feedback - rowan universityusers.rowan.edu/.../electronics_ii_files/feedback.pdf · voltage...
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FeedbackRobert R. Krchnavek
Spring, 2017
Feedback
• Negative or positive. Generally, amplifiers will use negative feedback and oscillators will use positive feedback. Active filters may also use positive feedback.
• Feedback is used in numerous physical systems. The theory was developed by Harold Black (EE) in 1928.
• General concept: trade off gain for other desirable effects.
Why Feedback• Desensitize the gain – gain is less dependent on component values.
• Reduce nonlinear distortion.
• Reduce the effect of noise.
• Control input and output impedances.
• Extend the bandwidth of the amplifier.
• The above positive effects come at the cost of reduced gain.
Feedback – Block Diagram Approachx
o
= Ax
i
x
f
= �x
o
xs � xf = xi
x
o
= A (xs
� x
f
)
x
o
= A (xs
� �x
o
) = Ax
s
�A�x
o
If A � � then Af ' 1
�
A
f
⌘ x
o
x
s
=A
1 +A�
Af =A
1 +A�
Also, the following is useful:
xi
xs=
1
1 +A�
A𝛽 – The Loop Gain
If A � � then Af ' 1
�
Some terminology:– feedback factor – gain with feedback; closed-loop gain – loop gain – amount of feedback
�
Af
A�
1 +A�
• positive for negative feedback. • determines how close Af is to the ideal
factor of 1/𝛽. • determines the amount of feedback which
impacts several parameters (to be seen.) • is frequency dependent and can cause
problems (to be seen later.)
A
f
⌘ x
o
x
s
=A
1 +A�
Af =A
1 +A�
Calculating Loop Gain• Set xs = 0. • Break the feedback loop. • Apply test signal at xt and find xr.
xr = �A�xt
A� = �xr
xt
Reality: A nice concept, good for qualitative understanding, but generally this will not work for real amplifiers because the feedback network always loads the amplifier to some extent. The real analysis is considerably more work.
Example: Inverting Op Amp
Feedback – Summary
Feedback – Improving Amplifiers• The block diagram of negative feedback provides a
simple way of seeing how negative feedback and gain reduction can improve several other amplifier properties.
• The block diagram does not cover enough details of real feedback. In particular, it assumes the feedback network does not load the amplifier.
Feedback – Gain DesensitivityAf =
A
1 +A�
dAf
dA=
1
(1 +A�)2
dAf
Af=
dA
A
1
(1 +A�)
The change in Af with respect to Af is smaller than the change in A with respect to A by the desensitivity factor, 1+A𝛽.
Feedback – Bandwidth ExtensionRecall from our work on frequency response. The high-frequency response was given by
A(s) =AM
1 + s/!H
With feedback, we have Af (s) =A(s)
1 + �A(s)Af (s) =
AM
1+s/!H
1 + � AM
1+s/!H
Some algebra Af (s) =AM
1+�AM
1 + s!H(1+�AM )
AMf =) AM
1 + �AM
!Hf =) !H (1 + �AM )
Feedback – Bandwidth Extension
Af (s) =AM
1+�AM
1 + s!H(1+�AM )
AMf =AM
1 + �AM!Hf = !H (1 + �AM )!Lf =
!L
1 +AM�
Feedback – Interference ReductionFor certain situations, feedback can increase the signal-to-interference ratio.
Consider this classic example.
S/I = Vs/Vn
Feedback – Interference ReductionNow, consider this solution.• Add a second amp that
does not suffer from the interference.
• Apply negative feedback around both amps (global feedback.)
How does this improve the situation?
Feedback – Interference Reductionvo
= vi
A1A2 + vn
A1
vi
= vs
� �vo
vo
= (vs
� �vo
)A1A2 + vn
A1
vo
(1 + �A1A2) = vs
A1A2 + vn
A1
vo
= vs
A1A2
1 + �A1A2+ v
n
A1
1 + �A1A2
S/I =vs
A1A21+�A1A2
vnA1
1+�A1A2
=vsvn
A2
Feedback – Reduction in Nonlinear Distortion
Amplifier (a): Gain is highly nonlinear. Goes from 1000 to 100 to 0. Amplifier (b): Add feedback to (a). Gain is reduced and so is the nonlinearity.
