Controller for Switching current sink

IR is fairly famous for that sort of problem. I don't do a lot of high power stuff, so I haven't run into it myself, but I've heard it discussed here before.

Cheers

Phil Hobbs

It's all in the dynamics of the formation of a hot-spot - once the conditions are right for a hot-spot to start sucking current away from the adjacent silicon and get even hotter, you've got your failure mechanism. No matter how big the part, one incandescent point on the die is enough to render the rest of the die irrelevant and useless.

I'm having trouble understanding the significance of RL=240R in a 500 amp circuit?????

You only need to pulse Q1 ONCE. Make the pulse width so that the energy stored in L1 is sufficient to drag the voltage at the L1-RL junction to zero after you turn off Q1. You're probably gonna want another clamp diode to ground anyway. Make it a transistor and use the negative voltage to gate the turnon of the big FET.

You're disclosing little about what you're doing. Looking at the math... You have a load of 500A at 24V. The load can't stand the 250mV drop across the 1/2 mOhm FET(s). Adding a cap with a time constant of RC = .0005*.005 = 2.5us makes the load happy. Assumes zero ESR in the cap(s).

If it's that critical, I'm wondering if charging the cap is your biggest problem.

What quits working if you move the cap from the drain to the source of the big FET?

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It's not especially significant, but that's what bridges the load while it's off. I didn't do it, I'm just reporting reality.

For (L*i^2)/2 =3D 1.6J, I'd need 260A in a 47uH inductor. A 10mH iron- laminated inductor still needs 18A. That's a big part.

This all needs to be surface-mounted and fit in a few square inches-- that's all I've got.

I'm sorry if it's not clear. ISTM I've given an unusually detailed account of exactly the task and the parameters, right down to supplying circuit diagrams--I don't see what's missing.

That's nothing to do with me. The load is a commercial device. It has cap. bank CL in it for its own reasons. I'm interfacing to it. I can't change it.

I do!

-- Cheers, James Arthur

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The best parts I've found so far are NXP. The NXP STL140N4LLF5--of a similar class (40v, rdson=3D2.8 milliohms) to my main switch array device--specs 3.5A @ 25v for 100mS, which is 10x my original part's SOA rating.

That NXP part might be a precharge candidate. Its pulse-current rating is too wimpy to use in the main array.

-- Cheers, James Arthur

I've waded through a bunch more data sheets, and it seems like there are at least two things going on.

Most SMD devices are thermally limited (as you'd expect) by their low mass and limited ability to get rid of heat quickly. Those parts (including DPAK, surprisingly) can't dissipate much more power for a

Basically, 100mS is more than long enough to get the die as hot as it's gonna get at a given power level. So, no, absent highly effective heat-sinking, I can't dissipate 25W for 100mS with a 2.5W- rated part. Not allowed.

Monster FETs also have this surprising hot-spot thing limiting linear- mode SOA to just a few percent of rated dissipation. My original switch device (IRLB3034) is rated at 375W, but can only dissipate ~350mA @ 30vdc, or about 10W (!) safely under the SOA limits.

Perusing datasheets, that limitation wasn't noticably correlated with logic-vs-standard threshold, or 30v vs 40v. Mostly it seems related to who made the part.

BJTs generally allow max dissipation while linear--a 25w-rated part is allowed 1A @25v, etc. That assumes heatsinking, naturally.

-- Cheers, James Arthur

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Which is to say, how the part was made.

That's not what I remember - albeit from a long time ago. My impression was that "second breakdown" was first discovered in BJT's, and limits their SOA in much the same way as for MOSFETs although the breakdown mechanism doesn't work in quite the same way. Since a little googling dated the discovery to 1962, before power MOSFETs were commercially available

it seems that my memory wasn't misleading me - at least on this point.

-- Bill Sloman, Nijmegen

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Thanks for the correction Bill. I should've said more clearly that I was reporting on the few devices' datasheets I had scanned, not on BJTs in general.

Going linear seems more trouble than it saves--I'd have to add protection circuitry for fault conditions, and I have to push parts close to their thermal limits. That bugs me.

I considered (ab)using ordinary ICs, such as using a linear regulator as a thermally-protected current limiter. That saves the need to protect the power device, but interacts with and complicates the other circuitry. For example, drop-out voltage--a precharger that guarantees precharging to < 0.25v simplifies my overload protection circuitry.

Ordinary switching IC's weren't suitable.

I've rolled a controller with a constant-ripple control scheme, otherwise the initial pulse times are impossibly short. It uses an LM393 and a handful of discretes, calculates out well and simulates as expected. That looks like the best, most rugged overall solution.

-- Cheers, James Arthur

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I'm also struggling to see why you cant use a linear device to charge it. To protect the fet against all possibilities, if its P_diss isnt inherently sufficient, you could multiply the fet Vd-s and the V across the current sense resistor and use that power product to limit the gate voltage to stay within the SOA at all times.

NT

The linear method does work, but by the time I'm done protecting it it's more complicated than the switcher, that's all.

The big switch, for example, is now fault protected. That means it can conceivably 'open' while the load is active. That would result in continuous application of +24v to the pre-charge circuit, forcing a 1A circuit to dissipate 24w, continuous.

That's impossible in an SMD device with essentially no heat sink, so I'd have to thermally protect it, or use your clever limiter that calculates Vds x Id. But, that'll be slower--initially only sinking ~100mA at high Vds--and more complicated.

So, reluctantly, I conclude the switcher is best. It'll sink 1A (or

It automatically turns itself off. When the cap is charged or the main switch closes it shuts down, so no control or protection circuitry is needed. That's a significant savings.

So yes, it's possible to go linear, but the switcher happens to be simpler in total parts and support circuitry, this time. And tougher. YMMV.

-- Cheers, James Arthur