# Calculating gain in this op amp stage

Discussion in 'General Electronics Discussion' started by Davewalker5, Sep 26, 2014.

1. ### Davewalker5

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Sep 20, 2014
The input voltage is 8.8 vdc going to R117
U12 pin#6 output voltage is 4.4 vdc
How do you calculate the gain?
What does C38 do?
What kind of gain stage is this called? it's weird how they arranged Rin and the feedback RF
Plus the signal and voltage is going to both inputs , so it's some kind of differential gain stage?

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3. ### Davewalker5

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Sep 20, 2014
Is R117 and R115 = R in?
R116 = Rfeedback?
or is R116 and R115 = R feedback?
Or is R114 and R115 = R in?
C38 with R117 is a filter, but some of the voltage is going to pin#3 and Pin#2
C38 voltage divides the input signal to R117
So this is a differential gain stage?

4. ### davennModerator

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Sep 5, 2009
R 115 is going to have an effect on both feedback and Vin

No, why do you say that ?

no, R117 and C38 form a low pass filter

No, not that I see, as there is no voltage going to the other input of the op-amp

this below is an op-amp using diff. mode

you have no V2 / R2 in your diagram

Dave

5. ### Davewalker5

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Sep 20, 2014
I'm confused what resistors are Rin and what resistors are Rf?

R114 does nothing for the gain formula?

6. ### Davewalker5

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Sep 20, 2014
R116 is 2.15 megs, why use a Meg ohm resistor in the feedback?

One tech said here that Meg ohm resistors are used as stablizers when the are "coupled" to other resistors or caps

This R116 must be a stablizer? to stable the feedback or op amp

7. ### davennModerator

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Sep 5, 2009
R116 provides the feedback path from the output to the input of the op-amp
there is NO other path

I'm not up on doing feedback gain calculations, but I would have to logically assume that a high value feedback resistor would set a low gain value

8. ### Laplace

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Apr 4, 2010
Are you sure of those measured voltage values? With 8.8VDC at the input and 4.4VDC at the output the gain would seem to be 1/2; however, the DC gain is calculated as R116/R117 which is a little more than 2. So it would make more sense that the output is 8.8V when the input is 4.4V. And the calculated gain would be exactly 2 if the 2.15MΩ resistor were 5% low and the 909KΩ resistor were 10% high. Note that for gain calculation at DC the capacitors can be removed from the circuit, and R115 can be zeroed because there is no signal current flowing through it due to the high impedance input of the op amp.

Check those measurements again. It is highly suspicious that the measured result is the inverse of the calculated performance.

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9. ### davennModerator

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Sep 5, 2009
Thanks Laplace

As said, I'm not up on the finer details of that stuff

D

10. ### LvW

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Apr 12, 2014
This circuit is one of the classical second-order active lowpass filters: Lowpass with multi-feedback in MFB topology.
Googling these keywords will give you a lot of information and hints how to find the transfer functiuon ("gain") of the circuit.
The resistor R114 is not so important - it just provides a kind of "dc input current balancing" (input current compensation for a "better" operational point).

11. ### Davewalker5

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Sep 20, 2014
So R114 sets up a biasing operational point? or how does it balance the input current?

Why would they want a 2.15MΩ resistor in the feedback loop? this makes the gain stable?

I don't see Meg ohms resistors in the feedback loop often in gain stages, I'm not sure why it's used and for what reason

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

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Jan 21, 2010
Dave, you're still talking about Meg resistors like they're something special. They're not. They're just resistors.

They've been chosen because they are the correct values for the particular circuit.

You keep showing parts of the circuit and you don't tell us what it is. So w really can't help you.

All we can say is that there is nothing intrinsically different between the purpose of a "meg resistor" and any other resistor you care to name.

13. ### Davewalker5

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Sep 20, 2014
I would think putting a resistor in the Meg ohms would be a lot of gain, like a open loop gain, not a close loop gain

14. ### davennModerator

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Sep 5, 2009
have you yet read any information on how gain Vs resistor values are calculated ?

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

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Jan 21, 2010
It depends on the other resistances. I could use a pair of 10M resistors and get a gain of 2. It's all about the ratio of resistors. In this case, the impedance of the capacitors would probably be a significant factor, but since we don't know what you've got, we can't say.

But it is a closed loop gain.

LvW has given you VERY IMPORTANT information. But I bet you didn't do as he suggested, right?

16. ### davennModerator

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Sep 5, 2009
Last edited: Sep 27, 2014
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17. ### LvW

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Apr 12, 2014
1.) Although the opamp input resistances are neglected during calculations, there is a small DC current into both input terminals. And it is the task of R114 to develop a small DC voltage (caused by the small DC input current into the non-inv. terminal). This DC voltage should be approximately equal the DC voltage developped at the inv. terminal). Thus, the remaining DC differential voltage is rather small (it is a kind of input offset). As a consequernce, the influence on the desired operating point is rather small. This method is called "input current compensation".

2.) It is not possible to isolate the Meg-ohm resistor from all other elements and to ask "why so large"? This resistor is part of the overall feedback network (which - in fact - it is a bridged-T topology). All elements of this RC-feedback network determine the filter response. If you want to verify all the component values, you have no other chance than to analyse the filters transfer function. If you are interested, I can write down this function. From this function one can derive gain, pole frequency and quality factor of the filter.