 In this video, we're going to be looking at a bridge rectifier unit and what makes a bridge rectifier unit different from a biphase rectifier unit is the fact that we use four diodes and that helps us set this up for a full wave rectification. Let's take a look here, negative to positive. If I go negative and block that way, you can go through, block that way. I'm going to go across here and out to positive and if we switch polarities, I have negative here now. So it's going to go down across here, through there and across down to there. So I end up with a full wave rectified circuit because I've got a hump, boom, hump, boom. Again, I'm going negative to positive this way and finally negative to positive. That gives me one, two, three, four and as long as it's just swapping back and forth, back and forth, back and forth, I will continue to get full wave rectification. So we have all our alternations are on the top. So let's take a look at how that's going to look with some numbers. I'm setting myself up with 140 volts, 60 hertz with a 15 ohm resistor. I'm going to try to determine my voltage peak, my voltage DC at the load, my current peak, my current at the load, the peak inverse voltage of the diodes and the power being dissipated at the load. So our first step is to take this 140 volts and take that up to peak. As before, we just go 140 divided by 0.707 and we get our peak voltage, 198 volts peak. There we go. We throw that right over there. Now, if we want to figure out what our DC voltage is, we take the peak voltage and we multiply that by 0.637 and that gets us our answer of 126 volts DC. So let's go ahead and throw that over in the side here. Next, we have our current peak. Well, we can take our peak voltage and divide it by our ohmic value of the load to get our peak current, which is 13.2 amps peak if you see that over here. So let's put that over here where it belongs. And the same thing to get my DC current, there's two ways. I can take this peak and multiply it by 0.637 or I can take the voltage that I already have, 126 volts DC, divide it by 15 ohms and I get the same answer, 8.4 amps DC. So let's put that over in the side. Now, the peak inverse voltage of these diodes we have to look at, they only see this voltage, this peak voltage, as opposed to the biphase which we saw, we had two windings that it was getting its voltage across because we had that center tap. Well, in this case, we only have one source. Therefore, it is only this 198 volts that we are dealing with. So these diodes have to handle at least a peak voltage of 198 volts. And last is our power being dissipated at the resistor. To get our power is just e peak times i peak or v peak times i peak divided by 2 because this is a full wave rectified circuit. That gives us our power. There's our formula. So we plug the numbers in and we get 1306 watts by taking 198 volts times 13.2 amps and dividing it by 2. And that's it. That is a full walkthrough on a full wave rectified circuit using four diodes, these diodes a.k.a. the bridge.