Maker Pro
Maker Pro

Those peak taming resistors in Jim Thomspon's gyrator

J

Joel Kolstad

Say Jim (or anyone else),

The resistor you add in your gyrator to "tame the peaking" related to the
finite GBW of the op-amps... I'd like to understand and/or derive the value
you use, R=2/(2*pi*GBW*C). Any pointers on how do go about doing so? I'm
thinking that if I go through the math for a regular integrator using an
op-amp with finite gain and a single pole (i.e., dominant pole assumption),
the resistor you add will have some "obvious" effect on the transfer
function that causes it to move somewhat closer to 1/sC.

I know at one time I read one of your comments about this resistor to the
effect that its usage was well known?

Thanks,
---Joel Kolstad
 
J

Jim Thompson

Say Jim (or anyone else),

The resistor you add in your gyrator to "tame the peaking" related to the
finite GBW of the op-amps... I'd like to understand and/or derive the value
you use, R=2/(2*pi*GBW*C). Any pointers on how do go about doing so? I'm
thinking that if I go through the math for a regular integrator using an
op-amp with finite gain and a single pole (i.e., dominant pole assumption),
the resistor you add will have some "obvious" effect on the transfer
function that causes it to move somewhat closer to 1/sC.

I know at one time I read one of your comments about this resistor to the
effect that its usage was well known?

Thanks,
---Joel Kolstad

It adds "lead" just at the point where excess phase kicks in.

More detail tomorrow... my wife says to get my ass out of the office
;-)

...Jim Thompson
 
T

The Phantom

It adds "lead" just at the point where excess phase kicks in.

Often in analyses where effects of GBW of op amps are considered, a single pole op amp
rolloff is used, namely A/(1+st), with A the DC gain and t the time constant of the pole.
If this is the model you use in your analysis, Jim, would you give us the values of A and
t you use? Or if your model is more complex, could you give us the op amp transfer
function in a similar form, such as A/p(s), with P(s) being a polynomial in s?
 
J

Jim Thompson

Often in analyses where effects of GBW of op amps are considered, a single pole op amp
rolloff is used, namely A/(1+st), with A the DC gain and t the time constant of the pole.
If this is the model you use in your analysis, Jim, would you give us the values of A and
t you use? Or if your model is more complex, could you give us the op amp transfer
function in a similar form, such as A/p(s), with P(s) being a polynomial in s?

My OpAmp model is more complex... it's modeled like a real device...
current driven capacitor forms the pole, and controls the slew-rate.
It also has an excess-phase element. It is excess-phase that causes
the need for the peak-taming resistors.

(Otherwise, most OpAmp models just have a constant ninety degree phase
shift.)

You can obtain a copy of the OpAmp model from my website.

...Jim Thompson
 
J

Joel Kolstad

Jim,

Jim Thompson said:
It adds "lead" just at the point where excess phase kicks in.

I believe that. I spent awhile reading Schaumann/Van Valkenburg last night,
and he goes through the analysis of "the effects of finite A(s)" on a bunch of
op-amp topologies. In the case of the regular old integrator, he ends up with
the transfer function being approximately -1/sRC * 1/(1+s/w_t) (where w_t is
supposed to be omega-sub-t, the unity gain bandwidth). Hence, by changing the
capacitor into a series resistor 'r' and capacitor, he changes 1/sRC into
(srC+1)/sRC. After some straightforward algebra, he uses this newfound zero
(phase lead) to cancel the 1+s/w_t pole. The end result is r=1/(w_t*C), which
is obviously similar to your value of 2/(w_t*C) ... but of course you're not
using a pure integrator, but rather a gyrator that's trying to produce
something more like 1-1/sC out of the "would be" integrator op-amp.

Interestingly, he goes through a similar analysis for the Antoniou gyrator,
but his analysis leads him in the direction of finding component value ratios
that ideally eliminate the loss term in the simulated inductor rather than
adding compensation to the op-amps themselves.

