The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6V and the load resistance is \$R_L = 4 k \Omega\$. The series resistance of the circuit is \$R_i = 1 k \Omega\$. If the battery voltage \$V_B\$ varies from \$8V\$ to \$16V\$, what are the minimum and maximum values of the current through Zener diode ?

At \$V_B = 8V\$\$i_L = \dfrac{6 \times 10^{-3}}{4} = 1.5 \times 10^{-3} A\$\$i_R = \dfrac{8 - 6 \times 10^{-3}}{1} = 2 \times 10^{-3} A\$\$\therefore i_{zener \, diode} = i_R - i_{load}\$\$= 0.5 \times 10^{-3} A\$At \$V_B = 16 V\$\$i_L = 1.5 \times 10^{-3} A\$\$i_R = \dfrac{(16 - 6) \times 10^{-3}}{1} = 10 \times 10^{-3} A\$\$\therefore i_{zener \, diode} = i_R - i_L\$\$= 8.5 \times 10^{-3} A\$

Two junction diodes one of germanium \$(Ge)\$ and other of silicon \$(Si)\$ are connected as shown in figure to a battery of emf \$12\ v\$ and a load resistance \$10\ k\ \Omega\$. The germanium diode conducts at \$0.3\ V\$ and silicon diode at \$0.7\ V\$. When a current flows in the circuit, then the potential of terminal \$Y\$ will be:

Ge conducts at \$0.3\ V\$ and \$Si\$ Conducts at \$0.7\ V\$. Both \$Ge\$ and \$Si\$ are connected in parallel. When current begins to flow, then the potential difference remains at \$0.3\ V\$, so no current flows through Si-diode.\$\therefore\$ Potential difference across resistive load \$= 12 - 0.3\$\$= 11.7\ V\$\$\therefore\$ Potential of \$Y = 11.7\ V\$.

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>>Class 12
>>Physics
>>Semiconductor Electronics: Materials, Devices and Simple Circuits
>>Introduction to Diodes
>> $Diodes and Their Uses$

Diodes and Their Uses

Q: An \$LED\$ is constructed from a pn junction based on a certain semi-conducting material whose energy gap is \$1.9\ eV\$. Then the wavelength of the emitted light is

Energy gap \$E_g = 1.9eV\$Wavelength of light emitted \$\lambda = \dfrac{hc}{E_g}\$ nm where \$hc = 1240eV\$\$\therefore\$ \$\lambda = \dfrac{1240}{1.9} \approx 650\$ nm \$ = 6.5\times 10^{-7}\$ \$m\$

Q: The colour of light emitted by LED depends on.

LEDs are light emitting diodes which are semiconductors that convert electrical energy into the light energy. The color of the emitted light depends on the type of semiconductor material. Depending on which LEDs can be classified into three wavelengths: ultraviolet, visible and infrared.

Q: A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA. When it operates with a 6 V battery through a limiting resistor R, the value of R is

Light emitting diode is a forward biased P-N junction which emits light. The voltage across it = 2 v Battery voltage = 6 v Hence total voltage in the circuit = V = 6 - 2 = 4 v since a LED has 0 resistance in the forward biased region, total resistance in the circuit = 0 + r = r current I = 10 mA = 10/1000 A = 0.01 A we know that, R = V/I => r = 4/0.01 = 400Ω Hence the value of the limiting resistor is 400Ω

Q: A red LED emits light at\$ 0.1\$ watt uniformly around it. The amplitude of the electric field of the light at a distance of \$1 m\$ from the diode is :

Sine we know that, intensity \$I=\dfrac{P}{4 \pi r^2}\$Also, \$I=\dfrac{1}{2}\varepsilon _0E_0^2\times c\$\$\dfrac{P}{4\pi r^2}=\dfrac{1}{2}\varepsilon _0\times E_0^2\times c\$\$E_0^2=\dfrac{2P}{4\pi\varepsilon _0r^2c}\$\$\dfrac{2 \times0.1 \times9 \times10^9}{1 \times3 \times10^8}=6\$\$\therefore E_0=\sqrt{6}=2.45 V/m\$

Q: The I V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes because

The \$V-I\$ curve for solar cell is available in the fourth quadrant, i.e. where current is negative, which means that current I is supplied by the solar cell and not drawn by it.

Q: Mention the important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED).What should be the order of band gap of an LED if it is required to emit light in the visible range?

1.The reverse breakdown voltage of LEDs are very low, which is around 5V. So enough care is to be taken while fabricating a pn-junction diode such that the high reverse voltages do not occur across them.2. There exist very small resistance to limit the current in LED. So a resistor must be placed in series with the LED such that no damage is occurred to the LED.The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 eV

Q: LEDs have the following advantages over conventional incandescent low power lamps:

LEDs have longer life as compared to incandescent lamps. LED require low operational voltage and less power as compared to incandescent lamps. LEDs also have fast on-off switching capabilities. So, the answer is (d) All of the above.

Q: A p-n junction diode cannot be used

To increase the amplitude of an AC signal, transistors are used. p-n junction diodes can be used as rectifiers and in photo-diode and LED.

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Types of diodes

The different types of diodes and their uses are listed below:

Diode	Uses
p-n Junction Diode	Half-wave rectifier, full-wave rectifier, switch, etc
Zener Diode	Voltage regulator
LED	LED bulbs, lasers, etc
Photodiode	Optoelectronic receivers, solar cells, etc

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Examples of optoelectronic devices

Some important optoelectronic devices are:
1. LED converts electrical energy into light energy. It is useful for low power bulbs.
2. LASER diode converts electrical energy into light energy. It is useful as a high intensity monochromatic light source and finds applications in optical communication systems.
3. Photodiode converts light energy into electrical energy. It finds applications in optical communication systems and DC isolator circuits.
4. Solar cell converts light energy into electrical energy. It finds applications in energy production.

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Semiconductors - LED and LDR

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