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AC Emitter resistance.

  1. Nov 19, 2011 #1
    Hello experts!!

    AC emitter resistance is given as,

    My question is that what does this 25mV tells us? and where does it come from?

    Thanks in advance.
  2. jcsd
  3. Nov 19, 2011 #2
    Vt = 25 mV is a thermal voltage at room temperature, about 20 degrees Celsius I believe. This value changes with temperature.

    It comes from semiconductor physics where you try to describe a PN junctions diode, you arrive at Shockley's equation which describes its IV characteristics. It turns that the Shockley's equation is highly non-linear.

    BJT transistors are composed of two such diodes which also behave non-linearly. So ultimately, when you try to describe the behavior of BJT transistor amplifier, it will be non-linear. But because it's difficult to deal with non-linear equations, the diode equations have been linearized by making some assumptions into what is called the small-signal model. As a results a couple of new parameters come into play, such as r_e, r_pi, g_m.

    When you work this out, Vt makes its way into the linearized model and that's why see it.

    For more info check out: Sedra/Smith.
    Last edited: Nov 19, 2011
  4. Nov 20, 2011 #3
    Thanks for reply.

    You mean it is thermal voltage at room temperature and it varies if temperature varies. Am I right?
    It is the voltage assumed on 20°C? Am I right?
  5. Nov 20, 2011 #4
    Thermal voltage temperature is at 300 K, so that is about 27°C (room temperature)
    Last edited: Nov 20, 2011
  6. Nov 20, 2011 #5
    OK. Thanks a lot. :approve:
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