Transcript PowerPoint

Chapter 8 - Feedback
1 - Desensitize The Gain
2 - Reduce Nonlinear Distortions
3 - Reduce The Effect of Noise
4 – Control The Input And Output Impedances
5 – Extend The Bandwidth Of The Amplifier
The General Feedback Structure
Basic structure of a feedback amplifier. To make it general, the figure shows signal flow as opposed to
voltages or currents (i.e., signals can be either current or voltage).
Source
xs

xi
A
xf
b
xo
Load
The open-loop amplifier has gain A  xo = A*xi
Output is fed back through a feedback network which produces a sample (xf) of the output (xo)  xf = bxo
Where b is called the feedback factor
The input to the amplifier is xi = xs – xf (the subtraction makes feedback negative)
Implicit to the above analysis is that neither the feedback block nor the load affect the amplifier’s
gain (A). This not generally true and so we will later see how to deal with it.
The overall gain (closed-loop gain) can be solved to be:
xo
A
Af 

xs 1  Ab
Ab is called the loop gain
1+Ab is called the “amount of feedback”
The General Feedback Structure (Intro to Simulink)
A
xo
A  xi
xi
xs  xf
xf
b  xo
b
Af
xs
xi
xf
xo
A
xs
1  Ab
xo
b
A b
1  A b
feedabck factor
loop gain
amount of feedabck
xf
Ab
xs
1 Ab
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  b xtst
xs=0

xi
xi  0  x f
A
xo  Axi   Ax f   bAxtst
xf
loop gain  
b
xtst
xo
xo
 bA
xtst
Negative Feedback Properties
Negative feedback takes a sample of the output signal and applies it to
the input to get several desirable properties. In amplifiers, negative
feedback can be applied to get the following properties
–Desensitized gain – gain less sensitive to circuit component
variations
–Reduce nonlinear distortion – output proportional to input (constant
gain independent of signal level)
–Reduce effect of noise
–Control input and output impedances – by applying appropriate
feedback topologies
–Extend bandwidth of amplifier
These properties can be achieved by trading off gain
Gain Desensitivity
Feedback can be used to desensitize the closed-loop gain to variations in the
basic amplifier. Let’s see how.
Assume beta is constant. Taking differentials of the closed-loop gain equation
gives…
Af 
A
1  Ab
dA f 
Divide by Af
dA f
Af

dA
1  Ab 2
dA 1  Ab
1 dA

1  Ab A
1  Ab 2 A
This result shows the effects of variations in A on Af is mitigated by the
feedback amount. 1+Abeta 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+Abeta
factor)
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 beta and the resulting closed-loop gain is:
As  
A f s  
AM
1  s H
As 
AM 1  AM b 

1  bAs  1  s  H 1  AM b 
•Notice that the midband gain reduces by (1+AMbeta) while the 3-dB roll-off
frequency increases by (1+AMbeta)
Basic Feedback Topologies
series-shunt
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, seriesseries, 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)
shunt-series
series-series
shunt-shunt
Examples of the Four Types of Amplifiers
RD
RD
vOUT
vIN
vOUT
vIN
Vb
Vb
iIN
Voltage
Amp
•
iOUT
iOUT
iIN
Transimpedance
Amp
Transconductance
Amp
Current
Amp
Shown above are simple examples of the four types of amplifiers. Often, these
amplifiers alone do not have good performance (high output impedance, low
gain, etc.) and are augmented by additional amplifier stages (see below) or
different configurations (e.g., cascoding).
iOUT
RD
RD
vOUT
vIN
Vb
RD
RD
Vb
vOUT v
IN
iIN
lower Zout
iOUT
iIN
lower Zout
higher gain
higher gain
Series-Shunt Feedback Amplifier
(Voltage-Voltage Feedback)
Samples the output voltage and returns a feedback
voltage signal
Ideal feedback network has infinite input
impedance and zero output resistance
Find the closed-loop gain and input resistance
The output resistance can be found by
applying a test voltage to the output
So, increases input resistance and reduces output
resistance  makes amplifier closer to ideal
VCVS
V f  bVo
Vi  Vs  V f
Vo  AVs  bVo 
Af 
Vo
A

