question about transistor

Discussion in 'General Electronics Chat' started by TAKYMOUNIR, Sep 3, 2011.

  1. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
    351
    1
    i was doing troubleshooting for circuits and the transistor was bad because Vbe was about 3vd insteade of being .7v my question is why when Vbe is high or very low this mean that the transistor is bad
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Transistors start conducting meaningful amounts of current when Vbe reaches around 0.6v, and by the time Vbe is around 1V or so, it's at around 60%-75% of it's maximum rated collector current.

    Basically, the base-emitter junction acts like a forward-biased silicon diode. They usually don't start conducting a meaningful amount of current until around 0.55v or so, and they're near or at their maximum current at ~2v.

    So, if a silicon transistor that is supposed to read between ~0.6v and ~2v from base to emitter reads outside of that range, it's either turned off deliberately (no current flowing through the base current limiting resistor), shorted if lower than ~0.6v, or open if higher than ~1.5v.
     
    TAKYMOUNIR likes this.
  3. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
    351
    1
    please can you explain phsically why when vbe is high the transistor is open
    thanks very much
     
  4. newb7777

    New Member

    Mar 6, 2011
    2
    1
    We are talking BJT.

    When Vbe is ~0.7V for a SILICON Transistor is ON state; (GERMINIUM will be lower)
    Depending on the amount of current through the base usually limited by a BASE RESISTOR (Rb), the transistor could be fully turned ON thereby acting like a turned ON switch or
    acting like an amplifier in the case where a controlled amount of current is flowing though it.
    The relationship between base and collector current is given by the formula Ic=Ib*hfe valid for NPN BJT (note: hfe=beta from datasheet or vice versa beta=hfe)

    Now if Vbe is far greater than ~0.7V then the base emitter junction is not a diode anymore!!!!!

    Now if Vbe is far less than ~0.7V then transistor is shorted on the base emitter junction!!!!!

    Hope that helps....... alternativly, you should try it on a breadboard take measurements with voltmeter then the concepts will be clear and you will be fully convinced. Would be good if you could jive it with some maths and make some sense of the datasheet for the specific device.
     
    Last edited: Sep 4, 2011
    TAKYMOUNIR likes this.
  5. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    All good points.

    Different makes of transistors act differently. If I were to see 1V or more on a 2N2222A BE (base emitter) I would declare the transistor bad. If I saw it on a 2N3055 I would take it for granted.

    Both are NPN BJT transistors, but the 2N3055 is designed for much higher currents, so the excessive BE drop is normal, for it. Most NPN transistors are much lower current, so 0.7 is the norm (for a silicon diode).

    As with ICs, transistors have datasheets. Datasheets are always your friend.

    As described, Vbe is a diode, it acts like a diode, and it drops current like a diode. If the voltage there is high then the diode junction is damaged or open.
     
    TAKYMOUNIR likes this.
Loading...