Controller for Switching current sink

Assuming your math is correct...

To charge the cap requires dissipation of energy. About all you can do is spread it out over time to reduce the peak. If you know all the circuit elements, you should be able to construct a drive waveform that limits the inductor current without sensing anything...depending on your other unstated requirements. Once the voltage gets to zero, turn it on hard.

If the end-state is 24 volts across the cap, with the fet fully on, the switcher looks like gross overkill to me.

I calculate 1.5 joules in the cap, so the fet has to dump the same 1.5 joules to charge it up. That's not much. A sot-89 or sot-223 wouldn't mind. Get a part with enough Rds to make the current whatever you like.

More partsy circuits could shift some of the dissipation into a resistor or so, but that hardly seems worth the bother.

John

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1/2 * 5,000uF * 25V^2 =3D 1.6 joules.

The job has to be done in

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My problem with the linear version is that, at say 5A and cap starting at 0v charge, dissipation across the linear element ramps from 125w to zero over 25mS, for an average dissipation of 62.5w over the 25mS.

I'd rather do that (that was my first approach), but it grossly exceeds my switch FETs' SOA. They're rated for 600mA x 25v for 10mS, single pulse, not 5A x 25v for 25mS/2 (kind of). I'm out by almost a factor of 10.

I haven't calculated the heat capacity of these devices, that just looks like an awful lot to ask of an SMD part with minimal heatsinking.

James

Charging that 0.005 F cap to 24 volts in 500 msec takes 240 mA. The fet absorbs 1.4 joules total. Its power dissipation starts at about 6 watts (at 240 mA) and tapers down. That doesn't look like a big deal to me. Am I missing something?

You could use a Miller cap to control the fet current. And/Or use a

100 watt-rated DPAK fet.

There are two sources of dissipation, the charging of the cap and the current in the resistor. The joules dissipated are always the same for charging the cap, but increase for driving the resistor as you switch slower. So there is a dilemma of sorts: minimum dissipation requires high peak current.

But the peak fet dissipation for driving the resistor is only 600 milliwatts, at the brief maximum-power-transfer point, so that hardly matters. And the cap only contributes 1.5 joules. The whole mess dumps less than 2 joules into the fet. I'd think that a cheap sot-223 or DPAK mosfet will take a lot less room than all that switcher stuff. And, as I noted, you can shift some dissipation into a resistor if you really need to.

The big advantage of a switcher would be if you need to recharge this thing often.

John

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Okay, I weighed the devices--2.0g each. If that's half copper, heat capacity works out to .385J / g * K, so even two joules would be just

5.2 degC rise. That's trivial.

I don't understand why the SOA rating of a 100's-of-amps big-ass FET is so miniscule.

I've screwed up something. None of that looks right. I wouldn't hesitate to slam a mounted TO-220, it's these new-fangled SMD parts I don't have a feel for.

Another small wrinkle is that the load might also be a dead short (pathologically possible).

James

No, that seems reasonable. I can only think I'm either misinterpreting the FET's SOA spec, or it's just plain wrong. The FETs are 1.7-milliohm 195A 375W monsters--it's absurd to think they can't handle more than 600mA x 25v for 10mS, yet that's what the graph says.

(Fig. 8)

(I had to drop that FET 'cause there wasn't room for the heatsinks. I've added s/c and overload protection to eliminate the heatsinks previously needed to withstand fault currents).

(I've switched to an SO-8-ish 1 milliohm device, in array, but don't have the part number handy.)

Originally I was soft-switching the big switch. I had a Miller ramp that did that over 50 or 100uS. On calculation, drawbacks included very high cap charging currents, and destruction of the FETs if the output load is operating, which is remotely pathologically possible.

...

The switcher is also safer--it won't overheat if there's a short--but I prefer simple. I'll put a timer on it so it can't run long enough to kill itself. I like your idea of using the FET's rds(on) as the current-limit.

1 amp charges the cap in 125mS--I s'pose that's fast enough.

