The P–N junction diode is one of the most important fundamental building blocks of modern electronics. From rectifiers to signal processing circuits, diodes play a crucial role in controlling the direction of current flow. It is mathematically described in the P–N junction diode equation.
This equation is the relationship between the diode current and the applied voltage and is known as the rectifier equation, the Shockley diode equation, or simply the diode equation.
In this article, we will explore
- The diode equation in depth—starting from its physical meaning, moving through its mathematical derivation,
- Analyze its behavior in different biasing conditions.
- Discuss its real-world applications, and
- Finally, apply it to real-world engineering problems.
By the end, you will not only understand the equation but also feel confident using it in practical scenarios.
Since the current in a p–n junction arises due to the diffusion and drift of charge carriers under an energy barrier. At equilibrium, diffusion current (due to the carrier concentration gradient) and drift current (due to the internal electric field) balance each other, and hence, the net current across the junction is zero.
When a forward voltage is applied, the barrier potential decreases; therefore, more majority charge carriers cross the junction, leading to an exponential increase in current as predicted by Boltzmann statistics.
In reverse bias, the barrier increases, blocking the majority carriers and allowing only a small reverse saturation current due to minority carriers.
To explain mathematically the current flowing across the junction, the pn junction diode equation is derived.
The diode equation describes how current varies with applied voltage across a p–n junction. The following formula represents it:
$$I = I_S \left(e^{\frac{qV}{\eta kT}} – 1\right) $$
This equation tells us:
- Current increases exponentially with voltage in forward bias
- Current remains very small in reverse bias
Now, let’s derive the diode equation.
Diode Equation: Derivation
According to Boltzmann’s equation, the hole concentration in the p-region (pp) and in the n-region (pn) are related as
$$p_n=p_pexp\left( -\frac{eV_B}{kT} \right)\quad……………. (1) $$
It represents the concentration of holes in the n-region when no voltage is applied. Suppose a forward voltage V is applied; then it reduces the potential barrier from VB to VB – V.
Due to this applied voltage, the majority charge carriers start diffusing in the depletion region, and after crossing the junction, holes reach the n-region and electrons reach the p-region. Therefore, minority carrier concentration increases in both regions.
Suppose Δpn is the increase in hole concentration in the n region; therefore, as per equation (1), we can write
$$p_n+\Delta p_n=p_pexp\left[ -\frac{\left( V_B-V \right)}{kT} \right] \quad…………. (2)$$
Subtracting equation (1) from equation (2), we get
$$\Delta p_n=p_pexp\left( -\frac{eV_B}{kT} \right)\left[ exp\left( \frac{eV}{kT} \right)-1 \right]\quad ………….(3)$$
Similarly, in the p region, an increase in electron concentration is given by
$$\Delta n_p=n_nexp\left( -\frac{eV_B}{kT} \right)\left[ exp\left( \frac{eV}{kT} \right)-1 \right] \quad………….(4)$$
Due to the diffusion of these charge carriers, current flows. And if A is the area of the junction, then the total current flowing across the junction is
$$I = JA = eA\left(\Delta n_p v_{d_n}+\Delta p_nv_{d_p}\right) $$
[Since J = nevd]
Putting the values of Δpn and Δnp from equations (3) and (4), we get
$$I = eA\left( n_n v_{d_n}+ p_pv_{d_p}\right)exp\left( -\frac{eV_B}{kT} \right)\left[ exp\left( \frac{eV}{kT} \right)-1 \right]$$
$$\Rightarrow I = I_S\left[ exp\left( \frac{eV}{kT} \right)-1 \right]$$
Adding an ideality factor η (≈1 for Ge, ≈2 for Si)
$$I = I_S\left[ exp\left( \frac{eV}{\eta kT} \right)-1 \right]\quad ……………(5)$$
Where $$I_S= eA\left( n_n v_{d_n}+ p_pv_{d_p}\right)exp\left( -\frac{eV_B}{kT} \right) $$
It is called reverse saturation current. And equation (5) is the required diode equation.
In forward bias: V is positive, hence exp(eV/kT) >> 1; therefore, $$I=I_Sexp\left( \frac{eV}{\eta kT} \right) $$
In reverse bias: V is negative, hence $$I=I_S$$
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Conclusion
- The diode equation explains current flow in a p–n junction
- It shows the exponential dependence of current on voltage
- Temperature and material properties play a major role
- It is essential for designing all semiconductor devices
Thus, the diode equation fundamentally captures how carrier movement over a voltage-controlled energy barrier produces the characteristic exponential I–V behavior of a diode.
Final Thought:
Understanding the diode equation is like learning the grammar of electronics—once you master it, you can design and analyze almost any circuit with confidence.
Important Questions for Exam
- Derive the diode equation.
- Explain the physical meaning of each term in the diode equation.
- Calculate the diode current for the given voltage and temperature.
- Discuss the temperature dependence of the diode equation.
- Sketch and explain I–V characteristics.
FAQs
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Q1. What is the diode equation?
It is a mathematical relation that describes current flow in a diode as a function of voltage.
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Q2. Why is the diode equation exponential?
Because carrier injection across the junction depends exponentially on applied voltage.
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Q3. What is saturation current?
It is the small reverse current flowing through the PN junction due to minority carriers.
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Q4. What is the current equation of a diode?
The diode equation is I = ISexp[(eV/kT) - 1].