# Gyrator for inductance measurement.

Discussion in 'General Electronics Chat' started by Wendy, Nov 7, 2008.

1. ### Wendy Thread Starter Moderator

Mar 24, 2008
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A gyrator will convert a capacitance into a pseudeo inductance.

Wikipedia article: http://en.wikipedia.org/wiki/Gyrator

I am under the impression that it is reversable, it will convert an inductance into a pseudeo capacitance.

Is this correct? What would be some good values (including op amp) to do this do you think?

My thought is it might be a good way to use the capacitance meter on a DVM to measure inductance.

2. ### steveb Senior Member

Jul 3, 2008
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Bill,

Yes, you can make an effective capacitance from an inductance. Your idea is creative. There probably is a way to use a capacitance meter to measure inductance with that circuit, but you would have to make sure that the method that the meter uses is compatible with the circuit. The frequency range and voltage biasing would have to be compatible. As you know, the impedance simulation is only accurate within the bandwidth limitations of the OPAMP.

I guess the argument would be that it could be more useful to design a circuit that converts inductance into a DC volts so you can use a voltmeter, since voltmeters are more common than capacitance meters. However, there are always special cases and exceptions.

I'm not an expert on this, but my understanding is that simulating inductors with capacitors is more useful than the reverse. My guess is that large coils have more problems with nonlinearity, frequency dependence and stray resistance, while capacitors are quite good nowadays. But again, there are always special cases.

It could be fun to run through the calculations to see what range of capacitance you can create with realistic OPAMPS, coils and resistors.

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The circuit shown on the wikipedia page isn't really a gyrator; it's an inductor simulator.

If you work out the input impedance (using the complex variable s instead of jw) you get:

s R RL C + RL
-------------
s RL C +1

An inductor should have an impedance of s L, so for that circuit you can see that the product R RL C is the value of the simulated inductor L, and RL is the value of a resistance in series with the inductance. Ideally RL should be zero, but since real inductors usually have some resistance, it's a useful simulation.

The denominator should be just 1, but the additional term of s RL C is small if RL is small, so, again, the simulation is useful.

But, if we replace the capacitor with an actual inductor L, the effective input impedance of the circuit is:

s RL L + R RL
-------------
s L + RL

A real capacitor should have an impedance of 1/(s C), and all those other terms in there prevent a good approximation. If the circuit were behaving as a real gyrator, the input impedance would be just 1/(s L), where the inductor L having a normal impedance of s L has its impedance inverted to 1/(s L), without all those extra terms.

I don't know of a really good gyrator having only one op amp. The good ones have at least two op amps.

4. ### Wendy Thread Starter Moderator

Mar 24, 2008
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I'm thinking of a one to one convertion, if it is practical. Since most coils are in the nano to micro range, which is a good fit for most DVM style capacitance meters.

Since most if not all DVMs use a dual slope A/D convertion technique I figure they adapt it to a constant current source doing the same.

If you know of a schematic that is better than the Wikipedia please post it.

5. ### steveb Senior Member

Jul 3, 2008
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Bill,

I recommend this circuit for impedance conversion. You can create a wide range of effective impedances with this. The circuit and formula is in terms of general impedances Z, you can then choose resistors, capacitors and inductors (or combinations thereof) for any of those elements.

If you place a coil in for Z2 or Z4 and put resistors everywhere else, you can simulate a capacitor.

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6. ### The Electrician AAC Fanatic!

Oct 9, 2007
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This is a slight variation on the circuit shown in that other thread:

There are 24 possible variations.

Bill said "I found gyrators facinating in college. I think I understood the math then, but it eludes me now. Such simple looking gadgets too."

A really fascinating question is the one I posed in that other thread:

Is there any reason to prefer any one of the variations over another?

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7. ### steveb Senior Member

Jul 3, 2008
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Yes, I would prefer the first one, so that I don't have to analyze the other 23!

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
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What if one of the others had some desirable property that the first lacked?

For example, if you simulate an inductor and then add another real capacitor and resistor at the top node to create a 2nd order filter, the well-known Q enhancement effect might be better for some particular variation.

How might one discover this fact?

Your two .pdf files appear to be from the days when you were learning to do SFG analyses.

You might find it interesting to do the analysis for the B2D variation which the OP in the other thread was using.

