# Voltage Divider feeding Op-Amp Confusion.

Discussion in 'General Electronics Discussion' started by Balrock, Jun 15, 2012.

1. ### Balrock

39
0
Mar 1, 2012
Hello everyone. I have been able to get back to my project again the last week. I have been struggling with the same problem for many days and I have finally run out of things to try.

I am trying to feed my Op-Amp Voltage Inverter with a Potential divider but with no accuracy.

The Potential Divider. - With a supply of 5 volts Point A gives me a ratio of 1:4 and correctly gives me 1.25volts.

The Op-Amp Inverter - If I supply say + 2 volts directly at point X from a power supply rail point Z correctly gives me - 2 volts. This is due to the fact that

Z (volts out) = - R4 / R3 x X (volts in)
so the real numbers are
- 1 / 1 x 2 = - 2

So far so good.

If I connect point A to Point X I do not see - 1.25v at point Z. I see -0.63 volts

If I remove R3 and replace R4 with a 3.9k resistor so that the Op-Amp maths looks correct I still do not see 1.25v at point Z. I see -5.02 volts

I have come back to this several times but I am stuck. I know that I am obviously missing fundamental knowledge here. Any and all help is very welcome.

Thanks kindly Paul.

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2. ### Harald KappModeratorModerator

11,280
2,583
Nov 17, 2011
Easy: Pin 2 of the OpAmp is effectively at ground (0 V).
Therefore R3 is effectively parallel to R2.
Do you see what follows and why?

There's no need to have R3/R4 at such a low resistance. Use 100 kOhm instead and the effect becomes almost negligible.

Harald

3. ### Balrock

39
0
Mar 1, 2012
thanks for the prompt replay Harald

I tried simulating it and it was failing past 20k. In the real world though 100k was fine, after I replaced a faulty Op-Amp that is!

Thank you for the sweet relief ! I will come back to this again later after a rest and try to work out why it works this way.

4. ### Harald KappModeratorModerator

11,280
2,583
Nov 17, 2011
A hint for understanding the operation of the circuit: The OpAmp will produce an output that forces V(+)-V(-) = 0 V (an approximation good enough for many purposes).

To be exact, it produces Vout = A*(V(+)-V(-)) where A is the open loop gain of the OpAmp. Since for a typical OpAmp A is on the order of >10^5 and Vout is limited to Vcc, you get V(+)-V(-) = Vcc/A ~ 0V.

Harald

5. ### BobK

7,682
1,688
Jan 5, 2010
Another way of looking at it is to consider the Thevenin equivalent of the R1/R2 divider. It is equivalent to a 1.25V source and a 975R series resistor. So that resistance is in series with the 1K R3 and therefore the gain of the amp is - 1K / 1.975K or 0.506. Multiply this by the 1.25V input and you get -0.63V.

Bob