EIE 211 Electronic Devices and Circuit Design II
ENE/EIE 211: Electronic Devices and
Circuit Design II
Lectures 9 &10 : Two-port Networks &
Feedback
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EIE 211 Electronic Devices and Circuit Design II
Two-Port Network Parameters
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EIE 211 Electronic Devices and Circuit Design II
Characterization of linear, two-port networks
Before we begin a discussion on the topic of oscillators, we need to study
feedback. However, in order to understand how the feedback works, we also
need to first learn the two-port network parameters.
A two-port network has four port variables: V1, I1, V2 and I2. If the two-port
network is linear, we can use two of the variables as excitation variables and the
other two as response variables. For example, the network can be excited by a
voltage V1 at port 1 and a voltage V2 at port 2, and the two current I1 and I2 can
be measured to represent the network response.
There are four parameter sets commonly used in electronics. They are the
admittance (y), the impedance (z), the hybrid (h) and the inverse-hybrid (g)
parameters, respectively.
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EIE 211 Electronic Devices and Circuit Design II
Two-Port Network (z-parameters)
+
V1

(Open-Circuit Impedance)
I1
z11
z12I2 +

z22
V
z11  1
I1 I 2  0
Open-circuit reversez  V1
12
transimpedance
I 2 I1  0
+
V1   z11
V    z
 2   21

V2  z 21 I1  z 22 I 2
V2
+ z21I1

At port 1
Open-circuit
input impedance
I2
z12   I1 
z 22   I 2 
V1  z11 I1  z12 I 2
At port 2
Open-circuit forwardz  V2
21
I1 I 2  0
transimpedance
V
Open-circuit
z 22  2
I 2 I1  0
output impedance
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EIE 211 Electronic Devices and Circuit Design II
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EIE 211 Electronic Devices and Circuit Design II
Two-Port Network (y-parameters)
(Short-Circuit Admittance)
I1
+
V1

I2
1/y11
y12V2
1/y22
I
Short-circuit
y11  1
V1 V2  0
input admittance
Short-circuit reversey  I1
12
transadmittance
V2 V1  0

I 2  y21V1  y22V2
V2
y21V1
At port 1
+
 I1   y11
I    y
 2   21
y12  V1 
y22  V2 
I1  y11V1  y12V2
At port 2
Short-circuit forwardy  I 2
21
V1 V2  0
transadmittance
I
Short-circuit
y22  2
V2 V1  0
output admittance
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EIE 211 Electronic Devices and Circuit Design II
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EIE 211 Electronic Devices and Circuit Design II
Two-Port Network (h-parameters)
+
V1

(hybrid)
I1
h11
h12V2 +

1/h22
V
h11  1
I1 V2  0
Open-circuit reverseh  V1
12
voltage gain
V2 I1  0
+
V1   h11
 I   h
 2   21

I 2  h21 I1  h22V2
V2
h21I1
At port 1
Short-circuit
input impedance
I2
h12   I1 
h22  V2 
V1  h11 I1  h12V2
At port 2
Short-circuit forwardh  I 2
21
I1 V2  0
current gain
I
Open-circuit
h22  2
V2 I1  0
output admittance
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EIE 211 Electronic Devices and Circuit Design II
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Two-Port Network (g-parameters)
EIE 211 Electronic Devices and Circuit Design II
(inverse-hybrid)
I1
+
V1

1/g11
g12I2
g22
I2
V2
+ g21V1

At port 1
I
Open-circuit
g11  1
V1 I 2  0
input admittance
Short-circuit reverseg  I1
12
current gain
I 2 V1  0
+

 I1   g11
V    g
 2   21
g12  V1 
g 22   I 2 
I1  g11V1  g12 I 2
V2  g 21V1  g 22 I 2
At port 2
Open-circuit forwardg  V2
21
V1 I 2  0
current gain
V
Short-circuit
g 22  2
I 2 V1  0
output impedance
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EIE 211 Electronic Devices and Circuit Design II
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EIE 211 Electronic Devices and Circuit Design II
+
I1
I2
V1

6
z-parameter examples
+
V2

Z11  6 Z 22  6
Z12 
Z 21 
V1
 6
I 2 I1  0
V2
 6
I1 I 2  0
Z   
6 6

6 6 
+
I1
V1

I2
12
3
V2
Z11  12 Z 22  3
Z12 
Z 21 
V1
 0
I 2 I1  0
V2
 0
I1 I 2  0
Z   
12 0

 0 3
+

I1
+
12
V1

3
6
Z11  18 Z 22  9
Z12 
Z 21 
+
V2
I2

V1
6I
 2  6
I 2 I1  0 I 2
V2
6I
 1  6
I1 I 2  0 I1
18 6
Z   

6
9


Note: (1) z-matrix in the last circuit = sum of two former z-matrices
(2) z-parameters is normally used in analysis of series-series
circuits
(3) Z12 = Z21 (reciprocal circuit)
(4) Z12 = Z21 and Z11 = Z22 (symmetrical and reciprocal circuit)
12
EIE 211 Electronic Devices and Circuit Design II
I1
+
y-parameter examples
V1
0.05S

y21 
+
V2
y11  0.05S y22  0.05S
y12 
I2

I1
 0.05V2

 0.05S
V2 V1  0
V2
I2
 0.05V1

 0.05S
V1 V2  0
V1
0.05  0.05
 y   

 0.05 0.05 
I1
+
0.1S
V1
 0.025S
0.2S
+
V2
I2

1
 1

y11  

  0.0692S
0
.
1
0
.
2

0
.
025


1
1
 1

y22  

  0.0769S
 0.2 0.1  0.025 
1
y12 
I1
V2 V1  0
But I 2  y22V2  0.0769V2
 I1 I 2  I1

0.1 0.025
 I1  0.8 I 2  0.0615V2
y12  0.0615S
By reciprocal, y21  y12  0.0615S
 y   
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 0.0615