9. ### steveb Senior Member

Jul 3, 2008
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LoL, it was just a joke. Of course it would be good to know if one has advantages. Also, I did do the analysis for the other one and provided some direction to the OP based on that. It would take some work to figure out what advantages there might be for the best of the 24 circuits, and if the non-idealness of the OP-amp ties into the choice of the best circuit. I would be willing to do that work if I had an application that needed the circuit.

Another thing to consider is that analysis alone may not reveal the best circuit. It may be necessary to actually build all circuits and try different types of OPAMPS.

10. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The non-ideality of the opamps is crucial.

As it happens, every one of the 24 variations has exactly the same driving point impedance at the top of the resistor string when the opamps are ideal, so the only differences in performance are due to the opamp non-idealities.

I've built a number of them and my experience with these GICs has been that at near audio frequencies, typical opamp models (usually just a single pole roll off is enough) give results completely consistent with actual hardware.

11. ### steveb Senior Member

Jul 3, 2008
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Well, you seem to be the best one here to answer your question. You know it's the non-idealness of the OPAMP that will create difference in performance. And, you seem to know that the single pole role off model is enough and gives results completely consistent with the acutal hardware. So, just analyze all 24 circuits and tell us which is best.

Or, is the point of your question to request a division of labor and have each of us calculate a few circuits. If so, count me in.

12. ### The Electrician AAC Fanatic!

Oct 9, 2007
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There a few people on this forum who seem to enjoy (and are good at) circuit analysis.

The usual problems posted are relatively simple homework problems.

So, the point of my question is to stir up some curiosity about a more interesting problem.

When I first discovered that all 24 circuits had the same driving point impedance with ideal opamps, I was rather surprised. Len Bruton discussed some of these GICs in a paper in the IEEE Transactions on Circuit Theory in about 1969, but he didn't discuss them all.

For instance, here's a point of curiosity: the circuit in the .pdfs you posted can simulate an inductor with a capacitor in either the Z2 or the Z4 position. There is no difference in performance if the opamps are ideal. It's an interesting exercise to discover what differences there are if the opamps are assumed to have a simple one pole roll off.

13. ### steveb Senior Member

Jul 3, 2008
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I agree. It is an interesting question and exercise. However, I'm still confused. Is this a question that you have already answered for yourself? Or, are you really asking if someone can tell you the answer? If you do know the answer, I would ask that you please tell us simply because most of us don't have the spare time to figure this out. If you don't know, I could spare time to calclate 3 or 4 circuits, and if a few other people are willing to do the same, we could find out the answer.

14. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I know some of the answers. I'm not asking anyone to do the work for me. It's purely for the enjoyment of those who can do it, and want to do something more than a simple homework problem.

Only do it if you can afford to devote the time to it, and if you will enjoy it.

I can tell you that as far as I can determine, there is a definite difference in performance between the two capacitor positions when the opamp isn't ideal.

I don't want to spoil the fun by just giving away the answers. Besides, I'm interested in seeing if anyone else gets the same result I do; I could have made a mistake.

I'll be away from the computer for a few hours.

15. ### steveb Senior Member

Jul 3, 2008
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I would enjoy it, but can't afford to devote too much time right now. But, somehow you've got me intrigued, so I may do a portion of it.

16. ### Wendy Thread Starter Moderator

Mar 24, 2008
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What resistors should I use for the resistor, do you think? I'll try building something simple.

17. ### steveb Senior Member

Jul 3, 2008
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I think you can use the typical resistor range for OPAMPS. If you just want to start with basic values that demonstrate the circuit, maybe 10K is good. You can use the resistor ratios as gain terms to help control the scaling of the impedance value. For example, a variable resistor would allow you to make a tunable coil or inductor.

The simple formula for impedance Zin=Z1*Z3*ZL/Z2/Z4 helps you decide, but you may also want to look at the more complicated formula which includes the OPAMP gain (A). The resistor ratios show up there too and become important for the high frequency performance of the circuit.

18. ### steveb Senior Member

Jul 3, 2008
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I have a general question about the 24 circuits. Have you built any of the circuits which have the opamps installed with positive feedback? If so, were you able to get stable operation?

