On Aug 23, 9:55*am, "ian field"
wrote:
"Al" wrote in message

...

You can do a lot with a battery, LED and resistor.

I have all these parts....;-)

If you apply reverse voltage from drain to source, the "body diode" will
conduct and light the LED. (can be useful to identify leadout on small
TO92
MOSFETs).

I am pretty new to this, can you explain what you mean by reverse
voltage? +ve terminal to ? and -ve to ?

I assume the resistor is just for limiting the current to the LED?

If you need that explaining, you shouldn't be messing with power supplies! -
However, what do I care if you kill yourself so here goes.

Its best to use a 2mA ultra-efficient LED, the battery should be a MN21/A23
12V car keyfob battery, calculate the resistor to limit the current to suit
the LED.

Assuming an N-channelMOSFETconnecting the negative lead to the drain and
the positive lead to the source, the body diode will conduct and light the
LED.

Next, connect the positive lead to drain and negative to source - the LED
may partially light due to static charge on the gate capacitance.

Keep the negative lead on the source and touch the positive lead on the
gate, this charges the gate capacitance positive and makes the N-channel
conduct - so if you then connect the positive lead to drain the LED will
light.

Next, keep the positive lead on the drain and touch the negative lead to
gate, this charges the gate capacitance negative and should make the
N-channel completely non-conducting - so put the negative lead on the source
and the LED should not even glow the slightest bit - if it does theMOSFET
is leaky and will go bang in a PSU!

Hi Ian,

Thanks - and to put your mind at rest its on a motherboard, so its all
DC.

I stay away from AC power supplies :-)

I assume all this must be done out of circuit?

-Al