 0.0615 0.0769 
0.0692
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EIE 211 Electronic Devices and Circuit Design II
Example: figure below shows the small-signal equivalent-ckt model of a
transistor. Calculate the values of the h parameters.
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EIE 211 Electronic Devices and Circuit Design II
Summary: Equivalent-Circuit Representation
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EIE 211 Electronic Devices and Circuit Design II
ENE/EIE 211: Electronic Devices and
Circuit Design II
Feedback
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Xs
+
-
Xi
Xf
Feedback
Xo
βf
 What is feedback? Taking a portion of the signal arriving at the load
and feeding it back to the input.
 What is negative feedback? Adding the feedback signal to the input so
as to partially cancel the input signal to the amplifier.
 Doesn’t this reduce the gain? Yes, this is the price we pay for using
feedback.
 Why use feedback? Provides a series of benefits, such as improved
bandwidth, that outweigh the costs in lost gain and increased
complexity in amplifier design.
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EIE 211 Electronic Devices and Circuit Design II
Xs
Feedback Amplifier Analysis
+
Xi
-
Xf
X f   f Xo
X o  AX i
Xo
βf
where  f is called the feedback factor
where A is the amplifier ' s gain , e.g . voltage gain
Xi  Xs  X f
where X i is the net input signal to the basic amplifier ,
X s  the signal from the source
The amplifier ' s gain with feedback is given by
Af 
Xo
AX i


Xs
Xi  X f
1
A

Xf
Xi
1
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A
 f Xo
Xi

A
1  f A
A
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EIE 211 Electronic Devices and Circuit Design II
Summary: General Feedback Structure
Source
Vs +

-
V
Vf
V  Vs  V f
V f   Vo
V  VS   Vo
Vo  A V
A
V
Load
A : Open Loop Gain
A = Vo / V
 : feedback factor
 = Vf / Vo

Close loop gain : Af 
Vo
A
1 T

 (
)
Vs 1  A  1  T
Loop Gain : T  A  
Amount of feedback : 1  A  
1
Note : Af A 

The product Aβ must be positive for the feedback network to be the
negative feedback network.
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EIE 211 Electronic Devices and Circuit Design II
Advantages of Negative Feedback
* Gain desensitivity - less variation in amplifier gain with changes in
β (current gain) of transistors due to dc bias,
temperature, fabrication process variations, etc.
* Bandwidth extension - extends dominant high and low frequency poles to
higher and lower frequencies, respectively.
L
 Hf  1   f A  H
 Lf 
1  f A
* Noise reduction - improves signal-to-noise ratio
* Improves amplifier linearity - reduces distortion in signal due to gain
variations due to transistors
* Impedance Control - control input and output impedances by applying
appropriate feedback topologies




* Cost of these advantages:
 Loss of gain, may require an added gain stage to compensate.
 Added complexity in design
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EIE 211 Electronic Devices and Circuit Design II
Gain Desensitivity
Feedback can be used to desensitize the closed-loop gain to variations in the
basic amplifier. Let’s see how.
Assume β is constant. Taking differentials of the closed-loop gain equation
gives…
A
dA
Af 
dA f 
1  A
1  A 2
Divide by Af
dA 1  A
1 dA


2
Af
1  A A
1  A  A
dA f
This result shows the effects of variations in A on Af is mitigated by the
feedback amount. 1+Aβ is also called the desensitivity amount
We will see through examples that feedback also affects the input and
resistance of the amplifier (increases Ri and decreases Ro by 1+Aβ factor)
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EIE 211 Electronic Devices and Circuit Design II
Bandwidth Extension
We’ve mentioned several times in the past that we can trade gain for
bandwidth. Finally, we see how to do so with feedback… Consider an
amplifier with a high-frequency response characterized by a single pole
and the expression:
Apply negative feedback β and the resulting closed-loop gain is:
AM
As  
1  s H
As 
AM 1  AM  
A f s  

1   As  1  s  H 1  AM  
•Notice that the midband gain reduces by (1+AMβ) while the 3-dB roll-off
frequency increases by (1+AMβ)
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EIE 211 Electronic Devices and Circuit Design II
Finding Loop Gain
Generally, we can find the loop gain with the following steps:
–
–
–
–
Break the feedback loop anywhere (at the output in the ex. below)
Zero out the input signal xs
Apply a test signal to the input of the feedback circuit
Solve for the resulting signal xo at the output
If xo is a voltage signal, xtst is a voltage and measure the open-circuit
voltage
If xo is a current signal, xtst is a current and measure the short-circuit
current
– The negative sign comes from the fact that we are apply negative
feedback
xs=0

