# Do I understand OpAmps wrong?

Discussion in 'General Electronics Discussion' started by acecase, Aug 26, 2014.

1. ### acecase

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Aug 26, 2014
First of all, thank you all for such a nice forum. A lot of great info here.

I have a very simple OpAmp circuit that consists of two opamps. One in difference amp configuration and another in voltage follower. (Attached)

The idea is an isolated voltage reading from Vin. I built a simple difference amp with all equal value resistors and noticed that I was able to get the output higher than the opamp supply. The supplies are +/- 2.5V and if I connect Vin to a 10V supply I get 10V at the output.

Everyone here is saying, "duh." It didn't hit me immediately that I have a path around the opamp with the resistors.

When I realized this, I added the simple voltage follower (in the image attached it's wired as a follower, but it's actually an LM310N, which is internally wired as a voltage follower). I'm still getting more than my supplies at the voltage follower's output (Vout). If I connect Vin to a 10V supply, I get (approx)10V at Vout. Even though my suppies are +/- 2.5V.

I didn't think I should be able to get an output greater than the supply. Am I wrong, or am I overlooking something else?

Thanks

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

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Jan 21, 2010
Are your input voltages totally isolated from this circuit's power supply and ground?

What are you reading the output with respect to?

3. ### acecase

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Aug 26, 2014
The input is a separate bench power supply. I'm using a simple 2-resistor divider to get +2.5V, -2.5V, and GND from a bench supply at 5V. Those are the supplies and the one ground that are marked in the image.

I'm then connecting a separate bench supply to Vin, and measuring Vout relative to GND from the above supply (the GND created by the divider).

Does that make sense, or did I butcher that description?

4. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
What are the values of the resistors?

Do the +2.5V and -2.5V supply voltages change as you vary the input voltage to the circuit?

5. ### acecase

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Aug 26, 2014
They do.
Thank you.

I'm injecting voltage at my ground through the one resistor at the non-inverting input of the difference amp, and changing my supply voltage.

I don't know how I could prevent that easily.

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

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Jan 21, 2010

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7. ### BobK

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Jan 5, 2010
I think you are exceeding the allowable voltage at the inputs. These are normally limited to the supply voltages.

Please measure the voltages on the + and - input pins.

Bob

8. ### acecase

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Aug 26, 2014
I'm sorry. That was a little vague.

The first answer, under your question was to it, and my last post was in response to KrisBlueNZ.

My + and - 2.5V supplies are relative to the ground on the one side of R4, so Vin is actually changing the value of the supplies. I should have included my power supply in the drawing, and one of you guys would have probably immediately picked up on it.

My goal is to take an isolated voltage measurement that will be the voltage drop across a .2ohm resistor, where the voltage across it could be as much as 50V and the drop should be less than 2V. I want to isolate it so that I don't create a path for the current in the event that the sense resistor fails. The voltage drop reading will be used by a 5V AVR uC, so I need to protect it.

If I had used this circuit, the opamps wouldn't have survived.

9. ### KrisBlueNZSadly passed away in 2015

8,393
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Nov 28, 2011
This is called "high-side current sensing". There are several techniques, and many ICs available for this purpose, depending on your requirements and your budget. Start with Digi-Key's selection guide at http://www.digikey.com/product-sear...ulation-management/2556448?stock=1&quantity=1 (download a few data sheets to get an idea of what's available) and a Google search.