 In this video, we're going to be going over a center tap by phase rectifier. So as you've seen in our other videos, a rectifier takes AC and converts it into a pulsating DC. So we've got these humps here. We're going to take a look to see what happens to my lovely humps. What happens is, if this is negative and positive, let's just say that this is negative. It's going to go through here. It blocks there. It goes across my load, which gives me a hump that way, but on its way back, it gets on its way back here, sorry, it gets blocked. So you'll see that gets blocked. But then on the second phase, as we see, we've got negative here. So current's going to flow through here and go there. And as it goes back through, it's going to get blocked. So then we just get that one hump and that side's blocked. Then it happens again, away we go. Current goes that way, but on its way back, it gets blocked. So we get this hump, and then one more time. I think you're picking up what I'm putting down, goes through, but gets blocked. So we end up with these humps. So we get this pulsating DC going across here. No more AC for us. So to see how this really works, we're just going to assign some values. We're going to try to figure out what our peak voltage is at the load, what our DC voltage is at the load, what our peak current will be at the load, what our DC current will be at the load, the power being dissipated across the load, and what our peak inverse voltage needs to be for these diodes. Now let's just start seeing what we can do with the voltages here. I've got 1200 volts across, so I'm going to have 600 there and 600 there. So we've got 600 volts across each winding. I've got a winding to winding ratio, if you remember, from transformers of five to one. So if I have 600 here, that means I'm going to have 120 volts over on those windings on my lower voltage windings. So therefore, I have 120 volts that I'm dealing with here. Now that's a center tap by phase, which means that the AC, if these guys weren't here, this guy would see 120 volts in one phase, this guy would see 120 volts on another phase. So we've got a three wire center tap. Now in all these questions, the first thing I want to do is I'm going to take my RMS value, which is 120, and I'm going to drive that up to peak. 120 divided by 0.707 gives me 170 volts peak, which coincidentally we wanted to find out. So there we're going to put it over here, 170 volts peak, and we're on to the next. Now that we have 170 volts peak here, we can determine what our DC value is by taking our peak value times 0.637 to get our DC value. Remember this peak times 0.637 to get DC. So I end up with 108 volts DC across the load resistor. So we get that drawn in there. Our next thing is to work out what our current peak is going to be. In order to figure out what our current is, I had to assign a resist value. In this case, we're going to use 10 ohms of resistance. And if I take my peak value of 170 volts peak, divide that by the 10 ohms resistance, I end up with 17 amps peak, which we're going to put over on this side now. 17 amps peak. Next, we've got to take this to a DC value. Well, just like we did with the voltage, we're going to take 17 amps peak, and we're going to times that by 0.637 to get my DC current value, which in this case is going to be 10.8 amps DC. So we're going to put that over on our side over here. We have 10.8 amps DC. We're getting close to the end. Power. We need to figure out what our power is. So the power being dissipated, we have a voltage here, and we have current here. We can determine what the power being dissipated across this resistor is. In order to do that, we have to remember the formula. E peak times I peak divided by 2 equals power. Well, I have my peak voltage here, 170 volts peak. I have my peak current of 17 amps peak. So I just go 170 times 17 divided by 2 is the average power being dissipated across this resistor. And in this case, that works out to be 1,445 watts. Last thing to figure out is we need to figure out what the peak inverse voltage, what's the maximum voltage that this guy has to be at least to block. Now in this case, with a center tap by phase, normally I have just 170 volts across here. But if you take a look here, this diode here has to block not only that center tap, it has to block the entire voltage across here. So its peak inverse voltage is not just going to be 170 volts. It's going to be 170 for this coil and 170 for that coil because these two coils are in series. So these diodes need to be able to block the entire thing, which is 340 volts peak. So the peak inverse voltage of the diodes on a biphase rectified circuit is two times the peak value of the one winding or 340 volts. And that's it. That's how you take the steps you need to calculate everything you need to know in a biphase rectified circuit.