Engineering Analog electronics (BJT circuit analysis)

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The discussion centers on calculating the DC voltage at the collector of a BJT circuit with a beta of 100 and a large ro. The user initially calculates IB and IC based on the emitter current (IE) of 0.5mA, leading to an implausible IC value of 5.05 A. Participants suggest correcting the equation for IE and considering V_BE as approximately 0.7 volts for accurate analysis. Additionally, they note that capacitors block steady-state current, acting as open circuits in DC analysis. The conversation emphasizes the importance of proper circuit analysis techniques, including the potential use of a T-model.
adamaero
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Homework Statement


upload_2016-12-9_20-6-1.png

For beta = 100 and assuming ro is very large, what is the DC voltage at the collector?

Homework Equations


IB = IE(β+1)

IC = β*IB

The Attempt at a Solution


Since IE = 0.5mA, IB = 0.0505 A
Then, IC = 5.05 A (which doesn't feel right)
VC is the same potential across the 30kΩ load?
I tried doing KCL with those currents (to find the current thought the load resistor).
Should I not be analyzing the direct circuit--should I make an equivalent T-model?
 
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adamaero said:

Homework Statement


View attachment 110203
For beta = 100 and assuming ro is very large, what is the DC voltage at the collector?

Homework Equations


IB = IE(β+1)

IC = β*IB

The Attempt at a Solution


Since IE = 0.5mA, IB = 0.0505 A
Then, IC = 5.05 A (which doesn't feel right)
VC is the same potential across the 30kΩ load?
I tried doing KCL with those currents (to find the current thought the load resistor).
Should I not be analyzing the direct circuit--should I make an equivalent T-model?
Your equation should read ## I_E=I_B (\beta+1) ##. Try again. I think you also need to assume ## V_{BE}=.7 \, volts ## (approximately). One hint for you: The capacitors basically block any steady state current trying to go through them so the capacitor is an open circuit for the DC part of the analysis.
 
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