Simple DC Circuit Analysis with Transistor

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Discussion Overview

The discussion revolves around the analysis of a simple DC circuit involving a transistor, focusing on the application of Kirchhoff's Voltage Law (KVL) and Ohm's Law. Participants are attempting to solve a homework problem related to the circuit's behavior and current values.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a series of equations based on KVL for different loops in the circuit, calculating various currents and voltages.
  • Another participant questions the sign convention used in the equations, suggesting that the voltage across a resistor should be represented differently.
  • Some participants express confusion over the values of current and voltage, noting discrepancies with Ohm's Law.
  • A participant mentions that the given answers do not adhere to Ohm's Law, indicating a potential error in the problem statement or solution.
  • One participant later acknowledges a correction from the author’s errata, confirming a previously disputed current value.
  • There is a request for clarification on the working of another participant's calculations.
  • Multiple participants express uncertainty about the correct application of KVL in the context of the circuit's passive components.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct application of KVL and the resulting current values, with multiple competing views and corrections being proposed throughout the discussion.

Contextual Notes

There are unresolved issues regarding the assumptions made in the calculations, particularly concerning the sign conventions and the application of Ohm's Law in the context of the circuit's configuration.

Who May Find This Useful

Students and individuals interested in circuit analysis, particularly those studying DC circuits and the application of KVL and Ohm's Law in transistor circuits.

Dan97
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Homework Statement


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Homework Equations



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The Attempt at a Solution


All loops are clockwise;

KVL Loop around 1V, 120kOhm and VBE (I1 is assigned to the loop):

-1V + 120kOhm * I1 + VBE = 0
I1 = IB = 0.3V / 120kOhm = 2.5 * 10-6 A;

Loop around 10kOhm, vCE (I2 is assigned to the loop);

KVL Loop around 10kOhm, vo, and 20V (I3 is assigned to the loop):

10kOhm * I3 + 20V - vo = 0
vo = 10kOhm * (I2 - I3)
10kOhm * I3 +20V + (I3 - I2) * 10kOhm = 0 (1);

At the node connecting the collector part of the transistor, Io and 10kOhm:

I3 + (I2 - I3) + IC = 0
since Io = I2 - I3

IC = B * IB = 80 * 2.5 * 10-6 = 2 * 10-4 A

I2 = -2 * 10-4 A (2);

Subbing (2) into (1):

10kOhm * I3 +20V + (I3 + 2 * 10-4 A) * 10kOhm = 0

20kOhm * I3 = - 20V - 2 * 10-4 A * 10kOhm

I3 = -22 / 20k A = -1.1 * 10-3

I2 - I3 = 9 * 10 -4 A ! = 6 * 10 -4 WRONG

And I've been stuck redoing the same equations for a day now... Would be nice if someone could help find the problem in my "solution" so I can finally move on haha.
 
Last edited:
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Hi Dan97. :welcome:

clockwise Σ of voltages: vo + 10k × I3 + –20 = 0

EDITED
 
Last edited:
Hi @NascentOxygen ,

If loop 3 is clockwise, and it goes through the passive configuration, then it should be +20V - vo +I3 * 10kOhm?

I'm not sure what you're trying to say...
 
Dan97 said:
Hi @NascentOxygen ,

If loop 3 is clockwise, and it goes through the passive configuration, then it should be +20V - vo +I3 * 10kOhm?

I'm not sure what you're trying to say...
The given values of Vo and Io do not follow Ohm's law.
I am also getting Io=9*10-4 A.
 
Hi @cnh1995 ,

Could you show your working (if it's not too much of a hassle)? :)
 
Dan97 said:
Hi @NascentOxygen ,

If loop 3 is clockwise, and it goes through the passive configuration, then it should be +20V - vo +I3 * 10kOhm?

I'm not sure what you're trying to say...
Sorry, typo. Corrected.

Following your path, it will be +20V - vo – I3 * 10k = 0

Current through a resistor (in fact, through any element that's not a source) flows from its more-positive end to its less-positive end. If you draw an arrow showing this PD across the resistor, the arrow's tail is at the less-positive end and the arrow head is at the more-positive end.
 
@NascentOxygen ,

Why would it be - I3 * 10kOhm? Isn't that breaking Ohm's Law?
 
The given answers are 12V and 600uA, which are incorrect. They don't follow Ohm's law.
Dan97 said:
Hi @cnh1995 ,

Could you show your working (if it's not too much of a hassle)? :)
Collector current Ic=2x10-4A.

Using KVL you can write,
104*Io=20-104*(Io+2*10-4).

Solving this for Io, we get Io=9×10-4 A.
 
@cnh1995 ,

Oh actually I've made a big fuss for nothing! I've just look at the author's errata and the answer is indeed 9 * 10 -4!

1zlddt0.png
 
  • #10
what book is this anyway?
 
  • #11
e0ne199 said:
what book is this anyway?

Alexander's Fundamentals of Electric Circuits
 

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