To start I am very much a beginner at electronics and am very limited in my tools.

I bought a IRF3205 mosfet a few days ago hoping that I could use it in a circuit I have where a 12V power supply at 1.5A is capable of going from the source to the drain to turn on an LED light strip. The would be mosfet controlled by 4.5V supplied by three triple a batteries at it's base.

I was disappointed though to test it and see that this mosfet only outputted less than 9v at the drain(I don't have a way of finding its current). Is there a way to make this mosfet work or do I need to buy a new one? If I need a new mosfet can I get a recommendation?

Again my goal is to control an LED strip—running at 12V, 1.5A—with a 4.5v battery supply going on and off with as few parts as possible, preferably just a single mosfet and no resistors or capacitors

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    \$\begingroup\$ Does not look like a logic level MOSFET, it won't turn on very well with only 4.5Vgs. But why would you want to use batteries to drive the gate when you have 12V available anyway? Sounds like a XY problem. \$\endgroup\$ Dec 31, 2020 at 20:27
  • \$\begingroup\$ You're going to have to draw the schematic before anyone can help. \$\endgroup\$ Dec 31, 2020 at 20:27
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    \$\begingroup\$ Very unlikely you are turning the device on adequately with 4.5 V...that is barely over the threshold voltage. It will be enough to get some current, but the drain to source resistance is killing you. Try upping the voltage, or use a logic-level MOSFET. \$\endgroup\$ Dec 31, 2020 at 20:31
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    \$\begingroup\$ @MathKeepsMeBusy Reaching the threshold voltage doesn't turn the MOSFET on like a switch, the drain to source resistance will be quite high. \$\endgroup\$ Dec 31, 2020 at 20:34
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    \$\begingroup\$ @MathKeepsMeBusy The reason I must use the 4.5 power supply is that it comes from a pic16F15345 micro chip which can handle a max of 5V \$\endgroup\$ Dec 31, 2020 at 20:41

1 Answer 1


This circuit may do what you hope for. Note that the 12V power is "normally on". That is, if the logic circuit is off, the 12V power will be applied to the load. If you need it to be the other way, it will take more components I think.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ @Unimportant thanks. \$\endgroup\$ Dec 31, 2020 at 21:52
  • \$\begingroup\$ Just a note: This circuit will drain the batteries the entire time the lights are off. \$\endgroup\$ Dec 31, 2020 at 22:07

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