo Ii Iε Ifb βε == iAI/oI i i i o if A A I I A + = + == 11 β β
  • 33. Shunt - Series Configuration  Common-base circuit with RE and RC RCIoRE Ii V- V+ RCIoRE Ii i E i m E m i o if A R r A rg R r rg I I A +      + = +      + == π π π π 11
  • 34. Series – Series Configuration gg g gf A A A β+ = 1
  • 35. Series – Series Configuration  The feedback samples a portion of the output current and converts it to a voltage – voltage- to-current amplifier.  The circuit consist of a basic amplifier that converts the error voltage to an output current with a gain factor, Ag and that has an input resistance, Ri.  The feedback circuit samples the output current and produces a feedback voltage, Vfb, which is in series with the input voltage, Vi.
  • 36. Series – Series Configuration Assume the output is a short circuit, the output current: εVAI go = feedback voltage is: oz IVfb β= Input signal voltage (neglect Rs=∞ ): fbi VV += εV gz g i o gf A A V I A β+ ==∴ 1 where ßz is a resistance feedback transfer function
  • 37. Series – Series Configuration Assume Ii=0 and Ix applied to output terminal.  Rof with feedback Output Resistance, Rof [ ] ogzxx oxzgxx ogxx xz xzfb RAIV RIAIV RIAIV II IIII )1( )( )( 0 β β β β ε ε εε += −−= −= −= =+=+ ( )gzo x x of AR I V R β+== 1 Or  Input current  Rif with feedback Input Resistance, Rif )( εεε β VAVVV gzfbi +=+=V )1( gz i A V V β ε + = )1( gzi i i i AR V R V I β ε + == )1( gzi i i if AR I V R β+==
  • 38. Series – Series Configuration  Series input connection increase input resistance  Series output connection increase the output resistance Equivalent circuit of series - series feedback
  • 39. Series – Series Configuration  The series output connection samples the output current  feedback voltage is a function of output current.  Assume ideal op-amp circuit and neglect transistor base-current: Ei o gf Eofbi RV I A RIV 1 == ==V
  • 40. Series – Series Configuration  Assume IE≅IC and Ri≈∞ ( ) ( ) Egm gm i o gf Eoigmo Eoifbi gmbm E fb o RArg Arg V I A RIVArgI RIVVVV VArgIrg R V I π π π ε εππ + == −= −=−= === 1
  • 41. Series – Series Configuration
  • 42. Series – Series Configuration Em LC C m i o gf Emfbi Emfb LC C mo Rg r RR R g V I A Rg r VVVV RVg r V V RR R VgI       ++       + − ==               ++=+=       +=       + −= π π ππ π π π π 1 1 1 1 )(
  • 43. Shunt – Shunt Configuration zz z zf A A A β+ = 1
  • 44. Shunt – Shunt Configuration  The feedback samples a portion of the output voltage and converts it to a current – current- to-voltage amplifier.  The circuit consist of a basic amplifier that converts the error current to an output voltage with a gain factor, Az and that has an input resistance, Ri.  The feedback circuit samples the output voltage and produces a feedback current, Ifb, which is in shunt with the input current, Ii.
  • 45. Shunt – Shunt Configuration Assume the output is a open circuit, the output voltage: εIAV zo = feedback voltage is: ogVI fb β= Input signal voltage (neglect Rs=∞ ): fbi II += εI zg z i o zf A A I V A β+ ==∴ 1 where ßg is a conductance feedback transfer function
  • 46. Shunt – Shunt Configuration Or  Input current  Rif with feedback Input Resistance, Rif )( εεε β IAIII zgfbi +=+=I )1( zg i A I I β ε + = )1( zg ii ii A RI RIV β ε + == )1( zg i i i if A R I V R β+ == Assume Vi=0 and Vx applied to output terminal. Or  Input current  Rof with feedback Output Resistance, Rof 0=+=+ xgfb VVVV βεε xgVV βε −= o zgx o zx i R AV R VAV I )1( βε + = − = )1( zg o x x of A R I V R β+ ==
  • 47. Shunt – Shunt Configuration Equivalent circuit of shunt - shunt feedback circuit or voltage amplifier
  • 48. Shunt – Shunt Configuration  Basic inverting op-amp circuit is an example of shunt-shunt configuration.  Input current splits between feedback current and error current.  Shunt output connection samples the output voltage  feedback current is function of output voltage. 2 2 R I V A IIwhere RIV i o zf ifb fbo −== = −=
  • 49. Shunt – Shunt Configuration  Az is open-loop transresistance gain factor (-ve value) ( ) 2 2 1 / R A A I V A RVIwhere IIAIAV z z i o zf ofb fbizzo + − == −= −−== ε
  • 50. Shunt – Shunt Configuration
  • 51. Shunt – Shunt Configuration       −      +      +      +       −− == =      +      −+      +      + − += = − ++ F m FFFC F m i o zf F o i F m FFC o F o i F o m C o R g RRrRR R g I V A R V I R g RrRR V R VV r V I R VV Vg R V 111111 1 0 11111 0 π π π π π π π
  • 52. ( ) ( )       − + ≅=       +      −      +      +       ++ == −=             − = F z z i o zf F C z FFF C F C z i o zf Cm C m z R A A I V A R Rr A RR r R R R Rr A I V A Rrg rR g A 1 1 1 11 11 ππ π π π Shunt – Shunt Configuration  Open-loop transresistance gain factor Az is found by setting RF=∞  Multiply by (rπRC)  Assume RC <<RF & rπ<< RF
  • 53. Feedback Amplifier Input and output Impedances  Summary 1. For a series connection at input or output, the resistance is increased by (1+βA). 2. For a shunt connection at input or output, the resistance is lowered by (1+βA).