What is the maximum input voltage for valid analysis in the BJT hybrid pi model?

AI Thread Summary
The discussion focuses on the maximum input voltage for valid analysis in the BJT hybrid pi model, particularly concerning the role of R_E and the conditions for maintaining active mode and small signal operation. R_E is noted to set the DC operating conditions but is neglected in the AC analysis due to the bypass capacitor's effect. The maximum base-emitter voltage (v_be(max)) is approximately 10 mV, leading to a permissible input voltage (Vs) of 13.9 mV, which is crucial for avoiding harmonic distortion. The circuit's design lacks negative feedback, exacerbating distortion issues if Vs exceeds the specified limit. Understanding these parameters is essential for accurate BJT amplifier analysis.
theBEAST
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Homework Statement


Here is the example along with the solution:
dRXPP9d.jpg


What I don't get is what happened to the R_E value. So in this model does it neglect this value? It seems like R_E could be anything and you would still get the same model.I have another question, for this same example, it asks what is the max v_s that the previous analysis is valid? There are two criteria in which we have to look at, i) we have to make sure it remains in active mode and ii) the small signal condition. I understand part i) which requires us to look at the allowable swing of the output value. What I don't get is part ii) and here is the solution:
JcrvxSx.png


It might be a bit blurry but the first part says v_be(max) approximately equals to 10 mV. Here v_be is voltage across the be junction due to the small signal I think?
 
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This is an AC analysis. The answer is the note, "Large C = short circuit, so no R_E in equivalent circuit". You choose the value for the capacitor to make its impedance negligible compared with the resistors in the circuit, at the frequencies the amplifier is designed for.

The value of R_E sets the DC operating conditions for the amplifier - specifically, the emitter current.
 
And what is the value or R_\pi? It is used in calculation of R_{in} and when you have R_{in} you can solve resistor divider R_s and R_{in} and get v_s from v_{be(max)} (equation in second picture).
 
mr_pavlo said:
And what is the value or R_\pi? It is used in calculation of R_{in} and when you have R_{in} you can solve resistor divider R_s and R_{in} and get v_s from v_{be(max)} (equation in second picture).

Woops, I forgot to mention that I don't understand what the v_be max is 10 mV?
 
Does anyone know why?
 
theBEAST said:
Does anyone know why?

This is the limit imposed by the problem. It results in a max. permissible Vs per the given formula ("ruling condition") of 13.9 mV.

This is a poor circuit in that it results in massive harmonic distortion if the input Vs isn't kept to a small value. Reason is lack of negative feedback due to infinite R_E bypass capacitor.
 
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