[sdiy] JFET switch question

[Ian Fritz]

At 04:21 PM 4/25/2008, JH. wrote:
>You all know the configuration:
>D and S are used as the two switch contacts.
>G is used to control the switch. A diode is connected to G (in opposite
>direction as the FET's own gate diode).
>Speaking of a n-channel FET like the 2SK30, the anode of this diode goes to
>G. If you connect the cathode of this diode to the negative supply, the
>switch is open. A positive voltage, or even no external voltage at all,
>closes the switch, because the FET conducts.
>So far so good.
>
>Now, in most applications we see a high impedance resistor added from G to
>either D or S of the FET.
>In some other applications, this resistor is missing, and we find a
>capacitor in the range of 10nF ... 100nF from G to GND.
>The latter is obviously intended for smooth transitions rather than for hard
>switching.
>
>Now I wonder: We rarely see this kind of switch without this capacitor and
>without the resistor.
>Why is this? For a very simple switch, wouldn't just the FET and the diode
>be enough?

If I understand what you are asking -- I believe the G potential is 
indeterminate if it is only connected to a reverse biased diode. Just 
floating at a very high impedance.  A large value resistor to the channel 
is basically a short circuit.  This ensures that the device is on the Vgs = 
0 curve of the Id vs Vds characteristics, where you have the lowest channel 
resistance and the best channel linearity.

   Ian