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i am using a TC4424 mosfet driver which has a peak output current of 3A driving a irl3705 mosfet(which is driving a 400mA load), The mosfet driver along with rest of the circuit is being powered by a 5V voltage regulator (CW7805) which at max can dissipate (( 12V Supply -5V Output)x 1.5A Max Current)=10.5 Watts.

The mosfet driver is being given a signal via an esp32 gpio pin.

The problem is that the regulator is getting hot even with a 2.1x2.5cm heatsink and a fan. From my calculation the total circuit current consumption is around 1.2 Amps excluding the mosfet driver consumption.

Is mosfet driver consuming a large amount of current to drive the mosfet?. Can i connect the mosfet gate directly to the esp32 pin so that i could lower the current consumption?

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  • \$\begingroup\$ How fast are you switching the load? A MOSFET driver shouldn't consume a lot of average current, but you are dropping a lot of voltage across your regulator so it's not surprising that you're seeing it get hot. Is the load running from the 7805 too? How hot is "hot"? \$\endgroup\$
    – John D
    Aug 23, 2020 at 18:25
  • \$\begingroup\$ i wish i could tell but i dont have an infrared gun to check the temperature. But its enough to burn my finger if i touch it for a few seconds. i am pwimg at 38khz. \$\endgroup\$ Aug 23, 2020 at 18:38
  • \$\begingroup\$ It's the 1.2A that's the problem. That's a tiny heatsink for dissipating 10W. If you decide the regulator temperature is a problem, you want either a bigger heatsink, or better, a buck converter which will save about 9W. \$\endgroup\$
    – user16324
    Aug 23, 2020 at 19:49

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If your total consumption is 1.2A you are dissipating 8.4W which is a challenge even with a heatsink and fan. If your thermal impedance is as good as 10C/W your temperature rise would be 84C over ambient. This is a good use case for a buck regulator rather than a linear regulator.

Your FET driver should consume minimal average current unless you're switching the load very fast.

You don't say what your gate drive voltage is, but if it's 5V then the power consumed by driving the gate is P=QVf where Q is the total gate charge at 5V, V is your 5V gate drive, and f is the switching frequency. So power consumed is $$98nC*5V*38kHz = about 19mW$$

There's another couple of mA for the quiescent current drawn by the driver chip, so in total not much at all.

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  • \$\begingroup\$ yes i am around 38khz for pwming ir leds. \$\endgroup\$ Aug 23, 2020 at 18:37
  • \$\begingroup\$ That's not fast enough to cause a driver current that's large in comparison with the 1.2A current consumption that you calculated. A FET gate is a capacitor, so there's no DC current, only charge current. \$\endgroup\$
    – John D
    Aug 23, 2020 at 18:43
  • \$\begingroup\$ maybe our ambient temperature is high in summer, its around 30 degrees at night and can go upto 45 degree on a very sunny day. The day average is around 35-40 degrees. \$\endgroup\$ Aug 23, 2020 at 18:53
  • \$\begingroup\$ OK, but this is really an application where you would be better off with a buck regulator. \$\endgroup\$
    – John D
    Aug 23, 2020 at 18:54
  • \$\begingroup\$ i am already using a buck converter on top of the laptop power supply to step it down to 12V. And further the linear regulator to move it down to 5V. I needed both 12V and 5V supplies for my project. \$\endgroup\$ Aug 23, 2020 at 18:55

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