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I'm choosing a switch for a power supply design. I would like to know the highest temperature the MOSFET can reach. The switch that I'm using is DMT3020LDV.

An input having 24 V and 8 A is supplied to the drain of the switch. The following are my calculations for the junction temperature:

\$T_j = T_a + (P_t \times R_{ja})\$

Where:

\$T_j\$ = Junction Temperature;
\$T_a\$ = Ambient Temperature;
\$P_t\$ = Total Power Dissipation = 0.9 W (from datasheet);
\$R_{ja}\$ = Thermal Resistance, junction to ambient = 138 °C/W (from datasheet)

So, \$T_j\$ = 25(room temperature) + (0.9 x 138) = 150°C.

But if the outside temperature is 85°C, will \$T_j\$ = 85 + (0.9 x 138) = 210°C?

Are my calculations correct? Because 210°C is too high in my opinion.

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    \$\begingroup\$ 210 is correct but too hot, solder a heatsink to the drain, see also figure 6 \$\endgroup\$
    – Jasen
    Jan 17, 2020 at 0:24
  • \$\begingroup\$ Thank You @Jasen for the quick response. \$\endgroup\$
    – J2018
    Jan 17, 2020 at 0:27
  • \$\begingroup\$ Thank You @JYelton for editing the question. \$\endgroup\$
    – J2018
    Jan 17, 2020 at 0:27
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    \$\begingroup\$ Your calculations are correct. You can look at the thermal section of the datasheet which lists the max operating junction temperature as 150°C. You'll either need to dramatically reduce your power consumption (which is probably not feasible), or dramatically reduce the thermal resistance. Rja = 138 is for minimal copper. Rja = 67 for 1 in sq copper plate. \$\endgroup\$
    – BEE
    Jan 17, 2020 at 0:36
  • \$\begingroup\$ @BEE I'm very limited on the space and 1 in sq plates are going to be very tough, but even if I do that, the Rja and Pt product is still around '125' (67 x 1.9). It would have been great if the product was around '75'. What I'm still wondering is, would the MOSFET temperature be related to the input of drain? Meaning whether the input is '12v, 1A' or '24v, 8A', will the MOSFET still reach 210 *C if the outside temperature is 85 *C? Forgive me for newbie questions, I'm still learning. \$\endgroup\$
    – J2018
    Jan 17, 2020 at 0:54

1 Answer 1

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Work backwards.
For desired power level and known other parameters you can calculate maximuj allowed ambient.
Rearrange as desired.

1 ... Tambient_max = Tjmax - Power(Rth_ca + Rth_jc)
or
2 ... Power_max = (Tamb-Tjmax) / (Rth_ca + Rth_jc)
or
3 ... Rthca = (Tjmax - Tambmax)/ Power - Rthjc

So (1) - Tambient_max = Tjmax - Power(Rth_ca + Rth_jc) tells you how hot the ambient can be for a given desired Power level, given Rthjc (fixed by manufacturer) and Rthca - set by heatsinking.

eg at P = 10 Watts, Tjmax = 150 c, Rjc = 3 c/W, Rthca = 10 c/W
(1) gives ... Tambient max = 150 - 10W x (3 + 10) = 150 - 130
= 20c :-(.

Change to a 2c / W heatsink and you get
Tambmax = 150 - 10 x (3+2) = 150-50 = 100 c.

ie IF you can get the case within 2c of the water then you can cool it with boiling water :-)

Adjust assumptions to suit.


You need to understand power dissipation.
If you have a low Rdson MOSFET and it is usually hard on or off then dissipation at 8A is "lowish". eg I^2R at say 30 milliOhm = 64 x 0.030 ~= 2 Watts.
Add to that some switching loses and it is still liable to work well with sensible minimal heatsinking.

If the FET operates partially NOT in on/off mode then it will be different or much different.

Show us your circuit

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