xi
xf
x f  xtst
xi  0  x f
A

xo  Axi   Ax f   Axtst
xtst
xo
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xo
loop gain  
 A
xtst
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EIE 211 Electronic Devices and Circuit Design II
Basic Types of Feedback Amplifiers
* There are four types of feedback amplifiers. Why?
 Output sampled can be a current or a voltage
 Quantity fed back to input can be a current or a voltage
 Four possible combinations of the type of output sampling and input
feedback
* One particular type of amplifier, e.g. voltage amplifier, current amplifier,
etc. is used for each one of the four types of feedback amplifiers.
* Feedback factor βf is a different type of quantity, e.g. voltage ratio,
resistance, current ratio or conductance, for each feedback configuration.
* Before analyzing the feedback amplifier’s performance, need to start by
recognizing the type or configuration.
* Terminology used to name types of feedback amplifier, e.g. Series-shunt
 First term refers to nature of feedback connection at the input.
 Second term refers to nature of sampling connection at the output.
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EIE 211 Electronic Devices and Circuit Design II
Basic Feedback Topologies
Depending on the input signal (voltage or current) to be amplified
and form of the output (voltage or current), amplifiers can be
classified into four categories. Depending on the amplifier
category, one of four types of feedback structures should be used.
(Type of Feedback)
(Type of Sensing)
(2) Series (Voltage)
Series (Current)
(1) Series (Voltage)
(3) Shunt (Current)
(4) Shunt (Current)
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Shunt (Voltage)
Shunt (Voltage)
Series (Current)
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EIE 211 Electronic Devices and Circuit Design II
Figure 8.4 The four basic feedback topologies: (a) voltage-mixing voltage-sampling (series–shunt) topology; (b) current-mixing currentsampling (shunt–series) topology; (c) voltage-mixing current-sampling (series–series) topology; (d) current-mixing voltage-sampling
(shunt–shunt) topology.
EIE 211 Electronic Devices and Circuit Design II
Basic Feedback Topologies
Depending on the input signal (voltage or current) to be amplified
and form of the output (voltage or current), amplifiers can be
classified into four categories. Depending on the amplifier
category, one of four types of feedback structures should be used
(series-shunt, series-series, shunt-shunt, or shunt-series)
Voltage amplifier – voltage-controlled voltage source
Requires high input impedance, low output impedance
Use series-shunt feedback (voltage-voltage feedback)
Current amplifier – current-controlled current source
Use shunt-series feedback (current-current feedback)
Transconductance amplifier – voltage-controlled current source
Use series-series feedback (current-voltage feedback)
Transimpedance amplifier – current-controlled voltage source
Use shunt-shunt feedback (voltage-current feedback)
series-shunt
shunt-series
series-series
shunt-shunt
EIE 211 Electronic Devices and Circuit Design II
Series-Shunt Feedback Amplifier - Ideal Case
Basic Amplifier
*
Assumes feedback circuit does not load down the basic
amplifier A, i.e. doesn’t change its characteristics
 Doesn’t change gain A
 Doesn’t change pole frequencies of basic
amplifier A
 Doesn’t change Ri and Ro
*
For the feedback amplifier as a whole, feedback does
change the midband voltage gain from A to Af
Feedback Circuit
Equivalent Circuit for Feedback Amplifier
Af 
A
1  f A


*
Does change input resistance from Ri to Rif
*
Does change output resistance from Ro to Rof
*
Does change low and high frequency 3dB frequencies
Rif  Ri 1   f A
Rof 

Ro
1  f A

 Hf  1   f A  H
 Lf 

L
1  f A

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EIE 211 Electronic Devices and Circuit Design II
Series-Shunt Feedback Amplifier - Ideal Case
Midband Gain
V
A V
AVf  o  V i 
Vs Vi  V f
Input Resistance
AV
AV
AV


Vf
 f Vo 1   f AV
1
1
Vi
Vi

Vs Vi  V f Vi   f Vo
Rif 


 Ri 1   f AV
Ii
Ii
Vi 
 R 
i


Output Resistance
It
Vt
V  AV Vi
It  t
Ro
But Vs  0 so Vi  V f
and V f   f Vo   f Vt so
It 

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
Vt  AV  V f

Ro
Vt 1  AV  f
Ro

  Vt  AV  f Vt 

V
Ro
so Rof  t 
It
1  AV  f
Ro

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EIE 211 Electronic Devices and Circuit Design II
Series-Shunt Feedback Amplifier - Ideal Case
Low Frequency Pole
For A  
1
L
A fo 
where
then A f   
Ao
s
Ao
1   f Ao
 Lf 
High Frequency Pole
For A  
where
then A f   
Ao
s
1
H
A fo 
Ao
1   f Ao


 A 
o 


 1 L 


s 



Ao


 1   f Ao 


A fo
Ao
A 




1   f A  
  L
 
 L  1  1   Lf 
1



A
f
o

 1 
 
Ao  
s
  1   f Ao  s  
s 
1   f





1 L 
s 

Low 3dB frequency lowered by feedback.
L
1   f Ao




A

o 

s 
 1

H 



Ao


 1   f Ao 


A fo
Ao
A 




1   f A  
 
 
 

