# BJT with Zener diode I got the picture from my prof's notes, and I don't quite understand why there can be two different currents at the given node. If there are only two currents shouldn't the currents' values be the same

• What about the current arrow that is not present but is the current entering that node from the far left? – jonk Nov 20 '17 at 2:49
• That's what I'm not sure about, from what I see, I believe it's just a voltage source to ground, but on the right side, I was able to use a KCL equation and there was no current going towards the circle to the right – Richard Chang Nov 20 '17 at 2:59
• Have you considered that the voltage source (ideally) can supply whatever the circuit is providing? – ThreePhaseEel Nov 20 '17 at 3:23

## 2 Answers

On the top node you can see that there is one incoming current that is equal to the sum of the two outgoing currents there. At the bottom node, you can see that the two incoming currents are equal to the outgoing current. And at the middle node (base of BJT), there is one incoming current and two outgoing currents and the sum of the two outgoing equals the incoming. Finally, you can see that the node at the BJT emitter receives the collector current and the base current as incoming currents and that the load current equals the sum of those. It all works out just fine. simulate this circuit – Schematic created using CircuitLab

• I'm very confused by the convention used in the diagram, because for the unshaded circle on the right, it was a power source, but on the right the unshaded circle was just telling me the voltage at that node. Is there any way to know whether or not it is a power supply? – Richard Chang Nov 21 '17 at 17:54
• @RichardChang In general, if a node is labeled as "15 V" then it is reasonable to assume it was meant as a "low impedance power supply rail, referenced to ground" (or the equivalent of some kind.) – jonk Nov 21 '17 at 21:05

The current going into the node from the left (the power supply) is 1.0209A, determined by KCL.

It leaves (splits into two) as 39mA, of which about 10mA flows into the load through the base-emitter junction and 0.99A which flows into the load through the collector-emitter junction.

The signed total of all three currents at that node is 0.0 as it must be.