The band diagram below is for a silicon sample that is subje

The band diagram below is for a silicon sample that is subjected to illumination. The quasi-Fermi levels provide detailed information on the doping distribution, carrier densities, and illumination intensity across the sample (i.e., their dependences on position). What is the doping concentration at x = 0 and x = L? Which side of the sample is illuminated more strongly? What are the values of the carrier densities at x = 0 and x = L under the shown conditions? What would the values the carrier densities be at x = 0 and x = L if the light were turned off? Sketch log (n) vs. x and log (p) vs. x under illumination.

Solution

Solution for A)
Quasi fermi level is the one which tells us about the concentration of electrons in both the bands and if the silicon was undoped E(i) would have been at the centre of covalent and valance band:
               E(i)= {E(v)+E(c)}/2,
But due to the donor type imputity added this band is not at the centre and is shifted upwards. Amount by which it is shifted up/down can be easily known by calculting the work function:
Work function:-       q(i)=E(i)-F(p),

To calculate the concentration of the doping inside the silicon we need to make calculation for two points x=0 and x=L

               Q(f)= (kT/q)*log(N(b)/ni)
here k is the Boltzman constant = 8.617*10^-5,
T is the temperature = 300 K,
q Charge on the elctron = 1.602*10^-19C,
ni intrinsic carrier concentration of silicon at 300K = 1^10
Q(f)= 0.3 for X=0 and 0.1 for X=L,
N(b) will give the concentration of the doping

case 1: X=0, Q(f) =0.3

0.3 =( 8.617*10^-5* 300/ 1.602*10^-19C)*Log{N(b)}/Log{1^10}

[0.3*Log{1^10}]/( 8.617*10^-5* 300/ 1.602*10^-19C) = Log{N(b)}


Similarly for case 2: X=L, Q(f)=0.1
0.1 =( 8.617*10^-5* 300/ 1.602*10^-19C)*Log{N(b)}/Log{1^10}

[0.1*Log{1^10}]/( 8.617*10^-5* 300/ 1.602*10^-19C) = Log{N(b)}

Solution for part B:-
Upper side will illuminated more strongly if the source of light is facing that surface or the surface facing the incident photons will illuminate more strongly as the charge carrier flow will be more that direction
          
Solution for part C:-

To calculate the carrier density from the given figure:-

a) Q(f)= 0.3 for X=0 and 0.1 for X=L,

we know that:-       n=n(c)*n(v)*exp(Q(f)/kT)
where n is carrier distribution
n(c), n(v) is concentration of covalent and valance band,
Q(f) is given at X=0 and X=L
here k is the Boltzman constant = 8.617*10^-5,
T is the temperature = 300 K,
For Silicon At 300 k values for n(c)*n(v)=n(i)= 1.45*10^16


So putting these values in the above equation we get

case 1: X=0, Q(f) =0.3

           n= 1.45*10^16 m^-3*exp(0.3/8.617*10^-5 *300)
Similarly for case 2: X=L, Q(f)=0.1

   n= 1.45*10^16 m^-3*exp(0.1/8.617*10^-5 *300)

Solving this further will give us the value for the carrier concentration.

D) Using the above formula in the similar fasion will give us the value for the carries density if light is turned off.