p-channel JFET calculation

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What confused me is that you drew the FET upside-down from how it's usually shown on a schematic (drain on top). So you do have a negative voltage on the drain as needed. Drawing circuits in a unconventional manner, makes it more difficult to analyze. That may be why you made the drain resistor smaller than the source resistor, which is opposite of the typical circuit.

You equation is correct. But it means Vs must be less (more negative) the Vg.
 

Hey.

I have finally got the cct to work.

I recalculated the eqns.

I have: (power supply 12 and -12 rails)

Vg=4.94v

Id= 1mA
Rs= 2k, Vs= (12v-[1mA*2k]) = 10v

This gives a Vgs = -5.05v

So in the end what i did for Vd was play around with the resistor value until i got an output. Not very technical but it worked.

Can you help me witht the equation for Vd, seeing though Rd has -12v on one side of it i used this calculation to find Rd, but it didnt really work. Rd 3k seems to work good.

Vd,max= Vg + |Vp| (to keep FET in saturation)

so Vd,max= 4.94 + 2.7 = 7.64

Rd = [7.64 - (-12)]/1mA = 19640 ohms. But that didn't work.

And also Rd being 3k would that give a Vd of (3k * 1mA)-(-12) = 9v?
 
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I think you missed it.

I get it.**broken link removed**

I don't think Eric got it though.


Yes they depletion mode JFETs, I've used them before, they're nothing like depletion mode BJTs or an enhancement mode JFETs though.
 
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