I am trying to measure an AC voltage using an ADC. However, the ADCs I know only measure the positive voltage. Hence, I am thinking of converting the AD to DC with an offset using the following circuit.

Voltage offset circuit

To offset the voltage, I need a voltage reference. I want it to be exactly half of 3.3V so that the ADC full scale can be used.

I don't know what the correct procedure is to generate the exact voltage reference. I am seeing there exists some voltage reference ICs but not for 1.65V.

I produced the following circuit, is there any better way than this?

Circuit to generate 1.65V reference

I have following questions:

  1. Is my offset circuit is correct? I have taken it from here.
  2. If it is correct what are the recommended values for R and C?
  3. Is there an ADC which can measure negative voltages?
  4. Is my circuit to generate 1.65V is good? Is there a better option?
  5. Is there an IC which generates 1.65V?
  6. Is there an IC which provides offset of the input voltage?
  • \$\begingroup\$ What are you using as your reference voltage? Usually you would take that and divide it in half (some ADCs will do this for you by the way and provide the exact voltage to use at an output pin). Using an external reference for this wouldn't make sense since its relative to your ADC reference voltage. \$\endgroup\$ Oct 14 at 2:28

For an Op-Amp input, a voltage divider output is fine, but you need to check whether your regulator requires a minimum load to operate correctly.

For anything that might be loaded, like a bias voltage that is applied to the signal itself, an Op-Amp configured as a follower can buffer that voltage.


You've asked far more than the one allotted question. I'll address the reference question(s).

Your circuit will generate a nominal 1.65V without a load, however many ADC reference inputs require a low-impedance source. You can use a low-voltage shunt reference such as a TLV431 and a couple resistors (plus the series resistor) to get a 1.65V reference of modestly decent accuracy accuracy and stability, and an output impedance of ~33\$\Omega\$ typical compared to 5k\$\Omega\$ for your circuit (an improvement of 150:1). Design equations are in the datasheet.

You can also use an op-amp to buffer the output of a voltage divider, however it may require more complexity and/or a fancier op-amp to get a low DC and AC impedance, combined with good phase margin.


I’m not sure if you’re intent is to rectify the input AC signal and convert it to a DC signal that’s proportional to amplitude, or if you want to preserve the AC signal and just introduce a DC offset to keep it in range with you’re converter.

If you want to preserve the AC signal, then I suggest you use one of the configurations in this reference design, as these are a more tried method for single supply applications and are less likely to have issues with amplifier stability: https://www.ti.com/lit/ug/tidu871/tidu871.pdf?ts=1634181235585&ref_url=https%253A%252F%252Fwww.google.com%252F

If you are using an embedded ADC in a MCU, then the best thing to do is to just use the same 3.3V signal being supplied to the MCU/ADCs AVDD/REFIN, as this will account for any inaccuracies or drift in the reference voltage and keep your AC signal truly locked in at the mid-point.

If you have the freedom to use an external ADC, than you can look into using an ADC with differential inputs in combination with a fully differential amplifier because these allow you differential signal to go from -Vref to +Vref. That means if you had a 2.5V Vref, you could actually apply a differential signal with a peak-to-peak voltage of 5V.

  • \$\begingroup\$ I can go for an external ADC. Do you know any part number where I can measure negative voltages? \$\endgroup\$
    – SKGadi
    Oct 15 at 15:21

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