Back to the regular old (Miller) integrator, he also shows the technique of
sticking a unity gain op-amp in the feedback loop of the integrator ("active
compensation") with guidance that this approach allows some pole/zero
cancellation (assuming matched op-amps) which can be preferable in that w_t is
a widely varying parameter and therefore you might find yourself having to
tweak passive compensation resistors a little. What do you think? Worth the
effort?

Thanks,
---Joel

P.S. -- Sorry for the typo in your last name in the subject line! :)
 
T

Tim Wescott

Jim said:
It adds "lead" just at the point where excess phase kicks in.

More detail tomorrow... my wife says to get my ass out of the office
;-)

...Jim Thompson
Does she want you to pick up the donkey crap as well?

"Passive and Active Network Analysis and Synthesis" by Aran Budak is a
very helpful book. It was the text for a very helpful class I took as a
grad student -- it was at the same time a lifesaving gut class because
the Estimation and Detection class I was taking was murder, and at the
same time it was very informative with in-depth information that I've
used ever since so it wasn't a waste of time. At any rate, the book
discusses how to analyze filters with non-ideal op-amps. The basic
technique is to make a feedback loop of it, and substitute a high-gain
integrator (or other appropriate model) for the op-amp.
 
J

Jim Thompson

Jim,



I believe that. I spent awhile reading Schaumann/Van Valkenburg last night,
and he goes through the analysis of "the effects of finite A(s)" on a bunch of
op-amp topologies. In the case of the regular old integrator, he ends up with
the transfer function being approximately -1/sRC * 1/(1+s/w_t) (where w_t is
supposed to be omega-sub-t, the unity gain bandwidth). Hence, by changing the
capacitor into a series resistor 'r' and capacitor, he changes 1/sRC into
(srC+1)/sRC. After some straightforward algebra, he uses this newfound zero
(phase lead) to cancel the 1+s/w_t pole. The end result is r=1/(w_t*C), which
is obviously similar to your value of 2/(w_t*C) ... but of course you're not
using a pure integrator, but rather a gyrator that's trying to produce
something more like 1-1/sC out of the "would be" integrator op-amp.

Interestingly, he goes through a similar analysis for the Antoniou gyrator,
but his analysis leads him in the direction of finding component value ratios
that ideally eliminate the loss term in the simulated inductor rather than
adding compensation to the op-amps themselves.

Back to the regular old (Miller) integrator, he also shows the technique of
sticking a unity gain op-amp in the feedback loop of the integrator ("active
compensation") with guidance that this approach allows some pole/zero
cancellation (assuming matched op-amps) which can be preferable in that w_t is
a widely varying parameter and therefore you might find yourself having to
tweak passive compensation resistors a little. What do you think? Worth the
effort?

Interesting... I'll look into it. Reminds me... I have a paper by
Geiger and Bailey, "Integrator Design for High-Frequency Active Filter
Applications", IEEE Transactions on Circuits and Systems, Vol. CAS-29,
No. 9, September 1982, that addresses using multiple OpAmps as
compensation.
Thanks,
---Joel

P.S. -- Sorry for the typo in your last name in the subject line! :)

No problem, I do it myself... typing two fingers on each hand... one
hand gets ahead of the other ;-)

...Jim Thompson
 
J

Joel Kolstad

Hi Tim,

Tim Wescott said:
"Passive and Active Network Analysis and Synthesis" by Aran Budak is a
very helpful book.

I have a copy (hooray for re-issues!), I'll take a look. From your
description, it sounds like he's analyzing non-ideal op-amps the same way as
Van Valkenburg; it'll be interesting to compare results.
The basic
technique is to make a feedback loop of it, and substitute a high-gain
integrator (or other appropriate model) for the op-amp.

Van Valkenburg must have at least a dozen repetitions of "the op-amp is really
an integrator!" in his book.

---Joel
 
T

The Phantom

My OpAmp model is more complex... it's modeled like a real device...
current driven capacitor forms the pole, and controls the slew-rate.
It also has an excess-phase element. It is excess-phase that causes
the need for the peak-taming resistors.

(Otherwise, most OpAmp models just have a constant ninety degree phase
shift.)

You can obtain a copy of the OpAmp model from my website.

Would that be the OP27 model?

