# BJT Gyrator

Discussion in 'General Electronics Chat' started by Jony130, Feb 9, 2010.

1. ### Jony130 Thread Starter AAC Fanatic!

Feb 17, 2009
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I have a schematics of a bjt gyrator

And I start matrix analysis with help of a Mathematica 7, but I but stuck.
I wont to find Zin and then find equivalent inductance.
But the expression of Zin is is awfully complicated and its very faraway from LC1*R1*R2.
So I have a question.
Maybe somebody has idea how to analysis this kind of circuits. Or maybe any kind of document that
describes this or similar circuit.

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Jul 17, 2007
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3. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I would proceed as shown in the attachment. This circuit doesn't work very well with just a single BJT. A wide band opamp would work better.

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4. ### Jony130 Thread Starter AAC Fanatic!

Feb 17, 2009
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OK, THX.
Can I plot impedance Zin(s) in Mathematica?
And how can I find Rs = ( (re+R1/β) || R3 ) + R2 for Fo
Fo=1/(2*pi*√(L*Cr))
Or maybe to find Rs I need to add Cr ?

5. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Here's the Mathematica solution to your request.

This is a good problem to show the power of using a symbolic algebra program.

The special symbol for SQRT(-1) in Mathematica, along with some other special symbols, can be conveniently be obtained from a Palette. Click on "Palettes" in the menu bar and select "Basic Math Assistant".

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• ###### BJT_Gyrator4.png
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Last edited: Feb 10, 2010
6. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The "gyrator" on the Wikipedia page (http://en.wikipedia.org/wiki/Gyrator) is shown as equivalent to an inductor with a series resistance. Strictly, this resistance should be shown as varying with frequency.

If another passive circuit is used as the equivalent to the opamp circuit, having a parallel resistance as well as a series resistance, then the two resistors can be constant with frequency.

This is shown on this web page:

7. ### Jony130 Thread Starter AAC Fanatic!

Feb 17, 2009
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THX The Electrician for help and tip.
So if we have Zin then we can find equation for L?
Zin=√ (Rs^2 + ωL^2)
I'm stupid XL is imaginary part of Zin, right?

Last edited: Feb 11, 2010
8. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Yes, XL is the imaginary part of Zin, all right.

You might think that the command Im[] should extract the imaginary part of an expression, and in fact it works well if the expression is not symbolic. But for a symbolic expression, Im[] may not work because Mathematica thinks the symbolic variables may be complex themselves. You could probably use the "Assume" option to tell Mathematica that the variables are only real.

But, there's a better way. The ComplexExpand command allows you to extract the real and imaginary parts of a symbolic expression, and it already assumes the variables are real.

I've attached an image showing how to extract the real and imaginary parts of the simplified expression for Zin. You should try this technique on the exact, un-simplified, expression for Zin. You will get a very large, complicated, expression.

I've also attached a second image showing the use of the real and imaginary parts of the simplified Zin to compare in a plot with the exact expression.

You can use the command Abs[] to do the same as √ (Rs^2 + ωL^2).

I've added an expression for the magnitude of Zin using Abs[] to the plot from post #5.

Looking at this last plot, we can see that Rs is so large that it dominates the impedance at low frequencies. It's only when the frequency is greater than about 100 Hz that the circuit behaves like an inductor, and then only for about one decade of frequency.

By the way, there's another type of inductor simulator at:

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• ###### BJT_Gyrator7.png
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Last edited: Feb 11, 2010