AC Emitter resistance.

  • #1

Main Question or Discussion Point

Hello experts!!

AC emitter resistance is given as,
rE'=25mV/iE

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

Thanks in advance.
 

Answers and Replies

  • #2
1,482
3
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:
  • #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?
 
  • #4
948
2
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?
Thermal voltage temperature is at 300 K, so that is about 27°C (room temperature)
 
Last edited:
  • #5
Thermal temperature is at 300 K, so that is about 27°C (room temperature)
OK. Thanks a lot. :approve:
 

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