Protecting a power supply

Mar 02, 2009 19 Replies

I've got a new power supply that I will be using to anodize aluminum. The supplier says to protect it when using it in that fashion by putting a diode inline with the positive terminal. The purpose is to prevent reverse current flow the instant the supply is turned on. The power supply is a Mastech HY3020e switching supply (30volts, 20amps) and if it sees even a small voltage when turned on, it does bad things to itself. Radio Shack has 50v,



6amp diodes. Can I put 4 of these in parallel with each other to get what I need? I'm just not sure if you can parallel diodes to increase the current rating. Thanks in advance.

Generally its not best practice to put diodes in parallel as there will be small differences in forward volt drop between diodes. The one with the lowest drop will hog more than its share of the current and heat up more - as temperature rises the drop falls even further causing the hottest diode to hog an even bigger share of the current.

If you give a big enough safety margin in current rating, you can get away with stud mounted diodes mounted close together on the same heatsink.

It seems very likely that a single stud mounted diode rated to take the current would work out cheaper and less hassle.

Thank you, guys. Your answers probably kept the smoke in, because that's not how I would have thought they would behave. (Electronically illiterate here) I'll look for a single 20+ amp diode to do the job.

_However_, if you're in a bind you can ballast the diodes with low-value resistors -- something that gives a few tenths of a volt at the diode's rated current should do. Be sure they've got a high enough power rating.

That'll keep the diode to diode current balance constant, at the cost of a few more watts puffing up into the air.

http://www.wescottdesign.com

At six amps per diode: --+--[DIODE>]--+-- | | +--[0.05R]---+ | | +--[DIODE>]--+ | | +--[0.05R]---+ | | +--[DIODE>]--+ | | +--[0.05R]---+ | | +--[DIODE>]--+ | | +--[0.05R]---+ || That comes out to 0.0125 ohms across the diodes which takes away their diodeness. JF

I suspect that Tim meant to place a resistor in series with each diode, not in parallel.

Peter Bennett, VE7CEI peterbb4 (at) interchange.ubc.ca GPS and NMEA info: http://vancouver-webpages.com/peter Vancouver Power Squadron: http://vancouver.powersquadron.ca

Aaarghhh!!! I\'m sure you\'re right. What _was_ I thinking? JF

Or could that be - what were you smoking?!

I'm pretty sure that's what he was thinking, too:

o---o---/\\/\\/---->|---o----o | | o---/\\/\\/---->|---o | | o --- etc --- o

But what do I know?

http://www.wescottdesign.com

All Electronics has some 40 amp stud mounts for $3.50 each:

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Shipping is a flat $7.

John

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NO Because their forward resistance will not be the same and the diode of lower resistance become more hot & burns therefore use single diode.

If RS is the only electronics supplier you can find nearby Go to an auto-electrician and get a rectifier diode from them, they'll have diodes rated at 50V 50A and higher for use in automotive rectifiers.

Come to that, suitable rectifiers can be had for free out of a scrap car alternator.

ian field wrote:

Aren't those suckers all press-fit? Getting them out of the unit's housing is a bit of a chore. Right? (...and half are polarized one way; half the other.)

Connecting to them is another issue: high thermal mass and no actual second terminal.

Not if you leave them in the heatsink and use the hole for a terminal lug that's already there.

Why turn an easy job into an ordeal?!!!

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Off the top of my head, how about a grunty MOSFET on the PS output to delay the PS connecting to the load? The mosfet turn on would be delayed with a RC charging circuit...

Cheers

depends, some are soldered to the heatsink eg:

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if from a scrap altenator leave them in situ or cut the heatsink to a suitable size leaving a bolt hole or stud for connecting the other terminal.

(remeber to mount the heatsink in a non-conductive way,)

Thanks for all the suggestions, I think I'll go this route.

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Paul

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