s
s
1  s 
1
  f Ao  1 




Ao
   H 1   f Ao    Hf 
1   f
  H
s 

1

 H 


 Hf   H 1   f Ao



Upper 3dB frequency raised by feedback.
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EIE 211 Electronic Devices and Circuit Design II
Practical Feedback Networks
*
Vi
Vo
Vf
*
*
* How do we take these
loading effects into account?
*
*
Feedback networks consist of a set of resistors
 Simplest case (only case considered here)
 In general, can include C’s and L’s (not
considered here)
 Transistors sometimes used (gives variable
amount of feedback) (not considered here)
Feedback network needed to create Vf feedback
signal at input (desirable)
Feedback network has parasitic (loading) effects
including:
Feedback network loads down amplifier input
 Adds a finite series resistance
 Part of input signal Vs lost across this series
resistance (undesirable), so Vi reduced
Feedback network loads down amplifier output
 Adds a finite shunt resistance
 Part of output current lost through this shunt
resistance so not all output current delivered to
load RL (undesirable)
King Mongkut’s University of Technology Thonburi
12/8/2016 32
EIE 211 Electronic Devices and Circuit Design II
Equivalent Network for Feedback Network
*
*
*
*
*
*
*
*
*
Need to find an equivalent network for the
feedback network including feedback effect
and loading effects.
Feedback network is a two port network
(input and output ports)
Can represent with h-parameter network
(This is the best for this particular feedback
amplifier configuration)
h-parameter equivalent network has FOUR
parameters
h-parameters relate input and output
currents and voltages
Two parameters chosen as independent
variables. For h-parameter network, these
are input current I1 and output voltage V2
Two equations relate other two quantities
(output current I2 and input voltage V1) to
these independent variables
Knowing I1 and V2, can calculate I2 and V1 if
you know the h-parameter values
h-parameters can have units of ohms, 1/ohms
or no units (depends on which parameter)
King Mongkut’s University of Technology Thonburi
12/8/2016 33
EIE 211 Electronic Devices and Circuit Design II
Series-Shunt Feedback Amplifier - Practical Case
*
*
*
Feedback network consists of a set of resistors
These resistors have loading effects on the basic
amplifier, i.e they change its characteristics, such as
the gain
Can use h-parameter equivalent circuit for feedback
network
 Feedback factor βf given by h12 since
*
Rin  Rif  Rs
Rout  1 /(
1
1
 )
Rof RL
1
Feedforward factor given by h21 (neglected)
 h22 gives feedback network loading on output
 h11 gives feedback network loading on input
Can incorporate loading effects in a modified basic
amplifier. Basic gain of amplifier AV becomes a new,
modified gain AV’ (incorporates loading effects).
Can then use feedback analysis from the ideal case.
AV '
Ro
AVf 
Rif  Ri 1   f AV ' Rof 
1  AV '  f 
1   f AV '

*
Vf
V
h12  1

f
V2 I 0 Vo
 Hf  1   f AV ' H
King Mongkut’s University of Technology Thonburi
 Lf 
L
1   f AV '
12/8/2016 34
EIE 211 Electronic Devices and Circuit Design II
Series-Shunt Feedback Amplifier - Practical Case
Summary of Feedback Network Analysis *
*
*
*
How do we determine the h-parameters
for the feedback network?
For the input loading term h11
 Turn off the feedback signal by
setting Vo = 0.
 Then evaluate the resistance seen
looking into port 1 of the feedback
network (also called R11 here).
For the output loading term h22
 Open circuit the connection to the
input so I1 = 0.
 Find the resistance seen looking into
port 2 of the feedback network (also
called R22 here).
To obtain the feedback factor βf (also
called h12 )
 Apply a test signal Vo’ to port 2 of
the feedback network and evaluate
the feedback voltage Vf (also called
V1 here) for I1 = 0.
 Find βf from βf = Vf/Vo’
12/8/2016 35
EIE 211 Electronic Devices and Circuit Design II
Summary of Approach to Analysis
Basic Amplifier
Practical Feedback Network
*
*
Evaluate modified basic amplifier
(including loading effects of feedback network)
 Including h11 at input
 Including h22 at output
 Including loading effects of source resistance
 Including load effects of load resistance
Analyze effects of idealized feedback network
using feedback amplifier equations derived
AVf 
Rif  Ri '1   f AV '
Modified Basic Amplifier
Idealized Feedback Network
AV '
1   f AV '
 Hf  1   f AV ' H
*
Rof 
Ro '
1  AV '  f 
 Lf 
L
1   f AV '
Note
 Av’ is the modified voltage gain including the
effects of h11 , h22 , RS and RL.
 Ri’, Ro’ are the modified input and output
resistances including the effects of h11 , h22 ,
RS and RL.
King Mongkut’s University of Technology Thonburi
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EIE 211 Electronic Devices and Circuit Design II
Example: Find expression for A, β, the closed-loop gain Vo/Vs, the input resistance
Rin, and the output resistance Rout. Given μ = 104, Rid =100 kΩ, Ro = 1 kΩ, RL = 2 kΩ,
R1 = 1 kΩ, R2 = 1MΩ and Rs = 10 kΩ.
King Mongkut’s University of Technology Thonburi
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EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
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EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
12/8/2016 39
EIE 211 Electronic Devices and Circuit Design II
Voltage feedback
to input
Series-Series Feedback Amplifier - Ideal Case
*
*
Output current
sampling
*
*
*
Feedback circuit does not load down the basic amplifier
A, i.e. doesn’t change its characteristics
 Doesn’t change gain A
 Doesn’t change pole frequencies of basic
amplifier A
 Doesn’t change Ri and Ro
For this configuration, the appropriate gain is the
TRANSCONDUCTANCE GAIN A = ACo = Io/Vi
For the feedback amplifier as a whole, feedback changes
midband transconductance gain from ACo to ACfo
ACo
ACfo 
1   f ACo


Feedback changes input resistance from Ri to Rif
Rif  Ri 1   f ACo


*
Feedback changes output resistance from Ro to Rof
*
Feedback changes low and high frequency 3dB frequencies

Rof  Ro 1   f ACo

 Hf  1   f ACo  H
King Mongkut’s University of Technology Thonburi
 Lf 
L
1   f ACo


12/8/2016 40
EIE 211 Electronic Devices and Circuit Design II
Series-Series Feedback Amplifier - Ideal Case
Gain (Transconductance Gain)
ACfo 
Io
A V
A
ACo
ACo
 Co i  Co 

