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

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Homework Help Overview

The discussion revolves around the maximum input voltage for valid analysis in the BJT hybrid pi model, focusing on the role of various components such as R_E and R_pi in AC analysis. Participants are trying to understand the conditions under which the model remains valid, particularly in terms of active mode operation and small signal conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the neglect of R_E in the hybrid pi model and its implications for the analysis. There are inquiries about the significance of v_be(max) and how it relates to the maximum permissible input voltage. Some participants are also exploring the calculation of R_in and its connection to the input voltage.

Discussion Status

The discussion is ongoing, with various participants raising questions and attempting to clarify concepts. Some guidance has been provided regarding the role of capacitors in the circuit and the implications of R_E on DC operating conditions. However, there is no explicit consensus on the interpretations of v_be(max) or the maximum input voltage.

Contextual Notes

Participants are working within the constraints of the problem, which specifies limits on input voltage to avoid distortion and maintain valid analysis conditions. The discussion highlights the importance of small signal conditions and the effects of component values on circuit behavior.

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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