Related Problem With Solution

Problem 1: Zener Diode as a Voltage Regulator#

Problem Statement:
A Zener diode with a breakdown voltage of 5V is connected in parallel with a load resistor of 1kΩ. If the input voltage is 12V, calculate the current flowing through the Zener diode when the load current is 4mA.

Solution:

1. Determine the total current (I_T):
   $ I_T = \frac{V_s - V_Z}{R} $

   Assume the series resistor (R) is 1kΩ. Therefore,

   $ I_T = \frac{12V - 5V}{1000Ω} = \frac{7V}{1000Ω} = 0.007A = 7mA $

2. Calculate the current through the Zener diode (I_Z):
   $ I_Z = I_T - I_L = 7mA - 4mA = 3mA $

Problem Statement:
Zener breakdown in a semiconductor diode occurs, when

Options:

(a) forward current exceeds certain value

(b) reverse bias exceeds certain value

(c) forward bias exceeds certain value

(d) potential barrier is reduced to zero

Answer: (b)

Solution:

Zener breakdown in a semiconductor diode occurs when reverse bias exceeds certain value, which is known as breakdown or Zener or Avalanche voltage.

Problem Statement:
Directions: Each of these questions contains two statements : Statement I and Statement II. Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below.

Statement I: A Zener diode is used to get constant voltage at variable current under reverse bias.

Statement II: The most popular use of Zener diode is as voltage regulator.

Answer: (a)

Solution:

Zener diodes are specially designed junction diodes, which can operate in the reverse breakdown voltage region continuously without being damaged. The Zener diode is used as a voltage regulator as constant voltage at variable current under reverse bias is obtained from it.

Problem 2: LED Forward Bias#

Problem Statement: An LED has a forward voltage drop of 2V and requires a forward current of 20mA. Calculate the required series resistor when connected to a 9V supply.

Solution:

1. Calculate the voltage across the resistor (V_R):
   $ V_R = V_s - V_{LED} = 9V - 2V = 7V $

2. Determine the resistor value (R):
   $ R = \frac{V_R}{I} = \frac{7V}{0.02A} = 350Ω $

Problem Statement:
Directions: Each of these questions contains two statements : Statement I and Statement II. Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below.

Statement I: Light Emitting Diode (LED) emits spontaneous radiation.

Statement II: LED are forward biased $\rho-n$ junctions.

Answer: (b)

Solution:

When a junction diode is forward biased energy is released at the junction due to recombination of electrons and holes. In the junction diode made up of gallium arsenide or indium phosphide, the energy is released in visible region. Such a junction diode is called Light Emitting Diode or LED. The radiated energy emitted by LED is equal or less than band gap of semiconductor.

Problem 3: Photodiode Output Voltage#

Problem Statement:
A photodiode generates a photocurrent of 5mA under illumination. If it is connected to a load resistor of 1kΩ, what is the output voltage across the load?

Solution:

1. Using Ohm’s Law to find the output voltage (V_out):
   $ V_{out} = I_{photo} \times R_L = 0.005A \times 1000Ω = 5V $