PN Junction

A PN junction diode is a crucial electronic component formed by joining p-type and n-type semiconductors. This junction permits current to flow in one direction (forward bias) and prevents it in the opposite direction (reverse bias). This crucial characteristic makes the diode indispensable in applications such as rectifiers, voltage regulators, and signal detectors, serving as the foundation of modern electronics.

1.0Definition of P-N Junction

• When a P-type semiconductor is joined to a N-type semiconductor such that the crystal structure remains continuous at the boundary, the resulting arrangement is called a P-N junction diode.

Symbol of P-N Junction Diode

2.0Process In P-N Junction

1. Diffusion

• It occurs for majority charge carriers due to concentration gradient. Holes diffuse from p-side to n-side of the junction. Electrons diffuse from n-side to p-side of the junction. Direction of diffusion current is from p-side to n-side.

2. Drift

• In a p-n junction, minority charge carriers are driven by electric fields within the depletion region. Electrons drift from the p-side to the n-side, while holes drift from the n-side to the p-side. This results in a drift current that opposes the diffusion current.
• Initially, diffusion current dominates, but as the electric field strengthens, the drift current increases. Eventually, the diffusion and drift currents balance out, forming a stable p-n junction with no net current under equilibrium.

$|I_{\text{drift}}|=|I_{\text{diffusion}}|$

3.0Depletion Width

• The depletion width is the area in a p-n junction where mobile charge carriers, such as electrons and holes, have moved away. This movement leaves behind fixed charged ions, creating a zone that is electrically neutral and devoid of free charge carriers. As a result, an electric field forms across this region.

• Width of Depletion region ≈ 0.1μm
• It depends on temperature, doping and type of material width ∝ 1Doping

4.0Barrier Potential

• An electric field is established from the n-side to the p-side of the junction.
• Consequently, the n-region has a higher potential compared to the p-region.
• Voltage between the p and n regions creates a barrier that restricts the movement of holes and electrons across the junction. This barrier is known as the "potential barrier."
• The magnitude of the potential barrier is influenced by factors such as temperature, doping concentrations, and the type of semiconductor material.
• While the potential barrier resists the flow of majority carriers, it allows the movement of minority carriers.

5.0Types of Biasing

1. Forward Biasing

• The p-side of the semiconductor is connected to the positive terminal of the battery and n-side to the negative terminal. The applied voltage (V) is in the opposite direction to the barrier potential (V₀).
• The depletion layer narrows, and the barrier height is reduced.. Under forward bias, the effective barrier height becomes V₀ - V.
• Current is mainly due to diffusion of charge carriers and it is of the order of mA.

$|I_{\text{diffusion}}|>|I_{\text{drift}}|$

• Forward bias resistance of diode is small and of the order of k. \( K\Omega \)
• Current is almost zero up to a certain value of applied voltage. This voltage is called knee-Voltage or threshold voltage VTH. \(V_th\)
• Current increases exponentially after the knee-voltage.

2. Reverse Biasing

• The p-side of the semiconductor is linked to the negative terminal of the battery, while the n-side is connected to the positive terminal. The direction of the applied voltage (V) supports the barrier potential (V₀).
• The depletion layer width increases and the barrier height also increases. The effective barrier height under reverse is V0+V. \(v_0+v\)
• Diffusion of majority carrier stops and current is mainly due to minority carriers. It is of the order of μA.

$|I_{\text{diffusion}}|<|I_{\text{drift}}|$

• Reverse bias resistance of diode is large and of the order of $M\Omega$.
• Current is very small & almost constant for any value of applied voltage, as it is limited by the concentration of minority carriers. It is called reverse saturation current $I_0$
• When $V=V_{BR}$ The reverse current in the diode rises sharply, with even a small increase in the bias voltage leading to a significant change in the current.

Conclusion:-

To check the biasing of diode or PN junction:

1. If $V_p-V_n>0$ then diode D is Forward Bias (F.B.) 

2. If $V_p-V_n<0$ then diode D is Reverse Bias (R.B.)

3. If $V_p-V_n=0$ then diode D is Unbiased. 

6.0Characteristic Curve of P-N Junction Diode

• In forward bias when voltage is increased from 0V in steps and corresponding value of current is measured,the curve comes as OB of figure.
• We may note that current increases very sharply after a certain knee voltage. At this voltage, the barrier potential is completely eliminated and the diode offers a low resistance.
• In reverse bias a microammeter has been used as current is very small.When reverse voltage is increased from 0Vand corresponding values of current measured the plot comes as OCD.
• We may note that reverse current is almost constant hence called reverse saturation current. It implies that diode resistance is very high. As reverse voltage reaches value VB \(V_B\)called breakdown voltage, current increases very sharply.

7.0Ideal Diode And Real Diode

Ideal Diode

Forward Bias : R = 0 (short circuit)   Reverse Bias : $R=\infty$ (open circuit)

Real Diode

• It allows the current only in forward direction.
•  There will be a finite potential difference, when current is passing through the diode.
•  For all practical purposes the current in Reverse bias is zero. It acts as very high resistance in reverse bias.

8.0Solved Example

1.Figure shows a diode connected to an external resistance and an e.m.f. Assuming that the barrier potential developed in diode is 0.5 V, obtain the value of current in the circuit in milliampere.   

Solution:

E = 4.5 V, R = 100 ,

Voltage drop across p-n junction =0.5 V

Effective voltage in the circuit V=4.5-0.5=4 V

Current in the circuit,

$I=\frac{V}{R}=\frac{4}{100}=0.04A=0.04\times 1000mA=40mA$

2. If current in a given circuit is 0.1 A then calculate resistance of P–N junction.      

Solution:

Let resistance of PN junction be R then $$\begin{gathered} I=\frac{5}{R+30+10}0.1=\frac{5}{R+40} \\ R=10\Omega \end{gathered}$$

3.What is the value of current I in given circuits?

Solution:

$I=\frac{20}{10+10}=1A$