I have a current source consisting of a 100Hz sine plus a DC offset: \$I=I_{DC}+I_{AC}\cdot\sin(2\cdot\pi\cdot 100Hz\cdot t)\$ with \$ I_{DC}\approx100uA\$ and \$I_{AC}\approx 1uA\$.

I would like to extract the AC signal and rectify it to get a digital output (rectangle) signal with a frequency equal to the modulated current. The phase may be shifted, but the shift should be constant. The circuit should consume as little power as possible and work from a single supply rail of \$ V_{SUP}\geq 1.8\$.

My idea was to have a shunt resistor (R_SHT) between the current source and GND to convert the current to a voltage. This voltage is then high-pass filtered to remove the DC component. Finally, an OPAMP comparator rectifies the AC voltage. I used a low power rail-to-rail OPAMP with low offset voltage (TI LPV821). Here's the schematics:


simulate this circuit – Schematic created using CircuitLab

After soldering the circuit, I don't see the expected rectangular output voltage. Instead, the output is drifting slowly between the supply rails, occasionally exhibiting the 100Hz input signal.

Is the proposed circuit generally suitable to achieve my goal of converting the modulated current to a digital signal?

Are there any obvious problems with the circuit, that may lead to the observed drifting behavior?

Are there better alternatives to achieve my goal while consuming similarly little power (\$ \approx1uW \$)?

Thanks for your input.


1 Answer 1


Are there any obvious problems with the circuit, that may lead to the observed drifting behavior?

  • The minimum supply for an LPV821 is 1.7 volts and you are expecting it to work above 1.6 volts.
  • Your AC current is 1 uA and this produces a voltage (across the 20 ohm resistor) of maybe 28 uVp-p. Given that your op-amp is running without feedback AND it has an input offset voltage of about 10 uV, you will certainly get poor performance.
  • Given that you are creating a voltage signal at the op-amp's +INPUT that rises above and falls below 0 volts by about 14 uV, the negative excursions exceed the common-mode input range for the op-amp as specified in the data sheet.
  • Guaranteed data sheet specifications are when operating from a supply voltage of 1.8 volts. The front page says 1.7 volts but there's no guarantees at this level.
  • The input voltage noise density is 215 nV/\$\sqrt{Hz}\$ and, without any filtering on the op-amp output this is likely to smudge your signal. Assuming a bandwidth of 5 kHz (GBWP = 8 kHz), the equivalent input noise is 15.2 uV RMS and and is a lot bigger than the signal you are trying to amplify.
  • Running open-loop (no feedback) means that any signal at the output will be shifted by about 90 degrees. See figure 11 in the data sheet. Given that you want the phase to be the same as the input signal this appears to be a showstopper.
  • \$\begingroup\$ Thanks for the helpful feedback. I got the supply voltage wrong, it's supposed to be 1.8V. Will edit the question. A phase shift is acceptable for my application as long as it is constant. \$\endgroup\$ Oct 14, 2019 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.