OP-Amp Configuration

What purpose does any capacitor have? To pass AC and not let DC through. SO when it's used as a coupling capacitor, it lets the AC signal pass between stages, but blocks the DC from the first stage from getting into the second stage where it would upset bias. When used as a bypass capacitor, it in effect shorts the AC signals to ground, without shorting out the DC voltage.

Of course, a capacitor's ability to pass AC varies with frequency and capacitance. A small value capacitor won't do well at passing low frequencies, while a high value capacitor will.

Now given that information, that should have been covered in the books, why do you think the capacitor is there? Think about why it might be different from having the resistor going directly to ground.

Michael

Reply to
Michael Black
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Just a swag based on your description but the impedance of the series cap will fall as frequency rises, so less negative feedback for higher frequencies and correspondingly more gain at higher frequencies.

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Rich Webb   Norfolk, VA
Reply to
Rich Webb

Hi,

I was recently in a job interview and was asked to describe an OP-Amp circuit.

The OP-Amp configuration consisted of negative feedback with a resistor and capacitor connected in series to ground from the inverting input.

I haven't seen this configuration before and I was asked the purpose of the capacitor?

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M.Joshi
Reply to
M.Joshi

No DC gain.

Graham

Reply to
Pooh Bear

Just to clarify a few points:

I have come across a capacitor on the input which is to block DC; and also in the feedback loop which gives an Integrator function I believe however, in this circuit, there was a resistor in the feedback loop and another resistor and capacitor in series to ground from the inverting input?

Thinking about it now, could it form some sort of filter (Low pass)?

Pooh Bear Wrote:

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M.Joshi
Reply to
M.Joshi

How about drawing the circuit? :)

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Reply to
Sam Goldwasser

On Thu, 1 Jun 2006 17:55:47 +0100, M.Joshi put finger to keyboard and composed:

Is this it?

+---- Rfb -+ | |\ | C | | \ | |---||-- Rin -+-|- \ | _|_ | \____|__ Vout = | / Vin ---|+ / | / |/

Av = Vout / Vin = 1 + Zfb / Zin = 1 + Rfb / (Rin + 1/jwC)

At DC, C=0 and Av = 1. At high frequencies, Av = 1 + Rfb / Rin

At high frequencies the circuit behaves like a normal non-inverting amp, while at low frequencies the gain approaches unity.

- Franc Zabkar

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Reply to
Franc Zabkar

I don't think you've been listening.

Graham

Reply to
Pooh Bear

I've redrawn it here:

R1 +----/\/\----+ | | | |\ | Input ---+-+---|- \ |

/ ??-|+ / R2 \ |/ / \ | _|_ C1 --- | GND

What about the + input? And the values of C1, R1, R2?

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Reply to
Sam Goldwasser

I have attempted an ASCII drawing which can be downloaded from the following location:

formatting link

Filename: Op-Amp Schematic.txt

I hope it makes things a bit clearer - It is the purpose of the capacitor (C) in the diagram that I am not too sure of?

Sam Goldwasser Wrote:

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M.Joshi
Reply to
M.Joshi

Isn't there a resistor on the input, too?

Assuming there is, you ought to be able to do a typical op-amp analysis -- the sum of currents into the virtual ground at the inverting input will be zero.

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Reply to
CJT

I do not remember the values of all the components as I only saw the circuit for a brief moment in an interview but, as far as I can remember, the positive input was connected to a potential divider consisting of two equal value resistors to +Supply and -Supply.

The invering input was being fed from a mic pre-amp circuit which I believe was capacitively coupled and there was another resistor in series with the input.

*CJT* - Could you give me some pointers as to how I would go about an Op-Amp analysis as it has been some time since I have done one.

Thanks.

CJT Wrote:

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M.Joshi
Reply to
M.Joshi

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