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Transcript
EIE 211 Electronic Devices and Circuit Design II
EIE 211 : Electronic Devices and
Circuit Design II
Lecture 9: Two-port Networks &
Feedback
King Mongkut’s University of Technology Thonburi
5/23/2017 1
EIE 211 Electronic Devices and Circuit Design II
Example: Design a 2nd order high pass active filter based on the inductor
replacement
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EIE 211 Electronic Devices and Circuit Design II
Second Order Active Filters based on the Two-Integrator-Loop Topology
To derive the two-integrator loop biquadratic circuit, or biquad, consider the
high-pass transfer function
We observe that the signal (ωo/s)Vhp can be obtained by passing Vhp through an
integrator with a time constant equal to 1/ωo. Furthermore, passing the resulting signal
through another identical integrator results in the signal (ωo2/s2)Vhp. The block diagram
on the next page shows a two-integrator arrangement.
King Mongkut’s University of Technology Thonburi
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EIE 211 Electronic Devices and Circuit Design II
From
It suggests that Vhp can be obtained by using the weighted summer in Fig b.
Now we combine blocks a) and b) together to obtain:
King Mongkut’s University of Technology Thonburi
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EIE 211 Electronic Devices and Circuit Design II
If we try to look at the Fig c. more carefully, we’ll find that
Ks 2
Thp 
 2
Vi
s  s (o / Q)  o2
Vhp
And the signal at the output of the first integrator is –(ωo/s)Vhp, which is a band-pass
function, with the center-frequency gain of –KQ,
Therefore, the signal at the output of the first integrator is labeled Vbp. In, a similar
way, the signal at the output of the second integrator is (ωo2/s2)Vhp, which is a lowpass function,
Thus, the output of the second integrator is labeled Vlp. Note that the dc gain of the
low-pass filter is equal to K. Hence, the 2-integrator-loop biquad realizes 3 basic 2nd
order filtering functions simultaneously, that’s why it’s called a universal active filter.
King Mongkut’s University of Technology Thonburi
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EIE 211 Electronic Devices and Circuit Design II
Circuit Implementation
We replace each integrator with a Miller integrator circuit having CR = 1/ωo and we
replace the summer block with an op amp summing circuit that is capable of assigning
both positive and negative weights to its inputs. The resulting ckt, known as the
Kerwin-Huelsman-Newcomb or KHN biquad.
5/23/2017 6
EIE 211 Electronic Devices and Circuit Design II
We can express the output of the summer Vhp in terms of its inputs, Vbp = –
(ωo/s)Vhp and Vlp = (ωo2/s2)Vhp, as
To determine all the parameters, we need to compare it to the original eq:
We can match them up, term by term, and will get:
King Mongkut’s University of Technology Thonburi
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EIE 211 Electronic Devices and Circuit Design II
The KHN biquad can be used to realize notch and all-pass functions by
summing weighted versions of the three outputs, LP, BP, and HP as shown.
Substitute Thp, Tbp and Tlp that we found previously, we’ll get the overall
transfer function
from which we can see that different transmission zeros can be obtained by
the appropriate selection of the values of the summing resistors. For instance,
a notch is obtained by selection RB = ∞ and
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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)
(Open-Circuit Impedance)
I1
+
V1

+
V1   z11
V    z
 2   21
V2
V1  z11I1  z12 I 2

V2  z21I1  z22 I 2
I2
z22
z11
z12I2 +

+ z21I1

At port 1
z12   I1 
z22   I 2 
At port 2
V
z11  1
I1 I 2  0
Open-circuit forwardz  V2
21
I1 I 2  0
transimpedance
Open-circuit reversez  V1
12
transimpedance
I 2 I1  0
V
Open-circuit
z22  2
I 2 I1  0
output impedance
Open-circuit
input impedance
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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
Two-Port Network (y-parameters)
(Short-Circuit Admittance)
I1
+
V1

I2
1/y11
y12V2
+
1/y22
y21V1
 I1   y11
I    y
 2   21
y12  V1 
y22  V2 
V2
I1  y11V1  y12V2

I 2  y21V1  y22V2
At port 1
At port 2
I
Short-circuit
y11  1
V1 V2  0
input admittance
Short-circuit forwardy  I 2
21
V1 V2  0
transadmittance
Short-circuit reversey  I1
12
transadmittance
V2 V1  0
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)
(hybrid)
I
I
1
+
V

h
1/h
11
1
hV +
12
2

V1   h11 h12   I1 
 I   h
 V 
h
 2   21 22   2 
2
+
22
V
hI
2
V1  h11I1  h12V2
21 1

At port 1
I 2  h21I1  h22V2
At port 2
V
h11  1
I1 V2  0
Short-circuit forwardh  I 2
21
I1 V2  0
current gain
Open-circuit reverseh  V1
12
voltage gain
V2 I1  0
I
Open-circuit
h22  2
V2 I1  0
output admittance
Short-circuit
input impedance
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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
Two-Port Network (g-parameters)
(inverse-hybrid)
I
+
I
1
22
11
V
1
gI
12 2

+
g
1/g
 I1   g11
V    g
 2   21
2
V
2
+ g21V1


g12  V1 
g 22   I 2 
I1  g11V1  g12 I 2
V2  g21V1  g22 I 2
At port 1
At port 2
I
Open-circuit
g11  1
V1 I 2  0
input admittance
Open-circuit forwardg  V2
21
V1 I 2  0
current gain
Short-circuit reverseg  I1
12
current gain
I 2 V1  0
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
z-parameter examples
I1
I1
I2
+
V1
6

