Gate driver power dissipation and temperature

Thread Starter

kalemaxon89

Joined Oct 12, 2022
389
Let's take this integrated power stage used with fsw=60kHz, V=48V and I=4A (it's an example!)

Using the formulas in the section "8.2.2.4 Power Dissipation" I've calculated:

Pgatedriver = 0.009W
Pcond = 0.063W
Psw = 0.4256W
Adding up the previous three I get Ptot = 0.5W should be < "max power dissipation limit package at the operating temperature"

Since Rja = 27 °C/W (and Rjc = 0.4°C/W)

Tj = Tambient + Pdiss*Rja = 25 + 0.5*27 = 38.5°C < Tj,max (that is 150°C)

1) Is this calculation right?
2) Should I use Rjc? Rjb?

The thing that does not convince me is that, assuming Pin = 88W and Pout = 84W .. on the driver there are 4W
..so Tj = 25+27*4 = 133°C
That seems too much to me ..or does it?
..I would like feedback from you, especially for formulas and calculations
 

BobTPH

Joined Jun 5, 2013
11,463
Rja tells you what happens with no heat sink. If it exceeds the max power dissipation of the package, you need a heat sink.
 

Thread Starter

kalemaxon89

Joined Oct 12, 2022
389
Rja tells you what happens with no heat sink. If it exceeds the max power dissipation of the package, you need a heat sink.
Thank you for your reply.

1) So the calculations are correct? ..I mean the Tj calculated

2) On the datasheet I find only Tj,max ... I do not find Pmax dissip of the package
 

BobTPH

Joined Jun 5, 2013
11,463
Thank you for your reply.

1) So the calculations are correct? ..I mean the Tj calculated

2) On the datasheet I find only Tj,max ... I do not find Pmax dissip of the package
You calculate it from the Rja, the max junction temp, and ambient temp.

For example:

Tc = Ta + Rja x Pd

125 = 20 + 27Pd
Pd = (125 -20) / 27 = 105/27 = 3.88W

Which sounds rather high to me. I think it is assuming some kind of thermal pad the PCB.
 
Last edited:

Thread Starter

kalemaxon89

Joined Oct 12, 2022
389
You calculate it from the Rja, the max junction temp, and ambient temp.

For example:

Tc = Ta + Rja x Pd

125 = 20 + 27Pd
Pd = (125 -20) / 27 = 105/27 = 3.88W

Which sounds rather high to me. I think it is assuming some kind of thermal pad the PCB.
Why Tc? You mean Tj? ..because there Tj,max in the datasheet, not Tc
 
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