# Op amp circuit help #2

Discussion in 'Homework Help' started by notoriusjt2, Dec 13, 2010.

1. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
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0

this is the first question I have received that does not state "Assume that all op amps are ideal". Should I assume that in this situation or is this all together different?

2. ### steinar96 Active Member

Apr 18, 2009
239
4
I think you can safely assume it's ideal. Otherwise you would be given parameters which make it non-ideal such as input and output resistances, open loop gain. Etc

3. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Greetings norotiusjt2,

What answer did you come up with? Be sure to show us your work so that we can determine any errors that you make.

hgmjr

4. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
I am coming up with answer (D) Vo = 4*Vin

Say we had 10V @ Vin. that would mean that +5V would have to be at +/- op amp terminals.

Since that network of resistors are all the same value (R), and based on their configuration they would basically divide the output voltage of the op amp by 4, effectively giving us 5V back at the - terminal on the op amp itself.

This is only true if Vo = 4 * Vin... hence the answer

am I correct in my reasoning?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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How did you arrive at that conclusion? I would think that for an ideal op-amp the voltage across the input +/- terminals would be 0V.

6. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
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thats correct...

for example if you put a voltmeter across the + and - of the op amp you would get 0V.

but if you put a voltmeter from + to ground or from - to ground you would have 5V

7. ### Ron H AAC Fanatic!

Apr 14, 2005
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In post #4, you used 10V as the input. In post #6, you say that the op amp +input will be 5V. Which is it?

8. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
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If Vin = 10V then wouldn't the positive side be +5V and the negative side be -5V? for a total peak to peak of 10V?

9. ### Ron H AAC Fanatic!

Apr 14, 2005
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The negative side is connected to ground. Ground is your reference, so is defined as zero volts.

10. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
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Right! so that means that the positive side of the power supply is automatically +10V?

11. ### Ron H AAC Fanatic!

Apr 14, 2005
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Yes.
Although, technically, it's not a power supply.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Hi notoriusjt2,

You seem to be confusing input voltage with the op-amp supply voltage.

In this case the op-amp supply voltage is not stated as it is irrelevant for the purposes of solving the problem.

If the op-amp were capable of running from either a single or dual polarity supply the outcome would be the same for this problem.

13. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
I understand... I was just assigning an arbitrary value for Vin for the sake of demonstration.

14. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
allright

if
node1 = 1V
then
node2 = 1V

now that I look at it, node3 and node4 are the same point
and node6 and node2 are also the same point

in order for this to work correctly
node4 would have to be 4V
node5 would be 2V
node6 = node2 = 1V

no?

15. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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No. See the attachment.

File size:
112.4 KB
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16. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
i see.... how did you calculate those voltages @ the resistors?

17. ### Ron H AAC Fanatic!

Apr 14, 2005
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I started with these facts:

1. The voltages at the op amp inputs are equal.
2. The op amp input current is zero.

Then I just used Ohm's law to find the currents in the resistors, and the voltage across each resistor. Do you see the currents that I added to the schematic?

notoriusjt2 likes this.
18. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
yes I see, that all makes sense now... Thank you!