Title: Chapter 8 Feedback
1Chapter 8 Feedback
Introduction 8.1 The general feedback
structure 8.2 Some properties of negative
feedback 8.3 The four basic feedback
topologies 8.4 The series-shunt feedback
amplifier 8.5 The series-series feedback
amplifier 8.6 The shunt-shunt and shunt-series
feedback amplifier 8.10 Stability study using
bode plot 8.11 Frequency compensation
2Introduction
- Its impossible to think of electronic circuits
without some forms of feedback. - Negative feedback
- Desensitize the gain
- Reduce nonlinear distortion
- Reduce the effect of noise
- Control the input and output impedance
- Extend the bandwidth of the amplifier
- The basic idea of negative feedback is to trade
off between gain and other desirable properties. - Positive feedback will cause the amplifier
oscillation.
3Three Parts
- PartI The basic concept and some Properties of
negative feedback - PartII The four basic feedback and analysis
- PartIII The loop gain, stability problem
- and frequency compensation
4PartI
The basic concept
Judgment and Properties of feedback
examples
5PartI The basic concept and some Properties of
negative feedback
8.1 The General Feedback Structure
This is a signal-flow diagram, and the quantities
x represent either voltage or current signals. In
electronic circuits, part of or all output signal
is fed back to input, and affects the input
signal value, which is called feedback.
6The feedback judgment for amplifier circuits
7Negative feedback and positive feedback According
to the effecting of feedback 1) positive
feedback increases the signal that appears
at the input of the basic amplifier 2) negative
feedback reduces the signal that appears at
the input of the basic amplifier
- DC feedback and AC feedback
- Feedback quantity only contains DC quantity,is
called DC - feedback
- 2) Feedback quantity only contains AC quantity,is
called AC - feedback
- Usually AC feedback and DC feedback are
concomitant
8The feedback judgment
- No feedback
- Feedback exists
- No feedback
9The judgment of feedback parity
Instantaneous polarity method
1) Regulate the polarity of input signal relative
to ground at sometime.
2) Decide all points parity step by step, at
last get the parity of output signal.
3) According to the parity of output signal
decides the parity of amount of feedback.
4) If amount of feedback increases the signal
that appears at the input of the basic
amplifier, the circuit inducts the positive
feedback. Otherwise, it inducts the
negative feedback.
10To integrated operational amplifiers,the input
quantity can be UD or iN(iP)
11To discrete components amplifiers,the input
quantity can be Ube or ib
12The judgment of DC feedback and AC feedback
DC feedback,no AC feedback
AC feedback,no DC feedback
13Example feedback?Positive or negative?DC or AC?
AC and DC negative feedback
14The General Feedback Equation
- Open loop gain A
- Feedback factor ß
- Loop gain Aß
- Closed loop gain Af
- Amount of feedback (1 Aß)
15The General Feedback Equation
- If Aß gtgt1, The gain of the feedback amplifier is
almost entirely determined by the feedback
network. - If Aß gtgt1, which implies that the signal Xi at
the input of the basic amplifier is reduced to
almost zero.
168.2 Some Properties of Negative Feedback
the percentage change in Af (due to variations in
some circuit parameter) is smaller than the
percentage change in A by the amount of feedback.
For this reason the amount of feedback, 1 Aß,
is also known as the desensitivity factor.
17Some Properties of Negative Feedback
Note that the amplifier bandwidth is increased by
the same factor by which its midband gain is
decreased, maintaining the gain-bandwidth product
at a constant value.
18Some Properties of Negative Feedback
19Some Properties of Negative Feedback
- 4. Reduction in nonlinear distortion
20Some Properties of Negative Feedback
- 4. Reduction in nonlinear distortion
21Homework
22PartIIThe four basic feedback and analysis
238.3 The Four Basic Feedback Topologies
- Voltage amplifier---series-shunt feedback
- voltage mixing and voltage sampling
24The Four Basic Feedback Topologies
- Current amplifier---shunt-series feedback
- Current mixing and current sampling
25Example
Figure 8.5 A transistor amplifier with
shuntseries feedback. (Biasing not shown.)
