Use of Nyquist plot for Analog IC design

8/21/2019 Use of Nyquist plot for Analog IC design

1/17

Analog Integrated Circuit Design

Prof. Nagendra Krishnapura

Department of Electrical Engineering

Indian Institute of Technology, Madras

ecture No ! "Negati#e $eed%ac& Amplifier 'ith Parasitic Poles and (eros) Ny*uist Criterian

Hello and welcome to another lecture of analog integrated circuit design, so far we have been

looking at what happens if you have a delay in a negative feedback amplifier, and where

those delays could come from? The delays come from parasitic poles in the system. We also

saw that pole acts like a delay for the integrator. So, today what we will do is elaborate a little

bit on these things and then see how to conveniently judge the stability of an amplifier and

how to design it, so that it is wellbehaved.

!"efer Slide Time# $$#%%&

So, this is our negative feedback amplifier and it could have a delay Tdin the feedback path

or anywhere in the system. While 'uantifying the amount of the negative feedback we have

computed the loop gain, and the loop gain of the particular system without the delay is

!(u)k&)s, which can be defined to be (u,loop)s. Here (u,loopis the unity loop gain fre'uency and

its given by !(u)k&, where (u is the unity gain fre'uency of the integrator used to reali*e the

negative feedback system.

+ow, how much delay can be tolerated depends on the value of (u,loop. So, our previous

analysis shows that for Tdless than or e'ual to )e times ) ( u,loop, that is by -./ times the


2/17

time constant of the integrator. We get no overshoot in the step response and essentially

behaves like an over damped system. That is for the Tdless than this number it is e'uivalent

to an over damped system. for Tde'ual to )e times the time constant of the loop gain, then

there is again no overshoot and this is the critically damped system. So, this corresponds to

the higher delays that you can have without having an overshoot in the step response. 0t also

corresponds to the fastest step response that you can have without overshoot. That is why it is

the critically damped response critically damped system.

!"efer Slide Time# $1#2/&

+ow, when Td is between )e times the )(u,loop and between pi)- times )(u,loop, what

happens is there is ringing in the step response, but the ringing eventually die3s out. So, the

system is stable. 0n that if your apply a step there will be some ringing and that thing

eventually dies out, and the same thing is true if you we have initial conditions in the system,

they will eventually die out, but there can be a lot of ringing. 4ou can tolerate a little bit of a

ringing, but usually not a lot of ringing and for Tdmore than5 )(u,loopwhat happens is, there

are sustained oscillations.

So that means that if you apply a step then there is ringing and it blows up also if you have

some initial conditions they won3t die down they will eventually blowup and give a response

even without having an input. So, this is an oscillatory condition and this is certainly

something that you do not want to have. 6nd for a good amplifier, that means wellbehaved


3/17

amplifier, you can tolerate a little ringing, it also depends a little on the application that you

looking at7 you can tolerate a little ringing.

So, typically Tdless than about )- of the )(u,loopis tolerable. +ow if your application says

that there has to be absolutely no ringing at all then you have to make Tdto be less than )e

times )(u,loop. So, this is what is tolerable and if you have Tdbeyond this, it will start ringing

more and more, and if you have Tdmore than pi)- times )(u,loop,you will have sustained

oscillation and certainly you do not want to have that. So, now, let us see what happened

when there is an e8tra pole and related to this one.

!"efer Slide Time# $#-&

What we said was there was e8tra pole in the loop somewhere7 we have not at actually looked

at e8actly where it can be and it can come from parasitic capacitances at various points in the

circuit. So, for now we will simply model it as our usual system, an integrator with a unity

gain fre'uency (u, a voltage divider of fraction ) k. 6nd here 0 will simply have an ideal

block which models the pole, just like the delay it does not matter where in the loop this pole

is. So, as long as it is in the loop it will affect the response of the system.

So, the first thing is, we see that, the step response of the loop gain without the e8tra pole will

be a ramp whose slope is (u)k or (u,loop. 0f you have an ideal delay of Tdthen the step response

will have the same slope, but is delayed by an amount Td, this corresponds to delay of Td. 6nd

finally, if you have a pole, if you have e8tra pole p-what happens is that you will also have


4/17

something like this and the response slowly rises. 9ut eventually it will settle to a slope

which is the same as for the other cases, it is e'ual to (u)k.

