Question

Which operating region of BJT enables Emitter-base and Collector-base junctions to undergo perfect short-circuit configuration?
A. Active region
B. Saturation region
C. Cut-off region
D. None of the above

Answer 1

A bipolar junction transistor (BJT) is indeed a transistor that utilizes charging carriers both for electrons and holes. In the cut-off region, the emitter-base junction and collector-base junction both junctions are reverse biased. In the active region, the emitter-base junction is forward bias and collector-base junction is reverse bias. The Power BJT drives from the quasi-saturation zone into the hard-saturation region by raising base current. This region is known as an area of extreme saturation.

Complete step by step solution:
Bipolar junction transistor (BJT): BJT is Bipolar Junction Transistor's short form. BJT is a three-terminal unit with an Emitter, a collector, and a base pin. The current movement between the emitter and the collector is regulated by the amount of current applied to the base.

In a transistor emitter-base junction, while the collector-base junction is reverse biased, forward bias is often observed. The input resistance, i.e. the resistance of the emitter -base junction becomes quite weak when the emitter-base is forward biased.

Active region: The area between cut-off and saturation is referred to as active area. Collector-base junction stays reverse biased in the active region whereas base-emitter junction remains biased forward. Additionally, in this area, the transistor would operate normally.

Saturation region: The point where the ${I_{\text{B}}} = {I_{\text{B}}}\left( {{\text{sat}}} \right)$ curve is intersected by the load line is called saturation. The base current is maximal at this point, and so is the current of the collector. Collector-base junction no longer stays reverse biased at equilibrium, and normal transistor operation is lost.

Cut-off region: The base-emitter junction is no longer biased forward when broken off, and normal transistor operation is lost. The voltage of the collector-emitter is exactly the same as that of ${V_{{\text{CC}}}}$ i.e. ${V_{{\text{CE}}}}\left( {{\text{cut}}\;{\text{off}}} \right) = {V_{{\text{CC}}}}$.

Hence, option B is correct.

Note: To answer the given question, we must have the proper knowledge of BJT, CB junction, EB junction, along with their behaviour in active, saturation, and cut-off region respectively.
In the cut-off zone all junctions are reverse biased in the emitter-base junction and the collector-base junction. In the active region, forward bias is the emitter-base junction, and reverse bias is the collector-base junction. The Power BJT drives by raising base current from the quasi-saturation zone into the hard-saturation field.