Very hot rectifier

[AvionicsMaster1]

According to what I read in the datasheet, at 103C the max output is 10A. That means to me you're doing something wrong. If you've got this thing mounted to the case made of metal you've probably have a good enough heat sink since it looks like it's operating at near max current. It doesn't specifiy the size or type of heat sink though I think it's necessary.

Me thinks, and I'm very often wrong, you might oughta look at mounting, wiring or assembly to find your excess draw. I'd also like to see your schematic if that's possible.

[The Master]

I've connected my multimeter in and the current draw is hovering around 3.7A with the LEDs on full power.

Just to clarify, I meant the metal casing around the rectifier itself. I haven't attached anything else to it that could act as a heatsink.

I haven't got a schematic available but it's a simple circuit. The rectifier feeds into 2 capacitors in parallel (2200uF) then through some transistors into the LEDs. There is a WS2801 chip with it's own 100nF ceramic cap for each LED but these chips draw very little current. The 3.7A I measured was taken from the output of the rectifier and represents the total load of the entire circuit.

[Charlie]

This is something so fundamental, the datasheet doesn't even bother to mention it, assuming no one would connect a power device without a heat sink.
If you look at the curve, it tells you you can do 25 A if you keep the core under 55 degrees C. It's up to you to do that with a properly calculated heat sink.
Here's a good paper to start: http://www.ti.com/lit/an/slva462/slva462.pdf

[amgen]

That's absolutely true about power devices having to have proper heat sinking....... but..... if current is only on a few seconds not so frequently, that situation may be ok without heatsinking and the 6 amp round axial diodes don't have heat sinking but still rated at 6 amp (don't know how they get away with that)? and finally going to shotkey diodes with .3 Volt drop instead of silicone diodes with .6 volt drop will consume 1/2 the power / 1/2 the heat.
don

[Archangel]

Consider the fact the average electric blanket is around 25 watts . . .

[AvionicsMaster1]

Rambling a bit here but just some thoughts.

Well, your location says UK so I'm assuming 220 VAC 50 HZ. The datasheet only specs a 60 Hz curve but I doubt that is the source of your heat.

If I understand you correctly, your powering this rectifier by 12VAC input to the rectifier. I'm also assuming the output has twice the input frequency so it should be 100 Hz before capacitor smoothing. If the bridge rectifier is good you should have a doubling of the frequency. You could check that to make sure the rectifier is good.

I'd think anything metal would be a good enough heatsink for your current load. A small chunk of metal with a little transfer compound on it should work. I'm curious if that will do it.

[Charlie]

Rambling a bit here but just some thoughts.

Well, your location says UK so I'm assuming 220 VAC 50 HZ. The datasheet only specs a 60 Hz curve but I doubt that is the source of your heat.

If I understand you correctly, your powering this rectifier by 12VAC input to the rectifier. I'm also assuming the output has twice the input frequency so it should be 100 Hz before capacitor smoothing. If the bridge rectifier is good you should have a doubling of the frequency. You could check that to make sure the rectifier is good.

I'd think anything metal would be a good enough heatsink for your current load. A small chunk of metal with a little transfer compound on it should work. I'm curious if that will do it.

I'm not sure why you mention frequency. It has absolutely nothing to do with the problem. If the circuit (no schematic shared) is a full wave bridge rectifier, then at any point in time 2 of the 4 diodes will be conducting.(except for the very small time the AC is within 0.9V of 0V).
From the data sheet, at 5A the forward volt drop will be 0.9V.
2 elements are conducting, so 5A x (0.9 * 2) = 9 Watts being dissipated in a 1/2 cubic inch of material = very toasty. Hence the giant hole in the middle of the device for a #8 screw to bolt it tightly to a heat sink. How big of a heat sink? Work out the thermal impedance to ambient air. How do you do that? Click on the link in my previous post.