The slope is the (u)k and it is shifted hori*ontally by an amount that depends on the pole. 6nd

0n fact, it is e8actly e'ual to )p-. So, a pole p-is appro8imately the same as a delay of )p-.

We can e8pect a similar behavior that we know that if the delay is very small, there will be no

overshoot in the response. 6s the delay becomes larger and larger there will be overshoot.

Similarly if the pole p- is very large which corresponds to small delay there will be no

overshoot. 6nd if the pole p-becomes very very small then you will start having overshoot.

6nd in fact, we have already analy*ed this through the damping factor and we have seen that

for this particular system, the damping factor will be half s'uare root of the unity loop gain

fre'uency divided by the parasitic pole p-. +ow, for no overshoot we need to have the

damping factor e'ual to one or more. So7 that means, that the pole p -as to be at least four

times greater than the unity loop gain fre'uency, 0 made a mistake in this e8pression, this as

to be p-divided by (u)k.

So, the damping factor will be more than or e'ual to if p -is greater than e'ual to four times

(u)k and this is consistent with the notion that, a pole is similar to a delay. 6 pole at p-

corresponds to delay of )p-. So, if the pole is at very highfre'uency the delays very small

and there is no effect on the step response, if the poles comes lower and lower in fre'uency,

then there will be an effect on the step response, in that it will start ringing. 6nd the critically

damped case : e'ual to one corresponds to p -e'ual to four times (u)k and you do not want to

have a pole that is substantially lower this, lower than this in fre'uency.


5/17

!"efer Slide Time# #2-&

+ow, one thing about the system with the pole that is different from the system with the delay

is with single e8tra pole p-, the system is unconditionally stable meaning there can be ringing

in the system, but there will never be a sustained oscillation. This is because it is a second

order system and that is what it turns out to be, we will see that when you have multiple e8tra

poles this is no longer the case. ;irstly, the system is unconditionally stable, the system is

critically damped for p-e'ual to four times the unity loop gain fre'uency and the system is

over damped for p-e'uals, p-more than four times unity loop gain fre'uency.

So, now again we can tolerate this condition, we can violate this condition a little bit, but not

by too much. it is ok to have, let us say p-is two times (u,loopand this case we will get a little

bit ringing, but certainly we do not want make p -at a lower fre'uency than this. 0n that case

you will have a lot more ringing and it will not be acceptable in an amplifier. +ow, this shows

that with an e8tra pole the system behaves as though there is an e8tra delay and the

conclusions that we drew from the original analysis with the delay holds good even now.


6/17

!"efer Slide Time# %#-%&

We will assume that the negative feedback amplifier as 0 assumed two e8tra poles somewhere

in the loop. So, what happens in this cases the step response without the parasitic pole of the

loop gain will be a ramp whose slope is (u) k, and with this parasitic delay, it will initially do

something, but finally settled to a ramp with the same slope and the delay of -)p-. 0n fact, if

you have multiple poles in the system each pole can the thought of us contributing a delay of

one over that pole. So, this is the units of step response of the loop gain and it again acts as a

delay.

So, for this particular case we will try to solve for the condition, when there will be sustained

oscillations just we see what happens when you go to higher and higher order systems. So,

=$)=iwill be e'ual to the gain of the forward path plus divided by plus the gain of the

forward path times the gain of the feedback path, this you can work out for yourself. >ven if

you do not know the formula by writing out the variables around the loop you will be able to

work out this.


7/17

!"efer Slide Time# #$/&

6nd this can be re rewritten as, in the standard form by multiplying by plus s)p- s'uare

times s)!(u)k&. So, we will get5 this is what we will get and for convenience, 0 will replace

this by variable (u,loop. 6nd now you can clearly see that as we go to higher and higher order

system, evaluating the step response in a closed form and finding out whether there is ringing

or not becomes very difficult, which is why we will later go to the more convenient condition

in which we do not have to solve for it analytically.

So, but we will look at we will look at the conditions for which this system is unstable. 6n

unstable system means the poles are on the j( a8is, here when 0 say poles, 0 am talking about

poles of the close loop system around the j( a8is or in the right half plane. 6nd it turns out

that it is rather easy to solve for the condition where the poles are on the j( a8is and that is

what we will do.