Vs 1  bA
Rif 
Vs
V
V
V  bAVi
 s  Ri s  Ri i
 Ri 1  Ab 
I i Vi Ri
Vi
Vi
Rof 
Ro
1  bA
The Series-Shunt Feedback Amplifier
The Ideal Situation
The series-shunt
feedback
amplifier:
(a) ideal structure;
(b) equivalent circuit.
Z of ( s )
Af
Rif
Vo
A
Vs
1  Ab
Vs
Vs
Ii
Vi
Ri
Vs
Vi
Ri
Ri
Rif
Ri  1  A  b 
Zif ( s )
Zi( s )   1  A ( s )  b ( s ) 
Vi  b  A  Vi
Vi
Z o( s )
1  A(s ) b (s )
Series-Series Feedback Amplifier
(Current-Voltage Feedback)
For a transconductance amplifier (voltage input, current
output), we must apply the appropriate feedback circuit
Sense the output current and feedback a voltage
signal. So, the feedback current is a
transimpedance block that converts the current
signal into a voltage.
To solve for the loop gain:
Break the feedback, short out the break in the current
sense and applying a test current
To solve for Rif and Rof
Apply a test voltage Vtst across O and O’
I
A
Af  o 
Vs 1  Ab
Rif 
I
A  o (also called Gm )
Vi
I
Loop Gain  Ab   out  Gm R f
I tst
Vs Vi  V f Ri I i  bI o Ri I i  AbVi



 Ri 1  Ab 
Ii
Ii
Ii
Ii
Rof 
Vtst I tst  AVi Ro I tst  AbI tst Ro


 1  Ab Ro
I tst
I tst
I tst
Iout
Gm
ZL
Itst
Vf
RF
Shunt-Shunt Feedback Amplifier
(Voltage-Current Feedback
• When voltage-current FB is applied to a
transimpedance amplifier, output voltage is
sensed and current is subtracted from the
input
– The gain stage has some resistance
– The feedback stage is a transconductor
– Input and output resistances (Rif and
Rof) follow the same form as before
based on values for A and beta
A
Vo
Ii
Vo
 bVo
A
V
A
Af  o 
I s 1  Ab
I s  Ii  I f 
Rif  Ri 1  Ab 
Ro
Rof 
1  Ab 
Shunt-Series Feedback Amplifier
(Current-Current Feedback)
• A current-current FB circuit is used for
current amplifiers
– For the b circuit – input resistance
should be low and output resistance
be high
• A circuit example is shown
– RS and RF constitute the FB circuit
• RS should be small and RF large
– The same steps can be taken to
solve for A, Abeta, Af, Rif, and Rof
• Remember that both A and b
circuits are current controlled
current sources
Iout
RD
Iin
RF
RS
The General Feedback Structure
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The General Feedback Structure
Exercise 8.1
4
A f  10
A  10
a)
b
b)
Af
c)
Amount_Feedback  20 log  1  A  b 
Amount_Feedback  60
R1
R1  R2
A
d)
1  Ab
Vs  1
Vo  A f  Vs
Vf  b  Vo
b  1
given
Vo  10
Vf  0.999
Vi  Vs  Vf
4
Vi  10  10
A
Af
1  Ab
e)
b  Find  b 
R1
R1  R2
0.1
4
A  0.8 10
b  0.1
R2
R1
9
10  9.998
10
 100  0.02
A f 
A
1  Ab
A f  9.998
The General Feedback Structure
Exercise 8.1
Some Properties of Negative Feedback
Gain Desensitivity
Af
A
1 Ab
deriving
dAf
dA
(1  A  b)
dividing by
2
dAf
1
Af
(1  A  b)

Af
A
1 Ab
dA
A
The percentage change in Af (due to variations in some circuit parameter) is smaller than the
pecentage cahnge in A by the amount of feedback. For this reason the amount of feedback
1 Ab
is also known as the desensitivity factor.
Some Properties of Negative Feedback
Bandwidth Extension
High frequency response with a single pole
A ( s)
AM
1
s
H
AM denotes the midband gain and H the upper 3-dB frequency
Af ( s )
A ( s)
1  b  A ( s)
AM
 1 AMb 
Af ( s )
1
Hf


s
H  1  A M  b
H  1  A M  b


Lf
L
1  AM  b
Some Properties of Negative Feedback
Noise Reduction, Reduction of Nonlinear Distortion
Read and discuss in class
The Shunt-Series Feedback Amplifier
Example 8.1
[ RL ( R1 R2) ]
A
Vo'

