In an npn transistor,current will flow with low resistance during the forward bias. Assuming the collector instead of being reverse biased is forward biased, will there be low resistance current flow, will there be current leaving the collector, will electrons still leave the collector to activate the dc power supply.
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hi rossi, Before answering the question will brief you about transistor structure. The base is physically located between the emitter and the collector and is made from lightly doped, high resistivity material. The collector surrounds the emitter region, making it almost impossible for the electrons injected into the base region to escape being collected, thus making the resulting value of α very close to unity, and so, giving the transistor a large β. refer fig1 attached. now answering your question The bipolar junction transistor, unlike other transistors, is usually not a symmetrical device. This means that interchanging the collector and the emitter makes the transistor leave the forward active mode and start to operate in reverse mode. Because the transistor's internal structure is usually optimized for forward-mode operation, interchanging the collector and the emitter makes the values of α and β in reverse operation much smaller than those in forward operation; often the α of the reverse mode is lower than 0.5. The emitter is heavily doped, while the collector is lightly doped, allowing a large reverse bias voltage to be applied before the collector–base junction breaks down. In this mode, the transistor has an emitter efficiency and base transport factor. Most transistors, however, have poor emitter efficiency under reverse active bias since the collector doping density is typically much less than the base doping density to ensure high base-collector breakdown voltages. In addition, the collector-base area is typically larger than the emitter-base area, so that even fewer electrons make it from the collector into the emitter.