Feedback – Reduction in Nonlinear DistortionNumerical example:• Amplifier (a) without feedback
has an initial gain of 1000, then 100.
• Assume a 𝛽 = 0.01.
Af =1000
1 + 1000⇥ 0.01= 90.9
Af =100
1 + 100⇥ 0.01= 50. Without feedback, (a), the gain changes by
a factor of 10. With feedback, (b), the gain changes by less than a factor of 2,
The Feedback Voltage Amplifer• Recall we can classify amplifiers as:
• Voltage amplifier
• Current amplifier
• Transconductance amplifier
• Transresistance amplifier.
• The feedback will vary for these different amplifiers,
• We begin with voltage amplifiers – the most common amplifier.
Voltage Amplifiers – Series-Shunt Feedback Topology
• Typically, in voltage amplifiers, the input resistance should be high and the output resistance should be low.
• On the output, you want to sample the voltage so you do not subtract from the output voltage. A SHUNT connection.
• On the input, a SERIES connection allows a subtraction to vs while maintaining high input resistance.
Voltage Amplifiers – Series-Shunt Feedback Topology: Examples
• Noninverting op amp configuration.
• Output is a shunt connection because the feedback voltage is taken with respect to ground.
• Input is in series because of the differential input.
Voltage Amplifiers – Series-Shunt Feedback Topology: Analysis
• The loop-gain analysis is not accurate because the feedback network loads both the input and output of the amplifier.
• We cannot accurately determine A or 𝛽, but we can determine the product, A𝛽, using the loop-gain method.
• When applying the loop-gain method, breaking the loop will likely change the circuit and this must be accounted for. It also may suggest a logical place to open the loop.
• The loop-gain method also does not help us determine input and output resistances.
Voltage Amplifiers – Series-Shunt Feedback Topology: Examples
� =R1
R1 +R2
Voltage Amplifiers – Series-Shunt Feedback Topology: Examples
� =R1
R1 +R2The ideal value for Af would be: 1
�= 1 +
R2
R1
Voltage Amplifiers – Series-Shunt Feedback Topology: Detailed Analysis
The ideal Series-Shunt topology.
The equivalent circuit of the ideal, Series-Shunt topology,
Voltage Amplifiers – Series-Shunt Feedback Topology: Detailed Analysis
The ideal Series-Shunt topology.
The equivalent circuit of the ideal, Series-Shunt topology,
Af
⌘ vo
vs
=A
1 +A�
Closed-loop gainfor open-circuit feedback amplifier.
Voltage Amplifiers – Series-Shunt Feedback Topology: Detailed Analysis
The ideal Series-Shunt topology.
The equivalent circuit of the ideal, Series-Shunt topology,
Feedback inputimpedance.
Voltage Amplifiers – Series-Shunt Feedback Topology: Detailed Analysis
The ideal Series-Shunt topology.
The equivalent circuit of the ideal, Series-Shunt topology,
Feedback outputimpedance.
Voltage Amplifiers – Series-Shunt Feedback Topology: Detailed Analysis
Voltage Amplifiers – Series-Shunt Feedback Topology: Detailed Analysis
• Need to find R11, R22, and the 𝛽 circuit.
• Note: the circuit to the right matches the ideal topology.
practical ideal
Voltage Amplifiers – Series-Shunt Feedback Topology: Detailed Analysis
Finding R11. Finding R22.
Finding 𝛽.
Voltage Amplifiers – Series-Shunt Feedback Topology: Detailed Analysis
The A circuit.
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
• Find R11. • Find R22 • Find 𝛽 circuit. • Find A circuit. • Calculate feedback parameters. • Calculate Rin and Rout.
