Story of RLC and pn-diode..

There is lot of hype on VLSI (what is VLSI? another topic please..).. before we start, let's understand most basic concept and it's application.. let's master this competely and solve few problems at end.

The p-n junction diode is one of the most important and fundamental semiconductor devices. It is a two-terminal device that consists of a p-type semiconductor and an n-type semiconductor. The p-type semiconductor has an excess of positive charge carriers (holes), while the n-type semiconductor has an excess of negative charge carriers (electrons).

When the p-type and n-type semiconductors are joined together, a depletion region is created at the junction. This is a region where there are very few charge carriers, and it creates a barrier that prevents current from flowing in the reverse direction. However, when the diode is forward biased, the barrier is lowered and current can flow.

The p-n junction diode has a number of important applications, including rectification, switching, and signal processing. It is used in a wide variety of electronic devices, such as power supplies, radios, and computers.

Resistance, capacitance, and inductance:

Resistance, capacitance, and inductance are three fundamental electrical properties. Resistance is the opposition to current flow. Capacitance is the ability to store electrical energy. Inductance is the opposition to change in current flow.

All electrical circuits contain resistance, capacitance, and inductance. These properties can be used to design circuits that perform a variety of functions. For example, resistors are used to control the current flow in a circuit, capacitors are used to filter signals, and inductors are used to store energy.

The p-n junction diode has all three of these properties. It has a resistance, which is determined by the doping concentration of the p-type and n-type semiconductors. It also has a capacitance, which is due to the depletion region at the junction. And it has an inductance, which is due to the leads of the diode.

The resistance and capacitance of the p-n junction diode are important factors in its operation. For example, the resistance of the diode determines the amount of current that can flow through it. And the capacitance of the diode can affect its switching speed.

Example:

A p-n junction diode is used in a power supply to convert alternating current (AC) to direct current (DC). The diode allows current to flow in one direction, but not in the other. This rectifies the AC waveform and produces a DC output.

The resistance of the diode determines how much current can flow through the power supply. And the capacitance of the diode affects how much ripple is present in the DC output.

Conclusion:

The p-n junction diode is a fundamental semiconductor device with a wide range of applications. It is important to understand resistance, capacitance, and inductance in order to understand how the p-n junction diode works and how it can be used in electronic circuits.

Please solve these problems and post answers as reply in linkedin...

  1. A p-n junction diode has a forward voltage of 0.7 V at a current of 1 mA. Calculate the dynamic resistance of the diode at this point.
  2. A p-n junction diode has a reverse saturation current of 10 nA. Calculate the reverse current through the diode when it is reverse biased with a voltage of 10 V.
  3. A p-n junction diode is used in a power supply to convert AC to DC. The input voltage to the power supply is 120 V AC, and the output voltage is 12 V DC. Calculate the efficiency of the power supply.
  4. A p-n junction diode is used in a switching circuit to turn on and off an LED. The LED has a forward voltage of 2 V and a forward current of 20 mA. Calculate the value of the resistor that needs to be connected in series with the LED to protect the diode from excessive current.

MOHAMED ISMAIL

An Enthusiastic VLSI Engineer, looking to explore the world of Electronics

1 年

1. 700 Ohm 2. 10^9 Ohm 3. Need input current and output current to find the efficiency. 10(output current / input current) % is the efficiency. 4. A Voltage V greater than 2 is needed (to accommodate some drop across resistor 50(V-2) Ohm resistor is needed. Pls correct me if I'm wrong

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