Vf
 f I o 1   f ACo
Vs Vi  V f
1
1
Vi
Vi
Input Resistance
Rif 

Vs Vi  V f Vi   f I o


 Ri 1   f ACo
Ii
Ii
Vi 
 R
i

Output Resistance
R of 

V
It
But V s  0 so V i  V f
The series–series feedback amplifier: (a) ideal structure and
(b) equivalent circuit.
+
V
-
and V f   f I o   f I t so V i    f I t
V  I t  ACo V i R o  I t  ACo   f I t R o
 I t 1   f ACo R o
so R of 
V
 R o 1   f ACo 
It
12/8/2016 41
EIE 211 Electronic Devices and Circuit Design II
Equivalent Network for Feedback Network
*
*
Feedback network is a two port network
(input and output ports)
Can represent with Z-parameter network
(This is the best for this feedback amplifier
configuration)
Z-parameter equivalent network has FOUR
parameters
Z-parameters relate input and output
currents and voltages
Two parameters chosen as independent
variables. For Z-parameter network, these
are input and output currents I1 and I2
Two equations relate other two quantities
(input and output voltages V1 and V2) to
these independent variables
Knowing I1 and I2, can calculate V1 and V2 if
you know the Z-parameter values
Z-parameters have units of ohms !
King Mongkut’s University of Technology Thonburi
12/8/2016 42
*
*
*
*
*
*
EIE 211 Electronic Devices and Circuit Design II
Series-Series Feedback Amplifier - Practical Case
*
*
*
Rin  Rif  R s and Rout  Rof  RL
*
*
ACfo 
Feedback network consists of a set of resistors
These resistors have loading effects on the basic
amplifier, i.e they change its characteristics, such as the
gain
Can use z-parameter equivalent circuit for feedback
network
 Feedback factor βf given by z12 since
Vf
V1
z12 

f
I 2 I 0 I o
Feedforward factor given by z21 (neglected)
 z22 gives feedback network loading on output
 z11 gives feedback network loading on input
Can incorporate loading effects in a modified basic
amplifier. Gain ACo becomes a new, modified gain ACo’.
Can then use analysis from ideal case
1

ACo '
1   f ACo '
Rif  Ri '1   f ACo '
 Hf  1   f ACo ' H
King Mongkut’s University of Technology Thonburi
 Lf 
L
1   f ACo '
Rof  Ro ' 1   f ACo '
12/8/2016 43
EIE 211 Electronic Devices and Circuit Design II
Series-Series Feedback Amplifier - Practical Case
*
*
*
*
How do we determine the z-parameters
for the feedback network?
For the input loading term z11
 We turn off the feedback signal by
setting Io = 0 (I2 = 0 ).
 We then evaluate the resistance seen
looking into port 1 of the feedback
network (R11 =z11).
For the output loading term z22
 We open circuit the connection to the
input so I1 = 0.
 We find the resistance seen looking
into port 2 of the feedback network
(R22 =z22).
To obtain the feedback factor βf (also
called z12 )
 We apply a test signal Io’ to port 2 of
the feedback network and evaluate
the feedback voltage Vf (also called V1
here) for I1 = 0.
 Find βf from βf = Vf/Io’
King Mongkut’s University of Technology Thonburi
12/8/2016 44
EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
12/8/2016 45
EIE 211 Electronic Devices and Circuit Design II
Series-Series Feedback Amplifier - Practical Case
Original Amplifier
*
Feedback Network
*
*
Modified Amplifier
*
Idealized Feedback Network
Modified basic amplifier (including loading effects
of feedback network)
 Including z11 at input
 Including z22 at output
 Including loading effects of source resistance
 Including load effects of load resistance
Now have an idealized feedback network, i.e.
produces feedback effect, but without loading
effects
Can now use feedback amplifier equations derived
ACfo 
ACo '
1   f ACo '
Rif  Ri '1   f ACo '
 Hf  1   f ACo ' H
 Lf 
L
1   f ACo '
Rof  Ro ' 1   f ACo '
Note
 ACo’ is the modified transconductance gain
including the loading effects of z11 , z22 , RS and
RL.
 Ri’ and Ro’ are modified input and output
resistances including loading effects.
12/8/2016 46
EIE 211 Electronic Devices and Circuit Design II
Example: A Feedback Triple
*
*
*
*
*
Three stage amplifier
Each stage a CE amplifier
Transistor parameters
Given: β1= β2 = β3 =100,
rx1=rx2=rx3=0
Coupled by capacitors, dc
biased separately
DC analysis (given):
I
IC1  0.60 mA, gm1  C1  23 mA/ V ,
VT
r1 
1
gm1
 4.3K
I
IC2  1.0 mA, gm2  C2  39 mA/ V ,
VT
Note: Biasing resistors for each stage are
not shown for simplicity in the analysis.
King Mongkut’s University of Technology Thonburi
r 2 
2
gm2
 2.6K
I
IC3  4.0 mA, gm3  C3  156 mA/ V ,
VT
r 3 
3
gm3
 0.64K
12/8/2016 47
EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
12/8/2016 48
EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
12/8/2016 49
EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
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EIE 211 Electronic Devices and Circuit Design II
Shunt-Shunt Feedback Amplifier - Ideal Case
*
*
*
*
 Hf  1   f ARo  H
Feedback circuit does not load down the basic
amplifier A, i.e. doesn’t change its characteristics
 Doesn’t change gain A
 Doesn’t change pole frequencies of basic
amplifier A
 Doesn’t change Ri and Ro
For this configuration, the appropriate gain is the
TRANSRESISTANCE GAIN A = ARo = Vo/Ii
For the feedback amplifier as a whole, feedback
changes midband transresistance gain from ARo to
ARfo
ARo
ARfo 
1   f ARo
Feedback changes input resistance from Ri to Rif
1   f ARo 
 L * Feedback changes output resistance
from Ro to Rof
 Lf 
Ro
1   f ARo 
Rof 
*
Rif 
Ri
1   f ARo 
Feedback changes low and high frequency 3dB
frequencies
12/8/2016 51
EIE 211 Electronic Devices and Circuit Design II
Shunt-Shunt Feedback Amplifier - Ideal Case
Gain
ARfo 
Vo
A I
A
ARo
ARo
 Ro i  Ro 

