Turn a MOSFET into a high Voltage IBGT type device?

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Hi guys, I want to make a motor drive, but the MOSFETs that can supply the current, dont have adequate Voltage rating (I want 1.2KV). And IGBTS are much more expensive than MOSFETs.

Can someone please critize the following circuit, do you think the MOSFET would be able to with stand the break over voltage of D1, or do you think the smoke will escape? :oops:

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Just a noob in this crazy world trying to get some electrons to obey me.

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If VDD is greater than the voltage rating of your FET, I think you're screwed.

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That's to put it mildly :lol:

fever2tel, if you're unsure on this one, I think that a 1.2kV motor-driver is a bit out of ... hmm ... range.

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Just a thought, can MOSFETs reliably be wired in series for an increase in operating voltage?

I've never seen it done, like I said, just a thought...

Michael

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Consider the circuit by assuming that every component except the one you're looking at, is a little leaky.

For example, if you had two diodes in series, each with 100V PIV rating, you might think that you'd get 200V PIV. But if one diode is more leaky than the other one, then almost all the voltage may express across one diode, at least for a short time.

So the result is that two is better than one, but not 2X.

Leakage and transients caused by capacitance are your unseen enemies here.

I'd look at supressing spikes at the motor. You could do a crowbar circuit, that would act like a high power zener, across the motor. Use a normal zener to tur on a transistor across the motor when the motor voltage exceeds the supply voltage by some siginificant number that is still within your FETs ratings.

Wire routing will make a difference too. Don't hang a supression network off the motor on 2' long leads and expect it to be effective. Route the wires from the control circuit to the supression network, and then to the motor.

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BTW: What is D1 doing for you here?

With an inductive load, you CANNOT stop the current instantly. When you turn off Q1, the voltage on the top of the load will go negative, as if it's trying to "suck" current out of Q1, so D1 isn't protecting you from anything.. If you put D1 across the motor (or inductor), such that it does not conduct when the motor is on, then when you turn Q1 off, D1 will allow the field to collapse by allowing the current to flow into the top of the load from ground.

This will stop the huge voltage spikes you're seeing when you turn off Q1. Just make sure that D1 can handle the max current for a short time.

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dbvanhorn wrote:
BTW: What is D1 doing for you here?

With an inductive load, you CANNOT stop the current instantly. When you turn off Q1, the voltage on the top of the load will go negative, as if it's trying to "suck" current out of Q1, so D1 isn't protecting you from anything.. If you put D1 across the motor (or inductor), such that it does not conduct when the motor is on, then when you turn Q1 off, D1 will allow the field to collapse by allowing the current to flow into the top of the load from ground.

This will stop the huge voltage spikes you're seeing when you turn off Q1. Just make sure that D1 can handle the max current for a short time.

Thanks for your wisedom, I now see how stupid it is to do what I was trying to do. :oops:
I was trying to use a forward biased diode to increase the MOSFETs reverse blocking ability... stupidest post ever on avr freaks for sure X_x

Just a noob in this crazy world trying to get some electrons to obey me.

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Are you switching big currents? The R_dsON (resistance between D and S while switched on) of FETs can cause big power dissipation. IGBTs don't have this problem.

btw: Could this problem be solved with two or more parallel FETs?

edit:
We do need a fly-wheel diode across the motor AND the FET, don't we?

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Nephazz wrote:

We do need a fly-wheel diode across the motor AND the FET, don't we?

Not necessarily, but the one across Q1 can't hurt.
Much depends on how fast the fet switches, and how much parasitic inductance is in the wiring.

There is such a thing as switching the fet too fast.
When that happens, the fet turns on, then turns off again due to the inductance of the wires opposing the current flow, and causing the source voltage to rise up near or above the gate voltage. This happens several times and generates significant bursts of RF.

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Thanks dbvanhorn.

@ all: Could the R_dsON problem be solved with two or more parallel FETs?

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Quote:
Could the R_dsON problem be solved with two or more parallel FETs?
Yes. But the problem that OP is facing is the 1.2 kV.

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I'm not sure where the 1.2kV is coming from.
With proper supression of the motor, that may be entirely a non-issue.

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dbvanhorn wrote:

Quote:
I'm not sure where the 1.2kV is coming from.

From the OP; fever2tel wrote in his first post:
Quote:
Hi guys, I want to make a motor drive, but the MOSFETs that can supply the current, dont have adequate Voltage rating (I want 1.2KV). And IGBTS are much more expensive than MOSFETs.

Nard

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As "nleahcim" Said:

Quote:
If VDD is greater than the voltage rating of your FET, I think you're screwed.

D1 does nothing except increase the flyback voltage when the load is switched off. No number of series diodes will be of use in this circuit as was pointed out by "dbvanhorn"

You are better off deleting this diode and using the integral drain to source diode to conduct flyback energy to the supply.

D2 is already incorporated within the mosfet, Same rating as mosfet usually, no improvement gained by adding D2.

How are you going to do your 1200V level shifted drive to your Mosfet ???

If you use series mosfets, how will you drive them ???

If there are say 3 in series, what happens when two turn on before the third ???

Just some things to consider before paying out the extra for the suitable IGBT.

Ron.

 

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I think that the 1200V was coming from trying to withstand the field collapse without snubbers.

I don't think we ever heard what "VCC" was, but I'm suspecting it wasn't a high voltage. Probably, with proper snubbers, a 30V part will work fine, assuming VCC is something like 5-12V-ish

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We will know about the 1.2 kV when OP responds.

A GIF is worth a thousend words   They are called Rosa, Sylvia, Tessa and Tina, You can find them https://www.linuxmint.com/

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