Carrier flow in an NPN transistor: When biased in the active region, most of the electrons injected from the emitter into the thin, lightly doped base will ultimately flow to which terminal?

Introduction:
Understanding carrier flow in BJTs clarifies why the device can amplify current. In an NPN transistor, the emitter injects electrons into the base. How these electrons travel determines the resulting collector current and overall device behavior.

Given Data / Assumptions:

• NPN BJT biased in active region.
• Emitter–base junction forward-biased; collector–base junction reverse-biased.
• Base is thin and lightly doped.

Concept / Approach:
In active operation, the forward-biased emitter–base junction injects majority carriers (electrons) from the N-type emitter into the P-type base. Because the base is thin and lightly doped, only a small fraction recombines with holes (creating base current). Most electrons diffuse across the base and are swept into the collector region by the reverse-biased collector–base junction's electric field, forming the collector current IC.

Step-by-Step Solution:
Electrons injected from emitter enter base.Base is engineered to be thin/lightly doped so recombination is minimized.Remaining electrons reach the collector–base depletion region.Reverse-bias field sweeps them into the collector, producing IC ≈ α * IE with α close to 1.

Verification / Alternative check:
The relation IE = IC + IB shows that IB is small. Therefore, most of IE must appear as IC, confirming that most injected electrons end up collected at the collector terminal.

Why Other Options Are Wrong:
Out of the base lead / into the base supply: would imply large base current, contradicting transistor design (IB is small).Into the emitter: reverse flow would contradict the forward-bias injection and device physics.

Common Pitfalls:
Overestimating base recombination. High transistor gain relies on making the base thin and lightly doped to minimize recombination and maximize collection.

Final Answer:
into the collector