Solved Examples on Solid and Electronic Device

Question 1:-

(a) What is the probability that a quantum state whose energy is 0.10eV above the Fermi energy will be occupied? Assume a sample temperature of 800 K. 

(b) What is the probability of occupancy for a state that is 0.10 eV below the Fermi energy?

Solution:-

(a) We can find P(E) from equation P(E) = 1/[e^{(E-E_f)/kT_+1].} But let us first calculate the (dimensionless) exponent in that equation:

E-E_F/kT = (0.10 eV)/(8.62×10^-5 eV/K) (800 K)

             = 1.45

Inserting this exponent into equation P(E) = 1/[e^{(E-E_f)/kT_{+1], we get,}}

P(E) = 1/[e^1.45 +1]

       = 0.19 or 19%

Therefore, the  probability that a quantum state whose energy is 0.10eV above the fermi energy occupied will be19%.  

(b) The exponent in equation P(E) = 1/[e^{(E-E_f)/kT_+1]} has the same absolute value as in (a) but is now negative. Thus from this equation,

P(E) = 1/e^-1.45+1

       = 0.81 or 81%

From the above observation we conclude that, the probability of occupancy for a state that is 0.10 eV below the Fermi energy would be 81%. For states below the Fermi energy, we are often more interested in the probability that the state is not occupied. This just 1- P(E), or 19%. Note that it is the same as the probability of occupancy in (a).

Question 2:-

Calculate the frequency of radio-waves radiated out by an oscillating circuit consisting of a capacitor of capacity 0.02 µF and inductance 8µH.

Solution:-

Here, C = 0.02 µF 

           = 2×10^-8 F

         L = 8 µH

           = 8×10^-6 H

We know that, frequency, f = 1/[2π√LC]

 

Substitute the value of L nad C in the equation  f = 1/[2π√LC], we get,

 f = 1/[2π√LC]

  = 1/[2π√(2×10^-8 F) (8×10^-6 H)]

 = [1/2π×4]×10⁷

= 397.8×10³ Hz

= 397.8 kilo Hz

From the above observation we conclude that, the frequency of radio-waves radiated out by an oscillating circuit consisting of a capacitor of capacity 0.02 µF and inductance 8µH would be 397.8 kilo Hz. 

 

Question 3:-

A triode has a mutual conductance of 3 mAV^-1 and anode resistance of 20000 ohm. Find the load resistance which must be introduced in the circut to obtain a voltage gain of 40.

Solution:- 

Here, g_m = 3 mA/V 

               = 3×10^-3 mho

         r_p = 20000 ohm

        µ = r_p×g_m 

          = 20000× 3×10^-3

             = 60

Since, voltage gain = µ / [1+ (r_p/R_L)]

                          40 = 60 / [1+ (20000/R_L)] 

So, 1+ (20000/R_L) = 60/40

                            = 3/2

Or, 20000/R_L = 3/2 – 1

                      = ½

So, R_L = 40000 ohm

Thus, from the above observation we conclude that,  the load resistance which introduced in the circut to obtain a voltage gain of 40 would be 40000 ohm.

 

Question 4:-

The junction diode in the following circuit requires a minimum current of 1mA to be above the knee point (0.7V) of its I-V characteristic curve. The voltage across the diode is independent of current above the knee point, If V_B = 5V, then what will be the the maximum value of R so that the voltage is above the knee point ?

Solution:-

Here,V_B = V_knee + IR

Substituting 5 V for V_B, 0.7 V for V_knee and 10^-3 A for I in the equationV_B = V_knee + IR, we get,

V_B = V_knee + IR

5 = 0.7 + [(10^-3) (R)]

Or, R = 4.3 KΩ

From the above observation we conclude that, the the maximum value of R so that the voltage is above the knee point would be 4.3 KΩ.