If
 f Vo 1   f ARo
I s Ii  I f
1
1
Ii
Ii
Input Resistance
Rif 

Is = 0
Io’
Vs
Vs
Vs


I s Ii  I f
I i   f Vo
Vs

V 
I i 1   f o 
Ii 


1   f ARo 
Ri
Output Resistance
_
+
R of 
Vo’
V o ' I o ' R o  A Ro I i
I

 R o  A Ro i
Io '
Io '
Io '
But I s  0 so I i   I f
and I f   f V o ' so I i    f V o '
  f Vo '
Ii

   f R of
Io '
Io '

R of  R o  A Ro   f R of
so R of 
King Mongkut’s University of Technology Thonburi
1   f ARo 
Ro

12/8/2016 52
EIE 211 Electronic Devices and Circuit Design II
Equivalent Network for Feedback Network
*
*
*
*
*
*
*
*
King Mongkut’s University of Technology Thonburi
Feedback network is a two port network
(input and output ports)
Can represent with Y-parameter network
(This is the best for this feedback amplifier
configuration)
Y-parameter equivalent network has FOUR
parameters
Y-parameters relate input and output
currents and voltages
Two parameters chosen as independent
variables. For Y-parameter network, these
are input and output voltages V1 and V2
Two equations relate other two quantities
(input and output currents I1 and I2) to
these independent variables
Knowing V1 and V2, can calculate I1 and I2 if
you know the Y-parameter values
Y-parameters have units of conductance
(1/ohms=siemens) !
12/8/2016 53
I1
EIE 211 Electronic Devices and Circuit Design II
Shunt-Shunt Feedback Amplifier - Practical Case
* Feedback network consists of a set of resistors
* These resistors have loading effects on the basic
amplifier, i.e they change its characteristics, such
as the gain
* Can use y-parameter equivalent circuit for
feedback network
 Feedback factor βf given by y12 since
If
I1
I2
y 


V1 y11
ARfo 
y21V1
y12V2
ARo '
1   f ARo '
1   f ARo '
 Hf  1   f ARo ' H
Rif 
Ri
12
Rof 

y22
Ro
1   f ARo '
 Lf 
V2

*
L
1   f ARo '


1
Vo
f
Feedforward factor given by y21 (neglected)
 y22 gives feedback network loading on output
 y11 gives feedback network loading on input
Can incorporate loading effects in a modified
basic amplifier. Gain ARo becomes a new,
modified gain ARo’.
Can then use analysis from ideal case
1
1
1
1
Rin  1 /(  )
Rout  1 /(
 )
Rif Rs
Rof RL

*
V2 V 0
12/8/2016 54
EIE 211 Electronic Devices and Circuit Design II
Shunt-Shunt Feedback Amplifier - Practical Case
*
*
*
*
I2
I1
V1
y11
y12V2
y21V1
y22
V2
King Mongkut’s University of Technology Thonburi
How do we determine the y-parameters
for the feedback network?
For the input loading term y11
 We turn off the feedback signal by
setting Vo = 0 (V2 =0).
 We then evaluate the resistance
seen looking into port 1 of the
feedback network (R11 = y11).
For the output loading term y22
 We short circuit the connection to
the input so V1 = 0.
 We find the resistance seen looking
into port 2 of the feedback
network.
To obtain the feedback factor βf (also
called y12 )
 We apply a test signal Vo’ to port 2
of the feedback network and
evaluate the feedback current If
(also called I1 here) for V1 = 0.
 Find βf from βf = If/Vo’
12/8/2016 55
EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
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EIE 211 Electronic Devices and Circuit Design II
Example: Analyze the circuit below to determine the small-signal voltage gain Vo/Vs, Rin and
Rout. The transistor has β = 100
King Mongkut’s University of Technology Thonburi
12/8/2016 57
EIE 211 Electronic Devices and Circuit Design II
*
*
*
*
Single stage CE amplifier
Transistor parameters. Given:  =100, rx= 0
No coupling or emitter bypass capacitors
DC analysis:
VBE ,active  0.7V
0.7V
 0.07 mA  70A I 47 K  I B  0.07 mA
10K
I C  I B I 4.7 K  I C  ( I B  0.07 mA)    1I B  0.07 mA
I10 K 
12V  I 4.7 K 4.7 K  I B  0.07 mA47 K  0.7V
12V    1I B  0.07 mA4.7 K  I B  0.07 mA47 K  0.7V
12V  0.33V  3.3V  1014.7 K   47 K I B
8.37V
 0.016 mA  16 A I C  I B  1000.016 mA  1.6 mA
522K
1.6 mA
IC

