In an NPN transistor, why isn't a typical C-E voltage sufficient to overcome the internal resistance of NPN junction directly? Since we only need 0.7V to overcome the PN junction from the Base to Emitter, isn’t that the same as applying a similar (or slightly higher) voltage from Collector to Emitter to achieve the same outcome?

Thinking of it an electronic level, the key thing you’re trying to do is ‘push’ electrons into an area of negative charge – i.e. the Collector. I don’t see why the existing potential difference from Emitter to Collector doesn’t do this without a base potential being applied at all.

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    \$\begingroup\$ For the same exact reason that you can’t get a reverse-polarized diode to conduct. \$\endgroup\$ Commented Jan 10, 2019 at 5:09
  • \$\begingroup\$ check this out .... electronics.stackexchange.com/questions/92131/… \$\endgroup\$
    – jsotola
    Commented Jan 10, 2019 at 5:26
  • \$\begingroup\$ Possible duplicate of Why can't two series-connected diodes act as a BJT? \$\endgroup\$
    – MatsK
    Commented Jan 10, 2019 at 5:33
  • \$\begingroup\$ BJT Transistors are defined as either current or voltage controlled current sinks or sources. So why are you asking why don't they amplify with just 2 leads and no input? You must have an Ib which determines the Vb vs Ic characteristic. At low current relative to max rated , yes you can operate at high gain with Vcb=0 but hFE drops sharply to 10% when Vce=Vce(sat). \$\endgroup\$ Commented Jan 10, 2019 at 5:36
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    \$\begingroup\$ to overcome the internal resistance of NPN junction directly? That's not how PN junctions work! PN junctions do not conduct because their internal resistance has been "overcome". The internal resistance is something entirely different. \$\endgroup\$ Commented Jan 10, 2019 at 6:43

1 Answer 1


I suggest that you consider the physical structure of an NPN transistor. It is composed of a pair of PN junctions connected as


simulate this circuit – Schematic created using CircuitLab

When the base is disconnected, the upper junction is reverse-biased, so very little current flows.

Before you start objecting, you need to be aware that the two diodes are not perfectly isolated, and as soon as current starts flowing in the base-emitter junction, it affects the collector-base junction in such a way as to allow more current to flow. The details require a whole lot more space and math than is appropriate here.

Nonetheless, the open-base model is accurate at this level of discussion. For a more complete discussion see the Wiki page as a start

All models are wrong. Some are useful.

Finally, as a nitpick your opening assertion is simply wrong. Some current will flow from emitter to collector, although the amount is usually quite small. The name for this current is the "collector cutoff current", and is normally only of much importance for small-signal transistors.

  • \$\begingroup\$ Thanks, but not getting it. At an electronic level, we have 2 regions of electrons, sandwiching a region of holes. When you apply a potential difference across this (without a 3rd connection being made in the middle, at the base), I don't see how this won't force a current in a particular direction. Further, I don't see how applying a base voltage from the positive region to one of the negative regions is any different to applying a voltage across the whole "sandwich". \$\endgroup\$ Commented Jan 11, 2019 at 11:02
  • \$\begingroup\$ @Bhajmeister - Sorry, but you're going to have to learn some solid-state physics before you can address your problem. Every part of your description is wrong at some level, so you have your work cut out for you. Good luck on your studies. \$\endgroup\$ Commented Jan 11, 2019 at 15:19

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