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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!

enter image description here

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    \$\begingroup\$ Please draw your schematic with higher voltages near the top and lower voltages near the bottom. That will make your issue much more obvious. \$\endgroup\$ – The Photon Apr 7 '18 at 20:19
  • \$\begingroup\$ 1950s schematics that used PNP transistors heavily and had negative VCC were funny. This is headache fuel. \$\endgroup\$ – rackandboneman Apr 8 '18 at 2:04
  • \$\begingroup\$ 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? \$\endgroup\$ – user3117759 Apr 8 '18 at 2:51
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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.

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

enter image description here

enter image description here

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:

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ 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. \$\endgroup\$ – user3117759 Apr 8 '18 at 3:10

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