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A 2N3906 Question this time

Discussion in 'Electronic Basics' started by Newbie, Feb 5, 2004.

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  1. Newbie

    Newbie Guest

    Hello All

    Thanks for all who replied to my 2N3904 question earlier. I
    have one about a 3906 this time. The set up is below

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

    When I apply voltage to this circuit the LED does not come on and also
    the 2N3906 seems to burn out. I'm pretty sure it does because the 0.7
    volt diode drop no longer apears across the BC and BE junctions when I
    take it out of the circuit and test it. If I put a 220 ohm resistor in
    the base the circuit does seem to work.

    I thought the base was pretty high impedance and the current going out
    through it would be limited internally as most would go through the
    collector. It would seem I am mistaken. Could anyone shed any light on
    this? Would a 3904 NPN burn out in a similar situation (base connected
    to ground with no resistor and 5V on the collector. I have tried this
    and it did not seem to burn out - admittedly the LED+220ohm was before
    the collector so this probably limited the current.

    Is the LED+220ohm resistor after the collector a good way to configure
    a PNP transistor to act as a switch (as in the diagram)? Or should I
    have it close to the supply before the emitter?

    Many Thanks - I would value your opinions
  2. Why? It's a diode! It's impedance is determined by what current you allow
    thru it.
    No, the 3904 is an NPN and grounding the base amounts to no on bias.
    (The transistor is off!)
    Use a 39K ohm from the base to ground in your circuit above.
    | __O Thomas C. Sefranek
    |_-\<,_ Amateur Radio Operator: WA1RHP
    (*)/ (*) Bicycle mobile on 145.41, 448.625 MHz
  3. The emitter-base junction is a conducting diode in this schematic. It used
    to have about 0.7V on its terminals. If you push 5V on it th cuurent will go
    skihigh and destroy the transistor.
    That's still to low. You may not destroy the transistor anymore, but the led
    will not light.
    A conducting diode has a very low impedance. Nevertheless, the base input
    of a transistor can have a pretty high impedance due to the feedback caused
    by the load on the emitter. In this circuit you have no load at all.
    A blocking diode has a high impedance. Moreover, a voltage <0.6V will make
    the transistor switched off as well. It does not harm but you will have no
    light either.
    I think the circuit you have in your mind is known an emitter-follower. So
    you have to move the load (LED+220ohm) to the emitter. Other side of the
    load to +5V and collector to ground. This way the load will limit the
    emitter-base current and the LED will light. The LED will be switched off
    when the base voltage rises above about 4.4V.

    Using a NPN transistor the load should be connected to the emitter and the
    ground, collector to +5V. A grounded base will keep the LED off, a plus 5V
    on it will light it.
  4. You have turned the NPN circuit upside down in every way but one. The
    base to emitter diode junction is now connected directly across a
    supply with nothing to set its current except for the supply
    capability. The base gets connected to the same voltage as the
    emitter to turn the transistor off, same as it was with the NPN
    version. To turn the transistor on, you have to bring the base
    emitter junction into forward bias. In the PNP case, this means
    getting the base to be about .6 volts more negative than the emitter,
    with enough current passing through this junction that beta times that
    current is at least equal to the desired collector current. Providing
    a pull down current of a milliamp or so should do this. So the base
    should be connected to ground only through something like a 4.7k
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