BJT Analysis Question: Solving for Vb with Voltage Divider Method

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SUMMARY

The discussion focuses on calculating the base voltage (Vb) in a BJT circuit using the voltage divider method. The formula provided is VB = (VCC * 30k) / (70k + 30k), resulting in a base voltage of 3V. Participants clarify that VB is equivalent to Vb and emphasize the importance of calculating Thevenin resistance for the resistors involved (30k and 70k). The conversation highlights the necessity of considering the loaded resistive voltage divider due to the transistor base current.

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Homework Statement


upload_2016-10-31_12-39-28.png


Homework Equations


KVL

The Attempt at a Solution


This is the solution given:
VB = (VCC*30k)/(70k+30k) = 3V
Why is a voltage divider being used to find Vb?
 
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Your question should be in the problem statement, not the attempt.

Are you sure that VB is the same as Vb? Looks to me like a determination of a Thevenin equivalent for the base voltage supply. Was there also a Thevenin resistance calculated (paralleling the 30 k and 70 k resistors)?
 
gneill said:
Your question should be in the problem statement, not the attempt.

Are you sure that VB is the same as Vb? Looks to me like a determination of a Thevenin equivalent for the base voltage supply. Was there also a Thevenin resistance calculated (paralleling the 30 k and 70 k resistors)?
VB is Vb.
Thevenin resistance was calculated for 30k and 70k.
 
What is your task? To find the DC voltage VB at the base?
Don`t forget that you have a loaded resistive voltage divider - loaded with the transistor base current.
 
This is what I imagine the author of the solution was going for:
upload_2016-11-1_6-26-6.png

With ##V_B## and ##R_B## being determined as the Thevenin equivalent of the base voltage divider.
 

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