Nodal Analysis- direction of currents flow

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Discussion Overview

The discussion revolves around nodal analysis in a circuit involving a transistor and its biasing. Participants explore the direction of current flow through resistors and the application of Kirchhoff's Current Law (KCL) at a specific node, addressing potential misunderstandings regarding current directions and the behavior of the transistor.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the application of KCL at the left-hand node, specifically whether the current through the 3kΩ resistor should be considered as H + IB or H - IB.
  • Another participant asserts that the current through the 3kΩ resistor is indeed H + IB, challenging the first participant's reasoning.
  • A participant raises a question about whether the transistor should be treated as a current source in the nodal analysis.
  • There is a discussion about the direction of current flow through the 3kΩ and 700Ω resistors, with a focus on the influence of the voltage source's polarity.
  • One participant suggests rewriting the equation for H to incorporate the emitter current (IE) instead of the base current (IB) for clarity.
  • Another participant confirms that the base current in an NPN transistor flows into the base, aligning with the direction indicated in the provided diagram.

Areas of Agreement / Disagreement

Participants express differing views on the correct application of KCL and the direction of current flow through the resistors, indicating that multiple competing views remain unresolved.

Contextual Notes

Participants have not reached a consensus on the correct interpretation of current flow and the application of KCL in this circuit analysis. There are assumptions regarding the behavior of the transistor and the treatment of its components that remain unaddressed.

user5
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The figure shows a transistor with a bias circuit. If Ic=50IB and if VBE=0.7V, find VCE.
b4sra8.jpg

My question is:
The current through 700Ω resistor is H=[0.7+(51IB(250))]/700,
then when KCL is applied at the left hand node, the current flowing through the 3kΩ resistor suppose to be (according to hint),H+ IB. However isn't it suppose to be minus(H+IB)? (since the currents drop through resistors equals minus IB, it flows out of the node to the source)?

Thank you for any help you can provide.
 
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Well the correct answer is that the current flowing through the 3kΩ resistor is equal to H +Ib.
Why you think that H - Ib is correct answer ?
 
-For the nodal analysis (at the left-hand node), does the part towards the current IB flows regarded as resistor? (By that I mean shouldn't the transistor part be regarded as a current source?)

- Does the direction of flow through 3kΩ and 700Ω resistors is downwards due to the polarity of the voltage source 9V?

Thanks in advance.
 
With the battery 'that way up', current will be flowing from top (+) to bottom (-) of the circuit. Thus, the current through the 3k will split through the lower resistor and the base.
The base of a transistor does not appear as a constant current - just a high resistance, when an emitter (voltage feedback) resistor is in place; it is the collector which approximates to a constant current source.
I think I would re-write your equation
H=[0.7+(51IB(250))]/700

as H=[0.7+(IE(250))]/700, initially, and develop it from there (if you haven't already done so).
 
The base current in NPN transistor always flow into the base. Flow in direction show by the arrow in your diagram.

user5 said:
- Does the direction of flow through 3kΩ and 700Ω resistors is downwards due to the polarity of the voltage source 9V?
Yes, simply current flow from "+" to "-". And we have "+" at top and "-" at the bottom.
So when looking at schematics, most people get comfortable with the general notion that things flow top to bottom and left to right.
 
sophiecentaur : Thank you very much!

Jony130: Thank you very much!
 

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