Part 1: Specify whether the following statements are true or false. In both case
ID: 3349731 • Letter: P
Question
Part 1: Specify whether the following statements are true or false. In both cases.briefly explain why (a) The BJT is a symmetrical device. (b) Whether connected in forward or reverse, across the Emitter-Base junction, the penetration of the space charge region into the Base is deeper than the penetration of the space charge region into the Emitter. (c) In the forward active mode the penetration of the space charge region into the Base at the EB junction is deeper than the penetration of the space charge region into the Base at the BC junction. Part 2: Answer the following questions: (d) The built-in voltage across a pn-junction at no bias is given by: VI where n, is the intrinsic carrier concentration in Silicon. Na is the acceptor concentration of the p-type semiconductor Nd is the donor concentration of the n-type semiconductor. At no bias, which junction will have a higher built-in voltage? Why? (e) What are the type of currents flowing through the BJT in the forward active mode? Sketch a BJT model to show the current types through the device. (1) What is the main structure difference that distinguishes the BJT from two back-to-back diodes? (g) Why is there a slight increase in the collector current when we increase Vce in an npn junction operating in the forward active mode?Explanation / Answer
Part 1:
a.) False, BJT has different doping levels in different regions, base, collector, emitter. And all regions have different widths.
b.) True, The penetration of space charge regions depends upon the doping levels, more the doping level, more will be the concentration of ions near the junction. Thus base will experience more penetration due to less doping level than the emitter.
c.) False, In the forward active mode, the BE junction is forward biased, therefore the space charge region is narrower in this case whereas in case of BC junction(Reverse Biased) the space charge region is broader. Thus the base will get more penetrated in case of BC junction.
Part 2:
d.) In accordance with the given formula. assuming temperature to be constant, the doping level of the different junction will decide the built-in voltage. And emitter is more heavily doped, thus the BE junction has more heavily doped