Intrinsic resistance of a transistor

AI Thread Summary
The discussion centers on understanding the intrinsic emitter resistance of bipolar transistors. The original poster seeks clarification on how to calculate this resistance, referencing their professor's explanation involving changes in collector and emitter voltages. A formula is provided, indicating that intrinsic emitter resistance (re) can be calculated using thermal voltage (Vt) and emitter current (Ie). Another participant suggests that the resistance can also be approximated by sweeping the base voltage and observing the resulting changes in collector and emitter voltages. The conversation highlights the need for a clearer methodology to determine intrinsic emitter resistance in practical applications.
adiman1984
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In my circuits class we were talking about bipolar transistors and our prof. mentioned that every bipolar resistor has an intrinsic emitter resistor. So, I tried modeling a couple of circuits in SPICE to see if I can replace the transistor with another circuit (voltage source, current source, and the intrinsic resistor). However, I never got to that point, because I could not find the correct value of this resistor. In class he just said that we need the change in collector voltage (delta Vc) and change in emitter voltage (delta Ve) to figure it out. That doesn't make any sense to me. Don't you need a current to find that?
Does anyone have an explanation of how to find the intrinsic emitter resistance of a bipolar transistor?

Thanks
 
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re = Vt/Ie.

Vt = kT/q; k = 1.38e-23 joule/Kelvin, T = temperature in Kelvin, q = 1.602e-19 coulomb.

Ie = dc or bias value of emitter current

Claude
 
Claude,

Thanks for your answer. This will help me to double check if I'm right with my approximation. However, there should be another way to get to the intrinsic emitter resistance.
According to the attached picture, the resistance can be approximated with delta Vc and delta Ve (the deltas are obtained by sweeping Vb over a certain range).
Any ideas how this is done?

Thanks
 
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