SOLUTIONS :

[3.1] Side1 Donor impurity (Phosphorous ) concentration =

Side 2 Donor impurity (Phosphorous ) concentration =

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/ ]