Q of a super-inductor

Hey, if you guys were plotting Q of a simulated inductor (e.g,. from a gyrator), would you plot Q as just staying at infinity once the resistance goes negative? Or would you just plot it as negative Q?

It's interesting to me that the resistance from gyrator inductors start heading south (drop below 0 ohms with a vengeance) generally well before the imaginary part starts to degrade similarly; anyone for an intuitive explanation as to why this is?

---Joel

Reply to
Joel Kolstad
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If you'll thumb back through the old threads about gyrators I think you'll find that I attributed it to excess phase in the OpAmps.

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Joel:

So called "Q-enhancement" is common in active RC networks, is a well known phenomena and was studied and written about ad infinitum back in the 70's.

As Jim [Thomson] points out Q-enhancement is usually caused by excess phase shift in the OpAmps.

The search for better active RC circuits [gyrators, gic's, etc...] always led to inventions of novel circuitry that somehow reduces that excess phase shift.

There are passive compensation methods that "tune out" the Q-enhancement, but...

The "best" circuits use active compensation of the OpAmps wherein the excess phase of one [of usually two] OpAmp in a circuit approximately cancels that of it's paired [matched] OpAmp.

The Akerberg-Mossberg circuit was the first such 'active compensation" circuit that appeared in the literature.

There is a lot of information on excess phase compensation, and a catalogue of the "best" circuits, including the celebrated Antoniou gyrator invented by Andreas Antoniou, now at University of Victoria, in Victoria, BC Canada, laid out, analysed and described in the book:

Adel S. Sedra and Peter O. Brackett, "Filter Theory and Design: Active and Passive", Matrix Publishers, Champaign, IL 1978.

In fact the "best" gyrator circuit, the one invented by Andreas, is depicted in the cover artwork on that book.

FWIW...

-- Pete

Peter O. Brackett Indialantic By-the-Sea, FL

Reply to
Peter O. Brackett

Peter O. Brackett wrote: (snip)

Another book that concentrates on active excess phase compensation is "Theory and Design of Linear Active Networks" by Natarajan. The performance of almost every opamp circuit described is compared to a similar one that has the excess phase shift compensated, to show how much more ideally it functions.

Reply to
John Popelish

I would plot the Q as going negative, or I would plot 1/Q as a nice straight line. Plotting the Q as staying at infinity would indicate a lossless, mild-mannered circuit rather than the threat to system stability that you'd really be dealing with.

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

the

So perfect, they seem a bit like Schroedinger's cat, always existing in a state of of crypto-stability. I've seen unplottable drifting Q's of maybe

1000-5000 one minute and then to infinity and beyond when the temperature changes. Despite excess phase, pos or neg inductor "loss" resistance, appears from normal practical circuit layout. Predictable enough to allow fixing Q at say a much less perilous 100X. Or conversely, guarantee (in oscillator format) a gentle startup. john
Reply to
John Jardine.

An oscillator in a filter's clothing, eh? :-)

Thanks, I think I will just plot it as going negative... maybe on a log scale (something like sgn(Q)*log(abs(Q)))...

Reply to
Joel Kolstad

Thanks for the advice Peter, I do have your book sitting on the bookshelf and will take another look at it shortly (it was actually the first book I read that went through the whole approach of using 1/s as an op-amps gain if you're doing hand calculations and are looking to come up with the pole-zero locations of a given op-amp circuit... funny how somehow this never made it into my coursework in college...)

---Joel

Reply to
Joel Kolstad

"Jim Thompson" wrote in message

OK... I'll have to drop your configurable op-amp into a simulation soon (since it of course includes that excess phasE) and see how it compares (using data sheet values) to simulations done with, e.g., Linear Tech. SPICE models.

---Joel

Reply to
Joel Kolstad

Doesn't 'active' mean 'might oscillate'?

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

Probably. :-)

That reminds me of a co-worker who is hesitant to use IIR filters since their stability is not guaranteed (as FIR filters are) nor completely trivial to predict once everything is quantizied (i.e., the Z-transform of the unquantized system don't tell you the whole story)... I tell him that he might as well protest any use of active continuous-time circuits containing feedback, since it's not completely trivial to predict oscillation given imperface real-world components either...!

Reply to
Joel Kolstad

Why are you using Q anyway? Just use the resistive (potentially negative resistance) and reactive components you know you have, if you're using a simple linear model, or better, put in a full model of the gyrator with decent models for the op amps, and you won't have to worry about "negative Q". As Tim was pointing out, remember that there are things we call "negative resistance oscillators." A "Q multiplier" (used in the old days to add selectivity to a receiver's IF section) could, of course, be made to oscillate by advancing the gain too much. Q is often an over-worked term (especially when people try to apply it to multiple-resonator filters, but that's a topic for another thread).

Cheers, Tom

Reply to
Tom Bruhns

Don't forget anything with noise.

I discuss predicting the effects of quantization in IIR filters in my book, by the way, including finding bounds on limit cycles (although I bet I don't say it quite so clearly -- hmm).

Of course, you can't completely characterize your FIR filter performance with the z transform, either -- you can just come a lot closer because the quantization effects are less severe.

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

Because sometimes plotting Q is more useful than plotting re(Z), im(Z) or whatever... My original question was prompted just by building a bunch of "probe" components for SPICE that you could drop down on a schematic to plot various "useful" things; I was using a gyrated inductor as the test case for them.

---Joel

Reply to
Joel Kolstad

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