Collector Current through BJT Circuit

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SUMMARY

The discussion centers on calculating the collector current (IL) through a BJT circuit using Thevenin's theorem and transistor equations. The user initially calculated IL as 4.15mA, while the correct value is 4.3mA, as indicated by their tutor. The discrepancy arises from the assumption that the base current (Ib) is negligible, which affects the voltage across the emitter resistor. The user learns that the correct approach involves considering the base current's impact on the emitter voltage.

PREREQUISITES
  • Understanding of BJT operation and characteristics
  • Familiarity with Thevenin's theorem
  • Knowledge of basic circuit analysis techniques
  • Proficiency in using transistor equations (e.g., Vbe, Beta)
NEXT STEPS
  • Study the impact of base current on BJT circuit calculations
  • Learn about Thevenin equivalent circuits in more complex scenarios
  • Explore advanced transistor biasing techniques
  • Review practical applications of BJT circuits in amplifier design
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or designing BJT circuits will benefit from this discussion.

curiousguy23
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Homework Statement



Consider the following circuit
http://img109.imageshack.us/img109/80/circuit1.jpg

IL, current through R3 (unknown resistance must be found)


Homework Equations


Vth=(Vcc/Rtotal)*R1 (V tThevenin)

Rth=R1//R2 (R Thevenin)

Vbe= 0.7 volts (silicon diode)
Beta (transistor constant)=100
Vcc=10v
IL=Beta*Ib
Ie=Ib + IL

The Attempt at a Solution



I used the following Equation to try and obtain an answer.

Vcc= (Ib*Rth)+ (Ie*R4) + Vbe + Vth

Vth=5v
Rth=2.5k Ohms

Substituting gives:

10=(IL/Beta)*(2.5k) + 1000*(IL*((Beta +1)/Beta)) + 0.7 + 5

This yields an answer of IL=4.15mA.

My tutor said this is a wrong answer, the correct one is 4.3mA
What is wrong in my reasoning/Working?
 
Last edited by a moderator:
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I like your answer better.

The tutor has just taken 0.7 V from the 5 V at the center of the two 5K resistors and said this must be the voltage across the emitter resistor.

But the base current pulls this voltage down a little and that makes the difference between your answers.
 
THanks,

I just realized what he did,


He assumed Ie=IL and Ib=0 since it is very small.
 

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