Let's now extend our equivalent Circuit Models that we derived in chapter 3 to include switching loss. And in particular what we're going to do is model the switching loss caused by the diode reverse recovery. Okay, so to obtain tractable results anytime we're modelling switching loss where we have complicated brain waveforms and things, we do have to idealize the waveforms. But we can at least include a current spike that comes from the diode switching loss and including the effects of the diode switching time. And this actually gives pretty good analytical results. there are many different things that cause switching loss as in some of the company lectures here that are discussed. What we're going to talk about here is the, the loss just from the [UNKNOWN] reverse recovery but we can include other things like, the, effective, simi conductor output compacidances and their energy stored in a similar manner. So the approach here is to correctly account for how the waveforms change. And in particular, when we apply inductor volts' second balance or capacitor charge balance or when we calculate the average IG input current, we include how the diode reverse recovery changes those waveforms. And the I think the simplest way to do this step is sketch the transistor in diode waveforms including the diode reverse recovery and then relate those waveforms to the inductor and capacitor and other waveforms. So I'm going to do an example of this with a buck converter. So here is a buck converter with a diode that has reverse recovery and with a MOSFET that switches quickly. so we're going to neglect everything, all the sources of loss except for the diode reverse recovery. Okay. So, here is the same buck converter and I've identified the transistor voltage and current BT and IT, the diode voltage and current VD and ID, and I'd sketched idealized switch waveforms for all of those here. And, they look like the usual rectangular waveforms except for the switching time here of the diode, where we have a current spike from reverse recovery and its effect on all four of the waveforms. Okay, when we do this, there's now somewhat of an, ambiguity in deciding how to define the duty cycle. So, if we look closely at the these waveforms, the switching period, oops, the switching period, TS. Goes from here, out to right here. And what's happening is that the gate driver and the controller is telling the MOSFET to turn on at this point here. But then we go through the switching transition at the diode and the voltage finally changes out here. Okay, I've also squared up this wave form, this is what we call a snappy diode, that where its voltage changes abruptly at the end of the, the, reverse recovery period to idealize that wave form. So that's a somewhat pessimistic assumption, as far as laws goes. But anyway, we have this ambiguity in, what is the duty cycle? So, actually, we can define the duty cycle howev, however we want and we could define the duty cycle based on when the driver starts to turn on the MOSFET right there. Or we could define it based on when the voltage of the mosfet actually changes, right here. And what's sketched here is the second choice. It's based on the voltage in the power stage of the converter. so if I extend the this light blue line, which is the transistor drain of source voltage, it's actually falling right there. And, if we extend the, the pink line, which is the MOSFET current. It has the current spike from reverse recovery, that is the same as this one that is happening right before t equals zero. So, what I'm going to do is define the duty cycle. Is, is being determined by this interval from here to here, so it does not include the diode switching time. the justification for that is that this voltage waveform is what drives the volt second balance of the inductor and what ultimately determines the effective. Conversion ratio of the ideal converter and once, when we had the switching loss of this converter as well. If we define the duty cycle as being determined by what comes out of our controller and, so, this period starts here. Duty cycle's defined starting there. we get equations that are valid but they are not as physically connected to the ideal converter. We predict a different effective conversion ratio and we get a lot of extra terms around the power stage model. So, in, effectively, the transistor switching, and the diode in transistor switching times, change the duty cycle from what comes out of the controller. And, the duty cycle that the converter effectively works at is based on this voltage waveform of when the voltage at the switch note is high, or, when the voltage across the MOSFET switches. Okay. So, with that definition, DTS, then is this interval that starts after the diode is switched off and ends when the MOSFET is switched off. And, then, the deep prime interval is the rest of the period, going from here out to there. Okay given those definitions, given these waveforms lets now work out the inductor voltage waveform and apply volt second balance, will work out the capacitor current waveform and apply charge balance. And will work out the input current waveform ig and calculate its average or DC component. So here's the inductor voltage waveform, and with the definition that I just gave you for the duty cycle, here's what it looks like and its the same waveform we've been drawing all along this semester. So that definition gives us back the original ideal inductor voltage waveform. And volt second balance gives us the usual equation. That the average inductor voltage is DVg minus V. The reverse recovery also doesn't change the capacitor current waveform. The capacitor's on other side of the inductor and we get the same equation from charge balance that we always have gotten. So so far, no change. And in fact, from these two equations, we can construct the usual equivalent circuit model, which is shown right here, with the