Thevenin Potential Homework: Simplifying a BJT Amp Frequency Model

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Discussion Overview

The discussion revolves around simplifying the frequency model of a BJT amplifier, specifically focusing on the Thevenin potential across a resistor in the circuit. Participants are examining discrepancies between their calculations and those presented in a textbook example.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion over the textbook's formula for the Thevenin potential, noting a small difference in calculated values compared to their own results.
  • Another participant suggests replacing the source and resistors R1 and R3 with their Thevenin equivalent to derive the textbook result, indicating that this involves a source of Vsig.R3/(R1+R3) and an impedance of R1 // R3.
  • A later reply confirms that using potential dividers with loading considered yields the same result as the Thevenin equivalent for the specified resistor values.
  • One participant expresses relief at finally understanding how the textbook arrived at its result after reviewing the calculations.

Areas of Agreement / Disagreement

Participants appear to be working towards understanding the same problem, but there is no explicit consensus on the correctness of the differing methods or results. The discussion reflects a mix of agreement on the approach but uncertainty regarding the exact calculations.

Contextual Notes

Participants mention different methods of analysis, including potential dividers and Thevenin equivalents, but do not resolve the discrepancies in their results. The discussion is limited by the specific values chosen for the resistors and the assumptions made in the calculations.

sandy.bridge
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Homework Statement


Hey guys,

There is an example in the textbook regarding simplifying the frequency model of a BJT amplifier. It ends up determining the potential across the highlighted resistor I attached. The formula that the textbook is rather different than the one I determined to be the Thevenin, and hence I am looking for some sort of explanation. I plugged in some values to determine if they were equivalent, and they were close, but not exact. I believe there was approximatedly 0.0016 V difference for the particular values I had chosen. The textbook gives
V'_{sig}=V_{sig}\frac{R_3}{R_3+R_{1}}\frac{R_{4}}{R_{4}+R_2+(R_{1}//R_3)}

I do not understand how they ended up with R_1//R_3. I analyzed the circuit two different ways. I used potential dividers and I also determined the current passing through that leg. Each time my results were vastly different; however, they do give the same results upon evaluation.
 

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To get the textbook result, replace the source and R1 and R3 with the Thevenin equivalent, giving a source of Vsig.R3/(R1+R3) and having impedance R1 // R3

This Thevenin equivalent is then loaded by the series resistances R2 and R4.

If I nominate some values, e.g.,
R1=4
R2=2
R3=5
R4=3

That expression gives Vsig‘ = 0.2307 Vsig

I'll recheck.
 
Last edited:
I checked using potential dividers with loading taken into account, the result with the above figures is also 0.2307
 
Finally! I was staring at that for a good hour attempting to determine how they got that. Thanks a bunch!
 

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