And the zenner diode between gate in source ?
:)

On 5/7/08, Sean Breheny <shb7@cornell.edu> wrote:
> 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
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