Spehro Pefhany writes:
> Ah, the classic n-channel JFET.
>=20
> The one in the middle is the gate. The other two are interchangeable.
>=20
> (for example, ON Semi and Fairchild show the reverse pinout--
> S-G-D and D-G-S respectively)

	Wow! I plan to use this as a current regulator for the
LED in an optoisolator so I guess I can just tie the gate to the
source (negative) and the load will be between the remaining
free pin (drain) and the positive terminal. The odd thing is
that in this particular circuit, I have the 2N3819 stuck in to 3
pin holes of an IC socket so I could flip it either direction
and it will probably behave the same way. Correct?

	My understanding of this circuit is that the gate is one
diode drop above the source when shorted in this manner and that
is what gives you the current regulation. The other question I
have is if you connected the positive side of the power supply
directly to the drain and put the LED between the source-gate
and the negative terminal of the power supply, would anything
really be any different? Would you still have 10 or 15 mills
regulated current between the source and negative terminal? If
that was a simple resistor instead of the JFET, you would have
the same current at any point in the series circuit except there
would no longer be any regulation. This is based on Norton's law
concerning current in a series circuit.

Anyway thanks for the information.

Martin McCormick
--=20
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