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Above is a npn transistor with a battery connected across emitter and collector.

My textbook and all online resources seem to say this: No current flows in above circuit because the emitter-base junction is not forward biased. The current will flow when another battery is connected across emitter-base junction.

But doesn't the existing battery across emitter-collector already make the emitter-base diode forward biased? Where would all the voltage of that battery appear?

Observations:
If emitter-base junction is not forward biased, then both the junctions will have large but same resistance. This means half the battery voltage appears across each junction. This makes the emitter-base junction forward biased.

Where is my error?

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Where is my error?

There will always be a leakage current from collector to base and emitter (\$I_{CEO}\$). This might be in the region of 100 nA and, at that current, the base emitter junction appears to be forward biased as if that 100 nA were flowing through a diode. There might be about 0.2 volts dropped across base emitter in this scenario.

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    \$\begingroup\$ If I get this correctly, almost all the voltage appears across the base-collector junction! Yes, the CB-junction is in reverse mode while the BE-junction is in forward mode. In reverse mode there can be a large voltage and little current. In forward mode there will only be a small voltage, less than 1 V, here only about 0.2 V as the current (set by the CB-junction which is in reverse and only leaking) is very small. \$\endgroup\$ – Bimpelrekkie Feb 25 '20 at 12:43
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    \$\begingroup\$ don't see why the emitter-collector battery fails to force the emitter electrons all the way into collector The battery can try but the CB junction is in reverse and only allows some electrons to "leak through". Most of the electrons cannot make it into the collector. \$\endgroup\$ – Bimpelrekkie Feb 25 '20 at 12:45
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    \$\begingroup\$ @beccabecca just review how much voltage is dropped across a forward biased diode at low currents. A rule of thumb is that the BE voltage increases or decreases by 0.1 volts for every decade change in BE current. So, if we say the BE drops 0.6 volts with 1 mA, it will drop 0.5 volts at 100 uA, 0.4 volts at 10 uA, 0.3 volts at 1 uA and 0.2 volts at 100 nA. It's a rule of thumb but if you look at (say) the 1N4148 diode you can see this. \$\endgroup\$ – Andy aka Feb 25 '20 at 12:55
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    \$\begingroup\$ @beccabecca your doubt is legitimate. Transistor action is mainly due to diffusion, not drift. The explanation you have been given is valid for two diodes back to to back. But two diodes back two back a transistor do not make :-). When in a transistor you add the B-E circuit, you are supplying charge from outside the C-E path. That tiny charge - linked to Ib - calls for compensation that can come from the emitter, but along with that tiny charge comes a lot of other charge that diffuse in base (against the field) and is collected by the collector. \$\endgroup\$ – Sredni Vashtar Feb 25 '20 at 12:56
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    \$\begingroup\$ Intuititively I still feel the emitter-collector battery should provide this tiny voltage How can it? The CB isn't cooperating. Only applying enough forward current to the BE-junction can make the electrons "go so fast enough" so that they can "jump over" the barrier of the CB in reverse. It takes a while to understand the BJT, we've all been there :-) \$\endgroup\$ – Bimpelrekkie Feb 25 '20 at 12:57

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