Single Transistor Overall Current Gain

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SUMMARY

The discussion centers on the equation for current gain in a single transistor circuit, specifically the formula Ai = α(R4/(R4 + Ri)). Participants clarify that the relationship between output current (Io) and emitter current (Ie) is defined by the parameter α, while the relationship between emitter current (Ie) and input current (Ii) is represented by the resistor ratio R4/(R4 + Ri). The confusion arises from the transition between these relationships, prompting further exploration of the circuit's behavior.

PREREQUISITES
  • Understanding of transistor operation and parameters, specifically current gain (α).
  • Familiarity with circuit analysis involving resistors (R4 and Ri).
  • Knowledge of basic electrical engineering principles, particularly Ohm's Law.
  • Ability to interpret and manipulate equations related to current and voltage in circuits.
NEXT STEPS
  • Study the derivation of the current gain formula Ai = α(R4/(R4 + Ri)).
  • Learn about the role of resistors in transistor circuits and their impact on current flow.
  • Explore the concept of input and output currents in transistor configurations.
  • Investigate the effects of varying resistor values on overall circuit performance.
USEFUL FOR

Electrical engineering students, circuit designers, and anyone seeking to deepen their understanding of transistor current gain and circuit analysis.

hisotaso
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Homework Statement


THe problem statement and given solutions are in this thread:

https://www.physicsforums.com/showthread.php?t=694054

I do not understand the equation for current gain:


Homework Equations


Ai = \alpha\frac{R4}{R4 + Ri}


The Attempt at a Solution


I have pages of work on this, but basically what I know is \frac{Io}{Ie} = \alpha

then \frac{Ie}{Ii} must equal \frac{R4}{R4 + Ri}, I don't see that.
 
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Why have you started a second thread for your own question?
 
I started a second thread because I thought I had figured it out. I stated so in that thread, and thus expected no one would reply. I asked another question about this same circuit in a third thread.
 

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