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BJT Question

Discussion in 'General Electronics Discussion' started by Raven Luni, Jan 1, 2013.

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  1. Raven Luni

    Raven Luni

    Oct 15, 2011

    I'm wondering specifically about the collector emitter path in a BJT when it is 'turned on'. Is this seen as a resistance or is it more like a diode voltage drop?

    Why I'm asking: I have a bunch of cheap digital thermometers that I want to use as part of a thermostat system. Normally they'd be independent of any other circuitry (used for display only) but I've noticed a feature that I can make use of. Open and short conditions on the sensor (a thermistor) cause 'Lo' and 'Hi' to be displayed on the LCD respectively which could be used to report the 2 possible fault conditions (heater failure or thermostat failure). In order to do this, the thermistor path would have to go through at least 1 transistor during normal operation (likewise it would also be shorted by another transistor). Can this be treated as a normal conductance path or does it introduce other variables?
    Last edited: Jan 1, 2013
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    The voltage drop across the EC terminals can easily be lower than a diode drop.

    In fact you'll find that saturation means that the potential of the base is higher than the collector.

    If you think about this too much it will warp your mind because the flow of current is reversed and you see negative voltage drops. It is at this point that you succumb to either madness or quantum mechanics.
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    If you use a BJT as a switch (sounds like you're going that way), you can coonsider it as a series switch with either "infinite" resistance or almost short circuit with a saturation voltage of ~0.1V (depending on the transistor and operating conditions).

    Whether this works in your circuit depends on the voltages and currents involved in the normal operation of the circuit. Just try it.

    One note: A BJT will conduct DC well in one direction (e.g. NPN: Vce>0V), but will be lousy, if it conducts at all, in the other direction (e.g. NPN: Vce<0V). So carefully observe the polarities of voltages and currents involved in the unmodified circuit and use suitable transistors.
    Another technique would use AC-rated optical relays (one brand name is PhotoMOS, but there are several brands on the market).

    Happy new year

  4. Raven Luni

    Raven Luni

    Oct 15, 2011
    Cool - thanks
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