Could you tell me what the *excess* phase shift is at a few frequencies, such as .1, .5,
1, 2, 5 MHz? This would help me make sure the model is working right.
 
J

Jim Thompson

Would that be the OP27 model?

Could you tell me what the *excess* phase shift is at a few frequencies, such as .1, .5,
1, 2, 5 MHz? This would help me make sure the model is working right.

The OP27 DOES have excess phase in it, but the better version is
"Op-Amp-Config.zip " on the Subcircuits & Symbols page.

...Jim Thompson
 
T

The Phantom

It adds "lead" just at the point where excess phase kicks in.

Are you saying that the peaking is caused by the excess phase shift? So that if we remove
the excess phase shift from the op amp model, the peaking will go away?
 
J

Joel Kolstad

Phantom,

The Phantom said:
Are you saying that the peaking is caused by the excess phase shift?

For the closed loop response of an op-amp you have the standard
T(f)=A(f)/(1+A(f)b), where A(f) is the open loop gain and b is the feedback
ratio, which I've assumed to be something like ideal resistors that aren't
frequency dependent. As f goes from zero to the unity gain frequency w_t of
the op-amp, A(f) goes from some large real value to magnitude 1, angle "180
degrees minus the phase margin," which is something starting to approach -1.
In other words, with zero phase margin, the op-amp would be an oscillator, and
with some phase margin you get peaking with the amount of peaking be inversely
proportional to the phase margin. One thing to note from the T(f)=...
equation is that the peaking is worst with the lowest gains (b approaches 1);
witness the availability (and desirability) of uncompensated op-amps require
gains >1 (or an external compensation network) to maintain stability.

This is explained in much more detail in many books, including "Intuitive IC
Op-Amp Design."
So that if we remove
the excess phase shift from the op amp model, the peaking will go away?

Yes it will!

---Joel
 
T

The Phantom

Phantom,



For the closed loop response of an op-amp you have the standard
T(f)=A(f)/(1+A(f)b), where A(f) is the open loop gain and b is the feedback
ratio, which I've assumed to be something like ideal resistors that aren't
frequency dependent. As f goes from zero to the unity gain frequency w_t of
the op-amp, A(f) goes from some large real value to magnitude 1, angle "180
degrees minus the phase margin," which is something starting to approach -1.
In other words, with zero phase margin, the op-amp would be an oscillator, and
with some phase margin you get peaking with the amount of peaking be inversely
proportional to the phase margin. One thing to note from the T(f)=...
equation is that the peaking is worst with the lowest gains (b approaches 1);
witness the availability (and desirability) of uncompensated op-amps require
gains >1 (or an external compensation network) to maintain stability.

I wasn't looking for an explanation of the theoretical basis for peaking, but rather
examination of the actual behavior of the Thompson gyrator bandpass as shown in the .pdf
file, "GyratorFilters-BP-LP-HP", on Jim's website.
This is explained in much more detail in many books, including "Intuitive IC
Op-Amp Design."


Yes it will!

It doesn't seem to. It would be interesting if you or someone else would run the
simulations I describe below and see what results are obtained.

I simulated Jim's bandpass filter with and without the excess phase block he uses in his
OP27 model. That excess phase block doesn't start to add excess phase until you get up to
about 1 MHz, so I was wondering how it could cause peaking down at 20 kHz, which is the
center frequency of the bandpass filter. After I got the model working, I decided to try
to duplicate the family of response curves shown on page 3 of the .pdf file. I swept the
GBW with smaller steps and noticed even more peaking at frequencies that Jim didn't show.
In fact, with certain op amp parameters, you can get peaking of greater than 130 db (I got
peaking of a factor of 3.5 million!). Just set the op amp DC gain to 1000000, the GBW to
279069.248 Hz and you should see a huge peak at 17437.0011 Hz (this is without excess
phase in the op amp model). With excess phase, the huge amount of peaking remains, but is
*slightly* changed in frequency.

I swept the GBW from 25 kHz to 100 MHz and plotted the results, with and without excess
phase; I used the excess phase block from Jim's OP27 model. I see two regions of
substantial peaking. I've posted the plots over on ABSE.
 
Top