100W LED with a switch

I would like to turn on a 100W LED using a switch. However, while the LED turns on the switch has no effect. What's wrong with my simple circuit?

My 12V and 30V grounds are connected to each other to be a common ground. I'm using an IRF4905 P-type MOSFET.

Does it have something to do with the Vgs?

Thank you!

• Please draw your schematic with higher voltages near the top and lower voltages near the bottom. That will make your issue much more obvious. – The Photon Apr 7 '18 at 20:19
• 1950s schematics that used PNP transistors heavily and had negative VCC were funny. This is headache fuel. – rackandboneman Apr 8 '18 at 2:04
• Apologies for the poor schematic. I'm a hobbyist and don't really have training in electronics. Perhaps I can get a recommendation on how to draw one using an electronic tool? – user3117759 Apr 8 '18 at 2:51

Your PFET has 30 V on it's source.

Depending whether the switch is open or closed, there's either 12 V or 0 V on the gate, so $V_{gs}$ is either -18 or -30 V. Either way, the FET will be switched strongly on.

Connect the pull up resistor to 30 V instead of 12 V and it should work.

Note: IRF4905 has a maximum (or rather, minimum) $V_{gs}$ of -20 V, so find a way (for example with a voltage divider) to have the switch connect the gate to something like 20 V, instead of to ground, when it is closed.

• I suspect the first time the OP closed the switch the MOSFET blew. It no longer comes on. – Sparky256 Apr 7 '18 at 20:22
• yea was about to comment the same lol @Sparky256 – Mitu Raj Apr 7 '18 at 20:24
• Thank you for the explanation! I will try the voltage divider on the input to the FET. – user3117759 Apr 8 '18 at 2:55

Its good to look into datasheet before you proceed things.

When switch is OFF, MOSFET is at $V_{GS} = -18 V$. And $-18V < -10V$ and hence it is sufficiently ON.

When switch is ON, MOSFET is at $V_{GS} = -30 V$. And its above limits $|-30V| > |- 20V|$. I wonder it blew off.

You can try this:

simulate this circuit – Schematic created using CircuitLab

• Thank you for this explanation. I didn't really have a good understanding of Vgs and when the FET is "ON". I will try this voltage divider solution where R1 = 10k and R2 = 20k to get 20V at the gate. – user3117759 Apr 8 '18 at 3:10