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Semiconductor Electronics - p-n Junction Diode

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

A pnp-n junction is formed when a pp-type semiconductor is joined with an nn-type semiconductor. Two processes occur: diffusion and drift.

The depletion region is a thin layer at the junction devoid of mobile charge carriers (electrons and holes), consisting only of immobile ions. Its thickness is approximately 10610^{-6} m.

Barrier Potential (VbV_b) is the internal electric field created by immobile ions that opposes the further diffusion of majority carriers. For Silicon, Vb0.7V_b \approx 0.7 V and for Germanium, Vb0.3V_b \approx 0.3 V.

In Forward Bias, the pp-side is connected to the positive terminal and nn-side to the negative terminal. This reduces the depletion layer width and the effective barrier height, allowing current to flow.

In Reverse Bias, the pp-side is connected to the negative terminal and nn-side to the positive terminal. This increases the depletion layer width and barrier height, resulting in a very small (negligible) current called reverse saturation current.

A Rectifier is a device that converts AC to DC. A Half-wave rectifier uses one diode and conducts during only one half-cycle of the AC input. A Full-wave rectifier uses two diodes (center-tap) or four (bridge) and conducts during both half-cycles.

The output frequency of a half-wave rectifier is the same as the input frequency (fout=finf_{out} = f_{in}), whereas for a full-wave rectifier, it is double (fout=2finf_{out} = 2f_{in}).

📐Formulae

VB=EdV_B = E \cdot d

rd=ΔVΔIr_d = \frac{\Delta V}{\Delta I}

I=Is[eeVnkT1]I = I_s [e^{\frac{eV}{nkT}} - 1]

fout=fin (Half-wave Rectifier)f_{out} = f_{in} \text{ (Half-wave Rectifier)}

fout=2fin (Full-wave Rectifier)f_{out} = 2f_{in} \text{ (Full-wave Rectifier)}

💡Examples

Problem 1:

A pnp-n junction diode has a dynamic resistance of 25Ω25 \Omega in forward bias. If the forward voltage changes by 0.020.02 V, calculate the change in forward current.

Solution:

Given ΔV=0.02\Delta V = 0.02 V and rd=25Ωr_d = 25 \Omega. Using the formula for dynamic resistance rd=ΔVΔIr_d = \frac{\Delta V}{\Delta I}, we get ΔI=ΔVrd\Delta I = \frac{\Delta V}{r_d}. Substituting the values: ΔI=0.0225=0.0008\Delta I = \frac{0.02}{25} = 0.0008 A.

Explanation:

The dynamic resistance is the ratio of small change in voltage to the corresponding small change in current in the linear region of the V-I characteristics. Therefore, ΔI=0.8\Delta I = 0.8 mA.

Problem 2:

What is the output frequency of a full-wave rectifier if the input frequency is 5050 Hz?

Solution:

For a full-wave rectifier, the output frequency foutf_{out} is related to input frequency finf_{in} as fout=2finf_{out} = 2f_{in}. Substituting the value: fout=2×50f_{out} = 2 \times 50 Hz =100= 100 Hz.

Explanation:

In a full-wave rectifier, the output reaches a peak twice for every single full cycle of the input AC, hence the frequency of the pulsating DC output is doubled.

p-n Junction Diode - Revision Notes & Key Formulas | CBSE Class 12 Physics