0
$\begingroup$

After moving onto some of the practice textbook exercises in the section covering Kirchhoff's current law and Ohm's law, I came across a problem which evades my best attempts at solving it.

problem text

circuit diagram

Using Ohm's law, it was relatively easy to determine a equation involving voltage drops around the right loop:$$(1000\,\Omega)I_C+(1000\,\Omega)I_E=V_2\\(1000\,\Omega)(I_C+I_E)=V_2$$Similarly, it was simple to write an equation for the node where $I_B,I_E$ meet:$$I_B+I_C+150I_B=I_E\\151I_B+I_C=I_E$$... but given just $I_B=100\,\mu A=10^{-4}\,A$ it seems impossible to determine a numerical value for $I_C,I_E$.

Nevertheless it's straightforward to solve for $I_C,I_E$ in terms of $I_B,V_2$:$$I_C+I_E=\frac1{1000}V_2\\-I_C+I_E=151I_B\\\implies\begin{align*}2I_E&=151I_B+\frac1{1000}V_2\\2I_C&=-151I_B+\frac1{1000}V_2\end{align*}$$

Substituting $I_B=10^{-4}\,A$ yields $I_C=\frac1{20000}(10V_2-151),I_E=\frac1{20000}(10V_2+151)$

Am I approaching the problem correctly? Am I justified in ignoring the unlabeled voltage source bridging the two loops along with the labelled elements in the left one (i.e. $V_1,R_1,R_2$)? Am I treating the CCCS properly in writing my nodal equation?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

You are massively overthinking the problem. The collector current is given (by the diagram) to be 150x the base current. The sum of base and collector current has to flow through the emitter... That's all you need to solve this.

In particular, a current source will look to a circuit like "whatever resistance" it needs to be in order for the correct current to flow. So you can't apply Ohm's law to the right hand circuit - instead you have to assume that the current source is doing its job, and that a current of 150 $I_B$ is flowing in the collector.

Incidentally - as drawn this circuit has a fatal flaw: there is no return path for the base current... Normally you would have a common ground between the left and right circuits, where the base current could return (and which would allow the emitter current to be $151 I_B$. So really - as drawn, this circuit doesn't quite work...

Improve this answer
$\endgroup$
6
  • $\begingroup$ what about the CCCS? what do I do about the current source supplying $150I_B$? $\endgroup$
    – obataku
    Sep 5, 2014 at 3:39
  • $\begingroup$ Since $V_2$ is not given, you can ignore it. That's how transistors work. They provide "whatever resistance is needed" to make the simple equation work. At least in the DC model they are assuming here. We have to assume that $V_2$ is big enough that the voltage drops across the load resistors in the emitter and collector are less than the supply voltage. Absent information to the contrary that is a reasonable assumption. $V_2$, through the 1k resistors, is providing the current. $\endgroup$
    – Floris
    Sep 5, 2014 at 3:44
  • $\begingroup$ the simple equation being $I_B+I_C=I_E$? what other equation do I use to solve for $I_C, I_E$ in terms of $I_B$? the loop equation? $\endgroup$
    – obataku
    Sep 5, 2014 at 3:47
  • $\begingroup$ You also have $I_C = 150 I_B$. But see my updated answer - normally $I_E=I_B+I_C$ but that is not quite possible in this configuration since there is no return path for the base current... $\endgroup$
    – Floris
    Sep 5, 2014 at 3:49
  • 1
    $\begingroup$ Yes - at any rate, that would be my answer to this question even though that is "strictly speaking" impossible because as drawn there is a net current coming into the right hand loop and no return current. But maybe the convention is that everything has a ground somewhere, even it it's not shown... $\endgroup$
    – Floris
    Sep 5, 2014 at 4:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.