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Wisdom states that the load is connected to the drain, and the source connected to pos or 0v dependant on P or N channel. However ...

I have a "2N3820" (a p-channel in TO92 package), so pin out (should) be DGS with the flat facing upwards.

If I connect in the correct configration as shown, I get weird results.

schematic

simulate this circuit – Schematic created using CircuitLab

Those readings don't seem right BUT ... if I put the load on the source pin, and ground the drain

schematic

simulate this circuit

So I wondered if the pinout was 'non-standard' (cheapo from ebay) so, using a diode tester on multimeter, and a continuity tester, I got the following results:

D-S reads 730 on my meter (a normal IN4001 also reads 730) S-D reads 688 (Don't ask what those figures relate too - I've no idea either!)

With neg of audio contin0uity tester to drain, + to source audio signal neg to drain, + to gate: no audio pos to drain, - to source: audio pos to drain, - to gate: audio

(The continuity tester, built to mainly test for broken leads is simply a 555 astable, but with a break between the neg of circuit board, and -ve of battery

schematic

simulate this circuit

I have NO IDEA how to interpret those results to find the correct pin config.

The second arrangement is what I want to happen, ie to be normally conducting when gate low, but switch off when gate taken high. But the pinout seems to go against all the diagrams I've seen where the load should be between drain and 0v.

Can anyone explain - IN SIMPLE TERMS (!) - what the heck is going on here. 'cos I'm totally bewildered

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    \$\begingroup\$ The 2N3820 is a P-type JFET but you're using the symbol for a MOSFET in your schematic, that is confusing. From the circuits, I guess that you treat the 2N3820 as if it is a P-channel MOSFET. As it is a JFET, that indeed will make it behave in a very different way. R2 is unconnected in the schematic yet you indicate that there's a voltage across it, that makes no sense at all! \$\endgroup\$ – Bimpelrekkie Feb 19 at 10:59
  • \$\begingroup\$ Thanks. I didn't know there was a JFET. I thought ALL FET's worked in the same way. (Will now research JFET) With respect to R2, in the first diagram, I wrote to the left (R2 shorted) Wrong wording, and what might have explained it better was a switch. What I meant to imply was when R2 was connected to the R1/gate junction, the outputs changed as shown. (On reflection "shorted" was deffo the wrong wording !) \$\endgroup\$ – Cristofayre Feb 19 at 16:54
  • \$\begingroup\$ Will redo diagrams lest anyone else gets confused \$\endgroup\$ – Cristofayre Feb 19 at 16:55
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What you have bought is a P-type JFET transistor, not a P-channel MOSFET.

Between JFET gate and JFET body a PN junction exists. But in JFET we want Ig = 0A (gate current).

So you must reverse bias this diode. And this is why Vg voltage cannot be smaller than the source (body) voltage.

enter image description here

For P JFET when the Vgs = 0V the P-JFET is full ON and \$I_{DSS}\$ current is flowing.

For 2N3820 \$I_{DSS}\$, will be in a range \$I_{DSS} = 0.3mA ....15mA\$

schematic

To turn off the P-JFET you need to force the gate voltage above the source voltage (\$V_{GS(off)} = 8V max \$).

So if a source is connected to 20V you need a positive voltage at the gate to turn off the transistor (28V max ).

schematic

simulate this circuit – Schematic created using CircuitLab

And JFETs are symmetrical device so you can swap the source with the drain terminal and vice versa.

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  • \$\begingroup\$ see, I told you was a noob. I didn't know there was a distinction of a JFet transistor and a MOSFET. Now I know that, I can look at researching the former, but if you say gate and drain can be transposed, and the example with load on high side, I'll stick with that. Very helful explainations \$\endgroup\$ – Cristofayre Feb 19 at 16:50
  • \$\begingroup\$ In practice, the voltmeter will be a 12v relay, which is around 390 ohms (it was about 30ma on 9v batt - given that the latter voltage drops under load, so obviously a little higher now using 12v) Using the load on the high side (obviously) creates a voltage drop. If the gate tied low keeps it on. would a high of just below Vcc - from 4017 output - cause the JFET to (partially) stop conducting? As long as it hits about 6v. I'll be happy as the relay should then 'drop out' \$\endgroup\$ – Cristofayre Feb 19 at 17:10
  • \$\begingroup\$ The maximum JFET current is IDSS when Vgs = 0V (gate short to source). Therefore I recommend you to buy a real MOSFET or a BJT. \$\endgroup\$ – G36 Feb 19 at 18:51

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