# Unknown voltage of a Op amp

Discussion in 'Electronics Homework Help' started by Offshore again, Jun 24, 2016.

1. ### Offshore again

11
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Jun 24, 2016
Hi,
Hoping for a prompt in the right direction regarding a unkown voltage on a Non Inverting Op Amp.
Reading through my notes i understand how the resistors R3 and R2 set up a potential divider which supply the Inverting input on the Amp.
And the fact that neither input is connected to Earth means the Virtual earth principle cannot be used.

My course notes show that:
V−=V+ because the op-amp is ideal and V+=Vin

So is this as simple as as saying V- =100mV ??? Any help would be greatly appreciated

11,442
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Nov 17, 2011
3. ### Offshore again

11
1
Jun 24, 2016
Thanks Harald,

I assume its the same course that you mention (its a distance learning course) but not sure how many or who the other people are.

The question threw me a bit as it seems to easy....so thanks for the re-assurance

4. ### LvW

604
146
Apr 12, 2014
The principle behind this simple answer is as follows:
As long as the opamp is operated in its linear region (and negative feedback, as in your circuit, enlarges this region considerably) the input differential voltage of a real opamp is in the µV range (because of the very large open-loop gain Aol).
During calculation of voltages and currents within such a circuit this small differential voltage can be neglected if compared with all other voltages. That means: We can treat the opamp as an IDEAL unit with infinite gain and (consequerntly) zero input voltage.Hence V+ = V- (V+ - V-=0) and V+=0 for V-=0 (virtual ground principle).

Offshore again likes this.
5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
one should also point out the voltage drop across R1 for a perfect op-amp would be...?

6. ### Offshore again

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Jun 24, 2016
In an ideal op amp if R1 is amp impedance then it should be near infinite resistance so therefore the full voltage would be dropped.....is that correct?

7. ### LvW

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Apr 12, 2014
In your circuit, R1 is an external ohmic resistor of - perhaps - som kOhms.

8. ### Offshore again

11
1
Jun 24, 2016
So the voltage drop across the resistor would be proportional to its ohmic value.
If it were a variable resistor (POT) then it would adjust the gain......
Am i on the right lines or going down the wrong track??

9. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
What is the input current for a theoretically perfect op-amp?

10. ### LvW

604
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Apr 12, 2014
Yes - of course. Ohms law must be applied.
However - what about the current ?

11. ### Offshore again

11
1
Jun 24, 2016
Current should be zero in a ideal op amp due to the impedance being infinite.

(*steve*) likes this.
12. ### LvW

604
146
Apr 12, 2014
So you can find the answer to your questions (post#6 and 8) by yourself?

(*steve*) likes this.
13. ### Offshore again

11
1
Jun 24, 2016
So because there is no current (ideal OP-Amp) then the full 100mV is dropped across the resistor.
And with V- = V+ this is why the Unknown voltage V (in original post) is 100mV

14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Reconsider this in light of V = IR

15. ### Offshore again

11
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Jun 24, 2016
V=IR
V=0xR
V=0

Seem to be confusing myself now.

604
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Apr 12, 2014
Yes.

17. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Possibly, but what does the equation say the voltage drop across the resistor will be?

18. ### Offshore again

11
1
Jun 24, 2016
If there is no current then the Voltage drop must be zero across the resistor
V=IR If I is zero then the V must be zero

19. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Be careful with what you say. "Voltage" and "Voltage drop" are two different things.

20. ### Offshore again

11
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Jun 24, 2016
I have done a bit more digging and think I see where we are going.
Using Kirchoffs Voltage laws
Vin (100mv)-V1 (volt drop across resistor)= V+

So therefore volt drop across resistor is 100mV and V+ = 0V

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