# 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.

How can we overcome the issue that current only flows during the positive half-wave? We need to gather more electric energy than we need to power our load during the positive half-wave and store it, so that we can use it during the negative half-wave. This way we can achieve a more stable voltage at the cost of higher current draws during the positive half-wave. For implementing this change, we just have to add a capacitor in parallel to the load: The capacitor provides the energy for the time in that no current flows through the diode. The output voltage is now nearly identical to the peak voltage of the AC signal. However, the result is still not a perfectly constant DC voltage. The capacitor can only store a limited amount of energy. During the negative half-wave it discharges, the voltage decreases slowly. In the next positive half-wave the capacitor gets recharged. The remaining voltage changes are called ripple. Depending on how sensitive components are they can tolerate more or less ripple. Microcontrollers with an integrated voltage regulator for example work with more ripple than those that depend on a specific input voltage.

How can we minimize the ripple? Well, the capacitance value needs to be chosen according to the load. The bigger the capacitance the less ripple. The bigger the load the higher the capacitance needs to be for the same amount of ripple. This obviously has its limits. For big loads it is better to use a full-wave rectifier.

## Conclusion

A half-wave rectifier is very easy to build and requires just one diode. With an additional capacitor we can stabilize the rectified voltage for use with our circuits. The main disadvantage of a half-wave rectifier is the high amount of ripple. It can disturb the components in the DC circuit. A half-wave rectifier is good for cost-efficient circuits with few components and a small load. For a bigger load and less ripple it is better to use a full-wave rectifier.