# Ops Amp

Discussion in 'Homework Help' started by forbi, Sep 20, 2012.

1. ### forbi Thread Starter Member

Sep 11, 2012
37
0

Hi,
how do you do an analysis on this circuit to get the gain = 0?

this is like a voltage follower but with a resistor at the input terminal.

i know the input resistance is 160k ohm and output resistance is 0 ohm.

cos there is a common ground between the input voltage, 20k resistor and Vo

----10k---100k---|--------100k-----|
|xxxxxxxxxxxxxxxx| xxxxxxxxxxxxxxxxx|
|xxxxxxxxxxxxxx | xxxxxxxxxxxxxxxxx|
Vi xxxxxxxxxxxxxx 20k xxxxxxxxxxxxxxx Vo
| xxxxxxxxxxxxxxx|xxxxxxxxxxxxxxxxx |
|-----------------------------------|

I did using superposition.
going around left side and right side.

Vi ( (20k//100k)/(110k + 20k//100k) ) = Vo (- (110k//20k)/(110k//20k + 100k))

but it doesn't give me the correct answer which is -70 gain

Last edited: Sep 20, 2012
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
If you consider the first circuit you'll hopefully recollect that the op-amp with feedback shown will have the output adjusting to make the -ve terminal potential equal the +ve terminal potential. Since +ve is tied to ground the -ve terminal will be at ground potential. So the output will also be at ground potential or 0V.

3. ### forbi Thread Starter Member

Sep 11, 2012
37
0
So for the first diagram is actually an "unless" ops amp since it doesn't provide any amplification and even "blocks" the voltage

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
That's correct. The circuit serves no useful purpose that I am aware of.

5. ### I have no education! Member

Sep 19, 2012
47
1
You are essentially providing 100% feedback!!

So no voltage gain out!!

A much simplier way is to think of it this way.

The gain equation for a an inverting amplifier is - rf/r1 x voltage in

Now in your case the rf is essentially equal to zero as you have no resistance and we will ignore the resistance of the wire!! So,

0/r1 x voltage in is equal to 0.

Also the other circuit is called a t junction! I'm sure there is analysis on the net somewhere for it!!

6. ### forbi Thread Starter Member

Sep 11, 2012
37
0
Wow this is interesting . How about the 2nd circuit . How do I go about analysising it

7. ### t06afre AAC Fanatic!

May 11, 2009
5,936
1,226
The superposition theorem here will be very useful. Have you tried that

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
For a clever shortcut one can make use of the star-delta transformation of the feedback network. The result is then quite obvious.

9. ### forbi Thread Starter Member

Sep 11, 2012
37
0
.

i did a star delta . what should i do with the 140kohm resistor? do i ignore them? if so the answer is correct with an inverting gain of
-700kohm/10kohm = -70

but why do i ignore the 140kΩ resistor ?

10. ### WTP Pepper New Member

Aug 1, 2012
21
6
T n K, you surprise me. There may not be any voltage gain, but there certainly is current gain. It's a classic circuit to act as a virtual ground that needs a decent impedance to work against.

11. ### I have no education! Member

Sep 19, 2012
47
1

if voltage out is equal to zero how can there be any current

12. ### t06afre AAC Fanatic!

May 11, 2009
5,936
1,226
The inverting opamp configuration is sometimes refferd to as a curreent to voltage conveyr/converter. I have used this skin conductance measurements systems. Using the skin conductance it self as the input impedance in a inverting opamp configuration, and a 3 electrode system

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Thanks for the correction. Goes to show you can teach an old dog new tricks.

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
The 140k at the negative input is essentially connected between virtual ground and physical ground. It therefore has no potential difference across its terminals and therefore has no current. As such it can be ignored. The other 140k connects the output to ground, thereby behaving as a load on the output. For an ideal op-amp this will have nill effect on the gain.

15. ### forbi Thread Starter Member

Sep 11, 2012
37
0
Wow this is interesting , my prof keep doing "magic" by relating things like no current flow here or is connected to virtual ground and by eliminating this resistor first. By doing all his "magic" the whole complicated circuit seems very easy to analys .
How do my prof or anyone able to deduce such conclusion such as no current flow through some resistor or elimination?
This is so "magicial" !