IGBT vs. MOSFET

There's a variety of well-known methods for controlling the voltage kick while getting a fast turn-off.

One way would be to position the catch diode so that the current-limit resistor is in the loop when it's on.

Another one (which I gather is fairly popular) is to either use a fairly high voltage zener to ground or +12V (like, put 75V or so on the solenoid), or power a rail with the catch diodes and then regulate that rail's voltage with a buck converter to +12V, to save power.

Just having a low-voltage coil with the current limit resistor should do some of this.

en

is

ignoring possible differences in pull-in and drop-out voltage, 2x supply aught to make the turnon and turnoff the same

I thought it was turtles all the way down?

How exactly do you couple the capacitor to the SCR you want to turn off? More badgers?

There are avalanche-rated mosfets; just let it fly.

No, the issue is that something has to dissipate the energy which is in the magnetic field. A clamp dissipates V*I where I is the solenoid current, so to speed up the dissipation, you need a higher clamp voltage - which is why zeners and TVS are suggested.

Clifford Heath

What turns it off?

There are cheap hundreds-of-peak-amps mosfets, easy to drive. Many are avalanche rated, so could absorb the flyback energy.

That too :)

I said cap, not badger...

Here's a half-bridge example:

Notice the supply is "squishy" because of the inductor. This allows shorting mode commutation: normally, one SCR is on; whe the other fires, the supplies collapse to ~0V (it remains balanced around GND, because it's a differential mode winding).

Because the 0.47uF cap had been charged to full supply voltage, it now discharges through the 9uH inductor and the two SCRs. A quarter cycle later, the inductor current reaches peak, and the SCRs carry a big gulp of extra current (kind of the opposite of when you have a MOSFET switching into a C load, and you get a current peak at the start of the waveform).

Another quarter cycle and the 9uH's current goes back to zero and the 0.47uF is now at negative peak voltage.

One more quarter cycle, and the 0.47uF discharges again through the 9uH, this time carrying negative current: both SCRs are reverse biased, and the MUR2020 conducts.

The driven SCR is still being driven, and turns back on when the series resonant tank swings back around one final time. Supply voltage overshoots (with a resonant frequency of 0.47uF into 9uH + 4*160uH), dampened by the load current (which should have a resistive and capacitive phase angle, to dampen this out).

If you do it in push-pull (two SCRs with cathode = common ground), you don't need the resonant tank: a cap from anode to anode does the same thing. (The cap still resonates with the PP output transformer's leakage, but you can demonstrate this with resistor loads, too.)

It's a lot like the old trick of NE-2 alternating blinkers, but with external trigger rather than self-breakdown.

Tim

I was actually thinking of inductive solutions: at 30A, inductors are cheap and reliable, capacitors less so. GTO is another possibility (but an ugly one).

That only gets to fail once.

Yep. That's where I aim my CMOS active clamps... if the process can stand the voltage. ...Jim Thompson

Maybe SCR to drive solenoid, then MOSFET from SCR anode to ground to commutate it off? ...Jim Thompson

Improve that by eliminating the SCR. A good mosfet will have a lot lower ON voltage drop, which could matter at startup time.

Nothing cheap or necessarily reliable about the solenoids in question, which are built into diesel injectors.

So you reckon I could get a fast turn-off using nothing more than an avalanche-rated MOSFET as the driver? That'd be pretty neat indeed.

Just a wild-ass idea...

Maybe pass some proportion of the current into a capacitor, forming a low-Q tuned circuit. When the discharge through the inductor makes its voltage fly past the accumulated voltage on the capacitor, it reverse biases the SCR and turns it off. Tune the resistance in parallel with the capacitance so that happens at the right point.

Clifford Heath.

Wrong way round. If the kickack diode is in series with the R, kickback current develops higher voltage across the solenoid coil, thus reducing current faster and releasing faster.

NT

At 30A? I would expect smoke. ...Jim Thompson

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Thinking outside the box... elegant solutions.

Maybe. It depends on the current, the inductance, and the rep-rate.

Just a part we have in stock, IRFP240:

No doubt there are better parts for your application. Or use a TVS, or the current-limit resistors, to dump the inductive energy.

The inductance will be sort of variable, so the flyback energy probably needs to be measured.

that's how it is done in cars (though the clamping is usually set with an internal zener so not really avalanche)

do keep in mind that the energy will be dumped in the fet so you'll have to figure out how much energy is stored in the inductor and check the datasheet to see if can handle it

-Lasse