[sdiy] JFET switch question

[Tim Stinchcombe]

> >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?

Well since I made that post I have been trying to get *my* head round it
too, and have even tried a few simulations, but I'm sure I'm not there yet
either!

However, to be clear, in the circuits in these books I'm looking at the cap
shunts the diode (i.e. is in parallel to it), and is *not* connected to
ground: JFET gate to anode + one end of cap; switching voltage to cathode +
other end of cap. Thus when the voltage switches, it is possible for a
'spike' to get through the cap. (I think I just about see the switching 'on'
case, but not 'off' yet...)

Tim
__________________________________________________________
Tim Stinchcombe 

Cheltenham, Glos, UK
email: tim102 at tstinchcombe.freeserve.co.uk