Why does it need to be p-channel?  An n-channel should work equally well, since the end result is still a roughly constant current over a wide range of terminal voltages.

TTFN

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Can't you use Jfet ?

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I need to turn on with a negative Vgs, so thus the p-channel.  

J-fet?  I need to device to be on (Rds=0) when no gate voltage is applied.  I didn't think it possible to use a J-fet in depletion mode.  

I'm confused.  You seem to be saying that you want to drive the gate.  Normally, a depletion FET simply has gate tied to source.  If that's the case, then it simply behaves like a roughly constant current source and it doesn't matter whether you're using p-channel vs. n-channel.  You simply keep the gate and source of the n-channel at the lower voltage and the drain at the higher voltage.  There is no body effect, since these are discrete transistors.

TTFN

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JFets only work in depletion mode.

Enhancing a jfet turns it into a forward biased diode.

Siliconix used to do depletion mode pmos, dunno if they still do.

Zeitghost is right. Siliconix still make p-channel and n-channel J-FETs but note that they are now part of the Vishay group:

http://www.vishay.com/fets-small-signal/

If you want descrete TO-92 package the J170 or J270 series look about right from your VDS stated requirements, but you might be struggling to get a RDSon very close to zero ohms in a small-signal device, more like low tens of ohms.

BrianG, I'll check out the Vishay options.  

Zeit - Whoa, what's that bright light!  Hmm, it looks like I'll have to dust off my FET books.  I used a 2N4416 N-channel JFET in a design long, long ago so you'd think I would remember these things.   

IRstuff - "Gate tied to source"?  Then Vgs = 0 for all cases.  Don't you mean "Normally, a depletion FET simply has gate tied to DRAIN"?  That way as Vgs/Vds increases the resistance increases and you have a constant current source.   

Thanks all!  

no, the whole point of a depletion transistor is that the threshold is such that a Vgs=0 still results in channel conduction. therefore, for a depletion transistor, Vgs must go below the source to actually shut off. otherwise, you'd might as well use an enhancement transistor.  

which raises the question of what you thought you were asking for ;)

TTFN

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That's correct for a N-channel depletion mode device.  Vgs goes above the source to shut off for a P-channel depletion mode device, right?  

I'm looking for a simple high-side 'constant' current source.  

No, a depletion-mode p-channel device has a threshold that's more positive than the source, so Vgs=0 still results in conduction.  Depletion mode, by definition means that the channel is conducting when Vgs=0.

Don't forget that p-channel source is supposed to be at a more positive potential than its drain or gate.  Therefore, a depletion mode p-channel has a positive threshold, so Vgs-Vt = -Vt with Vgs=0, therefore channel conducts.

TTFN

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Siliconix used to do Currrent Regulator diodes.

CR022 series and J500 series.

There may be a device in there that does what you want.

All they are is a depletion mode jfet with gate shorted to source, selected on IDSS.

Since it's a two terminal device, you can use it as a pullup if you want to.

For reference, I just found the 1N5283 thru 1N5314 series of current limiter FETs.  These are packaged in a two pin package with gate tied to source with a typical current limit (at 25C) of +/-10%.  

Of course my requirements have changed over the last month so now a 'current limiter' isn't sufficient for my application.  I just thought I'd post this to close this thread.  

Thanks for all your help.