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
Find R11
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
Find R11
R11 = R1||R2
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
Find R22
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
Find R22
R22 = R1 +R2
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
The A circuit
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
Find 𝛽
� =R1
R1 +R2
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
Using the A circuit
Find the open-loop gain:
vo
vi
= gm1gm2RD1
RD2|| (R1 +R2)
1 + gm1R1||R2
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
Using the A circuit and 𝛽 circuit
Find the closed-loop gain:
vo
vi
= Af
=A
1 +A�=
gm1gm2RD1
RD2||(R1+R2)1+gm1R1||R2
1 + gm1gm2RD1
RD2||(R1+R2)1+gm1R1||R2
R1R1+R2
Voltage Amplifiers – Series-Shunt Feedback Topology: Example
Using the A circuit and 𝛽 circuit
Find Rin:
Find Rout:R
outf
=R
out
1 +A�=
RD2|| (R1 +R2)
1 +A�
Rinf = Rin (1 +A�) ! 1
Other Feedback Topologies See Textbook
Feedback Analysis Techniques – Summary
Feedback Analysis Techniques Summary
The Stability ProblemNeed to consider the frequency response of the amplifier with feedback to determine the stability of the amplifier.
Af (|!) =A(|!)
1 +A(|!)�(|!)
Af =A
1 +A�=) Af (s) =
A(s)
1 +A(s)�(s)
L(|!) = A(|!)�(|!) = |A(|!)�(|!)|e|�(!)
• The stability of the amplifier is determined when. .
• If at �(!) = 180�
! = !180
|A(|!)�(|!)| � 1
|A(|!)�(|!)| 1
|A(|!)�(|!)| = 1
stable Af (|!) > A(|!)
oscillator and …
oscillator and …
Nyquist Plot|A(|!)�(|!)|e|�(!)
A polar plot of the magnitude of (radial component) and phase angle
(angular component).
A(|!)�(|!)
�(!)
• Frequency is not explicitly shown. • DC ( ) is on the real axis. • Negative frequencies mirror
positive frequencies. • is also on the real axis.
! = 0
!180
Feedback and Amplifier Poless = �0 ± |!nAn amplifier with the following pole:
Inverse phasor transform to put it into the time domain:
v(t) = e�0t⇥e+|!nt
+ e�|!nt⇤= 2e�0t
cos(!nt)
For an amplifier (or any other system) to be stable, the poles must be in the left half of the s-plane.
Feedback and Amplifier Poles Single-Pole Response
Af (s) =A(s)
1 +A(s)�(s)
A(s) =A0
1 + s/!P
Af (s) =A0
1+s/!P
1 + A01+s/!P
�(s)
feedback gain
single-pole amplifier frequency response
feedback gain with amplifier response
standard form
feedback network is frequency independent
Af (s) =A0
1+A0�(s)
1 + s!P (1+A0�(s))
Af (s) =A0
1+A0�
1 + s!P (1+A0�)
Feedback and Amplifier Poles Single-Pole Response
Af (s) =A0
1+A0�
1 + s!P (A0�)
The feedback changed the pole position
!P =) !P (1 +A0�)
Feedback and Amplifier Poles Single-Pole Response
Af (s) =A0
1+A0�
1 + s!P (A0�)
The feedback changed the pole position
!P =) !P (1 +A0�)
Let 𝜔 >> 𝜔P. Then
Af (s) 'A0
1+A0�s
!P (1+A0�)
'A0
1+A0�!P (1 +A0�)
s' A0!P
s' A(s)
UNCONDITIONALLY STABLE
Feedback and Amplifier Poles Two-Pole Response
Af (s) =A(s)
1 +A(s)�(s) feedback gain
two-pole amplifier frequency response
poles come from the denominator
closed-loop poles
A(s) =A0
(1 + s/!P1) (1 + s/!P2)
s2 + s (!P1 + !P2) + (1 +A0�)!P1!P2 = 0
s = �1
2(!P1 + !P2)±
1
2
q(!P1 + !P2)
2 � 4 (1 +A0�)!P1!P2
Feedback and Amplifier Poles Two-Pole Response
closed-loop poless = �1
2(!P1 + !P2)±
1
2
q(!P1 + !P2)
2 � 4 (1 +A0�)!P1!P2
UNCONDITIONALLY STABLE
Feedback and Amplifier Poles Three-or-More Pole Response
• Root locus diagram for a three-pole open loop amplifier response,
• Increasing A0𝛽 brings two poles together and moves the highest pole outward.
• After the two poles coincide, they become complex conjugates and head for the right side of the plane.
Stability: Bode Plots
Frequency Compensation
Dr. Schmalzel would like to do this!!!!
(but, you should also read it!)