I have to correct myself - I forgot the body diode, sorry:

2008/5/7 Sean Breheny <shb7@cornell.edu>:

>  Assuming you are using the MOSFETs as switches, then the power
>  dissipated is approximately:
>
>  P=I^2*R+f * (Tr+Tf) * 0.5 * V * I
>

If we assume that the FET carries I current with a duty cycle fraction
of D and the rest of the time it flows through the body diode, then:

P=D*I^2*R+V_d*I*(1-D)+f*(Tr+Tf)*0.5*V*I

If the body diode does not conduct the current during the off time,
then you can drop the V_d*I*(1-D) term.

V_d is the body diode forward voltage drop

Let me break down the reasoning a bit more here:

D*I^2*R

Here the FET dissipates normal I^2*R power for D fraction of the time

V_d*I*(1-D)

The body diode dissipates V_d*I for (1-D) fraction of the time

f*(Tr+Tf)*0.5*V*I

f times per second, the FET spends (Tr+Tf) amount of time in the
partially on state. We can roughly estimate the power dissipated
during this state as 0.5*V*I, where V is the full supply voltage and I
is the full on current. This is really an over-estimate but that is a
good way to do it (rather than an under-estimate). If driving a
resistive load, and you turned on and off linearly, the actual
dissipation would be 0.16666 * V*I (integral of the product of two
linear ramps, voltage ramping down from V to 0, current ramping up
from 0 to I)

Sean
-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist