# IRF 4905 heat sink (calculation)

1. ## IRF 4905 heat sink (calculation)

Do any1 knows the calculation of heat sink selection for IRF 4905?
What is the maximum operating power dissipation or 'Q' values that i should used.
Is it 200W as shown in the IRF4905 data sheet? isnt 200W too high?

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2. ## heat sink calculation

The TO-220 package is universally preferred for all commercial-industrial applications at power dissipation levels to approximately 50 watts.
And this 200W applies only to situations when you could maintain <25°C at its junction ..
Try this site for very interesting hints on heatsinks : http://sound.westhost.com/heatsinks.htm#3
Regards,
IanP

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3. ## heatsink calculate

Maximum Power Dissipation which can be handled by any device at a given ambient temperature is given by following formula

Pdallowed = (Tjmax - Ta)/Rthj-a

Where,
Pd - Power Dissipation allowed

Tjmax - Maximum junction temperature (from datasheet) = 175C max

Ta - Ambient temperature (system specification) - you will have to specify

Rthj-a - Thermal resistance from junction of IC to air This normally is submiison of various thermal resistances in the path of flow of heat from junction to ambient)

Rthj-a = (Rtj-c + Rtc-s + Rts-a) parallel Rtj-a(datasheet)

Rtj-c, Rtc-s, Rtj-a specified in datasheet

Rts-a is the radiation thermal resistance which I guess is difficult to calculate so we normally do by experimentation or define the maximum operating temperature at the sink itself rather than ambient temperature

If this allowed power dissipation is less then your caluclated power dissipation then you will need to use a heatsink and if you search on google you will find a lot of heat sink suppliers with applications notes on how to choose a heat sink

This is the detail calulation which we do

But if u just want a rough idea then you can check the datasheet for
Linear derating factor For power with temperature

so at particular case temperature how much power can be dissipated you can know
200W is for case temperature of 25C

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