I set up the basic circuit above as a way to test various LEDs and transistors with a 555 timer circuit in place of the switch. However, when I put an IRFP460 transistor there, even when the switch was open the LED remained lit. To say it in another way, there was no voltage going to the gate on the mosfet, yet I'm clearly getting current between the drain and the source. I'm pretty new to this, so what I'm wondering is what aspect of the specifications of the transistor allows this current flow to happen. Thanks in advance!

Attached Thumbnails

 

 

Simple, there is no discharge path for the gate capacitance. You close the switch once, charge the gate capacitance and turn the device on, then you open the switch, leaving the gate charged, and the device stays on. Connect a 10K resistor from gate to source and try again.

__________________
Mike ML.

 

You can not leave the gate of a MOSFET floating. Even though you stopped applying external voltage to the gate, there still is voltage on the gate. The gate of a MOSFET has a very high DC impedance (looks rather like one terminal of a capacitor) so when it is unconnected, it will tend to remain at whatever voltage was last applied. It does not automatically go to ground potential.

You need to add a resistor to ground at the switch output-R1 junction so that the gate voltage goes to ground when you open the switch. Resistor value is not critical.

Edit: Mike, you slipped in while I was writing.

__________________
Carl
Curmudgeon Elektroniker

 

Makes perfect sense. Thank you very much!

 

This is why CMOS circuits use pull up/down resistors. The potential presented to the gate of the MOSFET has to be deliberate.