"normally closed switch" semiconductor question

[Tim Godfrey]

Mark,

A JFET has the advantage of not requiring a power supply connection. It is
a simple 3 pin device, not a DIP IC package. However, it works best in
specific circuit configurations.

The circuit you need to use with a FET depends on whether your "normally
closed switch" has one side connected to ground or a power supply. If it is
in the middle of a circuit (both terminals of the switching element at an
arbitrary voltage with respect to the control voltage), then it gets more
complicated, and you'd be better off with the CD4066 solutions that have
been offered. 
 

For the case where the normally closed switch has one side tied to ground
try this:

                  -----   Output
                 |
                 | D
           G   |-
 i-----------<-|      2N5461 (or similar P-Channel JFET)
               |-  
                 | S
                 |
                -+-
                 -

If the input voltage is zero, the FET is on, thus the output will sink
current. If you raise the input to more than 2.5 to 5V above ground, the
FET turns off.



For the case where the normally closed switch has one side tied to the
power supply try this:


                 ^  +5V (or +12V, +15V, or whatever you're using)
                 |
                 | S
            G  |-
 i------------>|        2N5433 (or similar N-Channel JFET)
               |-  
                 | D
                 |
                  -----   Output

If the input voltage is equal to the supply voltage, the FET is on, thus
the output will source current (limited by the FETs on-resistance). If you
lower the input to more than 3 to 5V below the supply, the FET turns off. 


Don't forget that there are limits on current and voltage for FETs, like
all semiconductors. You should review the data sheets for the particular
part you select.

Hope this helps, and good luck.

Tim.

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