# Calculate Op Amp Gain

Discussion in 'General Electronics Discussion' started by steve harris, Jun 21, 2017.

1. ### steve harris

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0
Jun 21, 2017
Hi,

How do I calculate the gain of this attached circuit?

Basically do i have to take the R1 into account?

I am measuring the output to the right hand side of R1.

is it

A) A = 1 + (R2 / R3)

or

B) A = 1 + ( (R1 + R2) / R3)

Thanks

Steve

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2. ### davennModerator

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Sep 5, 2009
B would be the logical choice
but it really shouldn't be done that way. The feedback should be taken directly from the output pin

3. ### duke37

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771
Jan 9, 2011
I would go for A since the feedback comes from the output.

R1 throttles the output of the op-amp so the amp has to work harder. The frequency response and the output swing will be restricted.
An ideal op-amp has a zero output impedance.

4. ### steve harris

2
0
Jun 21, 2017
thanks all,

i would/should have mentioned.

this is a dc circuit thing.

this is from a board i am repairing and came across this part of the circuit which is not something i have seen before.

they are feeding a 5v reference voltage (low current) into the +ve input of the op-amp and using the output as a 15v output.

thanks

5. ### duke37

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771
Jan 9, 2011
R1 limits the output current so may be to protect the amp from a short.

6. ### davennModerator

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Sep 5, 2009
NO, the feedback comes after the resistor on the output
this will significantly change the gain value

7. ### ramussons

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Jun 10, 2014
A = [1 + ( (R1 + R2) / R3)] * (R2+R3)/(R1+R2+R3)

8. ### duke37

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771
Jan 9, 2011
The gain is determined by the feedback which comes from the circuit output, not from the op-amp output. The calculations you have shown deals with the voltage at the op-amp output which is of no concern, it will be greater than the circuit output voltage and will depend on the load.

9. ### LvW

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Apr 12, 2014
The voltage at the opamp output is of "no concern"??
I rather think, we have an active device (called opamp) which has a low-resistive voltage output and it is THIS output only which provides voltage feedback!
The only difference to the classical non-inverting circuit is that we do not further process the ouput voltage at the opamp output (for which reason ever) but the voltage between R1 and R2 (simple voltage division).
Referenced to this node (between R1 and R2) the gain is as given by ramussons (post#7).

And - yes, this output voltage (and the gain) will depend on a possible load resistor; however, thius was not part of the question.

Last edited: Jun 22, 2017
10. ### duke37

5,364
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Jan 9, 2011
No, no The feedback comes from the output so the output voltage will be controlled by R2 and R3 so long as the op-amp can drive it through R1.

Perhaps someone can spice it. I cannot do it at the moment due to a computer failure.

Harald Kapp likes this.
11. ### Ratch

1,093
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Mar 10, 2013
Why not just calculate it?

The gain is calculated from the loop equation. If you want the op-amp out, then two equations are needed. The op-amp will have to work harder to supply vout because of r1.

Ratch

duke37, Harald Kapp and davenn like this.
12. ### AnalogKid

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Jun 10, 2015
Referring to the original schematic, the opamp voltage gain from pin 3 to pin 6 is as in equation A, with a series feedback resistance of (R1+R2). Pin 6 is a zero-ohm point at voltage (Vin times the gain). Pin 3 is a zero-ohm point at voltage Vin, driven there by the opamp's closed loop. So R1 and R2 form a voltage divider between two zero-ohm points of two signals with identical frequency and phase, but different amplitudes. Enter Ohm's Law.

All of that assumes a perfect opamp and an infinite impedance load at Vout.

ak

13. ### Ratch

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Mar 10, 2013
I have a hard time understanding your explanation. Do you agree that my calculation are correct? If not, what are your calculations?

Ratch

14. ### Harald KappModeratorModerator

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Nov 17, 2011
R1 has no influence on Vout as long as the opamp can drive enough current, see post #10.
This kind of circuit is sometimes used with capacitive loads. Most opamps are limited in their capability to drive capacitive loads. The resistor R1 helps to stabilize the amplifier in that case while the feedback via R2/R3 from Vout is required to set the gain correctly.

As much as I deSPICE the use of a simulator when a thoretical calculation can give the result, I follow @duke37 's request:

As expected gain is 2 (1+R2/R3).
Changing R1 has a rather small influence on gain (in the 1 % range, depending on how much R1 is changed). This is a frequency dependend effect which at e.g. 10 Hz signal frequency is almost not visible. Possibly the opamp model and some implicit parasitic capacitances in the simulation cause this effect.
In real life the real parasitics and the real frequency response of the opamp will have even more impact.

duke37 likes this.
15. ### duke37

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Jan 9, 2011
Thank you Harald.
With 1V in, the voltage at the junction of R2,R3 will be 1V.
The output will be 2V.
U1 out will be 3V wih no load.

16. ### LvW

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Apr 12, 2014
Yes - of course, this is correct.
But this result was given in general form in post#7 already:

A = [1 + ( (R1 + R2) / R3)] * (R2+R3)/(R1+R2+R3)

Harald: How can you say that R1 has "no influence"?

17. ### Ratch

1,093
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Mar 10, 2013
It must be stipulated that the values of the output and U1 will be 2 v and 3 v respectively only if Vin = 1 v and R1=R2=R3.

Ratch

18. ### Ratch

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Mar 10, 2013
The above equation simplifies to A = (R2+R3)/R3 as I calculated in post #11. R1 cancels out.

I believe he meant no influence on the gain at vout.

Ratch

19. ### Ratch

1,093
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Mar 10, 2013
No, the gain is 1+R2/R3 . See post #11

Ratch

20. ### duke37

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771
Jan 9, 2011
Take R1 = R2 = R3
The simple equation A = (R2+R3)/R3 with 1kΩ, get 2
Complex equation A = [1+((R1+R2/R3]*(R2+R3(R2+R3)/(R1+R2+R3) = 2
Take R1 = 0.1k , R2,R3 = 1k
A= 2.1k*2k/2.1k = 4.2k/2.1k = 2

so it appears that R1 is cancelled and does not affect the answer.