VI Characteristics of a P-N Junction Diode

A P-N junction diode is a two-terminal semiconductor device formed by joining p-type and n-type semiconductors, creating a junction at their boundary.

• It allows electric current to flow easily in one direction (forward bias) and blocks current in the opposite direction (reverse bias).
• This unidirectional behavior arises from the formation of a depletion region and a potential barrier at the junction when two types of semiconductors are combined.

Formation of p-n Junction

To understand the formation of a p–n junction, consider a thin p-type silicon semiconductor sheet. If a small amount of pentavalent impurity (valency 5) is added to a portion of this sheet, that part of the p-type silicon is converted into n-type silicon. As a result, the same semiconductor crystal now contains both p-type and n-type regions. The boundary formed between these two regions is called the p–n junction.

• After the formation of a p–n junction, electrons diffuse from the n-side to the p-side and holes diffuse from the p-side to the n-side due to concentration differences, producing a diffusion current.

• When electrons and holes move, they leave behind ionized donor and acceptor atoms, which are immobile charges.
• These fixed charges near the junction form a depletion region that is free from mobile charge carriers.
• The positive and negative charges across the junction create an internal electric field.
• This electric field causes drift current, which flows opposite to the diffusion current.

Forward Bias

Forward bias is the condition in which a p–n junction diode is connected to an external voltage source such that the p-type region is connected to the positive terminal and the n-type region to the negative terminal of the battery. In this condition, the applied electric field opposes the junction's built-in potential barrier, reducing its strength. As a result, the depletion region becomes thinner, the junction resistance decreases, and current flows easily through the diode once the applied voltage exceeds the threshold value (about 0.7 V for silicon).

Reverse Bias

Reverse bias is the condition in which a p–n junction diode is connected to an external voltage source such that the n-type region is connected to the positive terminal and the p-type region to the negative terminal of the battery. In this arrangement, the applied electric field is in the same direction as the built-in electric field, which increases the potential barrier. As a result, the depletion region becomes wider, the junction resistance increases, and only a very small leakage current flows. If the reverse voltage is increased further, the depletion layer becomes even thicker and more resistive until breakdown occurs at a sufficiently high voltage.

P-N Junction Formula

The potential difference created by the electric field in the p-n junction is given by:

• E₀= junction voltage at no bias
• V_T = thermal voltage (≈ 26 mV at room temperature)
• N_D = donor concentration
• N_A = acceptor concentration
• n_i = intrinsic carrier concentration

Characteristics

• In forward bias, the p-type is connected to the positive terminal and the n-type to the negative terminal, reducing the potential barrier. For germanium diodes, conduction starts at ~0.3 V, and for silicon diodes at ~0.7 V.
• Initially, as the applied voltage overcomes the potential barrier, the current increases slowly, producing a non-linear portion of the I–V curve.
• Once the potential barrier is fully overcome, the diode conducts freely, and the current rises sharply with voltage, giving a nearly linear I–V curve.
• In reverse bias, the potential barrier and resistance increase, and a small reverse saturation current flows due to minority carriers.
• If the reverse voltage is increased further, minority carriers gain enough energy to trigger a large current, causing breakdown at the breakdown voltage, which may damage the diode.

Solved Problems

Question 1: A transistor has a current gain of 30. If the collector resistance is R_C = 6 kΩ and the input resistance is R_in = 1 kΩ, calculate the voltage gain.

Solution: Resistance gain =

Voltage gain = Current gain × Resistance gain

Question 2: A P-N junction has N_D = 5 × 10¹⁵ cm⁻³, N_A = 2 × 10¹⁶ cm⁻³, and n_i = 1.5 × 10¹⁰ cm⁻³.. Find the built-in potential E₀​ at room temperature (V_T = 0.026 V).

Solution: The built-in potential of a P-N junction is given by,

Substitute the values

Question 3: A diode has a forward resistance of 500 Ω. If a voltage of 0.6 V is applied, calculate the forward current.

Solution: Using Ohm's law

Substitute the values

Question 4: A silicon diode is forward-biased with a voltage of 0.8 V. The diode has a saturation current I_S = 10^-12 A and thermal voltage V_T = 0.026 V. Calculate the forward current. I using the diode equation

Solution: Substitute the given values

Unsolved Problems

Question 1: A P-N junction diode has a forward voltage of 0.65 V. The diode has a resistance of 250 Ω in the forward direction. Calculate the current through the diode.

Question 2: A silicon diode is reverse-biased with 10 V. The saturation current I_s = 5 × 10^{⁻¹³ A.} Find the reverse current flowing through the diode.

Question 3: The donor concentration in the n-type region is N_D = 10¹⁷ cm⁻³and the acceptor concentration in the p-type region is N_A = 2 × 10¹⁶ cm⁻³.. If the intrinsic carrier concentration is n_i = 1.5 × 10¹⁰ cm⁻³and V_T = 0.026 V, calculate the built-in potential E₀.​.

Question 4: A diode has a saturation current I_S = 2 × 10^⁻¹² A and thermal voltage V_T = 0.026 V. Calculate the forward current if the applied voltage is 0.75 V using the diode equation.

Question 5: A diode is forward biased with 0.7 V. If the diode’s forward resistance is 500 Ω and the saturation current I_S = 10^-12 A, estimate the total current through the diode, considering both the resistance and diode equation effect.