## Diode Bridge (Simple AC to DC converter)

Thunderstruck

A diode bridge is a simple arrangement of diodes that can be used to convert an AC signal into a DC signal.

### Circuit ### Explanation

A diode permits current to flow in a single direction only, from the anode to the cathode. Which is in the direction of the symbol shown below: The live and neutral from an AC power supply both have two diodes connected to them, one allowing positive charge to flow to positive out and the other allowing positive charge to flow from negative out. This is an example of full-wave rectification. This is animated below:   The problem is that the DC output isn’t constant, and will even switch off entirely when the AC input is at 0V. To solve this we use a capacitor, which discharges when the AC signal starts to dip, and is recharged when the AC signal starts to rise. This is an example of full-wave rectification with a smoothing capacitor. This is animated below:   ### Rectification Stages

The following shows different types of rectification, building up to the typical diode bridge circuit.

#### Normal AC This is just an AC signal.

#### Half-wave rectification The diode allows only the positive half of the signal at the output.

#### Full-wave rectification The output is the equivalent of the absolute value of the input.

#### Smoothed full-wave rectification (DC) The capacitor smooths the full-wave rectification, so that it doesn’t dip too far below peak voltage. The difference in voltage between the peak and the lowest point is called the ripple voltage.

### Calculations

To calculate the capacitor needed for the diode bride, use the following equation:

$$C\ge\frac{I}{2Vf}$$
Where:
$$f$$ is the frequence of the AC signal.
$$V$$ is the maximum ripple voltage allowed.

Example:
We Know:
$$I=100mA$$, the DC current we want to be able to provide.
$$f=50Hz$$, the frequency printed on the AC power supply.
$$V=1V$$, the maximum voltage wobble allowed.

To find $$C$$:
$$C\ge\frac{I}{2Vf}$$
$$C\ge\frac{0.1}{2\times1\times50}$$
$$C\ge1000μF$$

### Afterthoughts

When choosing a diode, ensure that it rated above the voltage of your circuit. Usually, the lower the voltage rating of the diode, the lower the voltage drop. The 1N4001 is perfect for beginners, as they will not break anything if used incorrectly.

When choosing a capacitor, the higher its capacitance, the more current the circuit can provided. The more current the the circuit draws, the quicker the capacitor discharges, which means a noisier DC signal. A 1000μF capacitor should be sufficient for almost any hobby-level current draw.

Written on the 24th day of January in the year of our Lord two thousand and seventeen