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:
E_0 = V_T \ln \left( \frac{N_D N_A}{n_i^2} \right)
- E0= junction voltage at no bias
- VT = thermal voltage (≈ 26 mV at room temperature)
- ND = donor concentration
- NA = acceptor concentration
- ni = 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 RC = 6 kΩ and the input resistance is Rin = 1 kΩ, calculate the voltage gain.
Solution: Resistance gain =
\frac{R_C}{R_in}
= \frac{6}{1}=6 Voltage gain = Current gain × Resistance gain
=30×6=180
=\text{Voltage gain} = 180
Question 2: A P-N junction has ND = 5 × 1015 cm⁻³, NA = 2 × 1016 cm⁻³, and ni = 1.5 × 1010 cm⁻³.. Find the built-in potential E0 at room temperature (VT = 0.026 V).
Solution: The built-in potential of a P-N junction is given by,
E_0 = V_T \ln \left( \frac{N_D N_A}{n_i^2} \right) Substitute the values
E_0 = 0.026 \ln \left( \frac{(5\times 10^{15}) (2\times 10^{16})}{(1.5\times 10^{10})^2} \right)
E_0 = 0.026 \ln \left( \frac{1\times 10^{32}}{2.25\times 10^{20}} \right)
= 0.026 \ln (4.44 \times 10^{11})
\ln (4.44 \times 10^{11}) \approx 26.8
E_0 = 0.026 \times 26.8
\boxed{E_0 \approx 0.70\,\text{V}}
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
I = \frac{V}{R} Substitute the values
I = \frac{0.6}{500}
= 0.0012\,\text{A} = 1.2\,\text{mA}
\boxed {I = 1.2\,\text{mA}}
Question 4: A silicon diode is forward-biased with a voltage of 0.8 V. The diode has a saturation current IS = 10-12 A and thermal voltage VT = 0.026 V. Calculate the forward current. I using the diode equation
Solution: Substitute the given values
I = 10^{-12} \left( e^{0.8/0.026} - 1 \right)
\frac{0.8}{0.026} \approx 30.77
I \approx 10^{-12} \left( e^{30.77} - 1 \right) \approx 10^{-12} \cdot 1.8\times 10^{13}
\boxed {I \approx 18\,\text{A}}
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 Is = 5 × 10⁻¹³ A. Find the reverse current flowing through the diode.
Question 3: The donor concentration in the n-type region is ND = 1017 cm⁻³ and the acceptor concentration in the p-type region is NA = 2 × 1016 cm⁻³.. If the intrinsic carrier concentration is ni = 1.5 × 1010 cm⁻³ and VT = 0.026 V, calculate the built-in potential E₀..
Question 4: A diode has a saturation current IS = 2 × 10⁻¹² A and thermal voltage VT = 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 IS = 10-12 A, estimate the total current through the diode, considering both the resistance and diode equation effect.