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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-Ef)/kT+1]. But let us first calculate the (dimensionless) exponent in that equation:

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

= 1.45

Inserting this exponent into equation P(E) = 1/[e(E-Ef)/kT+1], we get,

P(E) = 1/[e1.45 +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-Ef)/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, gm = 3 mA/V

= 3×10-3 mho

rp = 20000 ohm

µ = rp×gm

= 20000× 3×10-3

= 60

Since, voltage gain = µ / [1+ (rp/RL)]

40 = 60 / [1+ (20000/RL)]

So, 1+ (20000/RL) = 60/40

= 3/2

Or, 20000/R= 3/2 – 1

= ½

So, R= 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 VB = 5V, then what will be the the maximum value of R so that the voltage is above the knee point ?

Solution:-

Here,VB = Vknee + IR

Substituting 5 V for VB, 0.7 V for Vknee and 10-3 A for I in the equationVB = Vknee + IR, we get,

VB = Vknee + 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.