Heat sink for full wave rectifier? (metalworking content)

My 13" SB metal lathe has been retrofitted with a PM DC motor. It's not a treadmill motor, it's continuous duty according to the tag and can handle up to 180 V. I assembled a speed control from a treadmill-type board. Unfortunately it doesn't appear that the controller provides enough oomph.

What I'd like to try instead is a 120v 20amp variable transformer, output to a bridge rectifier, then filter the DC and send it to the motor. By happy coincidence I have all the parts, total cost will be simply a bit of time.

The rectifier I have is a GBPC1210W; data sheet says 12 amp, 1000 v. It is rectangular with a hole through the center. It looks like it ought to be mounted to a heat sink, but the data sheet I saw doesn't say anything about doing so.

The question is: should a bridge rectifier be heat-sunk (sinked?) in such an application? If so, how big (roughly) should the heat sink be? Would it be sufficient to bolt the rectifier to the 4" square metal enclosure? Thanks for your input.

-- Best -- Terry

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Does the data sheet say anything about junction temperature, thermal conductivity to ambient, or thermal conductivity to the case? Generally, power semiconductor device power ratings are made assuming pretty darn good heat sinking.

What current is the motor going to be pulling?

At 12A the rectifier will be dissipating something like 24 watts; that feels like something that would need a 4" square finned heat sink rather than a flat plate, but I do that sort of work so seldom that I always need to go do the math.

Reply to
Tim Wescott

In typical RCM fashion I'm ignoring your question and answering one you didn't ask. Speed regulation will be poor with a simple power supply. You'd be much better off with a DC motor drive similar to this:

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Reply to
Ned Simmons

I don't know how much current you're dealing with, but figure about 1-2 volts drop across such a module X how many amps you're pulling = how many watts it will dissipate.

since it's rated 12 amps, you're probably going to need larger than 4x4 of heatsink to keep it happy.

quite honestly though, it's going to be easier to just mount the thing and do the touch test than to try to calculate all sorts of stuff with guestimated values.

If it gets too hot, use a larger heat sink. big deal.

Reply to
Cydrome Leader

The voltage drop is approximately 1.5-2 volts, so at 20 amps it will be 30-40 watts. You need an aluminum plate or a heatsink for sure. I would also get a much larger rectifier, to account for starting current, at least 50 amps. Those bridge rectifiers are dirt cheap these days, I see no sense in getting something that just barely is adequate.

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Have a look at this datasheet:

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Figures 1 and 2, especially (they show some example heat sink dimensions). You can derive this by looking at figure 3, and at the degrees C/watt for various commercial heat sinks.

You'd probably be safe up to 5A continuous, for much more I would go to a much bigger heat sink and a 35A bridge rectifier (only a few dollars).

Reply to
Spehro Pefhany

The rectifier should have a heat sink.

Where are you? The local recycling place ( Actually scrap yard ) has aluminum finned heat sinks. Not especially cheap at a dollar a lb. They range from small to big.


Reply to

Yes, it needs to have a heat sink. if the enclosure is 1/8" aluminum, this may well be enough. If thin steel, then maybe not, but it will be better than just hanging in the air. You can figure that any time the bridge is conducting current, there will be two diodes in conduction, dropping about a volt each. So, at 12 A, you would have about 24 Watts to get rid of. If the cabinet is too light for removing the heat, a chunk of 1/8" aluminum would help. Also, you should put heat sink grease on the back of the bridge to conduct the heat better to the plate.


Reply to
Jon Elson

OTOH, if you've got some sheet aluminum, some heat sink goo, and no cash, you can bend up a bunch of ever-narrower 'U' shapes and stack them. I wouldn't do this unless I was _seriously_ short on cash, because if your scrapyard doesn't have heat sinks (mine doesn't) you can still find them cheap at electronics surplus places.

Reply to
Tim Wescott

Don't forget the silicon thermal compound between the rectifier and the mounting surface.

Pete Stanaitis


Terry wrote:

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Yes, you should heat sink it - I'd try for 100 sq inches of surface, more or less, in free air. Thats about half a sqare foot if both sides are exposed to the air.

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it's amazing how much just plain wrong information you are getting.

Vfd on a silicon diode is .7v. Fullwave rectifiers have two in the forward path conducting on each half wave. 1.4V drop. If you pull 12 amps, power is therefore 16 watts. But, it is AC, so you won't pull the

12 amps continually.