+
+


V2
Z11  6 Z 22  6
V1
I2
I1
+
12
3
V2
Z11  12 Z 22  3

12
+
V1

I2
3
+
6
V2

Z11  18 Z 22  9
Z12 
V1
 6
I 2 I1  0
Z12 
V1
 0
I 2 I1  0
Z12 
V1
6I
 2  6
I 2 I1  0 I 2
Z 21 
V2
 6
I1 I 2  0
Z 21 
V2
 0
I1 I 2  0
Z 21 
V2
6I
 1  6
I1 I 2  0 I1
Z   
6 6

6 6 
Z   
12 0

 0 3
Z   
18 6

 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)
19
EIE 211 Electronic Devices and Circuit Design II
y-parameter examples
I1
I2
0.05S
+
V1

I1
0.1S
+
+


V2
I2
0.2S
+
V2
V1

0.025S
1
y11  0.05S y22  0.05S
y12 
I1
 0.05V2

 0.05S
V2 V1  0
V2
y21 
I2
 0.05V1

 0.05S
V1 V2  0
V1
0.05  0.05
 y   

 0.05 0.05 
1
 1

y11  

  0.0692S
 0.1 0.2  0.025 
1
1
 1

y22  

  0.0769S
 0.2 0.1  0.025 
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
0.0692  0.0615

 0.0615 0.0769 
 y   
King Mongkut’s University of Technology Thonburi
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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
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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
Feedback
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Feedback
Xs
Xi
+
Xo
Xf
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
Feedback Amplifier Analysis
Xs
Xi
+
Xo
Xf
βf
X f   f Xo
X o  AX i
where  f is called the feedback factor
where A is the am plifier' s gain, e.g . voltage gain
Xi  X s  X f
where X i is the net input signal to the basic am plifier,
X s  the signal from the source
The am plifier' s gain with feedback is given by
Af 
Xo
AX i


Xs
Xi  X f
1
A

Xf
Xi
A
1
King Mongkut’s University of Technology Thonburi
 f Xo

A
1  f A
A
Xi
5/23/2017 26
EIE 211 Electronic Devices and Circuit Design II
Summary: General Feedback Structure
Source
Vs +
V

-
A
V
Load
A : Open Loop Gain
A = Vo / V
 : feedback factor
 = Vf / Vo
Vf

Vo
A
1 T

 (
)
Vs 1  A  1  T
V  Vs  V f
Close loop gain : Af 
V f    Vo
Loop Gain : T  A  
Amount of feedback : 1  A  
1
Note : Af A 
V  VS    Vo
Vo  A  V

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…
dA
A
dA f 
Af 
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:
As  
AM
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
x f   xtst
xs=0

xi
xi  0  x f
A
xo  Axi   Ax f   Axtst
xf

xtst
xo
King Mongkut’s University of Technology Thonburi
loop gain  
xo
 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)
(1) Series (Voltage)
Shunt (Voltage)
(2) Series (Voltage)
Series (Current)
(3) Shunt (Current)
Shunt (Voltage)
(4) Shunt (Current)
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
Af 
*
A
1  f A
Does change input resistance from Ri to Rif


Rif  Ri 1   f A
*
Does change output resistance from Ro to Rof
Rof 
Equivalent Circuit for Feedback Amplifier
*
Ro
1  f A
Does change low and high frequency 3dB frequencies


 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
AV
AV
AV


Vf
 f Vo 1   f AV
1
1
Vi
Vi
Input Resistance
Vi  V f Vi   f Vo
V
Rif  s 

 Ri 1   f AV
V
Ii
Ii
 i 
 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 





Ro
V
Ro
so Rof  t 
It
1  AV  f

King Mongkut’s University of Technology Thonburi

Vt  AV  V f
Vt  AV  f Vt

Ro
Ro
Vt 1  AV  f

5/23/2017 37

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 

 A
o

 1 L

s

A 

1   f A  

Ao
1   f


1 L

s
L
1   f Ao
then A f   
Ao
s
1
H
where
A fo 
Ao
1   f Ao


 Ao

s
 1
H







A 

1   f A  

Ao
1   f
s

1

H


 Hf   H 1   f Ao






Ao
 L

1  s   f Ao 





Ao


 1   f Ao 



 L  1 
1 
 
 1   f Ao  s 

A fo
  Lf
1 

s

Low 3dB frequency lowered by feedback.
High Frequency Pole
For A  














Ao


s
  f Ao 
1 
 H



Ao

 1   f Ao







s
1 

  H 1   f Ao 



A fo

1  s
  Hf





Upper 3dB frequency raised by feedback.
King Mongkut’s University of Technology Thonburi
5/23/2017 38




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)
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5/23/2017
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
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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
Vf
V
h12  1

f
V2 I 0 Vo
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 '

*
*
Rin  Rif  Rs
Rout  1 /(
1
1
 )
Rof RL
Hf  1   f AV ' H
King Mongkut’s University of Technology Thonburi
 Lf 
L
1   f AV '
5/23/2017 41
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’
5/23/2017 42
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 
AV '
1   f AV '
Rif  Ri '1   f AV '
Modified Basic Amplifier
 Hf  1   f AV ' H
*
Idealized Feedback Network
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.
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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
5/23/2017 45
EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
5/23/2017 46