Vi'
RL R1 R2
[ RL ( R1  R2) ]
RL  R1  R2
b
Af
Rof
Vf
R1
Vo''
R1  R2
Vo
A
Vf
1  Ab
Ro
1  Ab
ro
Ro
ro 
RL ( R1  R2)
RL  R1  R2
RL ( R1  R2)
RL  R1  R2
Rid

 ro Rid  Rs 
Rif
R1 R2
R1  R2
Ri  1  A  b 
Ri
Rs  Rid 
Rin
Rif  Rs
R1 R2
R1  R2
The Shunt-Series Feedback Amplifier
Example 8.1
3
6
R1 10
4
R2 10
5
Rs  10
Rid  10
3
ro  10
4
  10
[ RL ( R1 R2) ]
A  
RL R1 R2
[RL(R1  R2)]
RL R1  R2
b 
R1
R1  R2
Rid

R1R2

 ro Rid  Rs 
R1  R2
4
b  9.99 10
R1R2

Ri  Rs  Rid 
R1  R2
5
Ri  1.11 10
Rif  Ri1  Ab
Rif  7.766 10
Rin  Rif  Rs
Rin  7.666 10
Af 
A
1  Ab
3
A  6.002 10
5
5
Af  857.92
3
RL 210
The Shunt-Series Feedback Amplifier
Example 8.1
Ro 
RL ( R1  R2)
ro 
RL  R1  R2
RL ( R1  R2)
ro 
RL  R1  R2
Rout  10
given
Rout RL
Rof
Rout  RL


Find Rout  99.989
Ro  666.223
Rof 
Ro
1  A b
Rof  95.228
The Series-Shunt Feedback Amplifier
Exercise 8.5
The Series-Shunt Feedback Amplifier
Example 8.2
The Series-Shunt Feedback Amplifier
Exercise 8.6
The Shunt-Shunt Feedback Amplifier
Exercise 8.3
The Shunt-Shunt Feedback Amplifier
Example 8.4
The Shunt-Shunt Feedback Amplifier
Example 8.4
The Shunt-Shunt Feedback Amplifier
Exercise 8.7
Determining The Loop Gain
Exercise 8.8-9
The Stability Problem
The Stability Problem and Margins
Polar Plot
Af(s )
Nyquist
Closed-Loop
Transfer Function
Root-Locus
Bode
A(s )
1  A(s )  b (s )
The Nyquist Plot intersects the negative real axis at 180. If this intersection occurs
to the left of the point (-1, 0), we know that the magnitude of the loop gain at this
frequency is greater than the unity and the s ystem will be unstable. If the intersection
occurs to the right of the point (-1,0) the sys tem will be stable.
It follows that if the Nyquist encircles the point (-1,0) the amplifier will be unstable.
The Stability Problem
Nyquist
Closed-Loop
Transfer Function
Root-Locus
Magnitude and Phase
Bode
Gain Margin
Phase Margin - If at the frequency of unity loop-gain magnitude, the phase
lag is in excess of 180 degrees, the amplifier is unstable.
The Stability Problem
The Nyquist Plot
w  100 99.9 100
G( w) 
j  1
s ( w)  j w
f ( w)  1
50 4.6
3
2
s ( w)  9 s ( w)  30 s ( w)  40
5
Im ( G( w) )
0
0
5
2
1
0
1
2
Re( G( w) )
3
4
5
6
Stability Study Using Bode Plots
Bode plot for loop gain Abeta
– Gain margin is an indication
of excess gain before
instability
– Phase margin is an indication
of excess phase before -180°
phase shift at unity gain
Stability Study Using Bode Plots
w  .1  .11  2
G( w) 
K  2
K
j w ( j w  1)  ( j w  2)
j  1
Bode1( w)  20 log  G( w)

20
20
Open Loop Bode DiagramBode1( w)
Bode1( w)
0
0
20
20
0
0.5
1
w
T( w) 
G( w)
1.5
0.5
1
1.5
2
w
2
T( w) 
0
G( w)
Bode2( w)  20 log  T( w)
1  G( w)
1  G( w)
10
0
Bode2( w)
10
Closed-Loop Bode Diagram
20
0.5
1
1.5
w
2

The Stability Problem
Exercise 8.10
Effect of Feedback on the Amplifier Poles
Effect of Feedback on the Amplifier Poles
Exercise 8.11