# Opamp Circuit with Feedback Problem

Discussion in 'Electronic Basics' started by Guest, Dec 19, 2003.

1. ### Guest

8
0
Nov 7, 2014
Hello,
In the Electronics Club of our university, we are weekly solving
several problems at our meetings. But this time, noone could solve
this problem.

http://it.isikun.edu.tr/tango/electronics/circuit.jpg

If anyone is able to solve this, we would be grateful because many
friends of mine are on the job but still no solutions available.
(although there are speculations about the solution).

Thank you very much,
Electronics Club

2. ### BanGuest

|| Hello,
|| In the Electronics Club of our university, we are weekly solving
|| several problems at our meetings. But this time, noone could solve
|| this problem.
||
|| http://it.isikun.edu.tr/tango/electronics/circuit.jpg
||
||
|| If anyone is able to solve this, we would be grateful because many
|| friends of mine are on the job but still no solutions available.
|| (although there are speculations about the solution).
||
|| Thank you very much,
|| Electronics Club

it is very simple:
Vo= Vdc + A*V1 + B*V2
Vdc is 10V, thats the voltage on the non inverting input. The DC-gain is 1
A = ((10k||20k)/(10k||20k)+77k)*(100k/10K) = 0.797
B = -(100k/10k) = -10
assuming we are at at high frequency, the RC-time constants you should be
able to insert yourself, I'm not gonna do all the work.
Hey what kind of university is that where you are not even able to solve

Last edited by a moderator: Dec 11, 2014
3. ### Don PearceGuest

Almost right, but not quite(a bracket needs moving, and the last term
is wrong):

A = ((10k||20k)/(10k||20k+77k))*((100k+10k)/10K) = 0.876

d

_____________________________

http://www.pearce.uk.com

Last edited by a moderator: Dec 11, 2014
4. ### BanGuest

||||| Hello,
||||| In the Electronics Club of our university, we are weekly solving
||||| several problems at our meetings. But this time, noone could solve
||||| this problem.
|||||
||||| http://it.isikun.edu.tr/tango/electronics/circuit.jpg
|||||
|||||
||||| If anyone is able to solve this, we would be grateful because many
||||| friends of mine are on the job but still no solutions available.
||||| (although there are speculations about the solution).
|||||
||||| Thank you very much,
||||| Electronics Club
|||
||| it is very simple:
||| Vo= Vdc + A*V1 + B*V2
||| Vdc is 10V, thats the voltage on the non inverting input. The
||| DC-gain is 1 A = ((10k||20k)/(10k||20k)+77k)*(100k/10K) = 0.797
||| B = -(100k/10k) = -10
||| assuming we are at at high frequency, the RC-time constants you
||| should be able to insert yourself, I'm not gonna do all the work.
||| Hey what kind of university is that where you are not even able to
||| solve this simple task?
||
|| Almost right, but not quite(a bracket needs moving, and the last term
|| is wrong):
||
|| A = ((10k||20k)/(10k||20k+77k))*((100k+10k)/10K) = 0.876
||
|| d
||

THX for the correction Don. I was too fast!
A = (10k||20k)/(10k||20k+77k)*(1+100k/10K) = 0.876

ciao Ban
Bordighera, Italy
electronic hardware designer

Last edited by a moderator: Dec 11, 2014
5. ### GenomeGuest

20K and 10K off 30V puts the non-inverting (plus bit) at 10V. That's Mr Ohms
law. V = IR. 30 volts across 20K plus 10K iz 30 volts across 30K which gives
a whole 1mA of current. 1mA of current through a 10K resistor gives 10 of
the volts.

Mr 10uF capacitor blocks the DC path to Mr V1 so Mr 77K does not enter into
the DC party.

Mr Op-amp sticks his output so that his inputs (plus and minus bits) sit at
the same volts. Mr other 10uF capacitor blocks the DC yellow brick road to
Mr V2 so Mr Op-amp can stick his minus bit the same as his plus bit without
driving current through his 100K feedback resistor bit.

No current times 100K is no volts. So Mr Op-amp sticks his output at 10 of
the volts.

If Mr V1 goes for a wobble then his 10uF friend couples his wobble to Mr 77K
resistor.

If Bill is the 20K resistor and Ben is the 10K resistor then they find their
way to Ground and become parallel with a value of (Bill x Ben)/(Bill + Ben).
That's because Mr Bill might be connected to 30V..... but 30V is a voltage
source connected to ground. Voltage sources have no Bill or Ben.

Try it like this.

Call Martha a volt and let her wiggle her ass across Bill. The sexual
excitement, current, for Bill is Martha/Bill. Now we are probably talking a
bit of face sitting or getting into oral sex here.......

Since Martha is a woman she can multitask and wiggle her volt over Ben at
the same time. That gives a Martha/Ben.

The total sexual excitement is Martha/Bill + Martha/Ben. Cross pollinate and
you get.... Martha x Ben/Bill x Ben + Martha x Bill/ Bill x Ben which
makes....

(Martha x Ben + Martha x Bill)/(Bill x Ben) is the current

Extract a Martha and

Martha x (Ben + Bill)/(Bill x Ben) is the current......

Obviously Martha has a strapon and Bill and Ben are begging for it.

I think I've lost the plot here....

DNA

Last edited by a moderator: Dec 11, 2014
6. ### Walter HarleyGuest

Isik University, Istanbul. The palimpsest upon which he drew the schematic
is interesting - take the image, mirror and invert it to better read. Best
of luck to them!

7. ### John WoodgateGuest

I read in sci.electronics.design that Walter Harley
AIUI, they are much better at electrical power and BIG THICK cables.

8. ### Russell MillerGuest

That has to be the most creative and least informative method I've ever seen
for circuit analysis. Though I will admit to almost falling off my chair
laughing.
Think so?

--Russell

9. ### Dan DunphyGuest

Nobody picked up on the source impedance mismatch with the feedback
resistor. This will create a large offset on any op amp with any input
current. Yes, dcout is 10V with ideal op amp.

My advice may be worth what you paid for it.

Last edited by a moderator: Dec 11, 2014
10. ### Guest

8
0
Nov 7, 2014
Hello,
"Walter Harley" <> wrote in message news:<[email protected]>...
> "Ban" <> wrote in message
> news:I1CEb.13991\$...
> > Hey what kind of university is that where you are not even able to solve
> > this simple task?

>
> Isik University, Istanbul. The palimpsest upon which he drew the schematic
> is interesting - take the image, mirror and invert it to better read. Best
> of luck to them!

In fact, we are not the university Electronics club, I know nothing
about Electronics. I'm an IT student, as you may notice from the
I made up this Electronics club thingy because noone will answer such
a simple question, because it is obviously a homework for
non-electronics engineering classes.

The reverse page is a Requirement Analysis Document of a software
project, which is an assignment of a student. It's nothing serious,
but yeah, the English and the goals are funny

Sorry for the lies..

Thanks for the answers though, I have understood it.

11. ### Ross MacGuest

Very Funny!....and creative too!

Last edited by a moderator: Dec 11, 2014
12. ### Jim ThompsonGuest

See OpAmpCktWithFeedbackProblem.pdf on the S.E.D/Schematics page of my
website.

...Jim Thompson