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First things first: I'm very new to electronics

Getting right into it: I am trying to use an N-channel MOSFET as a LED strip light switch, with 12V at 1.5A supplied to the source and 4.5V from a battery supplied to the gate:

schematic

simulate this circuit – Schematic created using CircuitLab

However, I find that even without connecting the wires to my battery, my mosfet seems to allow flow from source to drain.

I'm hoping that someone may be able to tell me what is wrong and how I can fix it. The MOSFET I have is an FQU13N10L.

In addition I also have some particulars about the answers:

  • This circuit is just a test for a larger project which involves about 60 MOSFETs on a breadboard, which cramps my breadboard. I need a solution that does not involve resistors or capacitors that will cramp up a lot of space on my bread board;
  • Second, if needed I can change the 4.5V to be 3V or 3.3V by using different batteries;
  • Third, it may be important to note that the gate voltage of 4.5V is controlled by a GPIO pin on a PIC microcontroller. I do not know if that matters but I need nothing to go through the 4.5 lines that may hurt the microcontroller, which can not handle 12V.

I apologize if I sound demanding in this, I am just trying to be as clear in my purpose as possible, thank you for all your answers

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    \$\begingroup\$ A resistor is the obvious answer so maybe explain why your title says "without using resistors"? \$\endgroup\$
    – Andy aka
    Jan 8 at 13:18
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    \$\begingroup\$ @KidWithComputer "I do not want them" isn't really a good enough reason. Yeah, they may take up some space, but if they are required, then searching for shortcuts may well leave you with a project that doesn't work properly. Best to just do it correctly from the start and find a way to deal with the cramped space. You can always connect breadboards together to give yourself more space. But resistors will be the correct way to do it. \$\endgroup\$
    – MCG
    Jan 8 at 13:25
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    \$\begingroup\$ I’m voting to close this question because not having resistors from gate to source will cause the gate to float and once turned on, the MOSFET will take ages to turn off. \$\endgroup\$
    – Andy aka
    Jan 8 at 13:28
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    \$\begingroup\$ What happens when a MOSFETs gate is left open? \$\endgroup\$
    – Andy aka
    Jan 8 at 13:34
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    \$\begingroup\$ @KidWithComputer there is no proper solution that will give you consistent results. Resistors are the way to do it \$\endgroup\$
    – MCG
    Jan 8 at 13:35
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Leaving the gate floating (when the switch is open) can lead to what you are seeing; the gate is high-impedance and charge can accumulate on it, switching the FET.

It needs a pull-down resistor on its gate to make things predictable (explicitly driving the gate to 0 when the switch is open), space or no space on your breadboard.

I don't know about the PIC you are using; it may have internal pull-up and/or pull-down resistors you can activate in code, or actively drive its pins high and low.

You could use this instead of external pull-down resistors when you drive the gates directly from the PIC's IO pins. It would leave the gates in an undefined state until the PIC has booted up and has configured the pins, though.

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  • \$\begingroup\$ Thanks for the tip \$\endgroup\$ Jan 8 at 13:35
  • \$\begingroup\$ Note that resistors can come in packs, with 8 or more resistors in a pack, saving space. \$\endgroup\$
    – rdtsc
    Jan 8 at 13:39
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    \$\begingroup\$ Note that an internal pulling resistor will typically only work when the MCU is programmed and has completed booting up to the point in code where it is configured. \$\endgroup\$ Jan 8 at 19:55
  • \$\begingroup\$ @Chris-Stratton: Yes, added that to answer. \$\endgroup\$
    – ocrdu
    Jan 9 at 11:50
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The problem is that when using the MOSFET, you cannot leave the GATE open. It's a VERY high impedance input that controls a very high-gain device. So leaving it open allows anything and everything to happen to it.

This pin is the GATE:

enter image description here

Normally you will tie it to Vcc or GND via a resistor to keep it in the state you want when you are not driving it. In your case you have a simple switch that will drive it to 4.5V. So a resistor to GND is appropriate to keep the MOSFET OFF until you want it to turn ON.

Alternatively, if for some strange reason you really don't want to use a resistor that costs a tiny amount, you could use a SPDT switch that connects the GATE to GND when it's not connected to your 4.5V source.

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  • \$\begingroup\$ Thank you for the quick response, what is meant by leaving a gate open? Is that just not connecting anything to it? What would happen if I simply connected it via a wire to ground? \$\endgroup\$ Jan 8 at 13:31
  • \$\begingroup\$ I updated my answer with the GATE identified. \$\endgroup\$
    – jwh20
    Jan 8 at 13:33
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    \$\begingroup\$ To the OP: If the gate is wired directly to ground, and the source is also connected to ground, then the transistor will not turn on under any circumstances. In order to turn on the transistor you will have to remove the gate ground wire and drive the gate high. This may be obvious. It is hard to tell what people know and don't know. Please don't take offense. \$\endgroup\$
    – mkeith
    Jan 8 at 18:45
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    \$\begingroup\$ @mkeith I really appreciate the help, but mostly I appreciate the understanding, no offense taken my dude. \$\endgroup\$ Jun 1 at 21:46

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