Page No 511: - Chapter 14 - Semiconductors - Additional Exercise Solutions - Ncert Solutions class 12 - Physics

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Question 14.14:

In a p-n junction diode, the current I can be expressed as

where I₀ is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, k_Bis the Boltzmann constant (8.6×10⁻⁵ eV/K) and T is the absolute temperature. If for a given diode I₀ = 5 × 10⁻¹² A and T = 300 K, then

(a) What will be the forward current at a forward voltage of 0.6 V?

(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?

(c) What is the dynamic resistance?

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

Answer:

In a p-n junction diode, the expression for current is given as:

Where,

I₀ = Reverse saturation current = 5 × 10⁻¹² A

T = Absolute temperature = 300 K

k_B = Boltzmann constant = 8.6 × 10⁻⁵ eV/K = 1.376 × 10⁻²³ J K⁻¹

V = Voltage across the diode

(a) Forward voltage, V = 0.6 V

∴Current, I 

Therefore, the forward current is about 0.0256 A.

(b) For forward voltage, V^’ = 0.7 V, we can write:

Hence, the increase in current, ΔI = I^‘ − I

= 1.257 − 0.0256 = 1.23 A

(c) Dynamic resistance 

(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I₀ in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.

Question 14.15:

You are given the two circuits as shown in Fig. 14.44. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

Answer:

(a) A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in the following figure.

Hence, the output of the NOR Gate =

This will be the input for the NOT Gate. Its output will be  = A + B

∴Y = A + B

Hence, this circuit functions as an OR Gate.

(b) A and B are the inputs and Y is the output of the given circuit. It can be observed from the following figure that the inputs of the right half NOR Gate are the outputs of the two NOT Gates.

Hence, the output of the given circuit can be written as:

Hence, this circuit functions as an AND Gate.

Question 14.16:

Write the truth table for a NAND gate connected as given in Fig. 14.45.

Hence identify the exact logic operation carried out by this circuit.

Answer:

A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.

Hence, the output can be written as:

The truth table for equation (i) can be drawn as:

This circuit functions as a NOT gate. The symbol for this logic circuit is shown as:

Question 14.17:

You are given two circuits as shown in Fig. 14.46, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Answer:

In both the given circuits, A and B are the inputs and Y is the output.

(a) The output of the left NAND gate will be , as shown in the following figure.

Hence, the output of the combination of the two NAND gates is given as:

Hence, this circuit functions as an AND gate.

(b)  is the output of the upper left of the NAND gate and  is the output of the lower half of the NAND gate, as shown in the following figure.

Hence, the output of the combination of the NAND gates will be given as:

Hence, this circuit functions as an OR gate.

Question 14.18:

Write the truth table for circuit given in Fig. 14.47 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)

Answer:

A and B are the inputs of the given circuit. The output of the first NOR gate is . It can be observed from the following figure that the inputs of the second NOR gate become the out put of the first one.

Hence, the output of the combination is given as:

The truth table for this operation is given as:

This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.