compensation methods electronic engineering. the following presentation is a part of the level 5...

CompensationMethods

Electronic Engineering

The following presentation is a part of the level 5 module -- Electronic Engineering. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1 st year undergraduate programme.

The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.

Contents Instructions Gain Curve Compensation Methods Dominant Pole Compensation Frequency Compensation Lead Lag Compensation. Summary Credits

In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see:Clayton G, 2000, Operational Amplifiers 4th Ed, Newnes James M, 2004, Higher Electronics, Newnes

Compensation Methods

For our amplifier we are going to apply negative feedback to reduce the gain to 55dB.

Comment on the amplifiers stability.


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0Gain CurvefC1 fC2

Phase Curve

-40dB/dec

For our amplifier we are going to apply negative feedback to reduce the gain to 55dB.

Comment on the amplifiers stability.Unstable as it crosses the Gain Curve on the -

40dB/dec line.

The amplifier is therefore unsuitable for this application.

Can we do anything about it?



There are three methods which we will examine.

1. Dominant Pole compensation

2. Frequency compensation

3. Lead Lag compensation.


Dominant Pole CompensationIn this method the Manufacturer introduces an

artificial break frequency (pole) which causes the gain to drop to 0dB before the first natural one occurs.

fC1

fC2

Unstable

Stable

New fC Original Curve

New Curve

gain

frequency

Desired gain

Try this out on our plot and answer the following questions:

At what frequency does the dominant pole need to be placed?

What is the Bandwidth of the compensated amplifier?


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0Gain CurvefC1 fC2

Phase Curve

x 9.5dB – 5.1kHz

29.5dB – 510Hz

89.5dB – 0.51Hz

15dB – 2.8kHz

35dB – 280Hz

55dB – 28Hz




0.51 Hz

28 Hz


Notes This method is stable for all applications.

The user does not need to carry out the compensation exercise.

Bandwidths are limited in size using this method.


Frequency CompensationThis method is similar to the first in that a dominant

pole is introduced. This time its position is selected by the user and is positioned so that the gain drops to the desired gain at the point the first natural break frequency occurs.

fC1

fC2

Unstable

Stable

New fC Original Curve

New Curve

gain

frequency

Desired gain


A table supplied by the amplifier manufacturer allows the user to convert the new pole frequency measured from the plot into a capacitor value which is connected between two pins on the amplifier.


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0Gain CurvefC1 fC2

Phase Curve

x

69.5dB – 3kHz

89.5dB – 300Hz

fC1 – 15kHz




300 Hz

15 kHz (fC1)


Notes The user must compensate each amplifier

according to its use.

The Bandwidth, using this method is always at the first break frequency.

Once compensated if the gain is reduced the amplifier will become unstable.


Lead Lag Compensation.Let us examine the problem we have with our amplifier.

fC1

fC2

Unstable

Original Curvegain

frequency

Desired gain

The amplifier would be stable if either fC1 occurred at a lower frequency or fC2 occurred at a higher frequency Compensation Methods

We cannot move the two break frequencies as they are inherent parameters of the amplifier.

BUTWe can make fC1 “look” as though it started at a

lower frequency using the circuit below.

R1 R

2

C

VIN VOUT


Series combination of R2 and C

Now we can generate an equation for VOUT in terms of VIN

Cj

CRjRCj

2121

)( 211

21

121

21

121

21

RRCj

CRjV

CRjCRj

CRjV

RCjCRjCjCRj

VV NIININOUT

)( 211

21

RRCj

CRjGain


)( 211

21

RRCj

CRjGain

Quantative Analysis

At low frequencies ω parts << 1 so

the gain = 1 phase will be 0°

At high frequencies ω parts >> 1 so

the gain = R2/(R1 + R2) phase will be 0°

In between gain must roll off in some way.

e.g. R1 = 10k, R2 = 1k, C = 10nF

Plot over the range 100 Hz to 100 kHz

222

22

211

2122121

)(

)()(

RRC

RRRCRRCjCRjGain

)( 211

21

RRCj

CRjGain

222222

22

211

1

211

2121

)()(

)(

RRC

CRj

RRC

RRRCGain

222

2222

211

12121

)(

)(

RRC

CRRRRCGain

)( 2121

122

1

RRRC

CRTanPhase

Frequency Gain Gain dB Phase

100 0.99764 -0.0205249 -1.99651

200 0.990659 -0.0815145 -3.97231

400 0.964149 -0.3171164 -7.78572

700 0.901053 -0.9049962 -12.9442

1000 0.82426 -1.6787185 -17.2528

2000 0.590744 -4.572012 -26.0858

4000 0.350726 -9.1006474 -31.1146

7000 0.221129 -13.107079 -30.3226

10000 0.169115 -15.436358 -27.5696

20000 0.115878 -18.719977 -19.0967

40000 0.097777 -20.195268 -10.903

70000 0.093209 -20.610813 -6.45841

100000 0.092044 -20.720125 -4.5634

Lead Lag Network

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Frequency

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0


Referring to our equation for the network, we will have two break frequencies, one for the top line and one for the bottom.

Break frequency where roll off begins f1 is given by:

Break frequency where roll off stops f2 is given by:

Gain after roll-off is given by

)( 211

21

RRCj

CRjGain

kHzRRC

451212

1.

)(

kHzCR

91522

1.

dBRR

R83200910

21

2..


Lead Lag Network

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Frequency

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0x

x

f1

f2

-20dB/dec


NoteThe phase is not a concern as it returns to a low

value by the time the gain curve stops reducing.

Let us return to our amplifier.


fC1

fC2

Unstable

Original Curvegain

frequency

Desired gain

Introduce a lead lag network which starts before fC1 and stops at fC1

f1

f2

Stable

fC1 now looks as though it starts at f1



What component values do we need – assume C has a value of 100pF



Firstly f2 must equal fC1

Secondly what drop in gain is required

kHzCR

1522

1

k

pFkR 106

100152

12


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0Gain CurvefC1 fC2

Phase Curve

Required drop in gain = -10dB

Firstly f2 must equal fC1

Secondly what drop in gain is required

Finally, what is f1

kHzCR

1522

1

k

pFkR 106

100152

12

21

2316010

RR

RdB

. kR

RR 2292

3160

21

.

kHzkkpFRRC

7542291061002

1

212

1.

)()(


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0Gain CurvefC1 fC2

Phase Curve

x

xf1

f2

Bandwidth fC2


What component values do we need – assume C has a value of 100pF


229k 106k

100pF

VIN VOUT

fC2 = 300kHz


Notes The user must compensate each amplifier

according to its use.

The Bandwidth, using this method is always at the second break frequency.

Once compensated if the gain is reduced the amplifier will become unstable.


Summary

No compensation – Unstable

Dominant Pole Compensation – B.W. = 28Hz

Frequency Compensation – B.W. = 15kHz

Lead Lag Compensation - B.W. = 300kHz


compensation methods electronic engineering. the following presentation is a part of the level 5...

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