Got transformer for bench supply now...

[throbscottle]

Just found out my multimeter is a bit knackered. Real voltages are 35.7-0-35.7 and 12.8, so the smoothed DC is +/- 49v The amp's smoothing caps are 6800uF at 50v - cutting it a bit fine there! Since the amp says 120W on the back of it, I think this must be peak power. Going by the transformers core dimensions - centre leg is 41.7mm x 23.6mm, looks like it's under 100VA. I was under the mistaken impression that the VA of the transformer would be higher than the amp's power, but apparently not.

So going with the buck pre-regulator idea, with the voltage being dropped there should be more current available. The regulator's efficiency looks like 77% for the + rail and something less than 96% for the - rail. Errm, can anyone help me work out how much current I can get, supposing I have a max voltage of 35v? I'm hoping it'll do a couple of amps.

 

[dr pepper]

100va at 35v, just divide 100/35=2.85 amps, however you'd be better calling it 2amps, due to inefficiencies.

Your right about the caps working voltage rating thats pretty close.

My meter has messed up more than once, I'm working on a rf project right now and my 'scope is messing that up.

 

[throbscottle]

I thought I had to take the voltage as the full width of the winding, as crutschow pointed out?

 

[throbscottle]

I'm getting confused.
I get my smoothed DC voltage from Vrms / 0.7071 - 1.2. So if the transformer is 100VA, what happens to the current? Total ACV=71, after smoothing and rectification I get 99.2V. So for the AC side, 100VA / 71 = 1.4, but does the available current stay the same for the DC side? 1.4 x 99.2 = 138!
Or is this where the power rating of the amp and the smaller VA rating of it's transformer gets resolved? The amp says 120W on the back. Taking that, including the bit that's wasted in the bridge, I get 120 / 100.4V = 1.2A. If I use that with the AC voltage I get 1.2A x 71V = 84.8VA, which is more agreeable with the kinds of values I get playing with homo ludens' spreadsheet.

And then of course adjust it a bit to allow for the 12V winding if I use it.

Hmmmmm. Stumbling around in the dark here. Did I get this right?

 

[throbscottle]

*Sigh* talking to myself now. I split the secondary, so instead of 35-0-35 + 0-12, it is now 0-35 + 0-35 + 0-12. I thought I'd use the 0-12 winding to create a fixed 5v output.
That way I can have a dual supply rather than a split supply, no need for a negative regulator. I thought I'd borrow from this design: https://www.electro-tech-online.com/threads/dual-output-lab-supply-0-16v-3a.96187/ for the tracking and auto-parallel features (thanks bountyhunter, if you're reading ) Well OK in this case it would be triple supply.

I can bung a mc34063 on the 12v winding to get the 5v, then use LM2576's to pre-regulate the other 2 outputs to feed linear regulators. I can make a charge-pumped -V supply for those so they will go down to 0v.

I can put a foldback current limiter on each of the unregulated lines, which can disable the buck converters if the transformer current gets too high, then an adjustable current limit on the linear supplies. Should be able to make the maximum limit able to change with the output voltage, so a higher limit can be set at lower voltages.

 

[dr pepper]

You know your in trouble when you argue with yourself, talking is fine.

VA is volts x amps, transformers are rated this way as the windings are limited by current due to winding resistance, VA is used instead of watts, as the latter takes into account power factor, remembe the trans is limited by current so Volts x Amps is how they are rated.

So If your trans load has a power factor close to 1, you can work out the current by using the VA equation, if your using a switch reg then you can just use the o/p voltage as VA still applies at the transformer end.

I've confused myself a little now, thats kinda hard to explain, what I did was ball park by the way, power factor is never 1 and there are losses, but in practice its good to start with an approximation.

 

[throbscottle]

You'd be amazed at the arguments I have with myself. Anyway, thanks for the clarification. I was reading about this in a stackexchange thread, someone said watts doesn't count unless work is being done. Made sense when I read it. I did manage to get my head around power factor a while ago - it puts current and voltage out of phase is about as much as I can remember now though. I'm a suck it and see kind of person so I'll see if I can find out how much of a load it takes to get it hot(ish).

 

[dr pepper]

Yes, if you stick a cap across the mains it'd pull current, but hardly any power as volts and current would be out of phase and cancel.

This way its possible to blow a fuse without using much power, the trans windings act in a similar manner, current affects them whether its in or out of phase, so we rate trans's in VA rather than watts, VA is the same as watts when power factor = 1.

 

[throbscottle]

Oh ok, that is a further tiny nuglet of understanding for me then