I have an old Pekly multimeter that can be seen here. This is an analog multimeter, and only its Ohm-meter needs batteries. It can measure current intensities, DC and AC voltages, and resistances. The maximum resolution for the AC voltage is 0.01V (full deviation = 1.5V, 150 divisions). This means that the multimeter can measure AC voltages as small as 0.01V.
I can't figure out how they did that: obviously, you have to rectify the AC current to use a galvanometer; but the forward diode drop is at least 0.2V. It is of course possible to forward bias the diodes, but as I said, no battery is needed for the voltmeter.
I point out that my question concerns AC currents only, for there are of course very precise galvanometers working with DC currents.
I know there exist AC voltmeters that work with two coils (e.g. dynamometers), but nothing that correspond to this precision (in my opinion). Furthermore, the same mechanism is used for the other meters (ammeter, ohmmeter, DC voltmeter), which add even more to the cleverness of this apparatus.
I have managed to find a schematic of this multimeter here but I don't understand it.
Any insight on how this works?
Best Answer
From what I can tell (but the quality of the schematic is quite poor so this could be an incorrect interpretation) they use current sense resistors to turn the AC current into an AC voltage. This AC voltage is passed through the transformer (L1, L2 and L3). After this it is rectified and it's value is measured. Since the impedance of the voltage measurement circuit is so high, it doesn't influence the voltage drop due to the current.