Since the base in an n-p-n transistor is kept very thin, very few electrons get to recombine with holes and escape out of base, most of the electrons are injected from emitter into the collector. As a result, Base-current is very small. Whereas the Collector-current is almost equal to the Emitter-current.
Because a transistor's collector current is proportionally limited by its base current, it can be used as a sort of current-controlled switch. A relatively small flow of electrons sent through the base of the transistor has the ability to exert control over a much larger flow of electrons through the collector.
because in base region if we improve density of extra material There is more recombination of charge carriers so the current we want,collector current, is very much smaller. This is a big dis advantage.
By design. The whole IDEA of this type of transistor is for one current to control another one - thus, the transistor can be considered to be a type of amplifier, and it can indeed be used in amplifier circuits.
The current in a transistor is very due to the base of transistor is very thin of the order of 10/6m and it has very doping level as compare to the emitter or collector when electron and hole enter into the base region from the emitter the number of recommendation charges is very small reserve voltage VCC is greater than forward voltageVbb:
Emitter is heavily doped because to provide charge carriers to Base & Collector region, Base and Collectors are lightly doped because to accept those charge carriers.
base
hall coefficient of a lightly doped semiconductor will decrease with increase in temp as hall coefficient is inversely proportional to number density of charge carriers.
the starting material is either p+ or n+ substrate with a lightly doped epitaxial layer.
3: emitter, base, collectorThere are three regions but to be absolutely picky I think only two of them need be doped.Nope: they MUST be doped NPN or PNP. If any are undoped it will not function as a transistor.
Emitter is heavily doped because to provide charge carriers to Base & Collector region, Base and Collectors are lightly doped because to accept those charge carriers.
The width of the base is very thin to increase the majority carrier concentration gradient in the base region thereby enhancing the diffusion current and also to reduce the number of majority carriers lost due to recombination in the base.
base
Explain why the innerlayer two layers of an scr are lightly doprd and are
base
hall coefficient of a lightly doped semiconductor will decrease with increase in temp as hall coefficient is inversely proportional to number density of charge carriers.
Non degenerate semiconductors are those which: -are lightly doped -have less value of electron and hole concentration -violate Pauli's exclusion principle Degenerate semiconductors are those which: -are highly doped -have high value of electron and hole concentration -follow Pauli's exclusion principle
the starting material is either p+ or n+ substrate with a lightly doped epitaxial layer.
3: emitter, base, collectorThere are three regions but to be absolutely picky I think only two of them need be doped.Nope: they MUST be doped NPN or PNP. If any are undoped it will not function as a transistor.
phosphorus doped semiconductor will be N type.gallium doped semiconductor will be P type.There are also other differences due to the different size of the dopant atoms.
i. Constructional structures The UJT is made up of a lightly doped n-region known as the base region onto which is joined a small heavily doped p-region called the emmitter. The PUT on the other hand is a four layer device similar to an SCR except that the gate terminal of the PUT is connected to the n-region adjacent to the anode. ii.Operation The intrinsic stand-off ratio of a UJT is fixed hence operating characteristics can not be alterd. The PUT, on the other hand has operating characteristics that can be altered. These include base-base resistance,intrinsic stand-off voltage, valley current and peak current and all these can be altered by setting the values of two external resistors
Yes, petting above the waist is 2d base.