Guys i trying to step down voltage using mosfet.


  • Input battery supply 40 to 50 volt

  • Output voltage is 12 volt not more than 500 ma.

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Mosfet heating issue due to...

Since the mosfet is dropping voltage 37 volt (50v-13v), if output load consume 500ma. Then power dissipation will be 37*0.5= 18.5 watts. For this reason my mosfet getting too hot.

Can anyone suggest how to reduce the heat or what is the option for stepping down 50v to 12v without using buck converter.

  • \$\begingroup\$ Is this supplying the \$\approx 12\:\text{V}\$ that your other question mentioned as its input voltage source? And is it also supplying other loads, too? \$\endgroup\$
    – jonk
    Aug 28, 2019 at 6:56
  • \$\begingroup\$ Yes jonk, i need 12v, 5v and 3.3v from 50v battery supply. \$\endgroup\$
    – Bud
    Aug 28, 2019 at 6:58
  • \$\begingroup\$ Assuming that each supply is individually required for various devices, how much current compliance in each of them? (Don't include the current for the \$3.3\:\text{V}\$ into the other two, though. Keep it separated out, for now. Let's assume that you derive each of these separately from \$40\:\text{V}\le V_\text{BAT}\le 50\:\text{V}\$ and not from each other. We can always then work out other details, if we want to.) \$\endgroup\$
    – jonk
    Aug 28, 2019 at 7:07
  • \$\begingroup\$ As another option, you might want to look up web pages on Roman Black's buck converters and plan one to drop your battery supply down to \$12\:\text{V}\$ (you can use your zener for this) and then derive the rest from there using linear regulators or your zener methods. (You might cut your losses by a factor of 10, or so.) \$\endgroup\$
    – jonk
    Aug 28, 2019 at 7:44

2 Answers 2


Heat is not the problem. According to other things I've been told on this site, MOSFETs suffer a thermal runaway problem (hot spots on the silicon) when they are used as voltage controlled resistors.

Is it okay to use a MOSFET in its resistive region with a heat sink?

If you use a linear regulator which is made for the purpose, you will avoid the hotspot problem, but will still need a heat sink. However, in my cursory search, I couldn't find anything that could handle 50 V in and which would provide 500 mA output. The LM317 or LM217 might be able to handle it.

If you have your heart set on making a transistor controlled circuit using feedback, you can use a BJT (Darlington?) instead of MOSFET, however I am hesitant to give more specific advice since I'm a little out of the electronics game these days. You will need a heat sink for sure.

  • \$\begingroup\$ Thank-you for your reply, then sure i will try with bjt (pnp) as a voltage regulator. \$\endgroup\$
    – Bud
    Aug 28, 2019 at 7:47

If you want to avoid heat, a buck converter is the correct thing. Without that, you're going to be generating heat somewhere.

Sticking a sufficiently chunky wire-wound resistor in series with your circuit would lose some of the heat that would otherwise be generated in the mosfet. Perhaps a 40 ohm resistor, which would drop the 40V to 20V at 500 mA. But the resistor would dissipate up to 10 W.

  • \$\begingroup\$ Then can i use pnp as voltage regulator because its low voltage drop.But some how heat will be dissipates in Resistor, that i can manage by using high watts resistor. But My fet or transistor should be ok. \$\endgroup\$
    – Bud
    Aug 28, 2019 at 7:26
  • 1
    \$\begingroup\$ @Bud You can, but use a sufficiently high power one, and make sure it's on a big enough heatsink. \$\endgroup\$
    – Simon B
    Aug 28, 2019 at 8:11
  • \$\begingroup\$ PNP linear regulator will good think so, because of its Low voltage drop. \$\endgroup\$
    – Bud
    Aug 29, 2019 at 6:02

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