Electrical – Multisim – Struggling with a power supply circuit

Taking an educated guess here, I'd start from the origin of this \$V_{RMS} = V_{pp} \frac{1}{\sqrt{2}}\$

This formula holds only for the sine wave - it's actually the formula for the voltage that would deliver the same average equivalent power to the resistive load. It can be derived from a reasonably simple integration of the instantanous power.

Putting any non-resistive component (such as a diode) in the circuit, causes that integral to no longer hold. Specifically, around 0, the diode would isolate entirely (in both directions), and thus there would be no power contribution to the resistive load. For a simple model of a diode, you could probably work out the exact formula yourself with a little bit of integration.

1) The best way to reduce heat, assuming you insist on a linear regulator, is to use separate secondary windings to produce the raw voltages on your regulators. 20 VAC is OK for 25 volts, but 12 - 15 VAC for the 12 volt supply and 6-8 VAC for the 5 volt supply is much better. Of course, then you need separate rectifiers and raw DC capacitors, too.

2) If you must use the existing transformer, zeners are (technically) an excellent choice. The 1N3998A is available from Digikey, and you can use one for the 12 volt supply and 2 in series for the 5 volt supply. As long as output current is less than 1.5 amps you'll be fine. Of course, you'll need heat sinks for them, since each will dissipate 6 watts per amp. Of course, you may not be happy with the $25 price tag. That's per zener.

Power resistors are certainly possible, and about 10 ohms and 20 ohms will work. (They cannot be in series as the diodes are in your schematic). They'll need to be at least 10 watts and 20 watts for the two values I've given, and more is better. They'll also need cooling.

The problem with any linear regulator is that raw DC current must equal output current, and that current times the voltage drop from raw to final voltage is the amount of heat that must be dissipated in the supply.

3) In the charge circuit, it's true that the resistor will drop some voltage, but in direct proportion to current being drawn. So, as the battery charges, its voltage will rise, and the charge current will drop. At the very end of the charge, the battery voltage will essentially equal the charge voltage, the current will be very small, so the resistor drop will also be small. In fact, it is the difference in voltages divided by the resistance which determines what the current will be.