This is related to charge carries movements out of transistor terminals. Assuming a transistor in common emitter connection is biased properly where resistors are omitted for simplicity. Before asking the question here is what I know:

From V3's negative terminal the electrons will enter to the emitter terminal following the path A->B->C.

Most of these electrons will be drifted to the collector and from there they will follow X->Y->Z. The rest of the electrons will recombine with the holes in the base and will cause electron flow from the base to D.

My questions is:

Is there literally no charge movements(current) between B and K?

enter image description here

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    \$\begingroup\$ If there's base current it has to flow through the K-B segment of the wire. If there's a collector current (other than leakage) then there's a base current. See Kirchhoff's current law. \$\endgroup\$ – John D May 15 '19 at 12:16
  • \$\begingroup\$ Yes that is why I was confused. I used to believe that artificial current has to flow in loops. But the "electrons" forming the base terminal current seems are not passing through K and B. \$\endgroup\$ – cm64 May 15 '19 at 12:18
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    \$\begingroup\$ What’s an artificial current? You should get used to conventional current, which is in the opposite direction to electron current. \$\endgroup\$ – Chu May 15 '19 at 12:43
  • \$\begingroup\$ @Chu Artifical current is the current caused by an alternator or a voltaic battery. The current caused by static electric does not flow in loops afaik. Like in a lightning strike. \$\endgroup\$ – cm64 May 15 '19 at 12:46
  • \$\begingroup\$ I was trying to point the current flows in loops dogma. \$\endgroup\$ – cm64 May 15 '19 at 12:47

enter image description here

Figure 1. There are two current loops, Ib and Ic. Conventional current is indicated by the direction of the arrows.

There can't be any loss in current around the base loop or around the collector loop.

At one section of the circuit - between C and B - the current flowing is Ib + Ic.

From the comments:

The current caused by static electric does not flow in loops afaik. Like in a lightning strike.

It's still a loop. It's the same as discharging a charged capacitor. Once charge starts to flow it's static no longer.


simulate this circuit – Schematic created using CircuitLab

Figure 2. Lightning discharge equivalent circuit.

  • \$\begingroup\$ Are the electrons looping in Cathode-ray tubes? \$\endgroup\$ – cm64 May 15 '19 at 13:29
  • \$\begingroup\$ But can you also elaborate on the Iceo leakage current for the circuit in question? It originates at CB junction due to temperature. So the Iceo does not originate from the battery. Does it loop between C and B junctions? Thanks \$\endgroup\$ – cm64 May 15 '19 at 13:31
  • \$\begingroup\$ @cm64 Yes, they are. \$\endgroup\$ – Hearth May 15 '19 at 14:10
  • \$\begingroup\$ "Are the electrons looping in Cathode-ray tubes?" See the answers to Where do the electrons go in a CRT?. They have to get back to the supply somehow. \$\endgroup\$ – Transistor May 15 '19 at 17:24
  • \$\begingroup\$ Regarding \$ {I_{CBO}} \$ see if ICEO, ICBO physical interpretation in BJT answers your question. \$\endgroup\$ – Transistor May 15 '19 at 17:26

The current you have between Base in D is also flowing between K and B. The current there has the value of I/ß, with I the current from A to B and ß the amplification of the transitor.

  • \$\begingroup\$ To me the "electrons" forming the base terminal current are attracted by the positive side of V1 and ends their journey at D. How come they pass through K and B again I dont get it. \$\endgroup\$ – cm64 May 15 '19 at 12:22
  • \$\begingroup\$ A current always has to form a closed loop. If there is charge going into the positiv terminal, then there always has to be charging flowing out of the negativ terminal. \$\endgroup\$ – jusaca May 15 '19 at 12:32
  • \$\begingroup\$ Because if there is current going in or out of V1 at D then the same current must go out or into V1 at K. \$\endgroup\$ – Transistor May 15 '19 at 12:32
  • \$\begingroup\$ Yes that is what I memorized but in this case it is not clear. \$\endgroup\$ – cm64 May 15 '19 at 12:32
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    \$\begingroup\$ Your last comment is missing some punctuation so it is difficult to know what you are asking. Thinking in terms of electrons leads to great confusion. You are far better to think in terms of conventional current. As has been explained, current flows in closed loops. If electrons flow into the top of V1 then (different) electrons flow out the bottom of V1. \$\endgroup\$ – Transistor May 15 '19 at 12:39

Of course there is, but they are at the same charge potential so the net is 0. (disregarding any miniscule wire resistance)

  • \$\begingroup\$ What carriers are moving between K and B? \$\endgroup\$ – cm64 May 15 '19 at 12:23
  • \$\begingroup\$ Electrons of course. \$\endgroup\$ – SamR May 15 '19 at 12:28
  • \$\begingroup\$ The electrons flows from the base to V1's positive terminal and they end their journey there. Can you elaborate? I cannot reason any other electron movement. \$\endgroup\$ – cm64 May 15 '19 at 12:30
  • \$\begingroup\$ This answer is not correct - or at least not worded correctly. The not-insignificant base current flows from V1 through D -> C -> B -> K and back to V1. If you want to think in terms of electron flow then it runs the opposite direction. \$\endgroup\$ – Transistor May 15 '19 at 12:34
  • \$\begingroup\$ @cm64, they don't end their journey there. They participate in chemical reactions in the battery. From those reactions, electrons are produced on the other battery terminal, which flow through the wires KB and BC to the BJT emitter terminal. (Remembering also that no individual electron makes this whole journey) \$\endgroup\$ – The Photon May 15 '19 at 14:37

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