# Nodal Analysis- direction of currents flow

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

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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?

sophiecentaur
Gold Member
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.

- 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.

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