When the poles are on the j( a8is what happens is for a certain fre'uency s j(, this

e8pression =$)=ibecomes infinity. ;or this condition =$)=iis infinity for some s j(. +ow if

you had a single e8tra pole p- this would never happen, but for more e8tra poles this can very

easily happen. So, how do we solve for this. We will set the denominator to be e'ual to $, for

some s j(. So, what 0 have to do is, 0 have to substitute s j( in this e8pression and set the

result to $. So, as usual i encourage you to try this out before we look at my solution and find

the condition for which it happens.


8/17

!"efer Slide Time# -$#--&

This is $ for some s j( and this is nothing but again for s j(. So, with this means that -( -

by p-@(u,loopplus j( by (u,loopplus e'uals $. So, as you can see there are two variables here,

the value of ( for which this happens and the condition on p -at which this happens. 6nd

there are - e'uations because there is the real part of this e'uation and there is also there

imaginary part of this e'uation. 6nd each of them as to be e'ual to $ by e'uating them to $

you will get the two conditions. ;irstly, if you e'uate the real part to $ you will get.

!"efer Slide Time# --#--&


9/17

-(-divided by p- (u,loope'uals $ and if you e'uate the imaginary part to $, you will get A

(1)!p--(u,loop& plus this part () (u,loope'uals $. 6nd this gives you ( to be e'ual to p-and if

you substituted here, you will get, substituting in this first e'uation you will get p- to be

(u,loop)- for the condition to hold.


10/17

!"efer Slide Time# -B#1$&

So, for stability p- should be less than half the unity loop gain fre'uency of and when p-

e'uals the unity loop gain fre'uency by -, the poles of the closed loop system are on the j(

a8is, and when p- is more than (u,loop)-, close loop poles are in the right half plane, which

also means an unstable system. So, just as before when p- becomes very small the e'uivalent

delay becomes very large and this is becomes unstable.

So, we can carry this e8ercise for three e8tra poles and four e8tra poles and so on and it is

most convenient, if you assume that all the parasitic poles are in the same location. 9ut the

conclusion that you draw from this is, in general, even if you have poles at different

locations, the total delay contribution due to all the poles should be limited to certain value. 0f

the delay becomes too much the system will become unstable. 0 will 'uickly tabulate the

results.


11/17

!"efer Slide Time# -/#-%&

So, let us say we have the case where there they are no e8tra poles. 6nd this is the ideal case

that we had considered and in this case the loop gain is ( u,loop)s or !(u)k&)s. Cet me write it

more clearly, if you use an integrator of unity gain fre'uency (uand make an amplifier with

gain of k with it, it will have a loop gain like this which corresponds to (u,loop)s. 6nd this will

actually never ever become unstable. 0t is unconditionally stable, it is a first order system

with a pole at (u)k and it is unconditionally stable.

So, let us say you have one parasitic pole at certain fre'uency p-, the loop gain function will

be (u,loop)s by plus s)p-. 6nd actually even this never becomes unstable, but it does get

under damped or shows ringing if p- is less than four times the unity loop gain fre'uency.

6nd similarly you can have the two parasitic poles both identical at p-. 0n this case the loop

gain function is this and you can calculate when it becomes unstable. 6nd it becomes

unstable for p- less than $.2 (u, and you can do it for three parasitic poles at p- and this is

the function and this becomes unstable for p- less than .B (u.

Sorry .1 (uand finally, you can also try a % and 2, and how many were you wish to have.

6nd here also the system can become unstable, this is the function and this becomes unstable

for p- less than .B (u. So, to summari*e an e8tra pole is like a delay and as the delay

increases, we will tend to have ringing in the system and when the delay increases a lot, you

can even have instability. That means, the system can give you an output even without an


12/17

input, you can think of it as the gain becoming infinity at some particular fre'uency or the

poles being in the right half plane.

+ow, it turns out that for the special case of a single e8tra parasitic pole, the system is never

unstable. 0t can ring, but it never unstable. So, you still have to make sure that the parasitic

pole is at high enough fre'uency to ensure that the ringing is limited and that limit is usually

two times the unity loop gain fre'uency. 6nd if you do not want any ringing at all the

parasitic pole as to be at least four times the unity loop gain fre'uency. +ow, you can have -

or 1 or % e8tra parasitic poles and they can be at a various locations, but for analysis, it3s

easiest if you assume that all of them are at the same location. 6nd if you have two identical

parasitic poles p- what happens is the system can become unstable. 6nd it happens for p-

lesser than $.2 (u,loop.

