[sdiy] Roman Filippov power supply design

[Guy McCusker]

Can the LM25575 be used for 12V -> -15V ? The datasheet says that OUT
to GND is limited to Vin. That's why I guessed that it used the 15V
from the other converter as its input.

I didn't think about using the PCB layer as a heatsink though -- I am
a total novice at this! -- so that's a bit of very helpful insight.

On Tue, May 9, 2017 at 10:04 PM, Roman <modular at go2.pl> wrote:
> My guess is IC2 (LM27313) is used for +12 -> +15V conversion. IC3 (LM25575)
> is used as boost/inverter +12V -> -15V.
> Obviously IC1 (LM2575) is for 5V.
>
> If I was Roman F. I'd utilize most of the bottom layer as heatsink,
> especially that IC1 and IC3 have exposed pads made just for that. IC2 (+15V)
> has pretty good efficiency when supplied from 12V, about 90% according to
> datasheet, so might be just slightly warm using only a bit of top layer
> copper around it as heatsink.
>
> Roman
>
> Dnia 9 maja 2017 17:36 Guy McCusker <guy.mccusker at gmail.com> napisał(a):
>
> Roman Filippov is selling PCBs for a power supply which he says is
> enough for a Music Easel or 8 of his Buchla clone modules:
> https://electricmusicstore.com/products/power-supply-unit-model-211
>
> The board takes 12VDC input and produces +/-15V, +12V and +5V. Taking
> a look at the BOM, I can see a 5V switching regulator, a boost
> converter and a buck converter.
>
> Presumably the 12V input is passed straight to the output, and the 5V
> regulator does the obvious thing. For the +/-15V, do you think it
> likely that he's using the boost to produce +15V, and then using the
> buck converter to produce an inverted version of that for the -15v?
>
> If so, then could someone check my reasoning and calculations for the
> maximum current one could get out of this please? I am thinking this:
>
> the boost converter has a thermal resistance of 166 deg/W. I see no
> heatsinking and at 25C ambient and max operating temperature of 125 we
> have about 100 degrees to play with, so can afford to dissipate 0.6W.
>
> The device is up to 90% efficient, so that would mean we could get
> 5.4W out of it, dissipating the other 0.6W. So it can produce
> something like 360mA of current, and this has to feed both the +15V
> and -15V rails. You get a bit less because of the inefficiency of the
> buck converter for the negative rail, so we're looking at about 150mA
> per rail.
>
> This does not seem enough. On the other hand I don't exactly know what
> I am talking about here. Have I messed this up?
>
> Cheers,
> Guy.
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