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Topic: Measuring very very small voltage fluctuations? (Read 12314 times) previous topic - next topic

MikeX

majenko, that's pretty much what I planned from the begining, an op-amp. I'm wondering what purpose does the (not pictured, but mentioned) capacitor serve on the input?

majenko

It removes the DC offset, so the op-amp only sees the +/-0.000001 (or whatever) voltage variations and not the 1V offset.  It then uses the offset applied to the non-inverting input to place those variations at the mid-point of the ADC.


MikeX

majenko, I'm not sure I follow how the capacitor fits into the picture. So I have a voltage that fluctuates between 1.000v and 1.005 volts.



In this diagram I'll bring up the ground (non inverting on the op-amp) to 1.000v using a voltage divider. Now the op-amp will amplify the 0.005v fluctuation. Where does the capacitor go again?

Grumpy_Mike

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Where does the capacitor go again?

Between Vin and the voltage you want to measure.

MikeX

#20
Jul 02, 2012, 07:18 pm Last Edit: Jul 02, 2012, 08:05 pm by MikeX Reason: 1
Grumpy_Mike, wouldn't that do the opposite of what I want? That is the capacitor will dampen the 0.005v fluctuation that I'm trying to measure?

That capacitor will charge to 1.000v, correct? At the time when there are no voltage fluctuations. Then absorb, thus dampening any fluctuation which drives the Vin above 1.000v?

Grumpy_Mike

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That is the capacitor will dampen the 0.005v fluctuation that I'm trying to measure?

Only if you put it from Vin to ground. Not if you put it from Vin to the voltage you are trying to measure. It is called AC coupling.

MikeX

Grumpy_Mike, thanks, I read up on capacitive coupling and it makes sense now. But how large should the capacitor be if my AC frequency will be very low? Say 0.1 hz.

Magician

C = 1/ (2 x 3.14 x F x Rin ),  for F = 0.1 Hz  C = 1.59 / Rin.  To keep C below 10 uF, R should be > 100 k

Grumpy_Mike

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But how large should the capacitor be

I wouldn't disagree with the above.

MikeX


Keep your ref at 5V.  Connect the incoming signal to a capacitor, then that to an Op-amp.  Set the op-amp up so that it swings around a 2.5V virtual ground, and enough amplification so your tiny variations are now around +/- 2V.

Simplest arrangement is an inverting amplifier.  50/50 voltage divider on to non-inverting.  Signal via capacitor and resistor Rin to inverting.  Output through resistor Rf back to inverting.  The gain is -(Rf/Rin).



Note that the values then sampled by the ADC will be backwards, so subtract them from 1024 to get the real values.


I'm still not quite sure how this all comes together. So if I connect the signal via a capacitor, my inverting input will wobble very close to 0v. Then the non-inverting is 2.5v (50/50). How is this 2.5v differential going to be amplified?!

Magician

This schematic is not the best for your needs. If you put cap in input line, than OPA should have double power line (+ and -).  And it will leave DC offset question unanswered.  Easiest way in reply #17.

MikeX

Magician, can you please elaborate why that op-amp circuit won't work? I already have generic op-amps in my bin, so it would be nice to just one of them.

Magician

I tested a schematic linked in #17 with LM358, works just fine. Do you have them at your hands?

MikeX

Magician, I'm trying to understand how everything works, so I'm equally interested in the process as the end result. If you can, please help me understand why the source->cap->resistor->op-amp won't work...

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