100
gm 

 63 mA / V r 

 1.6 K
VT 0.0256 V
g m 63 mA / V
IB 
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EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
12/8/2016 59
EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
12/8/2016 60
EIE 211 Electronic Devices and Circuit Design II
Shunt-Series Feedback Amplifier - Ideal Case
* Feedback circuit does not load down the basic
amplifier A, i.e. doesn’t change its characteristics
Current sampling
 Doesn’t change gain A
 Doesn’t change pole frequencies of basic
amplifier A
 Doesn’t change Ri and Ro
* For this configuration, the appropriate gain is the
CURRENT GAIN A = AIo = Io/Ii
* For the feedback amplifier as a whole, feedback
changes midband current gain from AIo to AIfo
Current feedback
*


 Hf  1   f AIo  H
 Lf 
L
*
1   f AIo


*
AIfo 
AIo
1   f AIo
Feedback changes input resistance from Ri to Rif
Rif 
1   f AIo 
Ri


Feedback changes output resistance from Ro to Rof
Rof  Ro 1   f AIo
Feedback changes low and high frequency 3dB
frequencies
12/8/2016 61
EIE 211 Electronic Devices and Circuit Design II
Shunt-Series Feedback Amplifier - Ideal Case
Gain
AIfo 
Io
A I
 Io i 
I s Ii  I f
Input Resistance
Rif 

Is=0
AIo
AIo
AIo


If
 f I o 1   f AIo
1
1
Ii
Ii
Vs
Vs
Vs


I s Ii  I f
Ii   f Io
Vs

Io 


I i 1   f
I i 


Output Resistance
Io’
Vo ’
1   f AIo 
Ri

V o ' R o I o ' A Io I i 
Ii 


R of 

 R o  1  A Io
Io '
Io '
I o ' 

But I s  0 so I i   I f
and I f   f I o ' so I i    f I o '
  f Io '
Ii

  f
Io '
Io '

R of  R o 1   f A Io
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
12/8/2016 62
EIE 211 Electronic Devices and Circuit Design II
Equivalent Network for Feedback Network
*
*
*
*
*
*
*
*
King Mongkut’s University of Technology Thonburi
Feedback network is a two port network
(input and output ports)
Can represent with g-parameter network
(This is the best for this feedback amplifier
configuration)
G-parameter equivalent network has FOUR
parameters
G-parameters relate input and output
currents and voltages
Two parameters chosen as independent
variables. For G-parameter network, these
are input voltages V1 and the output
current I2
Two equations relate other two quantities
(input current I1 and output V2) to these
independent variables
Knowing V1 and I2, can calculate I1 and V2 if
you know the G-parameter values
G-parameters have various units of ohms,
conductance (1/ohms=siemens) and no
units !
12/8/2016 63
EIE 211 Electronic Devices and Circuit Design II
Shunt-Series Feedback Amplifier - Practical Case
*
*
*
I1
g22
V1 g11
g12I2
 Hf  1   f AIo 'H
1
1
Rin  1 /(  )
Rif Rs
I2
Feedback network consists of a set of resistors
These resistors have loading effects on the basic
amplifier, i.e they change its characteristics, such
as the gain
Can use g-parameter equivalent circuit for
feedback network
 Feedback factor βf given by g12 since
If
I1
g12 

f
I 2 V 0 I o
1
Feedforward factor given by g21 (neglected)
g21V1
 g22 gives feedback network loading on output
 g11 gives feedback network loading on input
* Can incorporate loading effects in a modified
L
basic amplifier. Gain AIo becomes a new,
 Lf 
1   f AIo ' modified gain AIo’.
use analysis from ideal case
Rout  Rof  RL * CanAthen
Ri
Io '
AIfo 
Rif 
Rof  Ro 1   f AIo '
1   f AIo '
1   f AIo '
V2

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*
*
*
I1
V1
g22
g11
g12I2
g21V1
I2
V2
*
How do we determine the gparameters for the feedback network?
For the input loading term g11
 We turn off the feedback signal by
setting Io = I2 = 0.
 We then evaluate the resistance
seen looking into port 1 of the
feedback network.
For the output loading term g22
 We short circuit the connection to
the input so V1 = 0.
 We find the resistance seen
looking into port 2 of the feedback
network.
To obtain the feedback factor βf (also
called g12 )
 We apply a test signal Io’ to port 2
of the feedback network and
evaluate the feedback current If
(also called I1 here) for V1 = 0.
 Find βf from βf = If/Io’
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Example: From this ckt, find Iout/Iin, Rin and Rout, assuming β = 100, VA = 75 V
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Xs
Xi
Summary of Feedback Amplifier Analysis
Xf
Xo
f
• Calculate low frequency poles and zeroes.
• Determine dominant (highest)
low frequency pole L including loading
effects of feedback network
• Calculate new dominant low frequency
L
pole Lf .
 Lf 
1   f Ao 
• Calculate high frequency poles and zeroes.
• Determine dominant (lowest)
high frequency pole H including loading
effects of feedback network
• Calculate new dominant high frequency
pole Hf .
 Hf  1   f Ao  H
*
*
*
*
*
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Identify the amplifier configuration by:
 Output sampling
 Io = series configuration
 Vo = shunt configuration
 Feedback to input
 Io = shunt configuration
 Vo = series configuration
Calculate loading effects of feedback
network
 On input
 On output
Calculate appropriate midband gain A’
(modified by loading effects of feedback
network)
Calculate feedback factor f.
Calculate midband gain with feedback Af.
A fo 
Ao
1   f Ao
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Summary of Feedback Amplifier Analysis
EIE 211 Electronic Devices and Circuit Design II
Feedback Amplifier Stability
A(ω)
*
Ao
Afo
Feedback analysis

Midband gain with feedback A fo 

New low and high 3dB frequencies
 Hf  1   f Ao  H  Lf 

 Lf  L
 H  Hf
i  *

A f  i  

1    i A i 
A  i 
11

Rif 
1   f AIo '
Ri


Rof  Ro 1   f AIo '