James

Aren't you going to be switching these? The SOA is for when you have Vds and Id simultaneously. You won't have that except during the switching period. That is probably less than 1us. You will therefore be well under the 100us curve. No problem.

The DC rating on the SOA looks like 24 volts, 8 amps maybe. That would of course need a heatsink if it were sustained.

I'd think that a little DPAK, with no heat sink at all, would do what you need.

John

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aren't you a factor of about 10 off? I see DC crossing 1A at 20V

-Lasse

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I can't hard switch it. Total resistance, with wiring, is about 2 milliohms. The supply is quite muscular. At 25v, that'd be 12kA. Bang.

All the resistive options generate about the same amount of heat, so the question is of doing it too quickly and frying the FET(s).

If I precharge the cap or ramp it slowly, that works.

-- Cheers, James Arthur

I must be missing something. Are you using an inductor and a switcher, or not?

Oops, sorry. I was looking at that wrong.

Yes, at 24 volts this 195-watt mosfet is good for around 12 watts. It's clearly headed for some sort of second breakdown.

A higher-voltage mosfet would have a better SOA up around 24 volts. This one is optimized for super low Rds-on at low voltages.

Something like an IRFR120N, rated 100 volts and (optimistically) 48 watts, has a lot more dissipation capability at 24 volts. That would work fine if you limit the drain current to, say, half an amp.

John

What are your estimates for L and Rs for L1? What are your requirements for DC load regulation at the load over what range of load currents? Assuming dead short is a fault condition, you can fix that with a fuse. Your schematic shows that RL is completely isolated and connects to NOTHING else anywhere. One can only infer that that's actually the case. Sometimes a simplified model masks BIG issues.

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surprising how being made for logic level makes that irlb3034 so wimpy, I just picked the first hit on the irf site for a 40V, TO220AB nfet - AUIRF1404 its rated for half the current (202A) and twice the rdson (4mOhm) but it's almost 20A,20V @ 10ms

-Lasse

-Lasse

James -

You are getting worked up over almost nothing.

Assume your inductor is about 3mH. Assume it has .01 ohms of inherent resistance.

You can turn on your superFET for 5ms, then turn it off and the inductor will dump energy into the C such that it will accumulate 27.6V by the time the dump is complete.

Peak current will be about 29A. I'm sure your device will handle that (while saturated, of course).

If you don't like the peak current or the size of the inductor, then use repetitive switching to reduce all that. Nevertheless, your devices are rated many times higher than what you need, if applied properly.

JohnS

But this assumes that the heat rise is uniform - that is, that the energy dissipation per unit silicon area is uniform. The maximum safe operating area curves tell you what happens when it isn't.

SOA is all about forming hot-spots somewhere on the switch. Once one point on the switch gets hot enough, all the current concentrates there and the silicon melts. Because it is localised, it doesn't take as much energy as it would take to cook the whole chip.

You might have to dissect the part to see the difference

--
Bill Sloman, Nijmegen

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Oh, you're missing the big switch. That doesn't appear in these schematics at all. The BIG switch is in parallel with the pre-charge circuit, and connects the (-) terminal of the load directly to GND.

I was mulling (above) using a small switcher with an inductor to precharge the cap before closing the (not shown) monster switch.

Does that make the situation clearer?

-- Cheers, James Arthur

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As I'm the designer they can be whatever I want them to be, naturally.

I'm afraid I've confused folks by diagramming only part of the system, while talking about the whole thing.

The load is a complex commercial device. I've shown the parts that matter for the time interval in question.

Here's an amended diagram that includes the main switch, the one the precharge-circuit is intended to support:

+24v | . . . .|. . . . . .--+---. . LOAD . | | . RL=3D240R . --- RL . CL=3D.005F . --- CL | . . | | . . '--+---' . . . . .|. . . . | Big switch | (FET Array) | ||-----------+

---|| | ||>-. | +24v | |_ | =3D=3D=3D L1 )| --- )| A ._)| --- Vcc | | D1 | +---------' .------. | | | ||--' | out|->--|| Q1 | | ||>-. | | | | Vs|-