I need to measure a low-power AC signal coming from a transformer and a voltage divider:
The above circuit provides a nice 5VAC (peak-to-peak) signal but then I need to avoid supplying negative voltages to my uC's ADC pin. I though in putting a clamping diode in parallel to the ADC pin as follows:
I expected to experiment a voltage drop in the negative side of the AC signal only. However, the whole rectified AC signal seems to be affected by an additional voltage drop of about 0.5V, as shown in the attached capture (yellow: original AC signal without rectification, blue: rectified signal).
I'm thinking at this point that the 750K resistor could be limiting the current through the line too much but this is just a weak supposition. I prefer to read from you before trying to get more conclusions.
It might be due to the capacitance of the diode. What is your input frequency, 50/60 Hz? Or, (more likely) it could be that your Schottky diode (assuming that's REALLY what you are using there) has a pretty crummy reverse leakage spec. You would just need 0.6uA to drop 0.5V across that 750K resistor.
It might be due to the capacitance of the diode. What is your input frequency, 50/60 Hz? Or, (more likely) it could be that your Schottky diode (assuming that's REALLY what you are using there) has a pretty crummy reverse leakage spec. You would just need 0.6uA to drop 0.5V across that 750K resistor.
That makes indeed sense. I've tried with multiple diodes (schottky and not) with more or less the same result. I even tried with the MCU in place and without the clamping diode and got again the same issue, probably due to MCU's internal protection diodes. In any case, I should maybe use lower value resistors in my voltage divider or a clamping diode with lower reverse current.
You might want to use an opamp between your divider and your ADC. Look at the spec for the ADC input impedance. Using an op amp rectifier would also eliminate any negative transition on the input signal.