Because a junction diode conducts in essentially one direction, its most important job is rectification — converting alternating current (AC) into direct current (DC). During the half-cycle that forward-biases the diode it conducts and current flows; during the reverse half-cycle it blocks. The output across the load is therefore one-directional, though pulsating.
A half-wave rectifier uses a single diode in series with the load. It conducts during one half of each AC cycle and blocks the other, so the output appears only during alternate half-cycles. It is simple but wasteful: half the input is unused, and the ripple is large. The DC (average) output is $V_{dc} = \dfrac{V_m}{\pi}$, and the maximum theoretical rectifier efficiency is about $40.6\%$.
A full-wave rectifier uses output during both half-cycles. In the centre-tap design two diodes share a centre-tapped transformer so that one conducts on each half-cycle; the bridge design uses four diodes and needs no centre tap. Either way the load current flows the same way in both halves, giving $V_{dc} = \dfrac{2V_m}{\pi}$ and a higher maximum efficiency of about $81.2\%$ — double the half-wave value — with a ripple frequency twice the supply frequency ($100\,\text{Hz}$ for a $50\,\text{Hz}$ mains).
Rectified output is still bumpy, so a filter smooths it. A capacitor placed in parallel with the load charges at each peak and discharges slowly through the load between peaks, filling in the gaps and greatly reducing the ripple. A larger capacitance and a larger load resistance give smoother DC; inductor and pi-filters do the same job more thoroughly.
A special diode designed to work in the reverse breakdown region is the Zener diode. It is heavily doped, so it has a thin depletion region and breaks down sharply at a defined Zener voltage $V_Z$. Beyond breakdown the voltage across it stays almost constant even as current varies widely, making it an ideal voltage regulator. Connected in reverse across the load with a series resistor $R_s$, it holds the output at $V_Z$: the series resistor absorbs the excess, $R_s = \dfrac{V_{in} - V_Z}{I}$, where $I = I_Z + I_L$. As long as the input stays above $V_Z$ and the Zener current remains within limits, the output is steady.
Other special diodes exploit the junction differently. A light-emitting diode (LED) is a forward-biased diode that emits light when electrons and holes recombine across the junction; the photon energy (and hence colour) is set by $E_g$, with $h\nu \approx E_g$. A photodiode works in reverse bias: incident light of energy $h\nu > E_g$ generates electron-hole pairs and the reverse current rises with light intensity, so it acts as a light detector. A solar cell is a large-area p-n junction that, with no external bias, converts light directly into electrical energy by separating photo-generated carriers across the junction field, giving a photovoltage.
The peak voltage of the AC across a half-wave rectifier is $V_m = 20\,\text{V}$. Find the DC (average) output voltage.
Solution- For a half-wave rectifier the average output is $V_{dc} = \dfrac{V_m}{\pi}$.
- $V_{dc} = \dfrac{20}{\pi}$.
- $V_{dc} = \dfrac{20}{3.14}$.
- $V_{dc} \approx 6.37\,\text{V}$.
Answer: The DC output of the half-wave rectifier is about $6.37\,\text{V}$.
A full-wave rectifier is fed from a $50\,\text{Hz}$ AC supply with peak voltage $V_m = 20\,\text{V}$. Find the DC output voltage and the ripple frequency.
Solution- For a full-wave rectifier $V_{dc} = \dfrac{2V_m}{\pi} = \dfrac{2\times20}{\pi}$.
- $V_{dc} = \dfrac{40}{3.14} \approx 12.7\,\text{V}$ — double the half-wave value.
- Each input cycle gives two output pulses, so the ripple frequency is $2f$.
- $f_{ripple} = 2\times50 = 100\,\text{Hz}$.
Answer: DC output $\approx 12.7\,\text{V}$; ripple frequency $= 100\,\text{Hz}$.
A Zener diode of $V_Z = 6\,\text{V}$ is used to regulate an unregulated supply of $10\,\text{V}$ through a series resistor. If the load draws $20\,\text{mA}$ and the Zener carries $10\,\text{mA}$, find the value of the series resistor $R_s$.
Solution- Voltage across $R_s$ is $V_{in} - V_Z = 10 - 6 = 4\,\text{V}$.
- Current through $R_s$ is $I = I_Z + I_L = 10 + 20 = 30\,\text{mA} = 0.03\,\text{A}$.
- $R_s = \dfrac{V_{in} - V_Z}{I} = \dfrac{4}{0.03}$.
- $R_s \approx 133\,\Omega$.
Answer: The series resistor must be about $133\,\Omega$.
An LED is made of a material with energy gap $E_g = 1.9\,\text{eV}$. Find the wavelength and colour of the emitted light. (Take $hc = 1240\,\text{eV nm}$.)
Solution- The emitted photon energy equals the gap: $h\nu \approx E_g$.
- $\lambda = \dfrac{hc}{E_g} = \dfrac{1240\,\text{eV nm}}{1.9\,\text{eV}}$.
- $\lambda \approx 653\,\text{nm}$.
- A wavelength near $653\,\text{nm}$ lies in the red part of the spectrum.
Answer: The LED emits light of wavelength about $653\,\text{nm}$, which appears red.
A semiconductor used in a photodiode has $E_g = 2.8\,\text{eV}$. Find the maximum wavelength of light it can detect. (Take $hc = 1240\,\text{eV nm}$.)
Solution- To create an electron-hole pair the photon must satisfy $h\nu > E_g$, i.e. $\dfrac{hc}{\lambda} > E_g$.
- The longest detectable wavelength is $\lambda_{max} = \dfrac{hc}{E_g}$.
- $\lambda_{max} = \dfrac{1240}{2.8}$.
- $\lambda_{max} \approx 443\,\text{nm}$.
Answer: The photodiode can detect light only up to about $443\,\text{nm}$; longer (lower-energy) wavelengths cannot free carriers.
In a Zener-regulated supply, $V_{in}$ varies from $8\,\text{V}$ to $12\,\text{V}$ while $V_Z = 5\,\text{V}$ and the series resistor is $R_s = 100\,\Omega$. Find the Zener current when the load is removed (open circuit) at $V_{in} = 12\,\text{V}$.
Solution- With the load removed, all the resistor current passes through the Zener: $I_Z = I$.
- Voltage across $R_s$ is $V_{in} - V_Z = 12 - 5 = 7\,\text{V}$.
- $I = \dfrac{V_{in} - V_Z}{R_s} = \dfrac{7}{100}$.
- $I_Z = 0.07\,\text{A} = 70\,\text{mA}$.
Answer: The Zener current is $70\,\text{mA}$; this worst case (no load, maximum input) sets the power rating the Zener must withstand.