From the course: Electronics Foundations: Semiconductor Devices

Rectify a signal

- [Instructor] One common application of diodes is to build a rectifier. Which is an electronic device that converts from alternating current to direct current. If you've ever broken apart a wall wart power adapter, you've probably seen a set of diodes inside. Those diodes are converting the 120 volts AC, that comes from a wall outlet, into a DC voltage for a device to use. Since an AC signal periodically reverses direction, it consists of both positive and negative voltages at different times. The goal of a rectifier circuit is to remove all of the negative voltages so that the output signal is entirely positive. Now, that can mean simply eliminating the negative voltages by making the output voltage zero during that time. Or, it can mean converting those negative voltage swings into positive output voltages. The simplest type of rectifier circuit is called a half-wave rectifier. It consists of a single diode placed in series between an electric load and the AC voltage source. When the input voltage from the AC source swings in the positive direction, the diode will act like a short circuit and allow current to pass through. So, the voltage across the load, which is the output of the half-wave rectifier, will be similar to the input voltage. Now, when the input voltage changes direction and swings to the negative side, the diode will act like an open circuit to block current from passing through it. Since the circuit has been effectively broken open, no current flows through the load. So the output voltage across the load will be 0 volts. These two plots illustrate the relationship between an AC input voltage and the corresponding output voltage from a half-wave of rectifier. When the input is positive, so is the output. But when the input is negative, the output goes to zero because no current flows through the load. Although it's common to think of the term DC as referring to a constant voltage, even though the rectifier's output is constantly changing, it is still technically direct current because the voltage is always positive, so the current will only ever flow in one direction. Now, that explanation of the half-wave rectifier is a bit of a simplification because the diode will not actually turn on and allow current to flow through it until the input voltage from the AC source is greater than the forward voltage of the diode. Also, due to the forward voltage drop across the diode, the output voltage from the half-wave rectifier will be slightly less than the input voltage. We lose some of that input voltage to the forward voltage drop across the diode. To demonstrate that, I've built the half-wave rectifier circuit on my breadboard using my function generator to provide a sine wave as the AC input voltage, and using a 10 kilo ohm resistor to serve as the electrical load. I've connected channel 1 of my oscilloscope to display the input voltage in yellow and channel 2 to display the output from the half-wave rectifier, which is the voltage across the load resistor, in blue. Looking at the scope, I can see that the output signal only contains the positive parts of the input signal. Additionally, the positive humps of the output waveform are slightly smaller than the input signal because I lose about 0.6 volts of the input voltage as the forward voltage drop across the diode. Although the half-wave rectifier circuit is simple to use, it's not very efficient for converting from AC to DC. Because I lose all of the energy from the negative half of my AC signal to the diode. If I want to preserve more of the energy from the input signal, then I should use a full-wave rectifier, which converts both the positive and the negative parts of the input AC signal into a pulsating DC output. The most common full-wave rectifier circuit consists of four diodes in a Wheatstone bridge configuration. The diodes are oriented so that when the input signal is positive, current will flow from the source through this diode and into the top of the load. And then current coming out of the load takes that path through the diode on its return back to the voltage source. When the AC signal switches polarity and turns negative, the current will take a different route through this diode into the top of the load, and then the return current from the load will pass through the bottom left diode back to the source. Regardless of whether the input voltage is positive or negative, the output of the full-wave rectifier is always positive and pushes current in the same direction through the load. This breadboard circuit is a full-wave rectifier that uses four diodes to convert the AC signal from my function generator into a DC voltage across a 10 kilo ohm load resistor. On the oscilloscope, I see that the output signal looks like a completely positive version of the input signal. And just like with the half-wave rectifier, the humps of the output waveform are smaller than the input signal due to the forward voltage drop across the diodes. Although the pulsating output of the rectifier circuit is technically a direct current signal, since the voltage is always positive, most of the time when we're converting from AC to DC, we really want a constant output voltage. So, to achieve that, most power supplies that convert from AC to DC, use a low-pass filter on the output of the rectifier to smooth out the bumps and make the output more steady. When building rectifier circuits, we need to consider the amount of power that will be dissipated by the diodes to choose sturdy enough components. For the example circuits in this video, I chose the common 1N4148 signal diode because I was rectifying relatively small signals, so the diodes wouldn't need to dissipate much power. However, if I needed my rectifier circuit to handle more power, then I should use power diodes, like the 1N4001, which have a much higher maximum current rating than basic signal diodes. The 1N4001 power diode is designed to handle up to 1 amp of current, operating as a rectifier. But with that capability comes the downside that it requires a larger forward voltage drop of about 1 volt.

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