DVg source, the inductor and capacitor, node and voltage loops and the load. Okay. The only thing in the buck converter, at least, that is changed by these switching times is the input current wave form, ig. And, what happens is that this inductor, or, this, current waveform of the transistor current is the same as ig. So when, so the transistor current waveform, which includes the reverse recovery current spike, gets drawn out of Vg and changes the Ig waveform. So we need to work out the average value or DC component of this current, and we do it in the usual way. Oops. Do it in the usual way, by integrating to find its average value. So the average value of ig is the area under the curve divided by the period. The area under this part of, its first pulse, [SOUND] this area would be the duty cycle, D times TS times the height IL, where IL is the DC inductor current. We have zero here, and then we have some area here during the switching time. So we have the area of this rectangle first, which is the diode reverse recovery time in width and in height, it's the inductor current IL. And we also have the area of this current spike, which is the recovered charge that has area qr. Okay, dividing two by ts gives us this expression. So, in the ideal case, we had just one term. Average IG was DIL, coming from this area but now the reverse recovery adds to more terms. And the two terms depend on the reverse recovery time and the recovered charge. So an equivalent circuit to go along that, is this. So, ig is the current coming out of Vg, that's equal to DIL which is the usual Dependent source that is part of the ideal transformer. Then we have two more terms, that I'm going to draw as a independent current sources that model losses from reverse recovery. So the recovery charge gives us one of the sources. QR over TS is the same as QR times the switching frequency FS, this is the the charge per unit switching period drawn out of the source by the recovered charge and it has an average value QR over TS. And then, the other term from the switching time, we draw an extra amount of current equal to the inductor current, during the switching period, or the, I mean sorry, the reverse recovery time. And that gives us another term that is this source. So we put them together and combine the ideal, or the ind, the dependent sources into a transformer. And we get this equivalent circuit. Okay? This looks, really like being a ideal buck converter with its 1 to D transformer turns ratio. plus though, it has these two extra sources which model the switch in loss. You can these that these sources consume power because they have a voltage Vg across them, and they draw a current out of Vg and so we get a switching loss equal to Vg times the sum of these currents. OK, this can be a considerable amount of power. We have to plug in the numbers to c. But its very easy for this to be more powered less than all of the conduction losses combined. It depends on the application, you can also see that this para depends on the switching frequency or a switching period. In fact its directly proportional to switching frequency. Since fs is 1 over Ts. [NOISE] So, if we turn up the switching frequency and switch faster, we turn up the switching loss, and we get more power loss. Which will reduce our efficiency. Kay, we can solve the model, so, The volt output bowl is easy. It's just the duty cycle times the input voltage. As far as efficiency goes, as usual, we can find the input power and the output power and then divide. The input power is, [COUGH] the input voltage multiplied by Ig, and we can solve these things defined Ig. and it's really from the average input current equation. The output power is the output voltage times the inductor current IL. and so, we can divide to find the efficiency and when we do we get an expression like this. it has a denominator that's one plus this, effective switching loss or terms that depend on the switching loss. And, this denominator depends on the switching frequency so, again, efficiency will go down. As the switching frequency goes up. Here's a plot, so I'm plugging in some numbers here and I think they're pretty innocent looking numbers, we were 100 killohertz switching frequency. Input voltage of 24 volts here are some numbers for the diode reverse recovery time of 75 nano seconds. And recovered charge of 0.75 micro coul ombs this is an okay diode, it's not a super great diode but it's not a real bad diode either, kind of an average diode. and that I've plotted those equations and I hadn't made any attempt to model how the recovered charge or the reverse recovery time change with the inductor current. it turns out that they do change with inductor current but it's actually a pretty weak dependence. And, modeling exactly how they change with inductor current is fairly complicated. They generally don't give you enough information on the data sheet, to, to have a valid model. And so we'll just keep this constant. But you can see, when we plot this, so here I am plotting efficiency versus duty cycle what happens is the, the switching loss degrades the efficiency at low duty cycle. And the reason for that is that, as we turn down the duty cycle, we're turning down the output voltage and turning down the output power, but we aren't changing the, the switching loss by very much. Switching loss does depend on the inductor current, and that is modeled, but, the dependence is not very strong. So you can see what happens is the efficiency becomes poor at low duty cycles. This is something we observe and practice as well. And in fact at low output power, the switch in loss makes the efficiency be, become more. So a converter may have good efficiency at full power, but when can turn down the power and efficiency goes down, for a variety of reasons. But, one of the major reasons is because we still have substantial switching loss without having a lot of output power.