Calculate the diffusion capacitance of a p-n junction diode.

Collegedunia Team logo

Collegedunia Team

Content Curator

The diffusion capacitance of a p-n junction diode can be calculated using the following formula:

Cd = (q / τ) * (W / A)

where Cd is the diffusion capacitance

  • q is the electronic charge 
  • τ is the minority carrier lifetime 
  • W is the depletion width 
  • A is the junction area.

The diffusion capacitance is a measure of the rate at which the diode can switch from a conducting to a non-conducting state, and is dependent on the characteristics of the diode material and the doping concentrations. The minority carrier lifetime is a measure of how long the minority carriers (electrons in p-type and holes in n-type) can survive before recombining, and is related to the doping concentrations and the temperature.

The depletion width is the width of the region at the p-n junction where the majority carriers have diffused away, leaving behind a region with a low concentration of free charge carriers. The junction area is the area of the p-n junction, which can be approximated by the area of the diode's metal contact pads. The diffusion capacitance of a p-n junction diode can be calculated using the above formula with the appropriate values for q, τ, W, and A.

Also check:

CBSE CLASS XII Related Questions

  • 1.
    An electric field $\vec{E}$ is established across the ends of a cylindrical conductor of length L and area of cross-section A. Discuss how electrons attain an average velocity, independent of time. Hence, obtain a relation between current in the conductor and this ‘average velocity’ of electrons.


      • 2.
        Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.


          • 3.
            Read the following paragraph and answer the questions that follow.
            A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.


              • 4.
                Capacitors are manufactured with certain standard capacitances and working voltages. However, these standard values may not be the ones that are actually needed in a particular application. Two or more capacitors can be grouped in series or in parallel to achieve desired capacitance and voltage. When connected in series, the total capacitance decreases while the voltage rating increases, whereas in parallel connections, the total capacitance increases and maintains the same voltage rating. A capacitor stores energy in the electric field between its plates and stored energy is proportional to the square of the voltage and capacitance $U = \frac{1}{2}CV^2$, where symbols have their usual meanings.
                Two capacitors, one of $3 \ \mu$F and the other of $6 \ \mu$F, are connected in series in the circuit as shown in the figure, for a long time. }


                  • 5.
                    A charged particle $+q$ in an electric field $\vec{E}$ experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field $\vec{B}$. But this magnetic force is perpendicular to both velocity $\vec{v}$ of the charged particle and the magnetic field $\vec{B}$, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses $m$ and $\frac{m}{2}$ having charges $-q$ and $+2q$ respectively. They are accelerated from rest through the same potential difference $V$ and acquire kinetic energy $K_1$ and $K_2$. Then they enter in a region of uniform magnetic field $\vec{B}$ perpendicular to their velocities.


                      • 6.
                        The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.

                          CBSE CLASS XII Previous Year Papers

                          Comments


                          No Comments To Show