p-n junction diode

I-V characteristics and use as a rectifier.

Physics5 questions

Formation of a p-n junction

When p-type and n-type regions are formed in one crystal, electrons diffuse from the n side to the p side and holes from p to n. Near the junction they recombine, leaving behind immobile positive donor ions on the n side and negative acceptor ions on the p side. This thin region with no free carriers is the depletion region. The ions create an electric field and a barrier potential (about 0.7 V for Si and 0.3 V for Ge) that opposes further diffusion. At equilibrium, diffusion current equals the opposite drift current.

Forward bias

The p side is connected to the positive terminal. The applied voltage opposes the barrier, the depletion region narrows and current rises sharply once the voltage exceeds the threshold (knee or cut-in) voltage, about 0.7 V for silicon. The current is in milliamperes and is due to majority carriers.

Reverse bias

The p side is connected to the negative terminal. The barrier increases, the depletion region widens and only a very small current (microamperes) flows, due to minority carriers. This current is nearly independent of voltage until breakdown occurs at a large reverse voltage.

Dynamic resistance

r_d = ΔV/ΔI. If a change of 0.1 V causes a change of 20 mA in forward bias, r_d = 5 Ω. A diode's resistance is low in forward bias and very high in reverse bias.

Diode as a rectifier

A diode conducts only in forward bias, so it converts AC into one-direction (DC) output.

  • Half-wave rectifier: one diode; output only during alternate half-cycles. Output ripple frequency equals the input frequency.
  • Full-wave rectifier: two diodes with a centre-tapped transformer; each conducts in alternate half-cycles, so current flows through the load in the same direction during both halves. The ripple frequency is twice the input frequency: 100 Hz for 50 Hz mains.

A capacitor filter across the load smooths the pulsating output.

Worked example

A silicon diode (drop 0.7 V) in forward bias in series with a 1 kΩ resistor across a 5 V supply carries I = (5 – 0.7)/1000 = 4.3 mA.

Key points

  • Depletion region: immobile ions, no free carriers.
  • Forward bias narrows it; reverse bias widens it.
  • Reverse current is tiny and due to minority carriers.
  • Full-wave output frequency = 2 × input frequency.

Common mistakes

  • Forgetting the 0.7 V drop when calculating circuit current.

Practice questions

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