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MOSFET Data sheet https://docs.rs-online.com/ec6b/0900766b811b5c5c.pdf

Taking the MOSFET driver from the attached data sheet (1.5A 30V), if the MOSFET wishes to remain on for a period of 24 hours does it require the following power to drive it:

1.5 A*30 V=45 W

45*24=1080 Wh

Or is this only if it is continuously switching over the 24 hour period? If the MOSFET wishes to only stay on for this 24 hour period how would I calculate the power consumption required?

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  • \$\begingroup\$ What is the supply voltage? What is the specified typical quiescent current? Where else do the 1.5A and 30V figures show up in the datasheet, and what does that suggest about how those figures apply, and whether they dictate steady-state power? \$\endgroup\$
    – Tim Williams
    Commented Mar 19 at 15:44
  • \$\begingroup\$ The current and voltage will not appear at the same time. The static consumption will be the quiescent current times drive voltage. What are you driving and how are you switching it? \$\endgroup\$
    – winny
    Commented Mar 19 at 15:48
  • \$\begingroup\$ How do you find the quiescent current from the data sheet \$\endgroup\$
    – Jasper
    Commented Mar 19 at 16:14
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    \$\begingroup\$ If the MOSFET is staying on or off for a long time, you may not need a dedicated driver, perhaps some 4000 series CMOS with near zero power consumption will do. \$\endgroup\$
    – Neil_UK
    Commented Mar 19 at 17:15

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The data sheet you’re showing isn’t for the MOSFET, but rather for a MOSFET driver. The driver sinks and sources a large dynamic switching current as it charges and discharges the big MOSFET gate.

The driver data sheet is boasting is that it has a large current output (+/-1.5A), so that it can switch a big power MOSFET very quickly. What isn’t so obvious is that when the output isn’t switching, the driver dissipates almost no power at all, only its standby (quiescent) current.

Considering the total circuit composed of this driver, a big MOSFET, and a load, and the fact that it’s ‘on’ continuously, almost all the power goes to the load.

You can estimate your power based mainly on the load and add a small allowance for the driver’s quiescent draw to come up with with the total power budget. You’ll find that quiescent draw spec in the data sheet DC Characteristics.

That said, do you really need a MOSFET driver? A simple level shifter will do for switching a p-FET. Also, you might consider a suitable integrated high-side load switch that could do the same thing, and have the added bonus of overcurrent protection.

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  • \$\begingroup\$ So in a situation where the MOSFET has to stay on for 24 hours powered by a battery, how would I size this battery? \$\endgroup\$
    – Jasper
    Commented Mar 19 at 15:54
  • \$\begingroup\$ @Jasper You size it by the current draw which will be the load current plus the driver supply current. The driver supply current is listed as 2.5mA with logic 1 input, 300uA with logic 0 input. You need to find your load current and the on and off times and then calculate the total current drawn over the 24 hours, if it’s always on it’s just (Iload+2.5mA)*24hrs. That will give you Ah, and you can size your battery from that. \$\endgroup\$
    – GodJihyo
    Commented Mar 19 at 16:54

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