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PIC and Transistor

Discussion in 'Electronic Basics' started by [email protected], Feb 28, 2006.

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  1. Guest

    Basically what I want to do is use the output pins of my PIC to
    activate some transistors. If i understand correctly, activating a PIC
    pin essentially grounds it, so an NPN won't work. I am assuming then I
    need to use a PNP, so when it is grounded, it activates the "switch".

    Now, does it matter what is going through the collector and emitter? I
    would like to have a Positive voltage sitting on one side of the
    transistor, and the other side of the transistor would go to an ECU in
    a vehicle. Under most circumstances, the transistor will be ungrounded,
    which should mean that there is no flow through the transistor, and no
    signal to the ECU (that is what I want). When instructed, the PIC will
    ground the base, allowing current to flow through the transistor and
    sending the positive signal from the other side of the transistor to
    travel to the ECU.
    Maybe I am missing something, but I cannot get this work no matter
    which I use PNP, or NPN.
    Is anyone able to shed some light on this for me?
     
  2. Noway2

    Noway2 Guest

    Without referencing a particular device or pin, generally speaking, you
    can drive the output of the PIC IO either high or low which means that
    you can use an NPN transistor. What you will need to consider, though,
    is how much base current you will need to flow in order to drive the
    transistor sufficiently and you need to be sure that you don't exceed
    the limits of the PIC's drive capability. If it turns out that the
    current drive is insufficient, you can use a FET instead and if
    necesary drive an NPN or PNP transistor with the FET.

    Again, generally speaking, using an NPN or PNP device determines wether
    you use it as a "source" or "sink" device when used as a switch. In
    this case, by source or sink, I mean is it switching on the power side
    or the ground side of the circuit (load).

    One thing to keep in mind if you do decide to use a PNP transistor, if
    you ground the base by turning on a PIC IO you could very well apply an
    extreme Vbe that could damage or destroy the device.
     
  3. Chris

    Chris Guest

    Transistors can be used as switches. You want a switch that connects
    your ECU to GND when it's active, and disconnects when it's inactive.
    You can use an NPN transistor to do this, as shown (view in fixed font
    or M$ Notepad):

    |
    | ------. .----o
    | | |
    | | |
    | | |
    | | |
    | | ___ |/
    | o-|___|-o-|
    | PIC | | |>
    | | .-. |
    | | | | ===
    | | | | GND
    | | '-'
    | | |
    | | ===
    | | GND
    | |
    | |
    | ------'
    |
    (created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

    I would assume your transistor will sink less than 20mA when on, so use
    2.2K resistors for R, and use a small-signal NPN like a 2N3904 or
    2N4401 for the transistor. These are commonly available.

    To turn the transistor ON, set the PIC output to "1".

    Good luck
    Chris
     
  4. No. You get to select which pins act as outputs and which act as
    inputs. If output is selected, you can order the pin to a logic high
    or logic low state. The output circuit has just a bit more current
    capability in the low state (sinking current) than it has in the high
    state (sourcing current), but the difference is not all that great.
    The easiest kind of transistor to drive is a logic level, N channel
    mosfet. You connect the gate to the active-high output pin and the
    source to ground. The drain connects to a load which is always
    connected to a positive supply. If the load is inductive, you may
    need a diode across it, to limit the turn off voltage.
    In this case, you need a resistor between the output pin and the base,
    to limit the output current to what can safely pass through that pin.
    And the emitter of the PNP transistor must be tied to the same
    supply rail as the PIC.
    Not to the PIC.
    The transistor will pull up to almost the supply voltage for the PIC,
    probably 5 volts.
    Simulate the ECU by connecting a 1k resistor between the collector and
    ground.
     
  5. Guest

    <No. You get to select which pins act as outputs and which act as
    <inputs.

    Yes, I know, sorry...I do have them selected as outputs.

    Just for clarification, this is what I am trying to do.

    http://wotid.com/tls/images/ssatre/transistor_tre.jpg

    The exception is that the signal coming through the gearbox, will be
    replace with the output of my PIC.

    Hope that helps.
     
  6. Guest

    <No. You get to select which pins act as outputs and which act as
    <inputs.

    Yes, I know, sorry...I do have them selected as outputs.

    Just for clarification, this is what I am trying to do.

    http://wotid.com/tls/images/ssatre/transistor_tre.jpg

    The exception is that the signal coming through the gearbox, will be
    replace with the output of my PIC.

    Hope that helps.
     
  7. John Fields

    John Fields Guest

    ---
    Yes, according to your later post, (the one with the link to the
    drawing) the problem I see is that you're expecting the PIC IO
    driving "N" to at least sink the incandescent lamp neutral
    indicator's current (and there's also funny going on with the
    side-stand relay as well) while at the same time trying to pull the
    transistor's input low. That won't happen.

    Neglecting all of the other IOs, you should wire up the "N" function
    like this:

    View in Courier:

    +12
    |
    +---+---+
    +5 |NI /SSR|
    | +---+---+
    [86] |
    | C
    +-----------B NPN
    | | E
    _ | C |
    N |---+--[860]----B NPN GND
    | | E
    PIC | | |
    | GND
    |
    |
    +--->To 47k resistor
     
  8. A little. Do you want the PIC to approximate the selector switch in
    the gear box (7 separate connections that each pull down to ground, in
    turn)? If so, I would use 7 mosfets. 6 of them could be small, but
    the one that represents the neutral contact has to also drive the
    neutral lamp, so it would be a bigger device.
     
  9. Guest

    actually I only need to approximate 1,2 and 3. Neutral and 4,5 and 6
    will actually turn the PIC pin the opposite way and disable the
    transistor.
    Make sense?
     
  10. John Fields

    John Fields Guest

     
  11. Guest

    <Do you want the PIC to approximate the selector switch in the gear
    box
    <(7 separate connections that each pull down to ground, in turn)?

    actually I only need to approximate 1,2 and 3. Neutral and 4,5 and 6
    will actually turn the PIC pin the opposite way and disable the
    transistor.
    Make sense?
     
  12. Guest

    Sorry

    <Do you want the PIC to approximate the selector switch in the gear box

    <(7 separate connections that each pull down to ground, in turn)?

    actually I only need to approximate 1,2 and 3. Neutral and 4,5 and 6
    will actually turn the PIC pin the opposite way and disable the
    transistor.
    Make sense?
     
  13. Nope. Please start back at the beginning and describe what circuit
    and signals you are dealing with. What part of the bike schematic is
    included and what additional stuff are you trying to add. I can't
    make any sense of your last several posts, put altogether.
     
  14. Guest

    Got it working, trial and error.
     
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