LET?

Yes. Bipolar transistors are three layers of silicon, forming two diodes (like the picture.) The base is in the middle. If you create a voltage between the base and emitter, it'll help charges cross over the reversed diode. Just connecting the anodes of a couple of diodes won't work, because the charge across the 'reverse biased' one won't be helped into jumping across the barrier.

As I discovered when I was, I think, 11 years old!

John

The difference between two diodes in series and a transistor with two junctions in series is that the middle node is physically very thin, so that when you forward bias the base emitter junction, the charge carriers injected from emitter into thin base layer move more by diffusion (random movements closely related to thermal Brownian movement) than by the very small electrical field across to the base material. So most of those charge carriers wander over to the collector side of the collector to base junction, where they suddenly experience the effect of the stronger electric field across this reverse biased junction. This field quickly sweeps those charge carriers out as collector current.

Rather a lot it seems ! Firstly it's important to understand that a bipolar transistor is a single device. The fact that it can behave like 2 diode junctions doesn't mean that's what its actually made of.

Graham

No, that was far too advanced for me. But the OP has us both beat, with a GaAlAs heterojunction light-emitting nonfunctional transistor.

John

"John Larkin" a écrit dans le message de news: snipped-for-privacy@4ax.com...

Oh, you too...

But then did you try SiGe (one Si diode and one Ge diode) as I did?

The left part of this diagram, showing diodes:

... really ought to be tossed into tue bit bucket. It does nothing to contribute to the understanding of bipolar transistors, and furthermore is misleading.

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