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I am trying to wire up a LED matrix using a FET transistor as a switch.

The AVR will be connected to the gate. To turn the LEDs on I have drive the pin connected to the transistor gate low, right?

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2 Answers

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The 2N3819 is an RF 'N' channel depletion mode JFET and is not really suitable for LED switching. Depletion mode means that the gate needs to be driven abut 7 volts more negative than the source to turn it off, and driven to the source voltage but no higher to turn it on. The maximum drain current is probably not high enough for your needs either.

Look for an N channel enhacement mode MOSFET instead.

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Yeah I think that's what my mate was recommending. Trouble is he tried to explain but me no comprehendo. So with the N channel MOSFET it's turn it low to switch on right? – Ageis Apr 7 '11 at 15:45
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No. MOSFETS are enhancement mode so the gate is high to switch on. – MikeJ-UK Apr 7 '11 at 16:06
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Just for the record: most normal (called "power") MOSFETs are enhancement mode (you will have a hard time finding a depletion mode power MOSFET). With enhancement mode the N-channel type expects high voltage to switch on and P-channel type needs low voltage (negative with respect to the source pin which you commonly connect to you power supply -- a high-side switch). – jpc Apr 8 '11 at 12:25
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The 2N3819 is a poor choice because of its limited current-handling capability and because it needs a negative voltage on the gate to turn it off. A BJT would be more suitable.

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It's an N channel Fet. 2N3819. ee.buffalo.edu/courses/elab/2N3819.pdf – Ageis Apr 7 '11 at 14:31
Dammit! you posted your edits just as I posted my answer :) – MikeJ-UK Apr 7 '11 at 14:55
Sorry about that! He originally just said that he was using a FET, type unspecified. – Leon Heller Apr 7 '11 at 15:18

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