I have an application where I want to pass a current of about 300mA through a load, which should be turned on/off via a 3V/0V signal. I thought of using a 2N7000 as the switch and have it set up on a breadboard. I'm able to turn the MOSFET ON (checked with an LED) and OFF with the 3V output, but the current is much lower than 300mA. What could be the reason? The load resistance is 10ohm and my battery is capable of driving the load.


According to the datasheet for the Fairchild version of this part, the maximum gate threshold voltage is 3 V. Think of this threshold as the 'threshold' of conduction.

If you scroll ahead a few pages, you'll see some curves that correlate gate voltage with drain-source current. At 3 V gate voltage, the curve predicts well below 250 mA for any drain-source voltage.

transconductance curves

You'll need to either reduce your load requirement, drive the gate with 4 V (or more), or choose a different MOSFET.

  • \$\begingroup\$ Thanks for the answer and the hint on what to look at in the data sheet! I'll try out another part \$\endgroup\$ – eGovind Mar 26 '13 at 2:18

Your basic problem is that the 2N7000 is inappropriate for this application. It's a pretty crappy FET all around. Think of it as being for signals only. It's not much good for handling power.

A IRLML2502 would work nicely, assuming your supply is 20 V or less, which you didn't say. This FET will be 80 mΩ or less with 2.5 V gate drive. 80mΩ * 300mA = 24mV, which is the worst case full on voltage drop accross that FET. That will only dissipate 7.2 mW, which you won't even notice get warm.

  • 1
    \$\begingroup\$ Thanks for the part recommendation! That's a tough one for a beginner like me. \$\endgroup\$ – eGovind Mar 26 '13 at 2:20

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