# Why are IC and IE different? Aren't the two 1k ohm resistors in series and thus the current through them should be the same?

Honestly I don't understand how this circuit works and that's why I'm thinking that IC and IE should be the same.

Note: This problem is copied from Engineering Circuit Analysis Hayt, Kemmerly & Durbin 8th Edition Page 69.

• That is a terrible model of a transistor. Ib will be non-existent, because it isn't part of a closed path. Commented Sep 1 at 2:48
• If you are showing us diagrams from a textbook, you need to cite the source and the context. As @MathKeepsMeBusy said, this model is, on its face, nonsensical since Ib == 0. But obviously the textbook is trying to demonstrate something and was presumably written by people who know what they are talking about. We can't discuss this diagram without its context.
– vir
Commented Sep 1 at 2:54
• (Just so it makes sense to future readers, the comment above by vir was written before the source textbook reference was added to the question.) Commented Sep 1 at 9:30
• Just to be totally clear, this is (as others have suggested) an error in the textbook. The diagram is a variant of a standard BJT biasing circuit. In the standard version (example: researchgate.net/figure/…), a single DC voltage supply would power both the R1/R2 divider and the 1k/controlled current source/1k path. It looks like someone added a separate voltage supply for R1/R2 but forgot to tie the two grounds together afterward. Commented Sep 2 at 4:30

By removing everything that's not important for considerations of current, you're left with this:

simulate this circuit – Schematic created using CircuitLab

The presence or absence of the resistors and all the other elements is irrelevant to Kirchoff's Current Law (KCL), which when applied to the above is:

$$I_E = I_B + I_C$$

$$\I_E\$$ and $$\I_C\$$ are only equal when $$\I_B=0\$$. Any non-zero $$\I_B\$$ will cause $$\I_E \ne I_C\$$, and there's no other way to interpret this system of currents.

In a twist of irony, if we treat the circuit in figure 3.54 exactly as drawn, with no closed loop around which $$\I_B\$$ can flow, then $$\I_B=0\$$. For a reason that has nothing to do with your question, it turns out that in this case you are accidentally correct: $$\I_E = I_B + I_C = 0 + I_C = I_C\$$.

Just for completeness, to resolve the problem that the shown circuit cannot possibly produce any non-zero base current $$\I_B\$$, I believe what the author meant to draw (and should have drawn) would be this:

simulate this circuit

I added the red wire representing a point of equal potential between the two halves of the circuit, and providing a loop around which $$\I_B\$$ can flow, and be non-zero.

Everything in the dotted box represents (very naively) the transistor itself, with $$\\beta=150\$$, and $$\V_{BE}=0.7V\$$. KCL still applies:

$$I_E = I_B + I_C$$

That is, total current entering the dotted box must equal total current leaving it. Otherwise there's an accumulation/depletion of charge inside the box, or a current leaking in/out somewhere that isn't being accounted for. Using the water analogy, it doesn't matter what that dotted box contains, as long as the plumbing inside doesn't have any leaks, and there's nothing in there to accumulate water (like a "ballooning" pipe), then for every litre of water entering the box (via any route) one litre must emerge somewhere else.

• Yeah, I was also thinking about how can Ib be non-zero. This diagram makes more sense. Thanks a lot for your answer Commented Sep 1 at 9:27
• though about treating it exactly as drawn, if that's a current source for 150 times Ib in the collector branch, then if Ib = 0, wouldn't also Ic = 0? Commented Sep 1 at 14:59
• @ilkkachu yes, that's correct. Commented Sep 1 at 15:07

The full text of the question related to Fig. 3.54 is (Emphasis added):

1. For the circuit of Fig. 3.54 (which is a model for the dc operation of a bipolar junction transistor biased in forward active region), $$\I_B\$$ is measured to be 100 μA. Determine $$\I_C\$$ and $$\I_E\$$

There's obviously a connection not shown (probably grounds missing from the left and right bottom nodes), but the point of the problem is to get you to apply KCL.

You can really ignore just about everything but the pink annotations and where the respective currents flow.

• Got it. Thank you !!! Commented Sep 1 at 9:27