Hi, the attached circuit picture is supposed to be a gyrator with an inductance of L = C.R1.R2. However, I am am having trouble deriving this equation.

What I've tried to do so far is calculate the input impedance Zin = Vin/Iin. But no value of Iin is giving me a Zin with an inductive impedance. What I'm mostly confused by is how the output of the top op-amp interacts with the opamp below it.

So far I have:
Top Op-amp:
e- = e+ = Vin
Therefore I1 = Vin/R

Bottom Op-amp:
e+ = e- = Vin
I2 = (Vout - e+)/R1 = (Vout - Vin)/R1 = Vin/R1

I'm not sure what I have to do to get a current in terms of the capacitor, and then calculate the full input impedance.

Thanks,
Jin

 

I tackled the derivation of a gyrator circuit some time ago and posted it here.

hgmjr

 

The circuit can be redrawn as shown on this page:

http://mysite.du.edu/~etuttle/electron/elect66.htm

It's a GIC. The referenced page shows the expression for the impedance at the top of the stack, but it doesn't show how to derive it. A search on the web may turn up something. If you can't find anything helpful, post here again.

 

I'm not sure what I have to do to get a current in terms of the capacitor, and then calculate the full input impedance.

Jin

It's definitely a little tricky to think about, but that formula looks correct using the simplified model of the OP-AMP. The relations can be worked out as follows:

Vin=0.5*Vout
Vout=I2*R1+Vin
I2=s*C1*(Vin-Vo) where Vo is the output voltage of the lower opamp
Vo=Vin-R2*Iin

If you substitute and do some algebra you get Vin/Iin=s*C1*R2*R1

This looks like the impedance of an inductor Xl=sL or L=C1*R2*R1

 

I just realized something else about this circuit.

The input impedance Vin/Iin may not be strictly the correct thing to calculate.

You may instead want to calculate (Vout-Vin)/Iin since Vout and Vin are the identified terminals for connection. You get the same answer because Vout-Vin=Vin .

 

The method I used involved introducing a resistor in series with the input signal and the terminal labeled Vin.

I obtained the same results as steveb by doing this. That answer being the virtual L = C1*R1*R2.

hgmjr

 

Actually, I shouldn't have said: "The circuit can be redrawn as shown on this page", because the 2 opamps (along with 5 admittances) of the GIC can be rearranged topologically as shown in the attached .gif file.

I should have said, "...can be redrawn as one of the 24 possible GICs..."

The topology of the OP's circuit is designated B2D in the image.

An interesting question is, "why should any one topology be preferred over any other"?

If the OP is still following this thread, I would like to know where he got the problem circuit. Did it come from a text, and if so, what is the title of the text?

 

Thanks to all for your replies. I am able to get the correct inductance value now.
The circuit was part of our university electronics notes, under the heading "op amp applications," without any mathematical derivation. Actually, the notes are based on a book by Sedra & Smith.

As for whether to measure between the terminals Vin -> Vout, or just Vin -> Gnd, all it says on the matter is: "The circuit appears to the input to be an inductor to ground of a value L= CR1R2."

 

As for whether to measure between the terminals Vin -> Vout, or just Vin -> Gnd, all it says on the matter is: "The circuit appears to the input to be an inductor to ground of a value L= CR1R2."

That definitely makes more sense. It's unclear to me why they identify Vout as an output. It seems unnecessary to me, but maybe I'm missing something.

 

Thanks to all for your replies. I am able to get the correct inductance value now.
The circuit was part of our university electronics notes, under the heading "op amp applications," without any mathematical derivation. Actually, the notes are based on a book by Sedra & Smith.

As for whether to measure between the terminals Vin -> Vout, or just Vin -> Gnd, all it says on the matter is: "The circuit appears to the input to be an inductor to ground of a value L= CR1R2."

Greetings dman,

Could you post your complete work in deriving your solution for the benefit of the other members that might be interested in seeing your analytical approach?

hgmjr

 

Actually, the notes are based on a book by Sedra & Smith.

What's the title of the book by Sedra & Smith?

 

I didn't want to hijack this thread, so I posted this in the electronics chat forum.

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

 

Here is my effort at analyzing the gyrator (inductor simulator).

Comments welcomed.

hgmjr

PS. This attachment has been edited to make the corrections that electrician identified.

 

What software do you use to produce these documents? It gives good looking schematics and mathematical expressions.

One problem I see right off is that you've reversed the plus and minus inputs of A2.

Just after Eq. 1b and 3b you have the word "equally"; I think "equality" would be better.

 

I use Microsoft WORD, PC-PAINT, and EAGLEPCB.

Thanks for reviewing the document. I will make the changes that you have indicated and repost.

Thanks,
hgmjr

 

hgmjr,

May I suggest the next time you print it to the pdf, select the pdf printer properties and tell it you want b&w. I found that the color diagrams tend to lack the clarity as a b&w ... although it beats hand drawing the diagrams.

I've gone as far as printing the diagrams to pdf (b&w) before copy and pasting them into word.

 

hgmjr,

May I suggest the next time you print it to the pdf, select the pdf printer properties and tell it you want b&w. I found that the color diagrams tend to lack the clarity as a b&w ... although it beats hand drawing the diagrams.

I've gone as far as printing the diagrams to pdf (b&w) before copy and pasting them into word.

Thanks for the suggestion. I will make sure to use b&w pdf settings in the future.

hgmjr

 

OP here.
Here's how I solved the circuit.

 

Hello, Dman

First off, sorry for posting on this old thread...
I was looking for some Gyrator circuits, and I found your derivation. Although you get the right expression for the input impedance, i'm not sure if the assumptions you made to proceed with the calculations are correct.
In your report, you mentioned this:

v feedback C---v
​

vin -- R2 --- opa + -------|- R1---- vin *2

- ---- vin
​

If the op-amps are assumed to be ideal, then no current will
flow into their terminals. All
the current must
flow from Vin, through R2, the capacitor, and R1.
Thus, Iin is the same as the current
flowing through R2 and is equal to ...

An opamp will have negligible current flowing into it's inputs (+ and - terminals). However, the opamp will most likely have current flowing from/into its output terminal.
So at the node connecting R2, C, and the Opamp Output, there will be a main current flowing into the opamp, and two branch currents of R2 and C adding together.
So then the correct observations before analysis would be that

1. There is a current flowing through both the R1, and the
feedback capacitor
2. There is a current flowing through the output resistor R2,
different from the current flowing through R1
3. The noninverting amp on the top stage has a gain of 2

I've attached my analysis of the circuit.

Which for some reason give the same result as when the assumption is made that the current into R2 and R1 is one and the same. What is happening here? lol ! Am I getting the math wrong?
Alright, i can smell it already, what am I doing wrong?

~mikeska

 

If you'll search for "generalized impedance converter" (also known as GIC), you'll find some additional information about gyrators, e.g.:

http://mysite.du.edu/~etuttle/electron/elect66.htm

As to your comments about dman's analysis and his assumption about currents, I'll have to look carefully at his posted analysis. It is rather strange that he got the correct answer anyway.