But, as others pointed out, you would be much better served with a SCR/Triac controller with feedback - the ones I like are made by Minarik. They are frequently available on ebay but peruse the minarik catalog first

Reply to
Bill Noble

For most full-wave bridge rectifiers with holes in them, the device should be mounted to a metal panel (minimally) or a heatsink with a proper heatsink compound to transfer heat.

Many datasheets will indicate a notation that recommends heatsinking of a device, when used in excess of a nominal power (watts) rating.

As Ned commented, speed regulation will be fairly poor with a variac. That means the motor will slow down considerably from a set speed (no load), during the actual metal cutting operation. A variac will provide variable speed, which is a lot "better than" (it get's a gold star) single speed.

Variacs (autotransformers) do not provide line voltage isolation, so the output voltage is hazardous. Proper insulating practices need to be followed to protect the equipment users from potential electrical shock hazards.

Most commercially-built 180VDC motor controllers are typically 240VAC input modules, and the DC motor controllers for 90VDC motors are typically 120VAC input modules. So, the existing 180VDC motor will be slightly under-powered from a DC supply derived from 120VAC.. the results may be similar to the performance from the previous attempt with a treadmill controller. Finding a commercially-built 180VDC motor controller would likely provide the best results.. variable speed with excellent speed regulation, plus other features.

Looking up the manufacturer's specs for the existing motor would be worthwhile, before buying a controller that may not be suitable for your application.

Another consideration could be the need to cool the variac to prevent overheating, which could be accomplished with a small fan.

Reply to

maybe you should check a datasheet for a bridge rectifier some day, or takes some actual measurements.

Vishay is even kind enough to tell you the dissipation in watts for such items.

Reply to
Cydrome Leader

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Hmm ... your transformer is capable of putting out 20A, and the bridge is only rated for 12A. How much will the motor draw worst case?

I would certainly use a heat sink -- and a local fuse between the output of the variable autotransformer and the bridge. One with

*lots* of fins to increase the surface area for better transfer -- and perhaps a fan to assist cooling if you are runing close to the 12A rating of the bridge rectifier.

And you want a heat sink compound between the bridge and the heat sink.

How much current will your motor draw worst case? (For a lot of PM motors, you should find a maximum current rating on the motor's data plate -- because more than that current will partially demagnetize the permanent magnets.) If you find such a rating, go for a smaller quick-blow fuse, because it does not take any significant time for the excess current to degauss the (im)permanent magnets.

Good Luck, DoN.

Reply to
DoN. Nichols

AC current is "dimensioned" in RMS amps. So the fact that it's not pulling them "continually" [sic], has already been taken into account.

12 is the correct number to use for figuring power.


Reply to
Bob Engelhardt

You probably don't even need to "filter" it - the inductance of the motor windings should provide all the filtering you need.

If you're planning on running 20 amps through the motor, this rectifier will die. If the motor current is guaranteed to not go over 12A, then no problem.

Yes, absolutely. And use some kind of thermal goo - NOT a "sil-pad" - they're crap.

It's hard to answer this sight unseen - you need to look up the thermal resistance of the diode in the data sheet, and the enclosure, enough to keep the rectifier chip itself below the rated max. temp. It's pretty simple arithmetic if you have those figures. Personally, I like as much surface area as I can manage; forced-air cooling (like a fan) helps a LOT, but unless it's a great big huge heatsink, you should mount some kind of guard so that nobody gets their hand burned. You can look up thermal resistance of various extrusions and stuff on websites like Aavid and such. Google "heat sink."

It also could do no harm to ask this same question at sci.electronics.design.

My Pleasure! Rich

Reply to
Rich Grise

Personally, I like as much surface

The heat sink ought to be big enough that no guard is needed. If the heat sink is hot enough to burn someone, the silicon junction is probably shot.


Reply to

While I kind of agree with all that, the lathe would be unlikely to be pulling 12 amps at 190 volts continuously. Anyway, aluminum plates are inexpensive

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and a 1/4" thick 8x12" plate can be had for $8.86. That is far more than necessary.

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Yes, from YOU.

The forward voltage isn't one value, it varies as the log of the current, which you should know if you ever curve-traced semiconductors. The Measured drop across a power rectifier near its rated load is 1 - 2V as others have posted. Some of that is Vf, some is IR drop, some possibly corrosion on the hardware, but the heatsink has to remove it anyway.

I'd plan for a fan to use if needed, like for a larger motor.


Reply to
Jim Wilkins

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