OP-Amp Configuration
 

M.Joshi ) writes:
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?

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

OP-Amp Configuration
 

On Wed, 31 May 2006 02:06:47 +0100, M.Joshi
wrote:


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?

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.

--
Rich Webb Norfolk, VA

Op-Amp Configuration
 

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?

OP-Amp Configuration
 

"M.Joshi" wrote:

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?

No DC gain.

Graham

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 (High pass)?

OP-Amp Configuration
 

M.Joshi writes:

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)?

How about drawing the circuit? :)

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OP-Amp Configuration
 

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


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)?

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
--
Please remove one 'i' from my address when replying by email.

OP-Amp Configuration
 

Franc Zabkar wrote:

Is this it?

I don't think you've been listening.

Graham

OP-Amp Configuration
 

Pooh Bear spake thus:

Franc Zabkar wrote:

Is this it?

I don't think you've been listening.

Eggs-ackley: I think you missed this from the OP's posting:

with a resistor
and capacitor connected in series to ground from the inverting input



--
Any system of knowledge that is capable of listing films in order of
use of the word "****" is incapable of writing a good summary and analysis
of the Philippine-American War. And vice-versa. This is an inviolable rule.

- Matthew White, referring to Wikipedia on his WikiWatch site
(http://users.erols.com/mwhite28/wikiwoo.htm)

OP-Amp Configuration
 

M.Joshi writes:

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

http://three.fsphost.com/Joshi

I've redrawn it he

R1
+----/\/\----+
| |
| |\ |
Input ---+-+---|- \ |
| | ---+----o Out
/ ??-|+ /
R2 \ |/
/
\
|
_|_
C1 ---
|
GND

Filename: Op-Amp Schematic.txt

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

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Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
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Important: Anything sent to the email address in the message header above is
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I have attempted an ASCII drawing which can be downloaded from the following location:

http://three.fsphost.com/Joshi

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?

OP-Amp Configuration
 

M.Joshi wrote:
I have attempted an ASCII drawing which can be downloaded from the
following location:

http://three.fsphost.com/Joshi

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?

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.



Sam Goldwasser Wrote:

M.Joshi writes:

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)?

How about drawing the circuit? :)

--- sam | Sci.Electronics.Repair FAQ: http://www.repairfaq.org/
Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
+Lasers | Sam's Laser FAQ: http://www.repairfaq.org/sam/lasersam.htm
| Mirror Sites: http://www.repairfaq.org/REPAIR/F_mirror.html

Important: Anything sent to the email address in the message header
above is
ignored unless my full name AND either lasers or electronics is
included in the
subject line. Or, you can contact me via the Feedback Form in the
FAQs.





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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.