Most electronic systems, like TVs, audio systems, and computers, need a DC voltage to work properly. Since the line voltage is alternating, we need to convert it to a relatively constant DC output voltage. The circuit that converts the alternating voltage (AC) into a continuous voltage (DC) are called Rectifiers.
As you know, a diode only conducts current in one direction from the anode to its cathode. This feature makes them ideal for Rectification.
The diodes are connected together to form various types of rectifier circuits such as “Half-wave”, “Full-wave” or “Bridge” rectifiers.
The simplest of all rectifiers is a Half-wave rectifier.
The Half-Wave Rectifier
Following image shows a half-wave rectifier circuit.
When an alternating voltage is applied across a diode, the positive half-cycle of source voltage will forward-bias the diode. In this case, the diode will appear as a closed switch, and the positive half-cycle of source voltage will appear across the load resistor.
During the negative half-cycle, the diode is reverse biased. In this case, the diode will appear as an open switch, and no voltage appears across the load resistor.
In the half-wave rectifier, the diode conducts during the positive half-cycles and not during the negative half-cycles. Because of this, the half-wave rectifier clips off negative half-cycles. Such waveform is called a half-wave signal.
If the diode is reversed, it will become forward biased when the input voltage is negative. As a result, the output pulses will be negative.
This half-wave voltage produces a load current that flows in only one direction making the circuit unidirectional.
DC Value of a Half-Wave Signal
The DC value of a half-wave signal is the same as the average value.
The average value of the signal over one cycle is calculated with the below formula:
This equation tells us that the DC value of a half-wave signal is about 31.8% of the peak value. For example, if the peak voltage of the half-wave signal is 10V, the DC voltage will be 3.18V
When you measure the half-wave signal with a DC voltmeter, the reading will equal the average DC value.
A Second-order Approximation
In reality, we do not get a perfect half-wave voltage across the load resistor.
Because of the barrier potential, the diode does not turn on until the source voltage reaches about 0.7V. So, the output voltage is 0.7V lower than the peak source voltage.
For example, If the peak source voltage is only 10V, the load voltage will have a peak of only 9.3V.
Therefore the more accurate formula to calculate the DC value of a half-wave signal is:
The variation of the rectified output waveform during positive and negative half cycles produces a waveform with a large amount of Ripple (the fluctuating part).
The resulting ripple has the same frequency as the input AC supply.
Therefore, we can write:
Filtering the Output of a Rectifier
The output we get from a half-wave rectifier is a pulsating DC voltage that increases to a maximum and then decreases to zero.
We do not need this kind of DC voltage. What we need is a steady and constant DC voltage, free of any voltage variation or ripple, as we get from the battery.
To obtain such a voltage, we need to filter the half-wave signal. One way to do this is to connect a capacitor, known as a smoothing capacitor, across the load resistor as shown below.
Initially, the capacitor is uncharged. During the first quarter-cycle, the diode is forward biased, so the capacitor starts charging. The charging continues until the input reaches its peak value. At this point, the capacitor voltage equals Vp.
After the input voltage reaches its peak, it begins to decrease. As soon as the input voltage is less than Vp, the voltage across the capacitor exceeds the input voltage which turns off the diode.
As the diode is off, the capacitor discharges through the load resistor and supplies the load current, until the next peak is arrived.
When the next peak arrives, the diode conducts briefly and recharges the capacitor to the peak value.
If the load resistor is small for a given capacitor value, a high current will flow through the load which discharges the capacitor more quickly (Because of the RC time constant) and results in increased ripples. As long as the RC time constant is much greater than the period, the capacitor remains almost fully charged, and we get a perfect DC output voltage. To have a greater RC time constant, we need a larger value capacitor. This is not practical because there are limits on both the cost and size of the capacitor.
Also there is no output during the negative half cycle hence half of the power is wasted which results in lower output amplitude.
Because of their major disadvantages the half-wave rectifiers are rarely used. It would be more practical to use a full-wave rectifier as discussed in the next tutorial.