# Calculating Op-Amp Gain on AC voltage

simulate this circuit – Schematic created using CircuitLab

Schematics above is a parts of AC Voltage Meter. I know there are several related questions such as "Determining DC gain of single supply op-amp (no AC component)" and "Calculating gain on Op Amp with AC voltage". Things that caused my confusion are:

1. Both question above use a single resistor instead of potentiometer (see R4)
2. The presence of double-anode zener (D2 and D1), as limiting circuit
3. Two variables (SW1 and R4)

Can someone help me on the formula to calculate the gain (such as when the wiper on R4 is on 0%, 10%, 20%.. etc) or point me to the subject that I should be looking? Thanks in advance :)

Edit 1

Thanks for inputs from @analogsystemsrf , I add ground to + input of Op-Amp to remove floating input.

Why is the OpAmp circuit "inverting"? Consider the circuit with 240K feedback. Put one volt DC on the far left node. Assume opAmp's pin- is virtual_ground, thus at zero volts. And as before, the POT's wiper is fully at bottom.

How much current flows thru the 1MOhm? 1volt/1Mohm = 1uA.

Where does that current go? Thru the 240Kohm and then into the OpAmp's output pin, to the -VDD. What voltage across 240K? 0.24 volts, with Zero on left end (the virtual ground at OpAmp's pin-) and -0.24v on right end of resistor, which is the OpAmp's Vout.

Positive on Vin, negative on Vout. Gain is -0.24x. OpAmp provides 1uA + 5uA.

Now set the Pot's wiper at 50%. We can model the Pot as producing Vout/2 with Rout of 25K || 25K = 12.5K ohms. This would be a Thevenin Equivalent circuit.

Our feedback circuit now is 240Kohm + 12.5Kohm, but seeing only half of the True Vout. Consider the case with 1uA flowing. The voltage drop will be 252.5 milliVolts. The 50% position causes the OpAmp to double that, because the top 50% of the Pot takes most of the current to Ground.

Approximately 90% of the OpAmp's current does to Ground (~10uA). And 1uA goes thru the 240K to the 1MOhm. Current total is 10 or 11uA, which flows thru the bottom 25Kohm of the Pot, causing another 250milliVolt drop. Thus the OpAmp, to handle the 1uA flowing thru 1MegOhm, has to provide 10+uA to bottom of the Pot.

Probably want to tie the 50Kohm resistor (from + input to - input) to ground, thus running from (+) to GND. Otherwise only very high frequencies will be amplified, as the Cin provides a bit of phase shift between Vin- and Vin+ and thus the OpAmp as some input signal to work with.

Now for the gain: if pot's wiper is fully to bottom, the gain is just 24.3Kohm/1Mohm == attenuation by 40:1.

If wiper is 10% from the top (10% away from GND), you still get attenuation, of 4:1.

If wiper is 1% from the top, you get gain of 1/0.4 or Av=2.5x.

How to think about this circuit? {thanks for asking it, because I faced a similar circuit a few years back, but did not need to extract the GAIN MATH). KEY to understanding: the virtual_ground at (_) input pin. What is the maximum input current, thru the 1MOhm, the OpAmp can control? With output clamped by back-to-back zeners to ~3.4 volts, and with pot wiper fully at bottom, the maximum current availabl from the OpAmp is 3.4v/24Kohm or 140microAmps.

Now assume that current flows past the Virtual_Ground, and into the 1MegOhm.

The I*R value is 140uA * 1Mohm = 140 volts. With only +-3.4 volts on OpAmp Output pin.

• thanks for your response, I've edit the circuit :D Can you please be more clear about the math used to get 40:1, 4:1 etc? – cureinside Jan 31 '17 at 6:15

Let's get that POT out of the circuit, to avoid confusion. Set the POT's wiper all the way to the bottom. The feedback resistors (240K or 24K) are connected directly to OpAmp's Vout. The POT is merely one of the loads on the OpAmp, and does not affect the feedback (until we start moving the wiper).

Have you worked with "virtual Ground" of OpAmps, as clean way to predict Gain? Well, here goes. Our circuit is 1MOhm from input to Pin- of Opamp, and 24Kohm from Pin- to Vout of OpAmp. Assume the OpAmp has infinite gain, thus even a tiny voltage on Pin- is adequate to make the OpAmp's Vout be whatever V is needed.

Assume +1volt on input to 1Mohm. Assume Pin- is at ZERO volts (good assumption). We get 1microAmpere flowing, the OpAmp's Pin- rises toward that same 1volt, and the OpAmp's Vout starts heading toward -infinity (we have an inverting circuit).

As Vout reaches -0.024 volts, the troublesome 1microAmpere now flows thru the 24Kohm, and the OpAmp stabilizes............VOut at 0.024 volts. The Pin- voltage is.......... 0.024 volts / infinite_gain, or ZERO volts, our Virtual Ground. Our gain is 24K/1M = 0.024, or approximately 40:1 (32dB attenuation).

Lets consider that POT again. Put the wiper at 50%. We still need the 1uA to flow, so the OpAmp's Vout must "double (close to)", to 0.048 volts. Of course, Ohms Law tells us there is lots of error voltage in this analysis, because 1uA * 1/2 of the POT's resistance is 1uA * 25K or 25milliVolts, or 0.025 volts, an error trying to hide somewhere.

• Thanks for the answer (again). I want to clarify several things about your answer, (1) why do you said "(we have an inverting circuit)"? Isn't the circuit non-inverting? (2) I've got the point while the wiper is on 0%, but it's still unclear for me to calculate while the wiper is not on 0%. Can you please elaborate? – cureinside Feb 1 '17 at 1:09