# Op-amp integrator gain from calculation doesn't correspond with real-world measurement

In my integrator circuit, I added one feedback resistor R2 and one bias compensation resistor R3. The values of the components are values I have measured before I applied them on this PCB Evaluation Board.

All components are somewhat of standard SMD components lying around my soldering iron. I calculated the gain with this formula:

$$\ \text{gain}_\text{calc} = \frac{1}{2\pi f R_1 C_f} \approx \underline{\underline{2.9}}\$$

The simulation with LTspice gives me the same value.  But on my PCB circuit I measured a different gain. $$\\text{gain}_\text{meas} = \frac{V_{\text{out_rms}}}{V_{\text{in_rms}}} = \frac{5.793\text{ V}}{2.135\text{ V}} \approx \underline{\underline{2.7}}\$$.

This leaves me with a voltage difference of Vpp ≈ 1.127 V, just a bit too much for my gut feeling.

What can cause this difference, and where should I revisit my doings so far to have the values from the simulation equal to the values in the real world?

• What is the tolerance of your 49n9 capacitor? Is its value constant with applied bias? Aug 11, 2022 at 15:12
• @ThePhoton It is a 47 nF capacitor with +-20 % tolerance. The value I used in the simulation is the value I measured before. I can't answer your second question, because I can't assign. Probably the datasheet can help us? 0805 47nF Aug 11, 2022 at 15:33
• It's shown in the upper right chart on page 53. But your capacitor is actually pretty good over your +/- 5 V operating range, maybe only +/- 3% due to voltage effects, where you're looking for about a 10% error source. Aug 11, 2022 at 15:38
• @Daniel In your previous question, when I said that a resistor is placed across the capacitor, I said 1 M$\Omega$ or higher. You have to account for that "higher" part when using components, otherwise you could have found out for yourself about the difference (aka: the answer below is right). Aug 11, 2022 at 17:04
• @Daniel Well, I also need to polish my glasses, because I don't know why I saw something about 3 Hz, where there clearly is a gain difference. Oh well, at least the problem is solved. Aug 12, 2022 at 12:37

You are neglecting R2 as part of your feedback impedance in the gain calculation equation. The reactance of Cf at 50 Hz is about 64 k ohms and so the 998 kohms parallel resistor is significant in calculating the total feedback impedance.

To get a rough estimate of the feedback impedance do R2.Xc/(R2+Xc) which gives an estimated feedback impedance of 60 k ohms. 60 k/21.978 k = about 2.7

To calculate the gain accurately you need to include a j term in the capacitor's impedance calculation and calculate the transfer function.

Gain = ((R2/jwC)/(R2+1/jwC))/R1

That is the transfer function and you then need to get that in the form a+jB and derive an expression for the magnitude.

But the quick and easy estimation of the way I mentioned first without including the j term seems to agree with your experimentally derived gain.

• What's about R3? May 15 at 9:21

You have stated that your cap is +/- 20%. You expected a gain of 2.9.

With Vin at 3 volts pk, that is an RMS of 2.12, which corresponds very closely with your scope reading of 2.13.

Now assume that Cf is 20% larger than nominal. From your gain equation, it should be obvious that the gain will be about 20% less, actually 0.833.

0.833 times 2.9 is 2.42, so you are well within tolerance.

To run it another way, a gain of 2.7 can be explained by Cf being large by a factor of 2.9/2.7, or 7.5%. That is eminently reasonable. If you like, you can get a capacitance meter and check.

• I have measured the capacitor. This is the reason the value of the capacitor in the schematic is 49.9 nF and not 47 nF. So I think the difference in gain isn't from the capacitive tolerance because I measured the capacitor before using and take the value for the simulation. I can imagine, that a tolerance with applied operating voltage affected also the difference in gain, like @ThePhoton mentioned in the comments directly under my question. Aug 12, 2022 at 6:56