Semiconductor Electronics: Materials, Devices And Simple Circuits 14.1 Introduction | Class 12 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 5 min read
Semiconductor Electronics: Materials, Devices And Simple Circuits 14.1 Introduction – this guide gives you a concise, exam-ready overview of Semiconductor Electronics: Materials, Devices And Simple Circuits 14.1 Introduction from Class 12 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
14.6 SEMICONDUCTOR DIODE
A semiconductor diode is essentially a p-n junction with metallic contacts at both ends, forming a two-terminal device. It allows current to flow primarily in one direction, making it useful for rectification and switching applications.
The diode symbol shows an arrow indicating the conventional current direction when forward biased.
The diode behavior depends on the applied external voltage (bias) across its terminals, which modifies the built-in barrier potential of the p-n junction.
14.6.1 p-n junction diode under forward bias:
When the p-side is connected to the positive terminal and the n-side to the negative terminal of a battery, the diode is forward biased. The applied voltage opposes the built-in potential, reducing the barrier height and narrowing the depletion region.
At small applied voltages, only a few carriers have enough energy to cross the junction, resulting in small current. As the voltage increases beyond a threshold (cut-in voltage ~0.7 V for silicon, ~0.2 V for germanium), more carriers cross, and current increases exponentially.
Electrons from the n-side cross into the p-side, and holes from the p-side cross into the n-side, a process called minority carrier injection. This increases minority carrier concentration near the junction, causing diffusion currents on both sides.
The total forward current is the sum of electron and hole diffusion currents, typically in milliamperes.
14.6.2 p-n junction diode under reverse bias:
When the polarity is reversed (n-side positive, p-side negative), the diode is reverse biased. The applied voltage adds to the barrier potential, increasing the depletion width and barrier height.
This suppresses diffusion current drastically. However, a small drift current due to minority carriers exists, typically in microamperes, and is almost independent of applied voltage.
At a critical reverse voltage called breakdown voltage, the reverse current increases sharply, potentially damaging the diode if current is not limited.
The diode's current-voltage (V-I) characteristics show negligible current under reverse bias and exponential increase under forward bias beyond threshold voltage.
Dynamic resistance of the diode is defined as the ratio of small changes in voltage to current (r_d = ΔV / ΔI).
This section explains the fundamental operation of the p-n junction diode and its asymmetric conduction properties.
📊 Diagram: Figure 14.12(a) shows a semiconductor diode with metallic contacts; (b) shows the diode symbol. Figure 14.13(a) depicts forward bias connection; (b) shows barrier potential changes with applied voltage. Figure 14.14 illustrates minority carrier injection under forward bias. Figure 14.15(a) shows reverse bias connection; (b) shows increased barrier potential under reverse bias. Figure 14.16(a) and (b) show experimental setups for V-I characteristics in forward and reverse bias; (c) shows typical V-I characteristic curve of a silicon diode.
🧪 Activity: Experimental study of V-I characteristics of a p-n junction diode using variable voltage source and measuring current with milliammeter and microammeter.
🔗 Connection: Leads to practical applications of diodes such as rectification in power supplies.
Frequently asked questions
In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.
Option (c) is correct: Holes are minority carriers and pentavalent atoms are the dopants in n-type silicon.
Explanation: In n-type silicon, pentavalent atoms (such as phosphorus, arsenic) are added as dopants. These atoms have five valence electrons, one more than silicon's four valence electrons. The extra electron becomes a free electron, making electrons the majority carriers. Holes, which are the absence of electrons, are minority carriers in n-type silicon.
Therefore, electrons are majori
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to $(E_{ ext{g}})_{ ext{C}}$, $(E_{ ext{g}})_{ ext{Si}}$ and $(E_{ ext{g}})_{ ext{Ge}}$. Which of the following statements is true? (a) $(E_{ ext{g}})_{ ext{Si}} < (E_{ ext{g}})_{ ext{Ge}} < (E_{ ext{g}})_{ ext{C}}$ (b) $(E_{ ext{g}})_{ ext{C}} < (E_{ ext{g}})_{ ext{Ge}} > (E_{ ext{g}})_{ ext{Si}}$ (c) $(E_{ ext{g}})_{ ext{C}} > (E_{ ext{g}})_{ ext{Si}} > (E_{ ext{g}})_{ ext{Ge}}$ (d) $(E_{ ext{g}})_{ ext{C}} = (E_{ ext{g}})_{ ext{Si}} = (E_{ ext{g}})_{ ext{Ge}}$
The correct statement is (a) $(E_{\mathrm{g}})_{\mathrm{Si}} < (E_{\mathrm{g}})_{\mathrm{Ge}} < (E_{\mathrm{g}})_{\mathrm{C}}$. Explanation: Carbon (diamond) has the largest band gap (~5.5 eV), silicon has an intermediate band gap (~1.1 eV), and germanium has the smallest band gap (~0.66 eV). Thus, the energy band gap decreases in the order Carbon > Silicon > Germanium.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above.
The correct answer is (c) hole concentration in p-region is more as compared to n-region.
Explanation: In an unbiased p-n junction, diffusion occurs due to concentration gradient. Holes diffuse from the region of higher concentration (p-region) to the region of lower concentration (n-region). The movement is driven by concentration difference, not by attraction of free electrons or potential difference.
When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lowers the potential barrier. (d) None of the above.
The correct answer is (c) lowers the potential barrier.
Explanation: Applying a forward bias to a p-n junction reduces the built-in potential barrier, allowing majority carriers to cross the junction more easily, thus increasing current flow.
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