The "diode and resistor" have been replaced by a "charge pump" (you can google this) for the reason of "why waste power in that resistor." This is not a bait and switch, it is merely an improvement on an earlier proposal.
If you remove the battery from Nick's charge pump circuit proposal, you have what's commonly called a "voltage doubler" (you can google this) and the open-circuit output voltage will peak at twice the AC input amplitude, which would be 340V for "normal" 120VAC-RMS input. Limiting the size of the capacitor limits the current output capability. Placing a battery across the output limits the output voltage by shunting the current flowing through the capacitor. Also remember the current=C*dV/dt, and dV/dt peaks at 170volts/sec. So to reiterate, current and voltage have been limited: as long as the battery is there and can accept the current, you do not have 60V, 120V, 340V or whatever, you have 1.5V regulated by the battery and no additional regulator is necessary.
I would modify this design by replacing the diode whose anode connects to ground with a zener rated for 2.5 to 5 volts. That way, if the battery should open-circuit, the downstream components won't see the
370V spike, it having been shunted by the zener. Similar "transformerless stepdown" circuits are used to supply regulated 12VDC to motion detectors.I think Nick was off in his original math because he may not have considered that the output diode (the one with the cathode connected to the battery) begins conducting roughly as the AC input wave rises from its -170V minimum, and continues to conduct roughly till the input reaches its 170V maximum (we're ignoring .6V and 1.5V voltage drops across diodes and batteries as insignificant compared to 170V). Therefore the total charge moved per cycle is Q=C*340volts over the full voltage swing of -170 to +170. (I now see and agree with his math ammended in a subsequent post.) Current in amps (coulombs per second) is Q*60 because there are 60 cycles per second. Substituting the suggested value of .0047 uF for C, and current comes out to be 96 microamps. This is not meant to recharge a battery that has been run flat, but rather to compensate for the self-discharge of the battery such that it does not discharge during storage.
The question of whether the negative terminal of the battery is connected to an identifiable "common" node in the VOM's internal circuit is left for further consideration.
%mod%