DEC H744 +5 supply
bhilpert at shaw.ca
Fri Sep 21 16:18:08 CDT 2018
On 2018-Sep-21, at 12:38 PM, Noel Chiappa via cctalk wrote:
> So there's something about the H744 I'm not sure I understand; hopefully those
> with more analog-fu will set me straight if I'm confused.
> This supply runs off 20-30V AC. It takes the input AC, rectifies it, and runs
> it through a cap to filter out the ripple. What's next is that it's an early
> switching supply; i.e. the electronics inside switches that newly-created main
> DC off and on very quickly to keep the output voltage at around +5V.
> My understanding is that, without using a transformer (which creates an
> independent circuit loop - more below), there's no way to increase the
> _amperage_ out of circuit over what's fed into it: since amps are
> electrons/second, the electrons/second out more or less have to equal
> electrons/second in, since one can't easily 'create' electrons - at least, in
> normal electonic gear! (Transformers, by creating a whole new circuit loop,
> can 'create' more electrons/second in the new loop; since they tie the 'out'
> of the new circuit back to its 'in', they can recirculate the 'extra'
> And to the extent that the output is at lower voltage, the energy differential
> has to be dumped; hence the huge heat sinks - a lot of the incoming power in
> that 30V AC has to be thrown away, in producing +5V. Right?
> My (possibly confused) understanding is that later switching supplies take the
> incoming 60Hz wall AC, transform it to a higher frequency, run _that_ through
> a transformer (which can be a lot smaller, since it's at a higher frequency,
> and the higher the frequency, the smaller the transformer you can use - hence
> the use of high frequency AC in airplanes, to allow use of smaller - and thus
> lighter - transformers). That then turns out a massive amount of amps in the
> output loop (since with a transformer, energy out is roughly energy in, modulo
> resistive losses; and with constant power, if V goes down, I goes up).
> So the losses are a lot lower - N amps at 110V in produce ~20N amps out at
> +5V. (Well, depending on all the losses, ~20.) And the whole works is a lot
> lighter, to boot.
> Did this programmer get all that analog stuff correctly?
No. You don't need a transformer to increase the current ('electrons/S') in the output above that of the input.
In typical "down-converters" there are additional current paths in the supply, paralleling the input path, that can provide the 'additional' electron flow rate.
There are the parallel filter capacitors, and in the typical 'bucking converter' - which, while I'm not familiar with the H744 design*, I suspect it probably is - there will be what looks like a reverse-biased diode across the (switched) input which allows 'arbitrary' current levels to flow out of a series inductor which previously stored energy from the source, and deliver that current through the load.
The energy differential from the voltage reduction is not being dumped as a power loss, it was converted by the supply to a different EI relationship between the input and output.
The forgotten factor in your analysis is time: the whole rationale of a switching supply is to use time (varying switching periods) and temporary energy storage to change that EI relationship from input to output without energy loss.
If the heatsinks seem huge compared to modern day supplies, that's more likely the result of technology improvements - faster devices, and moving from bipolar switching transistors to mosfets. Bipolar transistors have a near-fixed voltage drop which can't be reduced and thus present an increasing power loss as current goes up, while mosfets can be designed for lower 'insertion' losses.
* If you supply a link & location to a schematic I'll take a look, I don't feel like wading around in bitsavers pdfs to try to find it right now.
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