Try this John:
It sounds like you are wanting to do some switching. I don't know how many
watts the lamp is that you mention, but the '540 will switch a fairly large
load - even larger than an auto headlamp - so it may be overkill for your
application, but should still work.

The Source should be toward your lowest potential, or most likely your DC
ground. The load can be on either side - that is above the drain or between
the source and Gnd.
Unlike a bipolar transistor, this thing needs a positive voltage of
somewhere between 2 and 4 V, as related to the source(that's why the data
sheet labels it Vgs) for it to switch and conduct. it does not need current
at the gate the way a bipolar needs to be biased, just volatge, so think in
terms of a voltage divider tapped to bring the Gate to at least 4V if you
want all parts to work all the time. The full 12V on the gate shouldn't hurt
it, according to the data sheet, but may not be most desirable.

If you tried them this way they should have worked.


Now all you have to do is figure out if you have it connected correctly and
get the

The pinout in the IRF540 is G, D, S as viewed from the front (printed) side
with the tab away from you. The source, or pin 3

How many do you have, and what is your application?

C


>
> May Be I have it fixed now...
> John Ferrell
> 6241 Phillippi Rd
> Julian NC 27283
> Phone: (336)685-9606
> Dixie Competition Products
> NSRCA 479 AMA 4190  W8CCW
> "My Competition is Not My Enemy"
>
>

--
http://www.piclist.com hint: The PICList is archived three different
ways.  See http://www.piclist.com/#archives for details.