What is happening to the PN, NP, and MS junctions when using a BJT?

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SUMMARY

The discussion focuses on the behavior of PN, NP, and MS junctions in a Bipolar Junction Transistor (BJT) when external voltage is applied. Initially, an insulating depletion region forms at the junctions due to the diffusion of free carriers, which ionizes dopants like phosphorus and boron. When forward bias is applied to the base-emitter junction, the depletion region shrinks, allowing current to flow and charge carriers to diffuse towards the base-collector junction, which remains reverse biased. This process enables a larger current to be swept from the emitter to the collector, demonstrating the operational principles of BJTs.

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  • Basic principles of forward and reverse biasing
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With no external voltage applied there exists an insulating depletion region between all of the junctions of dissimilar materials. This is formed when the dissimilar materials first contact: the free carriers in one material (either positive or negative) diffuse into the second material and recombine with an opposite carrier. Since the free carriers have crossed the junction the dopants (e.g phosphorus & boron) which these carriers originally belonged to become ionised. This creates electric fields which prevent further diffusion. This area of ions is void of free carriers, and is thus an insulator. This insulating layer is known as the depletion region.

The electric fields (shown by green arrows) in a BJT with no external voltage are shown below:

DKdPlqx.jpg
This is the part I don't understand

However, if a voltage is applied between the base-emitter and collector-emitter terminals, a current can flow between these terminals. But why? what has happened to the depletion regions which originally stopped the current from flowing?

Thanks!

fc4C46m.jpg
 
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Depletion regions shrink and conduct when they are forward biased. Normally, they expand and block current (up to a point) when they are reverse biased.

When you forward bias the base-emitter junction the B-E depletion region shrinks until current begins to flows. This brings holes and electrons into the base. The base-collector depletion region is still there because it is reversed biased when the BJT is in triode or saturation mode. It has a real voltage and electric field. Charge carriers brought into the base by the B-E current will diffuse towards the B-C junction and get swept across by the electric field if they have the right sign.

In short, if you inject current across the base and emitter, a larger current will get swept across from Emitter to Base to Collector.

A picture is worth a thousand words. Someone made a nice looking wiki which goes into detail and has graphs which are really needed to understand how BJT's work. You can find it here.
 

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