feedback - rowan universityusers.rowan.edu/.../electronics_ii_files/feedback.pdf · voltage...

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!)