26The Four Basic Feedback Topologies
- Transconductance amplifier---series-series
feedback - Voltage mixing and current sampling
27Example
Figure 8.6 An example of the seriesseries
feedback topology. (Biasing not shown.)
28The Four Basic Feedback Topologies
- Transresistance amplifier---shunt-shunt feedback
- Current mixing and voltage sampling
29Example
Figure 8.7 (a) The inverting op-amp
configuration redrawn as (b) an example of
shuntshunt feedback.
30Homework
- May 11nd, 2008
- 8.14 8.15 8.17 8.19
318.4 The Series-Shunt Feedback Amplifier
- The ideal situation
- The practical situation
- Summary
32The Ideal Situation
- A unilateral open-loop amplifier (A circuit).
- An ideal voltage mixing voltage sampling feedback
network (ß circuit). - Assumption that the source and load resistance
have been included inside the A circuit.
33The Ideal Situation
Equivalent circuit. Rif and Rof denote the input
and output resistance with feedback.
34Input and Output Resistance with Feedback
- Input resistance
-
-
- In this case, the negative feedback
increases the input resistance by a factor equal
to the amount of feedback. - Output resistance
-
-
-
-
-
- In this case, the negative feedback reduces
the output resistance by a factor equal to the
amount of feedback.
35The Practical Situation
- Block diagram of a practical seriesshunt
feedback amplifier. - Feedback network is not ideal and load the basic
amplifier thus affect the values of gain, input
resistance and output resistance.
36The Practical Situation
The circuit in (a) with the feedback network
represented by its h parameters.
Omit the controlled source h21I1
37The Practical Situation
The circuit in (b) with h21 neglected.
38The Practical Situation
- The load effect of the feedback network on the
basic amplifier is represented by the components
h11 and h22. - The loading effect is found by looking into the
appropriate port of the feedback network while
the port is open-circuit or short-circuit so as
to destroy the feedback. - If the connection is a shunt one, short-circuit
the port. - If the connection is a series one, open-circuit
the port. - Determine the ß.
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40Summary
- Ri and Ro are the input and output resistances,
respectively, of the A circuit. - Rif and Rof are the input and output resistances,
respectively, of the feedback amplifier,
including Rs and RL. - The actual input and output resistances exclude
Rs and RL.
41Example of Series-Shunt Feedback Amplifier
42Example of Series-Shunt Feedback Amplifier
- Op amplifier connected in noninverting
configuration with the open-loop gain µ, Rid and
ro - Find expression for A, ß, the closed-loop gain
Vo/Vi , the input resistance Rin and the output
resistance Rout - Find numerical values
43Example of Series-Shunt Feedback Amplifier
44Example of Series-Shunt Feedback Amplifier
458.5 The Series-Series Feedback Amplifier
- The ideal situation
- The practical situation
- Summary
46The Ideal Situation
Transconductance gain
47The Ideal Situation
Tranresistance feedback factor
48Input and Output Resistance with Feedback
- Input resistance
-
- In this case, the negative feedback increases
the input resistance by a factor equal to the
amount of feedback. - Output resistance
-
-
-
- In this case, the negative feedback
increases the output resistance by a factor equal
to the amount of feedback.
49The Practical Situation
Block diagram of a practical seriesseries
feedback amplifier. Feedback network is not
ideal and load the basic amplifier thus affect
the values of gain, input resistance and output
resistance.
50The Practical Situation
The circuit of (a) with the feedback network
represented by its z parameters.
51The Practical Situation
A redrawing of the circuit in (b) with z21
neglected.
52The Practical Situation
- The load effect of the feedback network on the
basic amplifier is represented by the components
Z11 and Z22. - Z11 is the impedance looking into port 1 of the
feedback network with port 2 open-circuited. - Z22 is the impedance looking into port 2 of the
feedback network with port 1 open-circuited. - Determine the ß.
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54Summary
- Ri and Ro are the input and output resistances,
respectively, of the A circuit. - Rif and Rof are the input and output resistances,
respectively, of the feedback amplifier,
including Rs and RL. - The actual input and output resistances exclude
Rs and RL.