I made a general model for 4 circuits A1a, A1b, A1c and A1d and used two scale factors (for feedback polarity) to control which circuit was modeled (i.e. set the constants to plus or minus one). Perhaps I made a mistake, but the equation I got seems to predict unstable operation for A1b, A1c and A1d. In other words I get a pole in the right half plane unless both opamps are configured with negative feedback as in A1a.

This was my initial concern when I first saw these circuits, however, we know that positive feedback can be stable if configured correctly (with counteracting negative feedback). I just assumed this would be true here, but the equation seems to make stability impossible under typical conditions.

I dont' want to post my results with a mistake, so I thought I'd check first. Your posting of 24 circuits with 18 of them with positive feedback seems to suggest that I made a mistake somewhere.

Last edited: Nov 10, 2008
19. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Good that you noticed about the feedback. Even A1A has some positive feedback, as do they all, so it's a matter of the balance of negative and positive feedback, as you said.

Exactly what did you analyze? The circuits as they are shown can simulate an inductor with either Y2 or Y4 as a capacitor, but then you also have to decide if you want to examine open circuit stability, as the circuits are shown, or short circuit stability, with the top node shorted, or some intermediate condition, with a resistor or other load connected to the top node.

So, did you analyze the driving point impedance at the top node of the circuits as they stand? One should also look at the impedance at the other nodes which are not connected to an opamp output; it's possible to see a stable impedance at one node but not another.

What did you use for your opamp open loop gain? I used the expression:

A01/(1+s*t1), with A01 being the DC open loop gain and t1 being the time constant of the single pole roll off, and similarly for the other opamp, A02/(1+s*t2).

I encountered a band pass filter using A3A. There was a single resistor, R0, in series with a real capacitor (Cext), and finally in series with the inductor simulator. Input was applied at the loose end of R0, and output taken at the other end. The whole thing simulated a series resonant tank, and the response is just the typical peaked 2nd order bandpass response.

For that circuit, the component values were R0=78700, Cext=1 nF, Z1=7870, Z2=1 nF, Z3=10k, Z4=10k, Z5=8060.

Analysis showed that it was marginally unstable, with a very small region of instability depending on the gain bandwidth (GBW) of the opamps.

I distrusted the analysis, so I actually built this circuit (with LM101 opamps, so I could vary the GBW of the opamps), and found that it did indeed have this instability.

The worry is that upon power up, the opamp's gain may pass through the region of instability as the supply voltages come up. The circuit may begin oscillating, and stay locked in the oscillation. I didn't see this happen, but if it were my design, I would be worried that I couldn't prove that it wouldn't happen for some opamps. (I had to adjust the opamp GBW to the region of instability for full supply voltages to make it happen)

The region of instability is somewhat difficult to find, but if you use something like Matlab and plot a 3D plot of the real part of the transfer function in the Z axis versus frequency, and opamp GBW on the X and Y axes, you can find it.

The other circuits I have only simulated, but simulation can find the grossly unstable configurations. The interesting cases are the marginally stable ones.

If you do a SPICE simulation, be sure to try a short, an open, and some resistance such as perhaps 10k at the top node. Also, be sure to have an initial condition on the capacitor (I had the simulation start with .1 volt on the cap) to start any oscillation.

The OP in that other recent gyrator thread showed a circuit using B2D, which my analysis shows has considerable difference depending on whether the capacitor is at the Y2 or the Y4 position. You might try having a look at that one.

Last edited: Nov 10, 2008
20. ### steveb Senior Member

Jul 3, 2008
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Maybe we are thinking about different things when we say positive feedback. As far as I can tell A1A has only negative feedback. I base this on the SFG I showed previously. There are 3 loops shown and all have negative value. Similarly, the other examples you mention, A3A and B2D, seem like they would have only negative feedback.

Yes, that's a good question because it's difficult to study all possibililties. I derived the expressions for general impedance Z, but at the end make them all resistors (actually, Z1 can be anything). It's possible that if that assumption is not made, there may be stability, but it really looks difficult to achieve in the cases that have the positive feedback.

Yes, only the top node, but still if the top node is unstable, I would be uncomfortable using that circuit.

I basically used the same open loop gain as you did

I'll post my analysis a little later when I get home. Unfortunately, it's hand written and hard to follow, but based on what you said, it may be essentially correct, or at least on the right track.

Last edited: Nov 10, 2008