Could anyone tell me how to calculate Mosfet temprature rise

[gardenyu]

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[Sean Breheny]

Hi,

That is a very general question. There are three main regions of
MOSFET operation:

1) Fully off
2) resistive (fully on)
3) Partly on (also called "linear region")

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

The first part is simple on resistive losses (R is on resistance -
beware that this is temperature dependent).  The second part is the
switching losses, or the power dissipated by being in state #3 while
turning on and turning off. f is the switching frequency, Tr and Tf
are rise and fall times of the gate drive (taking gate charge into
account).

Once you have P, you have to compute the temperature difference
between ambient and the MOSFET junction. If you will not have great
variations in P, then you can simply add up the thermal resistance
from junction to ambient (usually junction to case + case to heatsink
+ heatsink to ambient) and multiply by P in Watts. If P can make
sudden jumps, then you need to make a dynamic thermal model. The
internal dynamic thermal model of the MOSFET may be included in the
datasheet. One way is when they give a "Transient Thermal Impedance"
In this case, you look up the length of the pulse in P and it gives
you an adjusted thermal resistance from junction to case. For some
duration and longer, it will be the same value as continuous. It will
be less for shorter durations.

Sean


2008/5/7 gardenyu <gardenyu2004@...>:

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[Sean Breheny]

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

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

>  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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[Vasile Surducan-2]

And the zenner diode between gate in source ?
:)

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

> I have to correct myself - I forgot the body diode, sorry:
>
> 2008/5/7 Sean Breheny <shb7@...>:
>
> >  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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> http://www.piclist.com PIC/SX FAQ & list archive
> View/change your membership options at
> http://mailman.mit.edu/mailman/listinfo/piclist
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