1
\$\begingroup\$

I am using a MOSFET as a pre-regulator.

enter image description here

When the voltage is too high, the temperature of the MOSFET is getting too high. (Even if I take 15 mA current.)

As a safety precaution, if the junction temperature of the MOSFET exceeds 100 degrees, I want it to enter the cut-off region. What are your suggestions?

(Note:I am using a iron cooler.But still, there is a rise in temperature.)

\$\endgroup\$
4
  • 2
    \$\begingroup\$ I see D1 and D2 are 27V, so assume supply is +/-35v or more. At 35v, 15mA though Q1 means it must have a dynamic resistance of 35/0.015 = 2.33k This is going to be much larger that whatever the load is, so the MOSFET is going to dissipate that power as heat. 35V*0.015A = 0.525W, definitely enough to warm quickly, even with a heatsink. What are you actually trying to do - create a variable power supply? \$\endgroup\$ – rdtsc Apr 2 at 14:54
  • \$\begingroup\$ @rdtsc yes.My voltage can be 90V.90*0.015A=1.35 W. I need just temprature protection for mosfet.Rds=10 ohm. \$\endgroup\$ – john_slv Apr 2 at 15:16
  • \$\begingroup\$ For the IRF540, Rds = 10Ω only when gate voltage >= Vgs, so near 20V. If Vgs is lower than ~10v, the drain-source resistance is going to be much bigger and thus cause overheat. I would consider alternatives rather than temperature cut-out. \$\endgroup\$ – rdtsc Apr 2 at 17:38
  • \$\begingroup\$ Frankly, I don't see the problem. Even with 90 V in and a 15 mA load, each MOSFET is only dissipating about 1 W. A bare TO-220 can handle this, although it will get quite warm, and even a modest heatsink should bring down the temperature considerably. \$\endgroup\$ – Dave Tweed Apr 3 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.