So, all of these have written here should be ( u,loop, $.2 (u,loopand for three e8tra parasitic poles

it is .1 (u,loopand four of them it is .B (u,loop. +ow, in the other thing you also notice is

that when you increase the number of poles, the permissible location for the pole is at a

higher fre'uency. ;or instance with two parasitic poles, it is at half of ( u,loopand with four

parasitic poles it is at .B (u,loop. This is because each pole contributes a certain amount of

delay and when you have multiple poles the delays due to each of the poles adds up. So, it

effectively constitutes a greater delay.

So, the delay due to each pole should be smaller, if you have to ensure stability. 6nd this can

be easily related to the first result that we have obtain that if you have an ideal delay, 6nd if

the delay becomes too much that is .2 times )( u,loop, the system becomes unstable. 0t is

e8actly, it is pi)- times )(u,loop, which can be appro8imated to .2 times )(u,loop.

So, the greater the number of poles the higher fre'uency they have to be if they come to

lower fre'uency you will risk instability. +ow, what we are interested in amplifier designer is

not merely avoiding instability because it can be ringing a lot, which dies down, the ringing

dies down and it is still a stable system. That is also not what we are interested in. What we

are interested in is also wellbehaved system where it does not ring too much.

+ow, what happens is, you have multiple poles, evaluation of the step response becomes

difficult. When we had only one e8tra pole it is very easy, we know the step response of a

secondorder system. 6nd we can classify it into critically damped, over damped and underdamped and we can easily identify the condition for which there is no ringing that is it has to


13/17

be critically damped or under damped. +ow, when you go to higher and higher order systems

this is not the case, the analytical e8pression become too complicated and it becomes simply

too difficult to judge even stability, or it is even more difficult to judge when the system is

wellbehaved.

So, what we will do is, we will not evaluate the transfer function and work with the

polynomials which is very difficult, what we will do is taking our hint from the ideal delay

case for which we have a solution, and the secondorder case for which we also have a

solution7 from their we will e8trapolate some conditions for higher order systems and these

conditions will not be based on calculating polynomial and the roots and etcetera. 9ecause

that is what is very complicated, it will be based on evaluating the polynomial for some

sinusoidal fre'uency. That is the evaluating the sinusoidal steady state response of the loop

gain, because of that it is much easier, and that is the preferred way of analy*ing negative

feedback systems.

!"efer Slide Time# 1B#-&

Dlosed form e8pression for the step response are too complicated for higher order systems, so

which basically means that we need an alternative methods for analysis. 6nd there is an

alternative method, but does not depend on calculating the denominator polynomial and its

roots and so on, and that is based on the +y'uist criterion. So, the +y'uist criterion would be

familiar to you from courses on control systems, in this course we are not going to deal with


14/17

the proof of the +y'uist criterion and so on. 9ut we will just state the +y'uist criteria and use

it for our purpose which is to be able to design wellbehaved amplifiers.

!"efer Slide Time# 1#%-&

So, just a start of with, we will take the classical representation of the feedback amplifier

which also includes the amplifier that we have derived and if you look at the control system

te8tbooks, this is how it as written. The forward path is denoted by E and the feedback pathdenoted by H. 6nd for our proto type case E is nothing but the integrator and H is nothing but

)k ideally. 6nd in reality there will be parasitic poles in either E or H or both. There will be

e8tra poles in the system. So, this =$)=ias we know as an e8pression E by plus EH, where

the EH is nothing but the loop gain, as we have seen the loop gain is something that we used

to 'uantify the amount of negative feedback.

So, the way we do it is by breaking the loop somewhere, applying a signal in the forward

direction of the loop and see what comes back. ;or instance in this particular case the we can

break it here, we set the input to $ for evaluation of the loop gain because we are not

interested in any particular input, but we are only interested in what comes back around the

loop. So, we can apply some =testhere and what comes back here we know is minus EH

times =test. So, this EH is defined to be the loop gain, the minus sign is omitted because we

assumed that we are making a negative feedback amplifier. So, we e8pect a minus sign, and

that is omitted and EH is the loop gain and as you can see from this e8pression the gain will

become infinity if EH becomes minus 7 if the loop gain becomes minus .


