# Can someone confirm the gain here please.

Discussion in 'General Electronics Discussion' started by JPU, Nov 25, 2012.

1. ### JPU

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May 19, 2012
Hi

Could someone please confirm the gain from the op amp here (U3:A). Pin 8 voltage is 15V, Pin 4 is grounded. The input at 3 ranges from 1 to 5V

What is the gain at pin 1 on U3:A

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

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Jan 21, 2010
Some drongo looked at the wrong op-amp.

Last edited: Nov 27, 2012
3. ### BobK

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Jan 5, 2010
That looks like negative feedback to me and a gain of 1.

Bob

4. ### JPU

281
1
May 19, 2012
Thanks Bob & Steve

"a gain of 1", that is what I thought. Im not sure why my local electronics guy has changed the circuit that I gave him that Chris originally devised for me, but that now confirms my thoughts. Im not sure about the negative feedback (in fact I am not sure what you mean by that). Its the gain I was worried about." Will the negative feedback be a problem ?

Thanks

Justin

5. ### BobK

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Jan 5, 2010
Negative feedback is what you want, it controls the gain.

I think Steve was looking at the wrong op amp (U3B). That one does appear to have positive feedback. What is the function of that op amp supposed to be?

Bob

6. ### JPU

281
1
May 19, 2012
Hi Bob

The output from the Pic is approx 0 - 5V. The original circuit was tocontrol a brush less motor controller that had a 0 - 5 V input. However when I eventually decided on the brush less controller its input needed to be 0 - 10V, so the fix was to use an op amp with a gain of 2.

I gave the schematic that Chris designed for me to a local company to produce the boards but they say this config shown here gives a gain of 2 but my workings say the gain is 1 and therefore pointless in this application.

Am I correct?

Thanks

Justin

7. ### duke37

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Jan 9, 2011
Gain = 1+R4/R3 = 2

8. ### BobK

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Jan 5, 2010
Except that the input is divided by 2 by R1 and R2, which is why I said 1.

Bob

9. ### BobK

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Jan 5, 2010
Justin,

The gain of your opamp circuit is indeed 2, but you have taken the output from the PIC and put it through a voltage divider that halves it, so the output of the op amp is the same as the output from the PIC. Is this a DAC output? I did not know there were any small PICs with a DAC? And why do you have a .1uF capacitor shorting the output? This makes a low pass filter controlled by the output impedance of the PIC pin, is this your intention?

Bob

10. ### JPU

281
1
May 19, 2012
Hi Bob

You might remember the thread in the Summer that you made some valuable contributions to that helped me allot. Chris AKA "CDRIVE" developed a circuit for me and this is the result of that circuit. I am still working on it. I have been unable to make contact with Chris for help but here is the link to the thread to remind you,

I combined the circuit at this link

https://www.electronicspoint.com/digital-pot-setup-t248126p11.html#post1469754

with the circuit at this link

https://www.electronicspoint.com/digital-pot-setup-t248126p17.html#post1472431

I am still very novice, but I have worked out today that the voltage divider is halving the input. Chris probably thought that when I combined the circuits I would have altered them in some way,,however I simply added one to the other as I needed the output from the DAC to be 0 - 10V. Chris designed the circuit with the 0 - 5V output in mind and I threw the spanner in the works towards the end that I needed 0 - 10V as I ended up using a different controller to that which I stated I would use initially.

Justin

11. ### BobK

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Jan 5, 2010
Justin,

I am still trying to figure out how you are getting 0-5V analog out of the PIC12f629? This chip has no DAC or even PWM module. Is it mislabled, and you are actaully using another part?

Bob

12. ### JPU

281
1
May 19, 2012
Hi Bob

Sorry, I thought I had put this down somewhere but I haven't. Its not a PIC, its an 08M2 Picaxe. I used the PIC footprint as it was the only one close to the 08M2 footprint.

Justin

13. ### BobK

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Jan 5, 2010
Which PIC is that based on?

Bob

281
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May 19, 2012
15. ### BobK

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Jan 5, 2010
Thanks, yep, that would be the one. So you are using the 5-bit DAC, giving you 32 levels between 0 and 5V?

Bob

16. ### BobK

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Jan 5, 2010
Returning to your question, to get a gain of two, eliminate the capacitor and two resitors in the circuit connecting the DAC to the + input. Just connect the pin directly to the + pin.

Not sure why the capacitor was there, if you want to slow the response to changes, you could place a resistor from the DAC output to the + pin and a capacitor from there to ground, forming a low-pass filter.

Bob

17. ### JPU

281
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May 19, 2012
Thanks Bob for looking at this for me. You have now confirmed exactly what the guy said who was looking at the circuit for me. As you can see, he has placed the word omit next to the resistor.

I spoke to him yesterday and he basically said what you have said. I have just ordered 30 of these PCBs with surface mounted components as its to be on a very narrow PCB. I was worried about the outlay, when I was unsure about the circuit.

I should have trusted my local guy as he has a good reputation and now you have backed it up.

Thanks

Justin