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Discussion in 'Electronics Homework Help' started by Lerior, Mar 14, 2018.

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  1. Lerior

    Lerior

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    Jan 20, 2018
    Hi, I have a project that I'm working on. The main idea is to construct a BJT/MOSFET circuit to turn an LED on or off. The project uses a 4.5V solar cell source connected to a BJT transistor. That BJT transistor will work as an open switch when there's no voltage detected from the solar cell, and as a short line switch when the 4.5V remains there (will reach the saturation). The collector of that BJT is connected to the gate terminal of an N-channel MOSFET (Enhancement mode). BJT will either ground the gate when it's saturated, making the MOSFET off. Or, turn the MOSFET on when it's open. I will attach my circuit so you can have an idea of what I'm talking about.

    My question here... I initially wanted to do that with a JFET, but the circuit did not work. Is that related to the fact that JFET gate terminal will not be off when the BJT transistor grounds its gate (again, once the transistor reaches saturation).? Also, by looking at the circuit I attached, it seems like the 10k resistor acts like it's not even there when the BJT transistor is open (You will notice that the voltage is 0, hence the BJT acts as open switch). Any idea why is that?

    Here's the video that I watched on youtube to do this project, it is pretty much what I wanted to do:


    Thank you,
     

    Attached Files:

  2. Alec_t

    Alec_t

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    856
    Jul 7, 2015
    Yes. The JFET would need a gate voltage below the ground rail to turn it off.
     
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  3. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Definitely. A MOSFET, as used in the video typically is a so called enhancement device. Off without Vgs, On with Vgs > Vtreshold.
    A JFET is a depletion device, On without Vgs, Off with Vgs <Vthreshold.

    Which voltage do you mean? The voltage at the base of Q1? If so, why should it be influenced by R1? R1 limits the current through Q1's collector-emitter path. There is no feedback to the base.
    On the contrary: R 1 acts like it is there, becaus otherwise the gate-source voltage of Q2 wouldn't be a neat 9 V with Q1 open.


    By the way: it is better to refer to circuit elements by their name, not their value. In a simple circuit like this the components may be identifiable, no longer so when multiple instances of components with the same value are present. Names are unique and unambiguos (or at least should be).
     
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  4. Lerior

    Lerior

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    Jan 20, 2018
    Oh sorry, I meant why isn't there any voltage drop at the 10k resistor (R1 in the simulated circuit I attached) once Q1 is open. But thanks, you did answer my question! I really appreciate your help.
     
  5. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    because there is no gate currrent into the MOSFET.
     
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