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Ebers-Moll Equation --- Transistors

Discussion in 'Electronic Basics' started by A.J., May 10, 2004.

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  1. A.J.

    A.J. Guest

    Hello All,

    I have been reading an electronics book that shows the following

    | +10V
    R1(10k) O
    Vin /
    --------| Q1

    If The base Bias is set so 1mA flows through R1 it will have 10V
    dropped accross. It then goes on to say that an 8 Degree rise in
    temperature will cause the voltage accross R1 to drop to 0.2V. I
    can't say I really unbderstand how this is working! I know that Vbe
    will drop -2.1mV/Degree but how does this affect the current flowing
    through the collector? Does someone with a better undersatnding of
    the Ebers-Moll equation know exactly what is happening here and why
    the 8 degree rise will cause this problem?

    Thanks so much for any assistance offered?


  2. Mantra

    Mantra Guest

    The Ebers-Moll has an exponential term (exp(Vbe/Vfo) - 1) for the Ie
    and Ic values, right. So fortunately this is high school algebra: the
    Vfo is proportional to kT/q so as the temperature increases, Vfo goes
    up also. This means (assuming linear operation where you can ignore
    the reverse terms) you will get a shift down in collector/emitter
    currents for constant Vbe bias (exponential argument gets smaller), or
    equivalently, you will get a shift up in Vbe for constant Ic (the
    exponential argument must remain constant). So if Vbe is constant, Ic
    decreases with temperature increase. There is a KVL loop: Vcc (=20V) -
    Ic*Rc - Vce = 0, so then the drop across Rc decreases as Ic decreases
    (and the difference is made up by Vce which is usually unknown except
    by this KVL requirement).

    As a side note, since the circuit has no current feedback, there is
    nothing to prevent this from causing the amplifier to debias. An
    emitter degeneration resistor (aka current feedback) will cause base
    voltage to be divided across Vbe and Re, thus an Ie increase causes a
    drop in Vbe due to the drop across Re (KVL: Vb - Vbe - Ie*Re = 0).
    This feedback to lower Vbe then reduces the Ie based on the Ebers-Moll
    equations (hence negative feedback). Some of the temperature effects
    can thus be eliminated.

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