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I read the reply to this question but I cannot really make sense of it: In an NPN transistor, why doesn't current flow without applying a base voltage?

if I have a npn-transistor and and apply a voltage between emitter and collector which is let's say 10 V , why is no charge flowing from emitter to the base already?

As I understand the pn junction between base and collector, it is in reverse bias due to the external voltage and will not allow any holes from the base or electrons from the collector to cross the barrier. If due to a voltage between base and emitter bigger than 0.7 V however electrons flow into the base they can drain through the collector.

Now, why can't the applied emitter-collector voltage cause flow of electrons into the base (in a sense of pushing the electrons from the emitter over the potential barrier just as in a forward biased diode?) Then we would have free electrons in the base that could then flow to the collector.

Thank you for your help!

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    \$\begingroup\$ why can't the applied emitter-collector voltage not cause flow of electrons into the base because the pn junction between base and collector, it is in reverse bias due to the external voltage. Note how I just copy-pasted two sections from your question. \$\endgroup\$ – Bimpelrekkie Mar 28 at 14:19
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    \$\begingroup\$ You might expand your question to the case where the emitter-collector voltage is small -- say 200 mV. It still won't conduct. The reason is that the base region quickly arranges itself so that there is no shared conduction energy band between collector and emitter and so no current can flow. You have to restore the situation by continually supplying a small recombination current at the base or else it almost instantly becomes a barrier (there will be a very very tiny charge flow for a moment before that happens) and the only way to restore things is to supply that recombination current. \$\endgroup\$ – jonk Mar 28 at 14:29
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    \$\begingroup\$ Both junctions are in series. \$\endgroup\$ – Andy aka Mar 28 at 14:30
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    \$\begingroup\$ the junction between emitter and base is not OK, so what keeps that BE junction in forward? It can't be the CE junction because it is in reverse mode. Note that the only current from the CE junction is a leakage current. Is that enough to make the BE junction operate properly in "forward mode" or would more current be needed? If I wanted to "pull" the BE junction into "proper" forward mode, what would I need to do? Hint: it has to do with the base connection. \$\endgroup\$ – Bimpelrekkie Mar 28 at 14:31
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    \$\begingroup\$ Without the base voltage explicitly applied, you have one diode going to the supply and another to ground. The one going to supply is reverse biased, and passes little current, so the other one will keep the base voltage at essentially ground. \$\endgroup\$ – Cristobol Polychronopolis Mar 28 at 16:13

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