Hi,
I read the article concerning using JFETs as an on/off switch for a circuit, and while I understand it in principle, the application is slapping me all about the head and shoulders...
The diagram shows the source and drain connecting between the Vss and ground of the circuit to be switched. It specifies a control voltage being applied to the gate, which I assume is going to be slightly above Vp. This indicates to me that the battery (in this case) will always be connected to the circuit and the transistor serves to either provide or cut the link to ground within the circuit.
I have about 30 J201's that have been measured for Vp, the values of which range from about 0.6v up to almost 1v. If I connected the Vss (9v battery) to a voltage divider for the gate of the switch that provided a bit over 1v and connected the source to ground and the drain to the positive rail of the circuit to be switched, would this be the way to get this to work? I'm assuming the ground leg of the divider could be used as the pulldown resistor to take the residual capacitance within the transistor to ground.
It just doesn't seem to be working out on the breadboard.
I read the article concerning using JFETs as an on/off switch for a circuit, and while I understand it in principle, the application is slapping me all about the head and shoulders...
The diagram shows the source and drain connecting between the Vss and ground of the circuit to be switched. It specifies a control voltage being applied to the gate, which I assume is going to be slightly above Vp. This indicates to me that the battery (in this case) will always be connected to the circuit and the transistor serves to either provide or cut the link to ground within the circuit.
I have about 30 J201's that have been measured for Vp, the values of which range from about 0.6v up to almost 1v. If I connected the Vss (9v battery) to a voltage divider for the gate of the switch that provided a bit over 1v and connected the source to ground and the drain to the positive rail of the circuit to be switched, would this be the way to get this to work? I'm assuming the ground leg of the divider could be used as the pulldown resistor to take the residual capacitance within the transistor to ground.
It just doesn't seem to be working out on the breadboard.