half bridge mosfet switching

[krishmaniac@hotmail.com]

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.

 

[Pooh Bear]

"krishmaniac@hotmail.com" wrote:

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.

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

 

[Pooh Bear]

Terry Given wrote:

Artem wrote:
IRFp640 is not a "logic-level" device.


I couldnt be bothered to look, so used the weasel words "but I doubt
that is your problem"

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

 

[Terry Given]

krishmaniac@hotmail.com wrote:

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.

say Vth = 3V

If you have a +12V gate voltage, then (carefully ignoring Cmiller and
nonlinear Cgs)a fully discharged gate will take -ln(1-3/12) = 0.29 time
constants to rise from 0V to Vth = 3V.

By contrast, discharging a fully charged gate from +12V down to Vth =
+3V (when the FET begins to turn off) takes -ln(3/12) = 1.39 time constants.

At Vg = 12V, Qg = 140nC so Cg = 11nF or so. Ignoring the IR2110 Rout
(which is small c.f. 25 Ohms), the gate time constant is about 250ns.

And that completely ignores the flat spot caused by high Rgate, which
will further increase the switching time.

So it doesnt sound like your 500ns dead time is really enough.

 

[Terry Given]

Pooh Bear wrote:

Terry Given wrote:


Artem wrote:

IRFp640 is not a "logic-level" device.


I couldnt be bothered to look, so used the weasel words "but I doubt
that is your problem"


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

its actually an IRFp460 too.

 

[Mook Johnson]

I had the same exact problem with building a low power switcher that needed
to operate off 200-400 VDC (half bridge). By low power I mean 20mA off the
vbuss at full load. Noload current draw was 4mA. The IRFP460 was too much
capacitance and suffered from the same problems you are describing. my no
load was 60mA.

I had better luck by pushing up the Rds on rating and getting parts that had
less D-S capacitance. (irfbg30, bg20, some Ixys fets also had high voltage
and low capacitance). Also drive the gates harder with < 10 ohm resistors.

If you're trying to do this on anything other than a PCB...forget it.



<krishmaniac@hotmail.com> wrote in message
news:1121864620.903044.245010@g43g2000cwa.googlegroups.com...

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.