55Example of Series-Series Feedback Amplifier
56Example of Series-Series Feedback Amplifier
57Example of Series-Series Feedback Amplifier
58Example of Series-Series Feedback Amplifier
598.6 The Shunt-Shunt and Shunt-Series Feedback
Amplifiers
Fig8.19. Ideal structure for the shunt-shunt
feedback amplifier.
60The practical Shunt-Shunt feedback amplifier
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62Example 8.3
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64The shunt-series configuration
Fig 8.22 Ideal structure for the shuntseries
feedback amplifier.
65A practical shunt-series feedback amplifier
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67Example 8.4
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69summary
70The method of finding A circuit
- h,z,r,g parameter method for two-port feedback
network. - Equivalent circuit method
- Find the feedback network
- Find the feedback network equivalent load
Resistance to amplifier input, for output voltage
sampling feedback, output short-circuit (Vo0)
for current sampling feedback, output Io
open-circuit (Io0). - Find the feedback network equivalent load
Resistance to amplifier output, for input voltage
mixing feedback, it should be disconnected from
input to feedback network (Ii0) for input
current mixing feedback, let input connect to
ground to disconnect from input signal to
feedback network.
71??????????
- ????????,????????
- ?????????,????????
- ??????????????????,????????
- ??????????????????,????????
72- ?????????????,????????
- ?????????????, ????????
- ?????????????,??????????(shunt-shunt)
- ?????????????, ??????????(series-series)?
73Example
1)???????????????????????? 2)????????????????????
???? 3)???????????????
???series-shunt (???????) 8?10,3?9,4?6
???shunt-series(???????)7?10,2?94?6
?shunt-shunt(???????)2?9,8?10,5?6
74Homework
- May 13th, 2008
- 8.20 8.25 8.30 D8.37
75The Stability Problem
76The Stability Problem
- The condition for negative feedback to oscillate
- Any right-half-plane poles results in
instability. - Amplifier with a single-pole is unconditionally
stable. - Amplifier with two-pole is also unconditionally
stable. - Amplifier with more than two poles has the
possibility to be unstable. - Stability study using bode plot
77Balance condition Magnitude condition Phase
Condition
Start up oscillation condition
Judgment method of amplifier stability 1) If
?180 is not exist, then the Amplifier is
stable 2) If ?180 is exist and ?180lt ?1, then
the amplifier is not stable 3) If ?180 is exist
and ?180gt ?1, then the amplifier is stable
78The Definitions of the Gain and Phase margins
- Gain margin represents the amount by which the
loop gain can be increased while stability is
maintained. - Unstable and oscillatory
- Stable and non-oscillatory
- Only when the phase margin exceed 45º or gain
margin exceed 6dB, can the amplifier be stable.
79Example
80Effect of phase margin on closed-loop response
- To see the relationship between phase margin and
Close-Loop gain, consider a feedback amplifier
with a large low-frequency loop gain.
81Stability analysis using Bode plot of A.
82Stability Analysis Using Bode Plot of A
- Gain margin and phase margin
- The horizontal line of inverse of feedback factor
in dB. - A rule of thumb
- The closed-loop amplifier will be stable if the
20log(1/ß) line intersects the 20logA curve at
a point on the 20dB/decade segment. - The general rule states
- At the intersection of 20log1/ ß (j?) and
20log A(j?) the difference of slopes should
not exceed 20dB/decade.
83Frequency Compensation
- The purpose is to modifying the open-loop
transfer function of an amplifier having three or
more poles so that the closed-loop amplifier is
stable for any desired value of closed-loop gain. - Theory of frequency compensation is the enlarge
the 20dB/decade line. - Implementation
- Capacitance Cc added
- Miller compensation and pole splitting
84Frequency Compensation
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86Frequency Compensation-Miller compensation
- A gain stage in a multistage amplifier with a
compensating capacitor connected in the feedback
path - An equivalent circuit.
- Miller compensation can reduce the value of C
87Homework
- May 20th, 2008
- 8.51 8.54 8.63 8.66