If p-type semiconductor and n-type semiconductor of a diode are equally doped, and if the diode is forward biased, then holes will move toward the n-type semiconductor and electrons will move toward the p-type semiconductor and they will diffuse with each other. Then will there be any electron that will go to the positive terminal of the battery if all of them have diffused with each other? I can't understand, please help me!
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FORWARD BIAS OF A P-N JUNCTION As the electrons move towards the positive terminal and the holes towards the negative, they will come to the depletion layer. This is a very narrow layer around the junction (i.e. around the interface of the two semiconductors.) In the depletion layer, electrons and holes can recombine, but the recombination rate is not high enough so as not to allow electrons to reach the positive terminal. This recombination effect takes place in the diodes of solar photovoltaic cells as well, and it is an interesting field of research in how to reduce its effectiveness. The recombination rate is smaller for larger energy gaps. By the way, these recombination processes are what generate the emitted light in LEDs, where the energy gap of the diode is arranged to be visible light of a desired colour. |
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Reverse Bias of P-N junction When the voltage is applied this way round it tends to pull the free electrons and holes apart, and increases the height of the energy barrier between the two sides of the diode. As a result it is almost impossible for any electrons or holes to cross the depletion zone and the diode current produced is virtually zero. A few lucky electrons and holes may happen to pick up a lot of thermal (kinetic) energy. This gives them enough 'go' to cross the barrier, hence the reversed biassed current is not zero, just very, very small. |
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