  A     A  e j  
  i A i     i A i  e j  i   e  j  1

L
1   f Ao
Modified input and output resistances, e.g.
Define Loop Gain as βfA
sin ce
A  i 

1   f Ao
Amplifier’s frequency characteristics
A( s )  AM FH ( s ) FL ( s )
* Feedback amplifier’s gain
A (s)
A ()
Af (s) 
 Af () 
1   f (s) A (s)
1   f () A ()
Amplifier becomes unstable (oscillates) if
at some frequency ωi we have
  i A i   1
 Im  i A i 
0
*
and   i   tan 1 
  180   radians
 Re  i A i 
so

Ao
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Magnitude    A 
Phase
 Im  A 

 Re  A 
    tan 1 
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Transfer Function of the Feedback Amplifier
The closed-loop transfer function is given by
Af (s) 
A (s)
1  f (s) A (s)
 Af ( j) 
A ( j)
1  f ( j) A ( j)
The loop gain L(jω) is a complex number that can be represented by its magnitude and
phase,
L( j )    j  A j     j  A j  e j  
It is the manner in which the loop gain varies with frequency that determines the
stability or instability of the feedback amplifier.
 The freq at which φ(ω) becomes 180o, or ω180, the loop gain A(jω)β(jω) will be a
negative real number. The feedback becomes positive. Thus, |1+ A(jω)β(jω)| < 1,
which makes |Af(jω)| > |A(jω)|. Nevertheless, the feedback is stable.
 If at ω180, |A(jω)β(jω)| = 1, then |1+ A(jω)β(jω)| = 0. The amplifier will have an
output for zero input; this is by definition as oscillator. The amplifier output produces
sustained oscillation.
 If at ω180, |A(jω)β(jω)| > 1, then the oscillation will grow in amplitude until some
nonlinearity reduces the magnitude of the loop gain to exactly unity, at which point
sustained oscillation will be obtained.
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Reviews of Bode Plots (from your past life)
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The Nyquist Plot
The Nyquist plot is a formalized approach for testing for stability based on the
discussion previously. It is simply a polar plot of loop gain with frequency used as a
parameter. Note that the radial distance is |Aβ| and the angle is the phase angle φ.
 The solid line is for positive freq.
 Since the loop gain has a
magnitude that is an even
function of freq and a phase that
is an odd function of freq, the Aβ
plot for negative freq is a mirror
image through the Re axis.
 If the Nyquist plot encircles the
point (-1,0), then the amplifier
will be unstable. (This is
sometimes referred to as the
Nyquist Criterion.)
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The effect of feedback on amplifier poles
1. Pole location and stability
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2. Poles of the feedback amplifier
The feedback amplifier poles are obtained by solving the characteristic equation of
the feedback loop:
1  A( s)  ( s )  0
3. Amplifier with a single-pole response
Let the open loop gain of an amplifier be given by
The closed loop gain will be given by
Thus, the feedback moves the pole along the negative real axis to a freq ωpf:
From the Bode Plot, we see that at low freq, the difference between |A| and |Af| is
20log(1 + Aoβ), but the two curves coincide at high freq. For
,
Practically speaking, at such high freq, the loop gain is much smaller than unity and
the feedback is ineffective.
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 Applying a negative feedback results in extending its bandwidth at the expense of
reduction in gain.
 Since the pole of the closed-loop amplifier never enters the right half of the splane, the single pole amplifier is stable for any value of β. Thus, this amplifier is
said to be unconditionally stable.
 This is because the phase-lag associated with a single-pole response can never be
greater than 90o. Thus the loop gain never achieves the 180o phase shift required
for the feedback to become positive.
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βfA(dB)
Feedback Amplifier Instability
1
0 dB
  
00
- 450
- 900
- 1350
- 1800
*
General form of Magnitude plot of βfA
General form of Phase plot of βfA
180
Amplifier with one pole
 Phase shift of - 900 is the maximum
A  
1 j


*

 HP1
Cannot get - 1800 phase shift.
 No instability problem
Amplifier with two poles
 Phase shift of - 1800 the maximum
Ao
A  

 
 
1  j
1  j




 HP1 
 HP 2 

 Can get - 1800 phase shift !
For stability analysis, we define two important
frequencies
 ω1 is where magnitude of βfA goes to
unity (0 dB)
 ω180 is where phase of βfA goes to -180o
 Instability problem if ω1 = ω180

*
Ao
A f 1  180  
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A 1 
1   1 A1 

A 1 
11

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Gain and Phase Margins
βfA(dB)
0 dB
  
- 30
*
Gain
margin
00
- 450
- 900
- 1350
- 1800
Phase
margin
1
1
*
180

180
Φ(ω1)
Φ(ω180)
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Gain and phase margins
measure how far amplifier is
from the instability condition
Phase margin
Phase margin   1    180 
Example
Phase margin   1    180 
 1350  (1800 )  450
* Gain margin
Gain margin  A1   A180 
 Example
Gain margin  A1   A180 
*
 0 dB  (30 dB )  30 dB
What are adequate margins?


Phase margin = 500 (min)
Gain margin = 10 dB (min)
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References
Microelectronic Circuits by Adel S. Sedra & Kenneth C. Smith. Saunders
College Publishing
Microelectronic Circuit Design by Richard C. Jaeger. The McGraw-Hill
Companies, Inc. 2011
Microelectronics Circuit Analysis and Design by Donald Neamen, The
McGraw-Hill Companies, Inc. 2010
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