SOLUTIONS :

[3.1]      Side1 Donor impurity (Phosphorous ) concentration  =

Side 2     Donor impurity (Phosphorous ) concentration   =

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Si intrinsic carrier concentration be

Side1:  For donor impurities              = 0.0259ln ( )

=0.0259* 20.03 = 0.5187ev

Side2:   For donor impurities              = 0.0259ln ( )

=0.0259* 13.592 = 0.352ev

[a]:

Work function is the potential difference between vacuum level and Fermi level …

And for Si

Side1: work function is    4.05+(0.55-0.5187)ev = 4.08137ev

Side2: work function is   4.05+(0.55-0.359)ev = 4.241ev

[b]: Potential difference between the 2 regions is the difference between work functions

4.241-   4.08137 =0.1597ev

 
[3.2]

[a] :

In figure 3.2(b) the  vacuum levels are not continuous

In figure 3.2(c ) the vacuum level of semiconductor does not  have same difference all

Through.

In figure 3.2(d) the Fermi levels should be in a straight line when metal and semiconductor     are  brought together, but this is not the case with figure 3.2(d).

[b]:

The correct diagram assuming the basic schottky theory applies is shown below..

 

 

 

 

 

 

3.3:(a ) We considered the work function of metal as4.5ev and electron affinity  =  4.05ev. then

(b):

Now if light is shined on the metal to generate electron hole pairs.

(i)                 There will be no flow of current when the two terminals are connected with a wire.

(ii)               The potential that is measured is still the built in potential

 

(c ):    Now the silicon is having a work function of 4.9ev . The band structure will look as follows

(d ) :  The semiconductor  in  problem 3.3(a) is Ntype and thus the majority carries are electrons where as the semiconductor in the problem 3.3(c) are Ptype whose majority carriers are holes and likewise the current  flow will be in opposite direction  for the two configurations.

 

3.5(a)

Let  5.3ev

 

For donor impurities  Nd=

= 0.0259ln ( )

=0.0259* 13.815 =  0.357ev

 

 

=  4.343ev

The potential barrier  is

The built in potential is

= 0.684ev

The small signal capacitance  C is given by the formula                       where                is the permittivity of the silicon  Nd is the doping concentration of the si and  Va is the applied voltage. Here the applied voltage  Va is 0.  Substituting all the above parameters in the equation we get

=

 

(b)   Now if a reverse bias Va   is applied and the effective capacitance is reduced by 25%. This Va   is calculated from the above relation

 

 

Va =0.38V     i.e. a reverse bias voltage of o.38 is to be applied.

REFERENCES:

[1] “Principles of semiconductor Devices” Brat Van Zegbroeck.

[ http://ece.colorado.edu/~bart/book/ ]

 

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