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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?

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  • \$\begingroup\$ The first division on 1.5VAC scale is 0.05V, not 0.01V. \$\endgroup\$ – Bruce Abbott May 10 '17 at 5:56
  • \$\begingroup\$ No. you have missed the other 150 divisions that are difficult to see on the image. I know this because I have this multimeter at home. \$\endgroup\$ – MikeTeX May 10 '17 at 11:44
  • \$\begingroup\$ On the red scale marked 0-20-40 etc. (1.5V AC) there is only one division below 10, and the scale is highly compressed at the low end (compare to the black scale with same range above it). This separate AC scale is needed to compensate for diode voltage drop, which is significant even after the voltage has been boosted with a transformer. \$\endgroup\$ – Bruce Abbott May 10 '17 at 16:29
  • \$\begingroup\$ Good point to be noted Bruce, thx. \$\endgroup\$ – MikeTeX May 11 '17 at 14:09
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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.

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  • \$\begingroup\$ So, they multiply the AC voltage with the transformer. Nice! I don't understand you last sentence. Do you mean: since the AC voltage is now higher, the voltage drop of the diodes is negligible ? \$\endgroup\$ – MikeTeX May 10 '17 at 6:04
  • \$\begingroup\$ The transformer is in parallel to the sense resistor - it thus forms a second path for current to flow. However, because it appears as a large resistor (high impedance), the current through it (and thus the error on the current measurement) is very small. \$\endgroup\$ – Joren Vaes May 10 '17 at 6:06
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This is a classic 20kΩ/V or 50uA current meter that has series resistor dividers to measure higher voltage, shunt resistors to measure higher current and diode half bridge to measure AC voltage. Ohms mode supplies some fixed uA/mA current and is converted to Ω on some V/mA using 50uA full scale.

The AC appears to use a centre-tapped transformer with labels for L1,L2,L3 in a planar EH type transformer.

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  • \$\begingroup\$ What about the 0.2V diode drop (at the least) ? \$\endgroup\$ – MikeTeX May 10 '17 at 6:05
  • \$\begingroup\$ There are 5 diodes, 2 for clamping meter coil over current and 2 for AC rectification with drop error, (0.1V at low current) and one I can't make out the connections \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 10 '17 at 6:09
  • \$\begingroup\$ Thanks. You're right to say that the other meter types are quite straightforward (as I knew). The main point of my question is related to your last sentence regarding the transformer. I had not hit on the idea that the AC voltage can be raised with a transformer. \$\endgroup\$ – MikeTeX May 10 '17 at 6:14
  • \$\begingroup\$ Looking at the turns ratio For L1 to L2:3 it appears to step up 1:10. I guess the drop error must be very low at this low current (Schottky. Diode can be like RF Hot carrier diode.) no spec here \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 10 '17 at 6:16

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