15/17

So, the stability criteria usually revolve around the loop gain and the value of minus , that is

you should not have loop gain to be e'ual to minus . So, how to avoid it and what to do with

the loop gain that is what the stability criteria work with. So, in a very crude way we can say

that stability criteria is about avoiding EH being minus , there is a little more to it than that,

but that e8plains the significance of minus in in the loop gain.

!"efer Slide Time# %#-$&

+ow, what is the actual definition of stability or instability rather7 instability means, that

transfer function =$by =iwhich is E by plus EH has some poles in the right half plane. So,

if this is the s plane, instability means, there are some poles in this half of the plane in the

right half plane or on the imaginary a8is. So, this includes the imaginary a8is. +ow, it turns

out that for this, to work out this thing you need to obviously, calculate the =$by =i and

there will be some denominator polynomial. 4ou have to find out the root of the denominator

polynomial and see whether they have positive real parts are not, if they have positive real

parts they will be in the right half plane and the system is unstable. 9ut that is what you said

was very difficult that is very difficult because you will end of with very higher polynomial

and for which the roots are not very easy to determine.

So, the alternative is to use the +y'uist criterion which states the following, So, what you can

do is instead of evaluating E by plus EH and working with the polynomials you can simply

evaluate EH. 6nd that to not everywhere, but on their imaginary a8is s j(. This essentially

means that what you are evaluating is the sinusoidal study state response of the loop gain


16/17

function. s j( corresponds to a sinusoid of fre'uency (. 6nd E!s&H!s& at s j( corresponds

to the loop gain for a sinusoidal fre'uency of (. So, you can evaluate E!s&H!s& and plot its

imaginary part versus the real part. So, instead of calculating the polynomial we simply

calculate E times H for sinusoidal fre'uencies s j( and then plot the imaginary verus real.

We later go into how this plot e8actly look like, like 0 said rather vaguely that stability criteria

involved avoiding the loop gain being minus . So, this !, $& is a critical point in this whole

techni'ue, and what the +y'uist criterion states is that the plot of imaginary part of the loop

gain verses the real part of the loop gain should not encircle !, $&. That is the plot can be

something like this and if it is like this for the negative fre'uencies, it will be the mirror

image. So, in this case it is not enclosing !, $&. We will make more concrete the definition

of what enclosing means, but in this case you can clearly see that !, $& is not inside this

picture.

Fn the other hand if the picture looks something like this then it looks like !, $& is enclosed

by this plot of imaginary part of the loop gain verses real part of the loop gain which is

evaluated on the j( a8is. So, in this case the system will be unstable. So, 0 still made some

vague statements, you had to make many more things more precise which 0 will do, but what

0 like you to do is to first appreciate the advantage of this particular techni'ue.

So, before if we had to ascertain stability you have to evaluate this function and evaluate the

denominator polynomial of that and find the roots, see if the roots are positive real parts. That

is a very difficult thing to do first of all the e8pressions become very combustion, even for the

baby cases that we took that is you have two identical poles, you have to do a little bit of

work. +ow, when you have multiple poles and that too not identical to each other it becomes

very very difficult. 0nstead what is done, is to evaluate only the loop gain and that too not

everywhere not on a general form, but on the imaginary a8is that is for sinusoidal input.

So, as you know it is easy to calculate as well as easy to measure if you are only interested in

the sinusoidal study state response. When you have a system you can apply a sinusoid see

what comes out and evaluate the magnitude and phase shift of what comes out compared to

what goes in. So, this +y'uist stability criterion is a way of ascertaining stability that is

evaluating whether the poles are in the right half plane by looking at only the loop gain and

this is a tremendous advantages. 6nd that3s why this and the techni'ues that are sort of

related to this are very widely used.


17/17

"oughly speaking it involves the sinusoidal study state response of the loop gain and the

critical point of minus , and the intuitive reason for minus being there is that if the loop

gain e8actly e'uals minus , you will have instability for sure because = $)=ibecomes infinity.

So, this +y'uist criterion says something about the plot of the imaginary part of EH verses

real part of EH and make some statement which is e8actly e'uivalent to the poles being in

the right half plane. >8actly what the statement is, we will elaborate in the ne8t class and see

with all the details.

Thank you.

Documents

Use of Nyquist plot for Analog IC design