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2N3904 Beginner Question

Discussion in 'Electronic Basics' started by Newbie, Feb 4, 2004.

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  1. Newbie

    Newbie Guest

    Hello All

    I'm learning about transistors and have a simple question to ask. I
    have set up a 2N3904 as below

    --- 5V
    V LED
    / 220 ohm
    | /
    220 |
    5V -/\/---|
    | Gnd

    The circuit works fine and the LED lights up. The question I have is
    when I take the 3904 out, reverse it, and put it back into the circuit
    so that the emitter is where the collector was and vice versa the
    circuit still works (the base does not change). The LED lights up,
    grounding the base shuts it off as you would expect. My grasp of the
    theory is not too good yet but why should it do this? I thought maybe
    it had something to do with the Emitter base breakdown voltage (listed
    at 6v) but I dropped the supply voltage to 3 volts and the effect
    persisted. Replacing the 220 ohm base resistor with a 47K resistor
    makes the circuit work in the first configuration but not in the

    Where is the current going? And how is it getting throught the
    transistor? Is this expected behaviour if a transistor is reversed? I
    would appreciate any light that anyone might shed on this.

    Many Thanks
  2. John Larkin

    John Larkin Guest

    If you swap the emitter and collector of an NPN transistor, it's still
    an NPN structure. Some very old germanium transistors were truly
    symmetric... emitter and collector were interchangable.

    A modern silicon transistor is not symmetric, but it still has
    "reverse beta", finite current gain if e-c are swapped. This beta is
    very low, on the order of 1, as opposed to normal beta (base-collector
    current gain) that is more like 100.

  3. The emitter and collector are both n type semiconductor and they lay
    on opposite sides of the P type base region. Either can act as
    emitter or collector. They have been doped differently (emitter
    highly doped, collector lightly doped, to optimize the current gain
    and breakdown voltage in one of these orientations. But it will work
    with poorer specs in the other orientation.
    That rating implies that if your supply was more than 6 volts, the
    emitter acting as collector would not be able to turn off and hold
    back the full supply voltage. The base to collector junction has
    something like 40 volts of reverse breakdown capability.

    You just proved that reverse beta (current gain) is not as great as
    forward beta. The high dopant concentration in the emitter is one of
    the big factors that make the forward beta high. When you use the
    lightly doped collector as emitter, most of the base current is holes
    from the base crossing to the emitter (the functional emitter), rather
    than electrons that would have been injected into the thin base layer
    by a highly doped emitter. It is these electrons dumped into the P
    type base layer that wonder into the collector region and get swept
    out by the collector positive voltage that gives the transistor
    current gain.
    I hope I have gotten you started.
  4. Newbie

    Newbie Guest

    Thank you for your kind and well thought out replys to my question.
    You have cleared up the issue considerably. I will go back to the
    theory and see if I can clarify in my own mind exactly what the
    electrons and holes are doing and where they are going.

    Looking at the diagrams your explanations now make a good deal of
    sense. One wonders why the textbooks make no mention of this. I guess
    a reversal of the supply voltage through the transistor just doesn't
    happen in "real life".

    Many thanks
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