# Half-Wave Rectifier

A popular application for diodes are rectifier circuits. Today, we are going to look at a simple half-wave rectifier.

## Half-Wave Rectifiers

A have-wave rectifier is probably the easiest rectifier circuit. For building it, we need nothing more than a single diode. Let's see how a half-wave rectifier works and what properties it has.

## The Rectifier Circuit

The circuit is really simple this time. The rectifier consist of a single diode. Additionally, I added a 10 kΩ load resistor over which we can measure the rectified signal with an oscilloscope.

How does this circuit work? Well, the diode only allows current to flow through it in one direction. The rectified signal only contains the positive half-wave of the AC signal. In the negative half-wave the diode is blocking the current flow. This is why this sort of rectifier is called a half-wave rectifier.

The rectified signal has roughly the same amplitude as the AC signal, half the peak-to-peak voltage and a root-mean-square voltage of roughly $$V_{RMS} = {V_{p}\over 2}$$. The reason the amplitude is not exactly the same are the losses over the diode. In my example I used a Schottky diode, which has a forward voltage drop of roughly 0.3 V at low currents. A normal diode has a voltage drop of roughly 0.7 V.

When rectifying a low voltage AC signal the forward voltage drop of the diode is easily noticeable. However, it becomes especially important if higher voltages are used or a high output power is required. The product of current and voltage drop define the power dissipation. A too small dimensioned diode can quickly overheat. Let's do the math for our example circuit.

The peak voltage of our rectified signal is:
$$V_{p} = {V_{pp}\over 2} - V_f = {9 V \over 2} - 0.3 V = 4.2 V$$

The RMS voltage over the 10 kΩ load and thus the RMS current our circuit are:
$$V_{RMS} = {V_{p}\over 2} = {4.2 V \over 2} = 2.1 V$$
$$I_{RMS} = {V_{RMS}\over R} = {2.1 V \over 10 kΩ} \approx 0.2 mA$$

The power dissipated by the resistor is:
$$P_{R_L} = V_{RMS} \cdot I_{RMS} = 2.1 V \cdot 0.2 mA = 0.42 mW$$

For the diode the power dissipation is:
$$P_{D} = V_f \cdot I_{RMS} = 0.2 V \cdot 0.2 mA = 0.04 mW$$

In this case the power dissipated by the diode is practically neglectable and below the limits of a standard diode. In case of a current of several amperes the forward voltage for a Schottky diode typically increase to about 0.7 V. The power dissipation increases to several Watts – more than enough to kill a normal diode.

The more serious problem in our case is what happens during the time no current flows. Most circuits will not be able to deal with this, they need a constant DC voltage. LEDs would have a noticeable flicker and most microcontrollers won't work at all. Just a diode is not enough in this case.