How do I calculate the maximum load on a center-tapped AC/AC transformer for a bi-polar linear power supply with full wave rectification?

Question

I have a 24 VCT AC to AC step-down transformer, which has three wire leads on the secondary. I'll call the secondary leads X, Y, and C, where C is the center tap. With a DMM, the voltage between X and Y is about 24 VAC (actually slightly higher), and the voltage between X and C is 12 VAC, and the voltage between Y and C is 12 VAC. The manufacturer rates the transformer as 24 VCT, 10 amps.

Putting aside safety factors, overhead, and load types for a moment (for clarity):

1. If we ignored the secondary wire, C, and connected an AC load across X and Y, we could have a maximum load on the transformer of 10 amps @ 24 volts, which is 240 watts. Correct?

2. Let's say we have an AC load that's designed to run at 12 volts. We should be able to run a maximum of 10 amps @ 12 volt load from X to C, at the same time as we are running a maximum load of 10 amps @ 12 volts from Y to C. In other words, we can now power 20 amps of 12 volt load because halving the voltage doubles the ampacity while power (240 watts) remains constant. Is that correct? This is assuming that we put no more than 10 amps on each 12 volt set. ie. two balanced 10 amp, 12 volt AC loads.

3. Now let's say we have a full-wave bridge rectifier on the secondaries X and Y, and we use the center tap C as circuit ground. We run both the + and - outputs of the rectifier past a smoothing cap(s), and through a 7912 voltage regulator (7812 on the positive rail). We now have three connections; the ground (which is the center tap), a +12 rail (relative to ground, ) and a -12 volt rail (relative to ground). Can we run a 12 volt, 10 amp DC load from the positive rail to ground at the same time as we run another 10 amp, 12 volt load from the ground to the negative rail? For a total of 20 amps load at 12 volt DC? Remember, this question is about the AC to AC transformer's rating when used in this context.

4. Now, for the title question. Let's say we have the same setup as question #3. But, instead of having perfectly balanced loads, let's say that we really only want the negative rail for op amps and there's not going to be much draw on the negative rail. Are we still limited to 10 amps @ 12 volts from the positive rail to ground? Or does using full-wave rectification allow us to "borrow" spare ampactiy from the negative rail to ground (that isn't needed) and use it on the positive rail instead? For example, could we have 12 amps @12 volts on positive to ground and 8 amps @ 12 volts on the negative to ground? Could we have 18 amps @ 12 volts on the positive to ground and 2 amps @ 12 volts on the negative to ground?

Or to put it another way, if a transformer is rated for 24 volts at 10 amps, can you use any combination of load on the positive and negative rail (in a bi-polar, full-wave rectifier scenario) as long as the total does not exceed 20 amps @ 12 volts?

For the purposes of this question, let's assume that "maximum load" on the transformer means either the point at which a fuse should blow (if fuses are sized properly) and/or the point at which the transformer would be damaged, become dangerous, or perform incorrectly due to overloading or unbalanced loading.

Answer 1

Since a bulk Cap with 10%V ripple needs a pulse of current to charge it for the 90% decay time with 10%Vpp ripple the current ripple is a pulse just 10% before peak to peak at 10x the load current, so the transfer of power is poor and can saturate the core, so the VA rating must be reduced significantly. A more efficient method is to use an active PFC and buck regulator and the negative side is same with a small cap. Old BJT LDO’s will drop 1.5 to 2V so you get 10V, maybe if the Vpp ripple is <10%. The XFMR should give you 12Vac at full load and 10% higher with no load per tap.
FET LDO’s are much less.

Answer 2

1.&2. are correct given load current proportional to voltage at every instance - resistive.

3.&4. are wishful thinking.

For a "balanced 2×half voltage load", the ideal centre tap current was zero.
In reference to this cheat-sheet, you can expect about 6 Amperes at (just incidentally!) about 12 Volts (with real diodes) at the charging capacitor for each rail when loaded symmetrically, 10 A from a single rail with the other loaded insignificantly.
Assuming the "negative" peak of ripple .5 V, allowing for 100-10% line voltage and a regulator drop-out of 2.5 V, it's "difficult" to get more than 7.5 V regulated.

Few loads operate from pulsating DC/unbuffered (single-phase) rectifier output just as well as from buffered DC.
By no means all DC loads profit from the supply voltage being stabilised/constant/defined exactly.
Some profit from supply voltage being well filtered, e.g. audio amplifiers (voltage amplification stages, which get additional filtering internally).

can you use any combination of load on [both rails of a bridge rectifier on a centre-tap secondary]
as long as the total does not exceed [double the secondary current rating]?

No, on two accounts:

• there is a derating to about 60% of the secondary's current rating with a "capacitor input filter" (90% with a (costly) DC choke input)
• total current decreases further with load(ing) imbalance - Hammond suggests down to (two times) 50% (72% with choke input).
  Note that, if the secondary was two-part, really, you could connect the halves in parallel instead of in series and use a regular bridge configuration and (2×) 60% of rated current (almost 190% (?!) with choke), but suffering one "diode drop" more.

In real life, things like (frequency dependant) ESR and "wiring"/PCB layout make a significant difference (e.g. in ripple voltage).