Sources of heat in a buck converter?

Why didn't you keep the board flat so the inductor wouldn't fall off?

It is pretty much impossible to know without seeing the schematic *and the part numbers*. But if the inductor fell off first, it could be that. Post the part numbers of the FETS and the inductor, at least.

His stepdown is 40:5 or 12.5%, makes for a heck of a high frequency component to dB/dt on top of a high repetition rate and ohmic losses in the coil. The component was not characterized well. But instead of looking at the specifics, the OP wants the distraction of a generalized treatise on buck converter loss factors...

The FETs are Vishay Si7850DP. The inductor was a Vishay IHLP5050CEER100M01 but has now been replaced with a Vishay IHLP5050EZER100M01 (as I hadn't expected current to be as high as it is). The schematic is essentially the first page of the datasheet - there wasn't too much creativity in the design of this board.

-Michael

It was flat, but upside down. The inductor didn't fall off till the board was tapped. Surface tension must have been holding it on.

-Michael

Well, those all look underrated for the job.

Remember those datasheet ratings are probably for 25'C ambient temperature, with individual infinite sheets of copper allocated to each part. In the real world everything will heat up everything else.

I can't predict which part is making the *most* heat (at this time of night). You do need to individually analyse the losses in each part. One thing you could try is mounting a part off the board with say 0.5 inches of wire. Normally a really bad idea but in this case this should isolate the part thermally so you can see how hot it gets!

Linear have a free simulator (switchercad) which would be *ideal* for this, but unfortunately I can't see a model in their library for this part.

Good diagnosis.

I was going to suggest that the OP spend at least a buck fifty for a better converter.

This is a synchronous rectifier so the mosfet will not get very hit if properly selected. The inductor needs to have very low DCR and with 8A of current flowing through it or it will heat up. ever 0.034 ohms would yeald 2.6 watts of power dissipation. if the inductor is not heatsinked, that will heat up until it falls off. you might want to look for an inductor with lowe DCR in tha range of 10m ohms or less.

Or you can heatsink the inductor with a Bergquist thermal pad.

What is you enclosure like?

LT3845 is in SWcad

So it is - I missed it for some reason.

In that case it would be well worth the OP setting up a simulation IMO.

One question: How do you get swcad to show power dissipated? I mean, you can alt-click to get electrical power - but that doesn't realy say much for an inductor, right?

Or am I missing something?

The circuit was designed with Linear's switchercad - so simulating it is not a problem.

Thanks,

-Michael

No, it works, the electrical power shown really is the power dissapated. And you can control-left click the power waveform to get average power.

I think you are missing that you have to include the parasitics in your circuit.

In the case of the inductor at a minimum you would include the DC series resistance (there is a parameter just for that in the standard inductor model). There will also be inductor losses due to frequency dependent effects (skin effect, core losses). These are much harder to model so you may have to make some guesses from the datasheets or some real measurements.

Holy crap Batman! That's a lot of step down for a buck. Usually inductor data sheets have some sort of data on heat rise due to current. Seems the coolest part are the power FETs these days.

At least the inductor self-desoldered. Easier to replace it with a beefier unit.

I once tried to slap together a big buck regulator (maybe 10-20 amps), with a one-layer coil on a plastic coil form (# 10 wire, about 3" dia. by about 5" long).

During testing, my coil form melted! IOW, my inductor was dissipating more power than my MOSFETS. Well, it seemed that way. So, what's the DC resistance of your inductor? It's subject to I^2*R losses.

And of course, take a good hard look at your heatsinking.

Good Luck! Rich

Probably solid wire at high frequencies. I became quite expert melting bobbins at GenRad before I put some thought to harmonic content, skin depth and Litz wire ;-)

...Jim Thompson

Watch for these:

1. The inductor's resistance increases with temperature. Resistance of most metals is roughly proportional to absolute temperature.

1. The inductor winding may have extra resistance seen by the AC component of the current through the inductor due to skin effect.

2. If the inductor has a core, there will be core losses.

1. If the inductor has a ferrite core, the core will be more prone to saturation as temperature increases. The buck converter may operate significantly less efficiently if the inductor starts saturating.

2. If the inductor overheats badly enough, a short may develop between turns. This may increase losses rather than causing the buck converter to not work at all. The short may be intermittent.

- Don Klipstein ( snipped-for-privacy@misty.com)

Would expect the fet (or other) current switches, and the en the inductor, as it has a physical series resistance its gonnna have i^2 r losses.

Marc