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

Question

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 V_pp ≈ 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?

Answer 1

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.

Answer 2

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.