mosfet speed versus power dissipation

[Simon]

I'm driving a power mosfet at 245 Hz, if has a combined rise and fall time of 184 ns and a channel resistance when full on of 6.6 mR, I'll be passing 2-3 amps through it, now I could easily work out the dissipation if it were just on, but how do I take into account the resistance during turn on and off ? clearly this will be of more impact than the actual current flow when fully on. is 1.6W allowance cutting it a bit fine ?

[scrat]

Rise and fall times really depend on the gate charging current. If you know those times, you can quite easily approximate dissipation from Vds and Id at the switching instants (they rise/fall with a ramp).
You have to look at the gate charge curve on the datasheet for the time lengths.

I hope that a look here will help:http://www.btipnow.com/library/white_papers/MOSFET%20Power%20Losses%20Calculation%20Using%20the%20Data-Sheet%20Parameters.pdf

At your frequency and with a quite fast gate driver (high max current), maybe switching loss is low.
Another loss, not mentioned in the above document (but should be negligible at a low switching frequency and high output power) is the gate charge: at each on-off cycle the gate driver dissipates (Qg @Vgs final) * Vgs final of energy (then realted Pdiss = Ediss*fsw).

[Simon]

Thanks scrat, I'll have a read,

The gate drive will not be that strong, I suppose I could do a test setup and and verify the dissipation buy observing the switch waveform with an oscilloscope to see how long it takes and calculating roughly the average resistance of the channel during the switch and use ohms law

[tecman]

With 184 ns out of 245 Hz, you can determine the duty cycle of the rise-fall, which is quite small.  If you assume the highest possible dissipation (I x E), which is highly load dependent, you could come up with a max dissipation, multiplied by the duty cycle, you will have an approximate loss during switching.

Paul

[Simon]

well my main unknown is the average resistance during the switching transition

[Zero999]

The peak power will be when the MOSFET is half on and half off.

The answer is to work out the amount of energy dissipated each time the MOSFET switches and multiply it by the number of transitions per cycle (double the switching frequency).

[Simon]

so what is half on and half off in ohms terms ? I mean considering that the off resistance is like 1+ MR and after a few hundred ohms it's of no more concern I'd need to know how long it takes to reach 1K in order to work out the average resistance over time

[scrat]

There is a main concern about what kind of load you're switching: I almost ever consider an inductive load, so waveforms are almost linear in places (as in the document I linked in the prev post).

For a resistive one things get more complicated, mainly because the datasheets of transistors meant for power switching usually don't report the necessary data to accurately calculate resistance variation.
If you can measure voltage and current waveforms, a simple integration of the voltage-current product gives the energy dissipated at each switching event (as Hero suggests). If you can download binary data from a scope, a program (for example Matlab) can be used.

[Simon]

as it happens this is an inductive load. Well I can download data from the scope, but I'm not that good at analyzing the data

[Zero999]

If the load is a constant current calculating the energy dissipated when switching on or off is easy:

E = 0.5*V*I*t

[scrat]

You can follow the formula Hero posted (easy to guess from the voltage-current waveforms), where power is a triangular peak.
You can apply it on the measured waveform (taking the important points only, mainly duration and maximum values), or post the downloaded data, too.
I remember you own a Rigol, for which there should be an already done import function written in Matlab. Then I can calculate the energy dissipated. It would be interesting to compare the results, I expect the approximation Hero wrote will be nearly correct.

When I will have some time (when?) I'd like to make a GUI for power analysis in Matlab, I see it would be an useful thing.

[Simon]

I expect some experimentation is needed, in the mean time I'll play safe

[scrat]

You can estimate rise/fall time from driver characteristics and MOSFET gate-charge curve.
Can you post the datasheet links?

EDIT: by the way, yesterday I generated some example waveforms, which are valid for an inductive load switching (with free-wheeling diode, of course), with MOSFET at the low-side. Until current reaches the max value (Vgs from Vth to plateau value), the diode (supposed ideal) clamps drain voltage at Vdd, then during the plateau charging of the gate voltage goes down.
Hope that it helps...

[Bambur]

Long time ago I used to simulate the circuit with PSpice and plot the power dissipated by the MOSFET. One more thing to keep in mind is to add a small series resistor to the gate wire to prevent any ringing -- otherwise, MOSFET might get really hot in practice.

[Simon]

Ringing ? please elaborate, what effect does frequency have ? My impression is that you want to supply as much current as possible to the mosfet gate in order to charge or discharge the gate capacitance

[Time]

The trace to the gate has an inductance and this can ring with the gate capacitance.  Adding some resistance will dampen any ringing.  I don't think you are switching this fast enough for it to be an issue though.  Its kind of a rule of thumb to have some resistance in there, atleast in the RF world I think.  I am sure someone can correct me if this is inaccurate.

[Simon]

so how does the "ring" manifest ? does the mosfet oscillate between fully on/of and partially before settling down ?

[Time]

Yes, depends on how the gate ringing behaves.  It would just cause erratic or erroneous behavior.

[Simon]

right, well as you say at my frequency it should not be an issue

[scrat]

Maybe your gate driver is sufficiently resistive not to make any ringing, but even if it could be not a problem for your circuit, I'll just plan to add one, since this ringing can cause many problems and the place for a resistance doesn't cost too much.