[sdiy] JFET switch question

[JH.]

>----- Original Message ----- 
>From: "Ian Fritz" <ijfritz at comcast.net>
>To: <tim102 at tstinchcombe.freeserve.co.uk>; "'JH.'" <jhaible at debitel.net>; 
><synth-diy at dropmix.xs4all.nl>
>Sent: Sunday, April 27, 2008 5:59 PM
>Subject: RE: [sdiy] JFET switch question
>

>At 09:43 AM 4/27/2008, Tim Stinchcombe wrote:
>>- however, with the diode reverse-biased for the switch to be ON, there is
>>now no path for a current to remove the charge in the reverse-biased 
>>channel
>>junction, which is needed to allow the JFET to actually turn on. The cap 
>>is
>>necessary to supply this path, the implication being that the JFET might 
>>not
>>actually switch on without it.
>
>That's exactly what I was trying to explain before. Guess I didn't say it
>very well.  You can't just leave the gate open-circuit because its
>potential is indeterminate.  It needs to be tied to something that will
>bias the device into conduction.

.. and I'm not sure I've completely understood it yet, even though I'm 
thankful to both of you for trying to explain.
The case with the resistor is clear.
The case with the capacitor ... well ...hmm.
Wouldn't an external capacitor which is tied to GND _preserve_ charge 
instead of removing it?

JH.