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

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In a BJT, the depletion regions at the PN, NP, and MS junctions act as insulators when no external voltage is applied, formed by the diffusion and recombination of free carriers. When a voltage is applied, particularly forward biasing the base-emitter junction, the depletion region shrinks, allowing current to flow as holes and electrons enter the base. The base-collector junction remains reverse biased, maintaining its depletion region and electric field. Charge carriers from the base diffuse toward the base-collector junction and are swept across by the electric field, resulting in a larger current flowing from the emitter to the collector. Understanding these dynamics is crucial for grasping BJT operation.
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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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