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Explain junction transistors (npn and pnp).
Junction Transistor:
This transistor consists of two p-n junctions combined in one crystal as demonstrated in figure below.
There are two major forms of junction transistors depending on whether the middle section is on a p material or an n material. The middle sector is termed as the base and the outer regions are termed as the emitter and the collector correspondingly.
The transistor is a 3-terminal device and its properties may be given in terms of characteristic curves involving the three currents and the three voltages demonstrated in figure below.
Within a transistor there is zero resultant current of electrons and holes across all junctions. If the collector is connected directly to the base and a small positive voltage is applied among emitter and base after that a current flows just as in a p-n junction diode in the forward direction. Likewise if the emitter is joined to the base and a negative voltage is applied in between collector and base, current flows in the reverse direction, demonstrating a saturation of collector current.
These voltages with similar polarity are applied simultaneously as demonstrated in figure below.
The potential barrier in between base and emitter is decreased and the flow of holes across the barrier greatly rises. There is an increased flow of electrons from base to emitter but because the hole density in the p- region is much higher than the electron density in the n- area, the current may be seem to be mainly because of the holes. The holes enter the n- region and diffuse by this, and combine with the electrons in that area as the base region is made sufficiently thin, many of holes reach the collector and base junction and very little arrive at the base terminal. Holes fall easily into the collector region on account of field at the collector base junction. Therefore the collector current is very nearly equivalent to the emitter current and the base current is almost zero. VCB has little effect on the current as long as its magnitude is above various minimum values and the base is adequately thin. A small change in VEB causes a change in IC resulting in an approximately equal change in IC.
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There are total 12 interrupts in 8085.
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