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I am doing power dissipation of a power mosfet. My design requirements are :-

  1. 20A @ 12-20V (Ambient temp will be around 50 deg Celsius)

The mosfet I am using is :-

http://aosmd.com/res/data_sheets/AOD425.pdf

My doubts are :-

In the datasheet , they have given 2 types of power dissipation as below :-

enter image description here

I understand that they are at different case & ambient temperature but what do they actually mean & how to use this data?

In one more table they have given :-

enter image description here

Rja is given & Rjc is given. Rja = Rjc + Rca. Rca is not mentioned anywhere because that might depend on the PCB layout but still, how do I calculate Rca to do my thermal analysis.

In general, what do I actually use for thermal analysis. If @ an assumed junction temp of 70 deg c, my Rdson is 20 mE & my current is 20A. calculating :-

P= (I2)Rdson = 400*.02 = 8W.

Is 8W okay through this mosfet? If not, how to prove that?If yes, how to validate?

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  • \$\begingroup\$ Made an edit from 'Junction' to 'Ambient' in the line under 1st table. \$\endgroup\$ – Oshi Jul 18 '16 at 17:49
  • \$\begingroup\$ Don't assume that if you manage to keep Tc at 25 deg C you'll be able to use 71 watts dissipation, based on table you quote above. If you look in the datasheet figure 9, Safe Operating Area, you'll find that the sloping lines marked with times that delimit the top right of the SOA, the DC line is at less than 71W. It can only achieve high dissipation in pulsed mode, almost all FETs have a thermal instability that restricts their high power operation to switching, which is the mode that almost all FETs are used in. \$\endgroup\$ – Neil_UK Jul 18 '16 at 18:20
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I understand that they are at different case & junction temperature but what do they actually mean & how to use this data?

Using the example of case temperature being controlled/limited to 25 degC, the spec says you can dissipate 71 watts. Now look at maximum junction to case thermal resistance - it says 2.1 degC/watt.

So, for 71 watts I would expect the junction to rise to 149 degC above the case. This ties nicely with the case being fixed at 25 degC because 25 + 149 = 174 degC - ok a slight discrepency with the stated max junction temperature.

8 watts would raise the junction by nearly 17 degC above the case. However using the "Maximum Junction-to-Ambient Steady-State" value, 8 watts will raise the junction up to 8x50 degC because now, the case isn't "held" at some value. This is why you need a heatsink.

If you went for a heatsink that was rated at 3 degC per watt (case physical connection to ambient), you use that figue (3 degC/watt) plus junction to case (2.1 degC/watt) to get a combined number of 5.1 degC/watt. Thus, for a junction temperature of not more that 175 degC in a controlled ambient of 25 degC, the temperature rise of 150 degC divided by 5.1 degC gives you the max continuous power i.e. 29.4 watts.

A heatsink of 16.65 degC/watt would be on the cusp of tolerating an 8 watt internal dissipation because 16.65 (external) + 2.1 (junction-to-case) = 18.75 and 150/8 = 18.75.

Does this help? Maybe these simplified pictures will: -

enter image description here

enter image description here

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  • \$\begingroup\$ @ Andy - Very nice explanation. I appreciate. But my doubt is that I actually cant dissipate this much. what is the role of power dissipation @ Ambient temperatures given in the datasheet which is only 2.5W @ 25 deg Celsius. \$\endgroup\$ – Oshi Jul 18 '16 at 17:45
  • \$\begingroup\$ @ Andy - Thank you once again for elaborating. My understanding is that the value 29.4 watt is the max power dissipation it will allow. Also, does it have to do anything with VDS? 20 A will be @ 12-20V. So, total max continuous power is going to be around 400W. \$\endgroup\$ – Oshi Jul 19 '16 at 7:31
  • \$\begingroup\$ Power is drain current x Vds. I'm not sure I understand the rest of your comment. \$\endgroup\$ – Andy aka Jul 19 '16 at 7:38
  • \$\begingroup\$ I am trying to follow these guidelines. I will update with the progress after simulation. \$\endgroup\$ – Oshi Jul 19 '16 at 10:48
  • \$\begingroup\$ @ Andy - We have shifted from using a P-Mosfet to using a N-Mosfet which gives very low RDSon values. \$\endgroup\$ – Oshi Jul 26 '16 at 12:22
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8 watts would give you 8W x 50 degC/W or 400 degrees Celsius above ambient temperature for no heat sink. (no good)

With a heat sink, you will have 8W x 2.1 degC/W or 16.8 degrees above ambient at the surface of the heat sink. You will have to add the heat sink RjA which is also stated in degC/W.

Mounting on the board will probably be inadequate for heat sink purposes, so plan on finding a heat sink for this part.

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  • \$\begingroup\$ Keep in mind that if you don't want to use a heatsink there are FETs available with much lower RDSon, you could pick one with maybe 1-2 mOhms and get away with mounting it on a PCB. \$\endgroup\$ – John D Jul 18 '16 at 17:49
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Since the maximum temperature is 174 degrees for the junction. So if the case is 25 degrees, which is different from the situation of the ambient is 25 degrees, then 71 watts is (174 - 25) degrees / 2.1 degrees/watt. That is where the 71 watts is from. By the way, you can use a sort of heat sink to keep the case at a temperature of 25 degrees.

And if let us have a look where the 36 watts (at 100 degrees) from by applying the same logic. If the case is 100 degrees, then 36 = (174-100)/2.1. 174 degrees is the maximum, 100 degrees is the case, not the ambient.

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