Op amp virtual ground confusion

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Hello Inter web

I am confused with virtual ground used for op amps and I cannot verify my calculations with measurment because of virtual ground.

My setup is as follows: LM358 connected in a non inverting amplifier configuration with a gain of 100. The power supply going to the op amp is 9v to the possitive power rail and -3v to the negative rail.

The input signal is 20mV. This signal is generated by running a variable current through a known resistor. The output should be 2v. I measure the signal relative to 0v but I am confused as to what ground the op amp see's the signal relative to? 0v or -3v?

If I measure the voltage drop over the known resistor I measure 20mV. When I measure the signal that goes into the op amp what ground do I put my negative tester lead to? -3v or 0v?

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In the non inverting amplifier, one of the resistors is connected to ground. That ground is the reference to where you should measure both the input voltage and the output.

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Please post the circuit.

Remember that input offser is also amplified, and a few mV times 100 can give a considerable fault.
An ultra low offser opamp coulis be needed.

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Yup, the LM358 probably wont cut it. Check this parametric search.

 

Probably some of those will be what you need. The INA199 with fixed x100 gain looks promising. It's not dirt cheap like the LM358 but still reasonable for a precision amplifier.

Last Edited: Sat. Feb 24, 2018 - 10:57 PM
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The power supply going to the op amp is 9v to the possitive power rail and -3v to the negative rail.

Wow, not only running the op-amp on a virtual Ground, but with an asymmetric power supply as well.

 

That is certainly doable, but is usually done where the signal is mostly positive, but one needs to capture a true zero crossing, so the negative rail is "slightly" negative.

 

Otherwise, one would generally feed the Op-amp with the V+ and Ground of the 9V battery, and use a resistor divider to generate 4.5 V.

That 4.5 V is then buffered with another op-amp, to be a "driven" 4.5 V line.

That 4.5 V line is + 4.5 V with respect to the battery's negative (Ground) terminal, but then becomes the Virtual Ground for the system.

 

One would also generally put a small cap across the resistor divider, to help filter noise out of the Virtual Ground.

 

Unipolar Op-Amp configurations with a component to Ground then are connected to the Virtual Ground.

 

JC 

 

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This Thread and comment #15 has an example of using an op-amp to create a Virtual Ground for the rest of the circuit.

It additionally shows a few other op-amp stages running on +3V, as a +/- 1.5 V system, with the op-amp generating the virtual Ground at 1.5 V.

 

JC

 

Edit:  Added the schematic, here.

 

 

 

Last Edited: Sat. Feb 24, 2018 - 11:14 PM
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The input signal is 20mV. This signal is generated by running a variable current through a known resistor

Well, you didn't say where the resistor was grounded to ...gnd or at -3V (should use gnd)...in any case, across the resistor is the desired "signal".  

 

The opamp is idealistically simple to think about: the output goes up & down (perhaps way up or way down), such that the input pins Signalin+ & Signalin- match up exactly.  if Sigin+ > Sigin-, the output goes up  & vice-versa.  The matching is examined under a high power magnifying glass, so any mismatch will cause a large reaction at the output.  Or conversely, a large swing of the output, corrects any mismatch by a very small amount.

 

The real opamp has a number of "reality checks", since the matching process is corrupted by an offset bias voltage, the pins draw a small current, the magnifier (error amp) has a finite gain that is also frequency dependent, etc, etc.  The output can (at most) only swing within the power supply limits & also has a slewing limit (max rate of change).

 

 

Remember that input offser is also amplified, and a few mV times 100 can give a considerable fault.

Any more, there are many many opamps that are way below 1mv offset, even for 30 cent parts.  Remember the good old days with a 4mv offset & $1.50 price tag?   The OP didn't say how "good"  his accuracy needs to be , frequencty response, etc.

When in the dark remember-the future looks brighter than ever.

Last Edited: Sun. Feb 25, 2018 - 01:11 AM