BJT Active mode - How does it work

I am finding it difficult to understand how exactly this works and would appreciate any help.

From my understanding:

The emitter base junction has to be in forward bias meaning that Vb is more positive than Ve ( ie Vb>Ve ?)

The base collector junction has to be in reverse bias hence Vc is more positive than Vb ( Vc>Vb ?)

From what I know current should go in from the emitter then through P-type and if it is thin enough it would reach the collector. Here the currents seem to be in opposite directions and I cant make much off it.

I understand the P-n junction theory between holes and electrons and how this can be seen as 2 back to back diodes however I am finding it difficult to analyse exactly how it works from the very start.

If we where to start straight from the begging where current flows from the cells how would it work?

I would appreciate the help. Many thanks.

• Are you asking about the physics of it? If so, what's your background in semiconductor physics, do you know the terminology of energy diagrams and such? Jan 13, 2019 at 22:18

Some very basic physics about transistor theory of operation http://hyperphysics.phy-astr.gsu.edu/hbase/Solids/trans2.html#c4

Explaining a BJT would likely take at least a couple of hours of instruction to get from wherever you current knowledge is to the point where you'd comfortably understand it from such a fundamental level.

You seem to understand the pn junction. The next step is probably to stop thinking of the collector-base junction as a pn junction, and instead look at the whole BJT as relying on diffusion of charge carriers - reverse biasing the collector base junction creates a depletion layer that permits (conventional) current flow from collector to emitter.

Start here, but try a variety of sources - BJT fundamentals are widely documented, but you'll need to find a source that suits your existing knowledge.

Decades ago I asked how a bipolar device worked; the answer came from a guy with PHD in device physics, who gave me a fine one-sentence explanation:

Base current enters the base region, and the charges flowing across the base-emitter junction do their best to annihilate the base-entering charges. Fortunately MOST of the charges provided by the emitter end up NOT INTERSECTING with the base-entering charges, instead being gathered up in the collector region. The ratio of miss-hit is the BETA.

OK --- that was 3 sentences.

• Three sentences - and no mentioning of the most important quantity that allows transistor operation as a precondition - the base-emitter VOLTAGE ?
– LvW
Jan 14, 2019 at 7:27
• It's a perfect explanation @LvW and much better than my cheese sandwich melted by the emitter hot sauce because the cheese melts more readily than the bread with the cool lettuce ground and drawing more heat from the collector and resistor. The voltages will rise with sauce accordingly when forced with current. But we know it can be also controlled directly from the bread-hot-sauce voltage. ;) Jan 21, 2019 at 5:12
• What I never will understand is the following: Everybody starts the design of a BJT stage by generating a bias voltage VBE of app. 0.7 Volts. And, of course, he knows that 0.7 volts will result in a larger IC than 0.65 volts. Nevertheless, some of them claim that it is the current IB that controls IC. In this context, I also have heard that the voltage-current relationship is something like the chicken-egg problem. But that`s wrong: Current is always the result of a driving voltage - the movement of charges needs an electrical field - not vice versa!
– LvW
Jan 21, 2019 at 7:35