Voltage Divider feeding Op-Amp Confusion

[Balrock]

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.

 

[LvW]

Hi Paul, I think your error is that you treat the node A as a voltage source (zero internall resistance) - thus forgetting that the divider resistors affect the feedback factor.

 

[FvM]

Or see it the other way around. The -1 gain of the op circuit is kept, but the voltage divider is loaded by an additional 1 k resistor to ground. You need to adjust the voltage divider for this load, by changing R1 and R2, Increasing R2 alone doesn't work.

 

[etmabreu]

Point Y is at ground potential. This means that, for computing voltage at Point A, R3 is in fact in parallel with R2 when you connect A to X. Do your calculations for the new voltage and smile...

 

[91divine]

hi barlock, you are ignoring the output impedance of your potential divider circuit.
If you take it in consideration, then u will find that its output impedance is ~0.975ohm (ie. 3.9||1.3).
now taking this in account your gain equation modifies as -R4/(R3+0.95). taking R4=R3=1k equation will became -1/(1+0.95) * 1.25 = -0.64v

So there are two solution to remove this impedance mismatch:

1. choose the value of R4 such that ratio of R4/(R3+0.95) will give you the required gain. (this solves a specific problem)
2. introduce a voltage follower circuit in between your potential divider and OPAMP circuit. This voltage follower will nullify the effect of impedance mismatch.

I will recommend you 2nd method if you are doing this project only on hardware(ie bread board). If you are doing only simulation then u can opt for 1st method

Hope you will find this answer helpful......

 

[LvW]

output impedance is ~0.975ohm (ie. 3.9||1.3).

0.975 kohms.