JFET as a on/off switch

Discussion in 'The Projects Forum' started by dpresley58, Sep 25, 2009.

  1. dpresley58

    Thread Starter New Member

    Sep 1, 2009
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    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.
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    A J201 Jfet is N-channel. Its datasheet shows that is is turned off when its gate is -0.3V to -1.5V negative to its source voltage. It is turned on when its gate voltage is the same as its source voltage. It does not conduct well when it is turned on since its current with a 20V supply is only 0.2mA to 1mA.

    You have its drain and source connected backwards but it is probably symmetrical and the drain can be the source. But you have it always turned off.
     
  3. Wendy

    Moderator

    Mar 24, 2008
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    A FET is a variable resistance device, which means it approximates a switch extremely closely. Look up CMOS multiplexers such as a CD4050, CD4051, or CD4052, they take full advantage of this.
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    A depletion-mode Jfet is completely different to an enhancement-mode Mosfet (inside a Cmos logic IC).
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Yes, but they are both FETs. The gates are different in theory and design, the Source/Drain are both fairly similar, which why they are both a type of FET.
     
  6. dpresley58

    Thread Starter New Member

    Sep 1, 2009
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    By the way, this is the article I'm referencing:
    http://www.allaboutcircuits.com/vol_3/chpt_5/2.html

    I see what you're saying about having it always turned off, and I should have specified the mechanical switching aspect of this. The plan is to have an SPDT switch that either provides or cuts voltage to the divider feeding the JFET, while having a constant voltage to the circuit being switched.

    Your point about the current supply is a good one that I hadn't thought of. The circuit in question has a quiescent of about 5mA, so if I understand your point, I'd need to find a transistor that provides at least that amount when on?

    The CMOS Inverter, while certainly a good suggestion, doesn't fit into the scheme of things. The circuit in question is a small JFET preamp, with space being a primary consideration. The boardspace available is going to require a minimum part count to accomplish the switching and will hopefully not be a significant addition to battery drain.

    Getting back to the item you brought up concerning the J201 being spec'd with a Vgss of -0.3v to -1.5v... Would the entire 9v supply go through the DS junction? If that's the case, them am I correct in assuming I'd need to be putting about 7v on the gate to ensure pinchoff?
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    The J201 is pinched-off with a gate to source voltage from -0.3V for some of them to -1.5V for some others of them. The max gate voltage is 20v so -9V is also fine to turn it off.

    If you load uses an extremely small current like 0.02mA (20uA) then the drain can be connected to +9v, the source connected to the load (with the other end of the load at 0V) and the gate will be +9V for the Jfet to turn on and 0V for the Jfet to turn off.

    A 2N3819 Jfet has a current of 2mA to 20mA. You could buy a few hundred and test them until you found one that consducts more than 10mA.
     
  8. dpresley58

    Thread Starter New Member

    Sep 1, 2009
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    I'm beginning to think that a MOSFET would be a more suitable device for this application. Am I even on the right track?
     
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