Michael Rigby-Jones writes:
>2) The important parameter is Vgs.  You need a certain potential on the Gate
>with respect to the Source in order to switch the MOSFET on fully.  By
>putting the load on the source side, as current flows through the load, the
>voltage on the source pin will rise.  As the voltage on the gate remains
>constant (i.e. the 5 volts supplied by your PIC pin) this obviously reduces
>Vgs, which partialy switches of the MOSFET.  What you have here is the
>equivalent of the emitter follower in the bipolar world, i.e. lots of
>current gain (virtualy infinite on a MOSEFT), but no voltage gain.

        Yup!  It's a source follower.  There are cathode followers in
vacuum tubes, also.  Since the current gain is very high, assuming
the voltage is correct, such a circuit makes a very good power supply
regulator.

        If the gate was correctly biased, you should see near 5 volts
on the source when the PIC pin is high and 0 when it is low.

        Now, I will let my ignorance shine.  I have used lots of
bipolar transistors like the 2N3055 in power supplies.  They show an
emitter voltage of the base voltage minus 1 silicon junction drop such
that the output is about .6 volts lower than the base bias.

        What kind of voltage drop does a properly-biased source
follower show?

        For that matter, do vacuum tubes have a certain voltage drop
between the grid bias and the cathode?  I haven't actually ever tried
such a circuit.

        One other little thing since I opened this can, while a
follower circuit can make a good power supply regulator, you must pay
close attention to the power dissipation of the device you are using.
Theoretically, one could put 100 volts on the collector of a
transistor, bias the gate at 5 volts relative to ground and get 4.4
volts out the emitter.  The trouble is that that transistor dissipates
the energy needed to run whatever is on the emitter as pure heat.  It
is like a big resistor dropping 95 volts across it so it won't take
much of a load under those conditions to get that transistor very,
very hot.

        When you use a FET or bipolar transistor as a switch and turn
it fully on, the voltage drop across it is extremely small like maybe
a volt or two relative to the power supply voltage.  That means the
power dissipation is orders of magnitude smaller than if it is
operating in linear mode.

        One thing I remember from DC circuits class that floored me
for a while was the fact that except for the case where the transistor
drops exactly half the power supply voltage, there are always two
impedances for any given power dissipation.  You must solve I^2*R for
R.  In other words, the quadratic equation.

        I remember making a fool out of myself in the class because it
just didn't take in my brain for a day or so.

Martin McCormick WB5AGZ  Stillwater, OK
OSU Information Technology Division Network Operations Group

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email listserv@mitvma.mit.edu with SET PICList DIGEST in the body