 In this video, we are going to be talking about a half-wave rectified circuit. And the way we achieve half-wave rectification is by putting a diode into the circuit. Now I'm going to assume that you know how a diode works, that a diode will allow current to pass through it one way, but not the other. It's like a check valve for electricity. So in this circuit what will happen is the current will flow through the circuit and the resistor here, but it won't be allowed to come back because it will be blocked by this resistor. So what's going to happen is for one alternation, so looking at it down here, with one alternation it is passing through, so it allows a hump, boom, but then on the way back it blocks it. So this part of the waveform is not going to be allowed through. And then we have again, it goes through, boom, and gets blocked. So we get what's known as a pulsating DC. You have a pulse of DC, then nothing, a pulse of DC, then nothing. And that is how we achieve halfway rectification with a single diode. I'm going to throw some numbers at the circuit and we're going to go through the calculations because I can find that could be a bit tricky for some. So let's move on and throw some numbers at this. So we're going to start out with with a very simple circuit. We have 120 volts, 60 hertz, and I've given it 20 ohms of resistance. Our first step is going to be to turn this voltage here. This is an RMS voltage. Our applied voltage is always an RMS or effective voltage. We want to take everything up to peak. So remembering what we've learned in our AC generation, we have to go 120 volts divided by 0.707 to get our peak voltage. And that peak voltage is what we're going to be working off of. I had one instructor tell me once that you have to go to the peak of a mountain because from the peak you can see everything, which always made sense to me. So let's take 120 up to peak. In this lovely green, we see that 120 volts divided by 0.707 gives me 170 volts peak. Now we have something to work with. So let's throw that over here and make sure that we have that written down. Okay, now I'm satisfied. I have 170 volts peak over there. Now what our next step is is to figure out what our DC voltage is going to be. Now when we learned about AC generation, we learned that DC voltage was going to be your peak times 0.637. However, we only have half of a cycle. We only have one hump, nothing, one hump, nothing, one hump, nothing. So instead of having an average of 0.637, we're going to cut that in half and we're going to have it as an average of 0.318 because you have that little part, you know, you have your hump and then nothing here. Let me just draw this out for you. You have a hump and then nothing, a hump and then nothing. It is during this nothing time that we have to take that into consideration. So therefore we're not going to use 0.637. We are going to use 0.318. Now I've got it written down here. E peak, which is 170 volts, times 0.318 is going to give us our voltage at the resistor. In this case, it's going to be 54 volts DC. Okay, we're getting there. Now we want to determine what our current peak current is going to be. The nice thing about current or any of these values is if you want to figure out what peak is, I can take the peak voltage and divide that peak voltage by the resistive element there and I will get my peak current. And that is going to be my peak current, 8.5 amps peak. And all I did to get that again was go 170 volts peak divided by 20 ohms, gave me 8.5 amps peak. So there we go. Now we have our peak current just as we did with our voltage. Over here I took 170 volts times 0.318 to get 54 volts DC. We can do the exact same thing with the current. I have 8.5 amps peak. I multiply that also by 0.318 and I get my current, which is 2.7 amps DC. Now we have our peak voltage. We have our DC voltage. We have our peak current. We have our DC current. Our next step is our power. When we were determining power in a regular circuit, it was E peak times I peak divided by 2. Now in this case again, let me just get my pen out here, we have a hump, nothing, a hump, nothing. So instead of having the peak voltage and the current peak current divided by 2, we don't have this portion happening in here. So instead of by 2, we are going to divide it by 4. So there we have it. E peak times I peak divided by 4 will give me my power in this circuit. So if I take 170 volts and I multiply that by 8.5 amps peak and I divide that by 4, I get a power of 361.25 watts. Now we've determined everything in this circuit. Let me just move that back. We've determined everything in this circuit except for the PIV, which stands for peak inverse voltage, which is the maximum voltage that this diode can block safely. Exceeding that, this diode will break down, we'll get a short circuit and then an open. And in this case, this diode here is going to see the peak voltage of 170 volts peak, which means it has to block at least 170 volts peak. Therefore, our PIV of this diode must be 170 volts. So there we go. We've determined our peak voltage 120 volts RMS divided by 0.707 gives us 170 volts peak. We've taken our peak voltage of 170 volts peak multiplied by 0.318 to get 54 volts DC. We've taken 170 volts peak divided by 20 ohms resistance, gives me 8.5 amps peak, 8.5 amps times 0.318 gives me 2.7 amps DC, and also FYI to double check that, take 54 volts DC divided by 20 ohms, you will get the same answer as 2.7 amps DC. And then we took our peak voltage of 170 volts, multiplied by our peak current of 8.5 amps divided it by 4 and got 361.25 watts. And our peak inverse voltage for this diode has to be 170 volts. And that is a single wave rectified circuit using one diode.