Structure of bipolar junction transistor, Electrical Engineering

Assignment Help:

Structure of Bipolar junction transistor:

 A BJT contains three differently doped semiconductor regions that are: emitter region, base region and collector region. These regions are p type, n type and p type correspondingly, in a PNP and n type, p type and n type correspondingly, in a NPN transistor. Every semiconductor region is connected to a terminal, properly entitled as: emitter (E), base (B) and collector (C).

The base is physically located among the emitter and the collector and is created from lightly doped and high resistivity material. The collector that surrounds the emitter region, creating it almost not possible for the electrons injected into the base region to escape being collected, so making the resulting value of α very close to unity, and so, providing the transistor a large β. A cross section view of a BJT points out that the collector-base junction has a much larger area than as compared to emitter-base junction.

The bipolar junction transistor, different from other transistors, is generally not a symmetrical device. Here this means that interchanging the collector and the emitter makes the transistor leave the forward active mode and begin to operate in reverse mode. Because the internal structure of transistor is generally optimized for forward-mode operation, interchanging the collector and the emitter makes the values of α and β in reverse operation much smaller than as compared to those in forward operation; frequently the α of the reverse mode is lower than 0.5. The lack of symmetry is primarily because of the doping ratios of the emitter and the collector. The emitter is heavily doped, whereas the collector is lightly doped, permitting a large reverse bias voltage to be applied before the collector-base junction breaks down. In normal operation the collector-base junction is reverse biased. The cause the emitter is heavily doped is to increase the emitter injection efficiency: the ratio of carriers injected via the emitter to those injected by the base. For high current gain, most of the carriers injected into the emitter-base junction have to come from the emitter.


Related Discussions:- Structure of bipolar junction transistor

Function generator, explain the waorking principle of function generator

explain the waorking principle of function generator

Explain the operation of 8279, Explain the operation of 8279.  Explain the ...

Explain the operation of 8279.  Explain the following terms: (i)  N key Roll over. (ii)  Key board debounce. (iii)  FIFO RAM. Ans The 8279 microprocessor i

Pn junction, the depletion layer in the pn junction is caused by

the depletion layer in the pn junction is caused by

Explain effect of frequency of applied electric field, Explain effect of fr...

Explain effect of frequency of applied electric field. If an external electric field is applied, the distance among charges that is related to chemical bonding keeps constant i

Current transformers., CT 10VA, 50A/5A, 6ohz, 2.4KV required nominal primar...

CT 10VA, 50A/5A, 6ohz, 2.4KV required nominal primary voltage?

Electrical circuits and systems, Electrical circuits and systems : Elec...

Electrical circuits and systems : Electrical systems allow energy to be conveniently delivered from the point of supply to the point of application - e.g. electric railways, ca

Zero flag - registers , Zero Flag - Registers If the  result of any  a...

Zero Flag - Registers If the  result of any  arithmetical  or logical  operation in the accumulator  is zero i ,e  all the bits  of accumulator  ( with  some exceptions )  are

For the low-pass filter configuration calculate cf, Q. For the low-pass fil...

Q. For the low-pass filter configuration of Figure, with R i = R f = 1M, calculate C f such that the 3-dB point is at 1 kHz.

Determine the bit-error probability for the two systems, Let both coherent ...

Let both coherent ASK and coherent PSK systems transmit the same average energy per bit interval and operate on the same channel such that E b /N 0 = 18. Determine the bit-error p

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd