in this lecture we're going to be talking about three-phase half-wave rectifiers with resistive loads so so far we've talked about single-phase rectifiers throughout the course but now we're going to talk about the three-phase equivalent of those rectifiers so similar to the single-phase half-wave rectifier that we looked at the half-wave three-phase rectifier is the simplest type of rectifier for a three-phase voltage source and also similar to the single-phase rectifier the three-phase halfway rectifier has one diode per voltage source and it's also the simplest type of three-phase rectifier there is and so the reason why you may want to use a three-phase rectifier is that let's say for example that you have an industrial application where three-phase sources are common but you need a dc voltage so you can use a three-phase rectifier to create that dc voltage now when we talk about three-phase voltage source we typically assume that they're what's called balanced meaning that the three-phase voltage signals are a third apart from each other every period so let's say for example that you have a period that's 2 pi in length then the voltage sources are going to be separated by 2 pi over 3 and 4 pi over 3 meaning that in one period of 2 pi you're going to have the three phase voltage sources equally separated from each other so let's go ahead and draw the circuit and the input voltages for this rectifier so you can see here that the three-phase voltage sources are separated from each other by two pi over three equally so they start every one third of a cycle so for our case one cycle is two pi so the b phase voltage starts at two pi over three and the c phase voltage starts at four pi over 3. and so let's write the equations for these three phase voltages so we can say that va is going to be equal to vs sine omega t so this is the same as we have looked in the single phase example so remember that the input voltage for that was also vs sine omega t but then vb is going to be equal to vs sine omega t minus two pi over three so this represents the shift of the b phase voltage with respect to the a phase voltage and then v c is going to be equal to vs sine omega t plus 2 pi over 3. so you may be wondering well why not minus 4 pi over 3 since it starts at it starts going positive at 4 pi over 3. well plus 2 pi over 3 and minus 4 pi over 3 is the same thing remember that sine waves repeat so if we were to say we could also say here minus 4 pi over 3 but saying plus 2 pi over 3 is equivalent of that and so i'm going to define it as plus 2 pi over 3. and so one important thing here is to define where the three phase voltages cross so essentially this point right here where for example the a phase voltage so remember that this is the a phase voltage va this is v b and this vc so we want to see where they cross each other so for example if we look at this point right here this would be the point where va crosses vc and you can kind of see that graphically in this waveform you kind of know that that's going to be pi over 6 in other words half of pi over 3. but let's calculate it so in order to do that we can say well va is equal to vc at that point so let's set va equal to vc and solve for omega t so this will be equal to vs sine omega t is going to be equal to vs sine omega t plus 2 pi over 3. and so of course the vs cancel out and if we solve for omega t and this obviously is going to have several solutions right because va and vc are going to cross in multiple points so they cross over here they cross again over here but we know that that crossing is symmetrical in other words if we take a look at the first point where they cross then we know that they're going to cross again at let's say this second point right here is going to be pi plus whatever that point was so if we calculate this point right here let's say omega t 1 then we will know that this point right here would be pi plus omega t one so we don't need to calculate every point we just gotta calculate one and then we know it's going to be symmetrical after that so if we solve for omega t in this equation at the bottom we would get that omega t is equal to pi over six and again you can kind of graphically see that from this waveform at the top because it looks like it's half of pi over three which is pi over six and so the reason why i wanted to calculate that is because that's going to tell us what the voltage is when they cross each other so now knowing what omega t1 is let me call this one omega t1 then we can plug it into either equation of va or vc and see what the voltage is at that point so let's use va so we're going to say that we're trying to calculate va which is v as sine omega t but omega t is pi over 6 so this is going to be equal to 0.5 vs in other words the voltage at which these three waveforms cross each other is 0.5 es so remember that pvs is the peak of the three-phase input voltages so this point right here is going to be 0.5 vs and again you can kind of see that graphically on this waveform because that looks like it's about half of vs now as far as the output voltage what ends up happening is that the output voltage is going to be equal to the greatest of any of the three phase voltages at any point in time so what i mean by that is let's say for example during this portion right here from here to about here so from pi over 6 until va crosses vb the a phase voltage is greater than the b and c phase voltages so what that means is that diode d1 is for bias so the voltage right here is going to be equal to va because d1 is on and so if that's the case then the voltage again the voltage on the cathodes of diodes d2 and d3 are equal to va remember that the cathode is this pointer here so cathodes and anodes on this side so the voltage at the cathodes of d2 and d3 is equal to va and the voltage and the anode of d2 is equal to vb and the voltage at the anode of d3 is equal to vc and so because vv and vc are lower than va then that means that diodes d2 and d3 are reverse bias and they're off so what happens is that for this portion of the waveform diode d1 is on and diodes d2 and d3 are off now after that when vb becomes greater than va so from here to about here then the same thing happens but for diode d2 so diode d2 becomes four bias so the voltage at the output is equal to vb which means that the voltage at the cathodes of diodes d1 and d3 is equal to vb and the voltage at the anode of d1 is va and the voltage at the anode of d3 is vc but because those two voltages are lower than vb then diodes d1 and d3 become reverse bias and they're off so for this part right here d2 is on and d1 and d3 are off and then the same happens after that so d3 would be on d1 and d2 would be off so the diodes can't take turn when the voltage for that phase is greater than the other two phase voltages then that diode the corresponding diode for that source comes on and the output voltage is equal to that and then when the next phase becomes greater then the next static turns on and so on and so forth so what ends up happening at the output it's that it's going to look like this it's going to be equal to whichever voltage is greater so if i draw the output voltage here in white then it's going to look like this and it's going to repeat after that so the ripple of the output voltage kind of fluctuates every third of a cycle because it's equal to the creator of the phase voltages at any point in time and it fluctuates between the peak of the voltage sources so vs and then the minimum is going to be where the voltages cross each other because that's when the transition happens from one diode to the next so that's going to be equal to 0.5 vs so the output voltage ripple for v out would be from here to here which is 0.5 vs now let's calculate one more thing let's calculate what the average of the output voltage would be so it would be probably somewhere over here would have drawn this dash white line so let me go ahead and erase a couple things to make room so remember that the average of the output voltage just like in the single phase examples it's going to be one over the period times the integral over the period of the waveform so v out d omega t and this would be omega t so we calculate that for this waveform we would get the v out is gonna be equal to one over the period so the period is two pi and we're gonna have to calculate several integrals so the first integral is going to be for this part right here which again is equal to vc so from 0 to omega t1 the output voltage is going to be equal to vc because that's greater than va and vb so we're going to calculate the integral from 0 to omega t1 which is pi over 6 of vc which is equal to vs sine omega t plus 2 pi over 3 d omega t plus then after that va is greater than vb and vc so the integral from omega t1 so pi over 6 to where they cross again so this point right here is going to be equal to and remember that i said that this is going to be symmetrical so if we know that this point right here um this point right here is 2 pi over 3 then where va and vb cross is going to be equal to 2 pi over 3 plus pi over 6 and so that's going to be equal to 5 pi over 6 of va which is vs sine omega t d omega t plus then after that vb becomes greater than va and vc so the output voltage is going to be equal to vb so from 5 pi over 6 until this point right here which is going to be equal to 4 pi over 3 plus pi over 6 which is equal to 9 pi over 6 of v s sine omega t minus 2 pi over 3 d omega t and then the last portion is going to be vc again so right here at the end so it's going to be from where we stopped so 9 pi over 6 until the end of the cycle which is going to be 2 pi of vc so vs sine omega t plus 2 pi over 3 d omega t so notice that we use vc for about a third at the beginning and then two thirds at the end so if you calculate all this this would be equal to so we can take vs at the beginning so we'll factor out vs so we'll say that is equal to vs over 2 pi times the first integral is going to be 0.366 o2 then the two next integrals are going to be actually the same and that makes sense because the width of the portion that we're using for va and vb so this and this are equal to each other so those two integrals are going to be 1.73 1.73205 plus again 1.73205 plus the last integral which would be for vc so this point right here and that's equal to 1.36603 and so all of this would be equal to zero or approximately equal to 0.82699 vs and this is also going to be equal to in more exact terms 3 times square root of 3 over 2 pi of vs so it's going to be the average of the output voltage for a three-phase halfway rectifier and i'm actually not going to calculate the output current because we know that for resistive load the output current is just going to look like the output voltage but scaled by the load resistance so you can see here that for a three-phase rectifier without having to add any capacitors for voltage filtering the output voltage actually looks better than for a single phase rectifier and so that's the nice thing about three phase rectifiers is that inherently without even having to do any filtering for voltage ripple we get a better output voltage just by the fact that we're using a three-phase voltage source so next we're going to take a look at the full wave three-phase rectifier with the resistive load