I'm a newbie at EE, and just I wanted to verify that the design I came up with for a MOSFET circuit that drives a 12V 500mA water pump will work. I'd also like to confirm I can run the N-Channel MOSFET in question (RUR020N02) without a heatsink.

MOSFET Circuit for Water Pump

P8 is a two-pole terminal block which the pump will be attached to. The MOSFET has a 1.5V Logic Level gate, which will be driven by a 3.3V I/O pin from an Atmega328p.

By what research I've done, I've learned that it's good to have a pull-down resistor (R9) in the event the digital I/O pin on the microcontroller (PUMP) is left floating, as well as a limiter resistor (R10) to keep the uC I/O pin within spec (20-40mA). I think this can also act as a voltage divider, but the voltage drop should be negligible in this case and still well over the gate threshold.

I've added a flyback diode just in case, though I'm not sure if one is truly needed. The pump is an impeller so I assume it has the same problems as running a DC motor--I'm not sure if the pump has its own internal protection circuitry.

So, my main question is whether or not this design is adequate for my needs.

However, I'd also like to know if I would need to use a heatsink for this particular SOT-23 MOSFET.

Based on the datasheet (linked above), it has a power dissipation of 0.54W at 25°C. Since the MOSFET has a RDS(on) value of around 105mΩ and my pump draws 500mA, I should be OK without a heatsink since it will only need to dissipate 52.5mW, correct?

0.105Ω x 0.500A = 0.0525W

Assuming the MOSFET turns on fully in an acceptable timeframe.

  • \$\begingroup\$ Your pump is a motor. Have you considered its start-up aka inrush current? \$\endgroup\$
    – Fizz
    Dec 23, 2015 at 20:03
  • \$\begingroup\$ Yes and no. I don't know what the actual inrush current would be as I do not have the physical pump yet to measure it, nor a datasheet for it. So I am assuming it will be < 2A until I can prove otherwise when I get the actual pump sometime this week. Here is the pump in question. \$\endgroup\$
    – Matt
    Dec 23, 2015 at 20:23

2 Answers 2


You have mostly thought this out correctly.

One problem I see in your figures is the power dissipation of the FET. You say the gate will be driven by a 3.3 V digital output. The only guaranteed on Rdson is 135 mΩ at 2.5 V gate drive. You can't use the 105 mΩ figure since that's for 4.5 V gate drive.

The power dissipation will be (500 mA)2(135 mΩ) = 34 mW. You will have difficulty noticing that get warm. It's not anywhere near any limits.

Inrush current will be much larger than the steady 500 mA operating current. However, it won't last long, so it looks like you'll be OK. Note that the maximum pulsed current this FET can handle is 6 A. That sounds sufficient at 12x the operating current. Of course the dissipation will be quite large then, but again, short lived. Without details on the motor, we can's say for sure, but I think you'll probably be OK.

It's good that you put R9 there, but I'd make is smaller. 100 kΩ will do a better job against noise that might be picked up, and is still of no consequence to your digital output.

If you plan to switch the pump on and off rapidly, I'd use a Schottky diode to get around reverse recovery time issues. If you always leave the pump off for more than a millisecond or so, then this doesn't matter.

I would also put a small cap across the pump, especially if you aren't going to switch it rapidly. That will help with RF emissions.

  • \$\begingroup\$ The pump will not be switched rapidly. It will only be turned off in the event the water level is too low for it to reliably operate, so as not to burn out the motor in the pump. I have already picked out a Schottky diode though, I'm assuming it won't pose a problem if I use one even though I don't truly need it. I wonder if maybe there is a better circuit design for a "usually on, seldom off" device? I have lowered the value of R9, and added a 100uF capacitor to the pump. Do you think 100uF will be adequate? \$\endgroup\$
    – Matt
    Dec 24, 2015 at 10:07
  • 1
    \$\begingroup\$ @Matt: The FET should be fine except possibly at startup. If not sure, you could use a relay. 100 uF is way too much. A 1 nF ceramic would be more appropriate. \$\endgroup\$ Dec 24, 2015 at 12:03
  • \$\begingroup\$ Just wanted to add that I built the circuit above with your suggestions and everything works perfectly. \$\endgroup\$
    – Matt
    Feb 3, 2016 at 22:24

It should be ok - read the data sheet and find this graph which I have red-marked: -

enter image description here

With a gate-source voltage of 2.5 volts and a drain current of 500 mA, the volt-drop across the device is typically about 50 mV.

Power dissipation is 500 mA x 50 mV = 25 mW static.

Driving it at 3v3 will improve this slightly. Always look for this graph and don't always rely on page 1 headlines although, for this FET the 0.105 ohms looks pretty much reflected by the graph.

BTW your calculation was wrong about power. Power is current squared x ohms so power = 0.5 x 0.5 x 0.105 = 26 mW.


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