![]() ![]() So the major charge carriers(electrons) move towards the base. Since the emitter side is -ve with respect to the base, the electric field is established from base to emitter. Taking the npn transistor, we simply connect the emitter side to 'x' volt, base to 'y' volt and collector to 'z'volt, where x < y < z must be strictly followed to forward bias the emitter-base region and reverse bias the collector- base region. The base region is very small in area in contrast to emitter and collector. The main cause of collector current in case of bjt is the size of base region. Of course the Base must be very thin, because it acts like a perforated grid electrode!) (Heh, see, the NPN transistors can be viewed as triode vacuum-tubes! Depletion regions are the vacuum, and doped silicon forms the electrodes. Instead it was more like a vacuum region, with the intense e-field repelling the usual carriers, keeping them on their proper side.īut when we have out-of-place charge carriers in the diode, that e-field in the reverse-biased depletion layer no longer behaves like a potential barrier. The depletion layer was never actually an insulator it doesn't pin charges in place like rubber or plastics do. Notice that the out-of-place carriers can easily pass through the depletion layer. ![]() Yet at the same time it also works normally, with the reversed junction preventing the free electrons of the Collector region from flooding into the Base region. In that case, a diode starts working backwards, and its reverse-bias voltage causes increased current, not decreased current. The extremely thin base-region in BJTs allows this to happen. For a PNP transistor, we reverse the carrier polarities, where free holes have been deposited in the n-doped base, and they tend to wander too close to the CB junction and get violently flung into the p-doped collector region.)Ĭonclusion: diodes don't work normally if we somehow moved some of the n-side charge-carriers over into the p-side, or vice-versa. (Fast charge-carriers crossing a large potential, that heats up the transistor because of Ic x Vcb.) However, if first they wander too close to the CB junction because of thermal diffusion, they'll be gripped by the large e-field within the depletion layer, and violently attracted into the n-doped collector region. If they remain there for long, they'll quickly fall into holes, which ends up producing an EB base current. Transistor CB diode: the p-doped base is full of free electrons! That's backwards. Normal diode: the n-doped side is full of free electrons, while the p-doped side is full of mobile conduction-band vacancies (holes.) A forward bias will produce e-fields which force the two populations together, turning the diode "on." A reverse-bias voltage will pull them away from each other, forming a potential barrier and turning the diode "off." In BJTs, charge-carriers are able to cross the reverse-biased CB junction because those charge-carriers have been placed in the "wrong" side. So what allows the electrons to flow through the reverse biased PN junction, as in the case of the collector-base junction of a BJT? However, in the reverse biased base-collector junction the external voltage will support the built in potential and cause a larger electric field (from the N-side to the P-side), which will stop positive charges flowing from N to P, and stop negative charges flowing from P to N.īut if you forward bias the base-emitter junction, and reverse bias the base-collector junction, electrons can still flow from the collector to the base, which is from P to N, which as I just explained in the previous paragraph should not be able to happen? I understand why the current can flow through the forward biased base-emitter junction: the external voltage (positive connected to P-side, negative connected to N-side) creates an electric field from the N-side to the P-side, which cancels out the built in electric field caused by diffusion of carriers across the dissimilar materials. ![]() This disagrees with my understanding of the PN junction, as I thought electrons cannot flow from the P-side to the N-side of the reverse biased junction, since there is a depletion region between them. If the base-emitter junction of a BJT is forward biased, then current can flow through the reverse biased base-collector junction (N-P junction). ![]()
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