half bridge mosfet switching

I have a simple half mosfet bridge. The mosfet i use has crss 350pF and Coss as 870pF. So the Cds would be 520pF. I have a gating pulse to these mosfet with deadtime of about 500ns, which more than enough to prevent shoot through. Suppose i have no load connected to the bridge and i just apply the gating pulse. The bridge keg should draw current that is charging the Cds. When i calculated it, it comes out to be 1A. But when i do simulation and experiment, i see it to be more than 6A. I am not able to track down the logic for it. Any MOSFET expert please explain me. kristo.

Reply to
krishmaniac
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I don't consider myself to be an *expert* but I happen to have exactly this on the bench right now. Vbus is 320 V. Similar specs to your example.

No load, it draws virtually nothing ( other than transformer magnetising current when that's connected )

What switching frequency ? Have you been purely simulating or actually got a working example ?

Graham

Reply to
Pooh Bear

i am switching at 80khz. i see both in experiment and simulation. still wondering why?

Reply to
krishmaniac
6A at 320v is

320*6

1920 Watt.

You mosfet should be a VERY hot (but only for a short time :) )

If you mosfet did not burn-out and cool. Than mean that you measuring current wrong.

Reply to
Artem

actually if 6A is pulsed current. Transistors could be cool. But measuring pulsed current is not simply task.

I this that if transistors is cool, it's ok. All right.

Reply to
Artem

Current transformer on the shielded cable may be more useful, even at DC current.

Gate current 15v / 10Om * 4transistors = 6A. But it's too simply. And current pulses must be much more shortest.

Reply to
Artem

hi, the mosfets is getting hot. i am not measuring gate current. I am measuring the dain current. my Vbus is 400V. i am using IRFp460 mosfet.

Reply to
krishmaniac

MOSFET with radiator or without radiator? If MOSFET getting hot WITH radiator, this is very bad.

Check common DC current. (Before condencer, after bridge, with oscilloscope and shunt, rms value).

  1. Are you shure that MOSFETs is completely close/open?
  2. How you make a gate driver?
  3. What about parasitic diode in MOSFET? Is this diode fast enough?
  4. Did you use a some kind of snubbers?
Reply to
Artem

since the current pulse occurs during turning on either mosfet on the half bridge... since Vbus=400V, the current pulse is 6A peak. the approx. average power dissipated = 200*3*1/20=30Watts...1/20 is current conduction time by the period it repeats. should the mosfet get hot with this power.

Reply to
krishmaniac

hi , intially i thought so much current will not flow through drain of the top mosfet so i did not put radiators. what i thought was: since maximum dv/dt across mosfet is 400/100*10-9 and cap of mosfet is maximum of 500pF. so worst case current would be 2A. is there any wrong in my fundamental calculations?

yes i checked the mosfet are compeletly open/close. i use ir2110 for driver with gate resistance of 25ohms. a slight increase in current is seen when gate resistance is increased. i assume mosfet diode to be fast enough as i see in datasheet the turn on/off delay and rise/fall time are in terms of tens/hundrends of nanoseconds. i have not used any snubber.

Reply to
krishmaniac

IRFp640 is not a "logic-level" device.

Reply to
Artem

terry, if they do not turn off, then i must get shoot through with which dc link voltage will be collapsing..but the voltage does not collapse. current is there only during the turn ON of the mosfets.

Reply to
krishmaniac

In my experience IRFp640 and IRF720 without load and radiator was cool.

------------- trr Reverse Recovery Time 680ns - max

------------- Try to increase death-time to 2000ns (at least for test).

You can connect 2 ultra-fast diodes for each transistors: one in serial between source and "-" and another between "-" and drain. By this way you can "disconnect" internal diode.

Maybe it's bad idea.

Reply to
Artem

its not matter of insanity..but learning..moreover its a voluntary discussion for your information the irfp460 has threshhold voltage minimum 2V and maximum 4V...i guess it not the reason.

Reply to
krishmaniac

If that much curent is being drawn, something must be getting *very* hot very fast !

So, what's getting hot ? What's your Vbus ?

Graham

Reply to
Pooh Bear

Oh yes ! 30W is plenty to get a mosfet *very* hot without any heatsink. In fact it'll fail very quickly. You shouldn't be seeing anything like that kind of dissipation though.

My own similar example ( also using IR2110 - but Rg = 10 R ) has no idle dissipation worth considering. Vbus is 320V and switching f = 100kHz btw.

I think you should consider a snubber though. The dV/dt at the junction of the fets will be very high I expect.

Graham

Reply to
Pooh Bear

It looks like an ancient device.

The datasheet on IR's site is simply a scan !

Vgs( thr) is as low a 2V.

A more modern device migh be more appropriate.

Graham

Reply to
Pooh Bear

It can be easy. As there is no load, slap a resistor in series with the

+DC bus. If there really is cross-conduction the voltage will collapse. use a small cap (just enough to charge up the expected FET capacitance) after the resistor.

he might be measuring gate charging current....if he has a low-side current sense resistor. If so, change the layout so gate current is not included (route gate driver V- to FET side of current sense resistor).

Cheers Terry

Reply to
Terry Given

If they are getting hot then your measurement is probably real.

What does your gate waveform look like? if it has a broad flat spot at about Vth then your gatedrive impedance is too high (and is losing the fight with Cmiller)

What value of Rgate?

Halve the switching frequency. that ought to halve the temperature rise.

then change the dead time to, say, 2us.

If the FETs stop getting hot, its a cross-conduction problem. Beware low Vth and high Vgate - it takes a small fraction of a time constant to charge from zero up to Vth, but many time constants to discharge from Vgate down to Vth, so if Rg is high then 500ns may not be enough. Reducing Rgate reduces the time constant, and hence the required dead time.

If the FETs still get hot then its not cross-conduction but rather switching loss, which can also be caused by gatedrive.

Also, FETs have positive thermal feedback. For a sufficiently high Rtheta a FET can suffer thermal run away, and because of the positive feedback (Rds gets higher with T, and usually I is constant so I^2Rds increases with T) doubling the heatsink Rtheta can more than double dT.

Cheers Terry

Reply to
Terry Given

Terry Given wrote: [snip]

Oh yeah, with ultra-low Vth FETs and a really bad gatedrive, its possible they dont ever turn off. I saw that once with a logic gate (0.4V output low) driving a complementary emitter follower (add a Vbe) driving a Supertex FET with Vth = 1V when cold. Vth dropped to about

0.8V when hot, at which point the damn thing wouldnt turn off.....

but I doubt thats your problem.

Cheers Terry

Reply to
Terry Given

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