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Why does most of the emitter current flow through collector base junction which is reversed biased?
The collector–base junction is reverse biased in normal operation. The reason the emitter is heavily doped is to increase the emitter injection efficiency: the ratio of carriers injected by the emitter to those injected by the base.
What is a reverse biased transistor?
If it is reverse biased, the transistor is off. This is just like a diode. If you forward bias a diode, the diode conducts. If you reverse bias the diode, the diode cuts off. Now, the base-emitter junction in a transistor is going to essentially turn the transistor on or off.
Why is the collector base junction reverse biased?
The collector base junction is reverse biased so that it attracts majority charge carriers and this jonction offers a high resistance to the current (as in rev. PN junc diode) Not the answer you’re looking for? Browse other questions tagged semiconductor-physics electronics electrical-engineering or ask your own question.
How does the reverse biased junction in a transistor work?
Here’s how it works: Current through the reverse biased junction in transistor. A transistor is a three terminal device. One terminal is called emitter, one collector and in between them is base. Now, during biasing the junction between emitter and base is made forward biased and the junction between collector and base is made reverse biased.
Why is the collector junction of a NPN always forward biased?
A transistor with a forward biased Collector-Base junction is said to to in “Saturation” Carriers flow from the collector to the base. That does nothing really useful for the average circuit. In a NPN transistor, the Base-Emitter junction is forward biased so that carriers (Electrons) flow from the Emitter to the Base.
Why does current not flow in reverse biased?
This increase in the number of ions prevents current flow across the junction by majority carriers. eg. electrons can’t complete the loop as they experience resistance of -ve ions on the p side. Thanks for contributing an answer to Physics Stack Exchange!