Ok guys I can't understand a thing of the Transistor here is the scheme:enter image description here

My teacher said that when I put 0,7V on the base the PN junction between Base and Collector is direct polarized so the whole structure of the transistor is ipotizable like a big N block, my question is if it is ipotizable like a big N block, I will have ever the maximum current (Since I have for example 20V on the Collector and 0,7V on the Emitter) regardless of the current that is flowing between the emitter and the base.But my hypothesis is wrong because the Transistor is an amplifier, my question is why it is wrong. I have also another question, why between the emitter and the base when the transistor is working I have a big voltage drop (For example if I have 20V on the Collector and 0,7V on the Base)

Thanks a lot for your time, hope that you will have an happy day :)

  • 6
    \$\begingroup\$ What does ipotizable mean? \$\endgroup\$ – Hearth Nov 30 '18 at 21:59
  • 1
    \$\begingroup\$ So many textbooks, websites, etc. have been written explain how a BJT works. We would be doing you a disservice to repeat only a limited section of that here. You mention of "a big N block" sounds like you think you're forming an inverted region in the entire P region like a MOSFET. That's not correct, and I suggest you forget that idea when trying to understand your textbook/professor. \$\endgroup\$ – Kevin Kruse Nov 30 '18 at 22:16
  • \$\begingroup\$ @Hearth ipotizzabile (Italian) = conceivable \$\endgroup\$ – Chu Dec 1 '18 at 1:30
  • \$\begingroup\$ Quote: "I will have ever the maximum current (Since I have for example 20V on the Collector and 0,7V on the Emitter) regardless of the current that is flowing between the emitter and the base". Perhaps you will be surprised - however, your statement is correct. This is because the base current is a small portion of the emitter current in form of an unavoidable "byproduct". That means: It has no controlling function - it is the voltage Vbe which determines the emitter resp. collector current. \$\endgroup\$ – LvW Dec 1 '18 at 8:41

Your right figure is misleading, because it's not paying attention to asymetric doping of emitter and collector, and neither the width of the base. The follwing picture is more accurate:


The base is only a thin layer between emitter and collector region. As soon the B-E diode becomes conducting, the thin base layer below the emitter is flooded with electrons so the C-E connection becomes conducting, too. The collector current cannot sustain this current by itself, however, because it runs the "wrong" direction through the C-B diode when the collector (as usual) is at higher voltage than the base.

That's why the function of this structure depends on a positive base-emitter voltage and a positive base current.

The doping assists the difference between collector and emitter. One could build a completely symetrical transistor but it would have worse properties.


I've been told (by a device physics guy) that the base charges are irksome to the charges coming from the emitter, those emitter charges trying to erase/annihilate what entered from the base.

Fortunately the base region is very thin, and most of the kill-kill-kill emitter charges miss and overshoot and end up in the collector-base-diode-depletion region where the relatively high voltage gradient moves the I-was-emitted-by-the-emitter to kill-me-some-base-charges but I-failed-to-fulfill-my-duty-in-life and now-I-am-just-a-worthless-charge being-swept-along-helplessly-in-the-collector-depletion-region and moved along into the collector-ohmic region.

The ratio, of the emitter-injected charges that miss the base charges, to the base-provided charges, is the BETA or the current-gain.

All of this illustrates the power of an electron to drive the behavior of matter.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.