To provide another example of modeling switching loss, let's drive the equivalent circuit model of a boost converter including the diode reverse recovery. So here's a boost converter. And we're going to, assume that all of the components are ideal, except for the diode and it's reverse recovery. And we'll, derive the averaged, or DC equivalent circuit model with terms to, incorporate the affects of reverse recovery and its loss. So here's the blues converter and the transistor and diode waveforms, they look qualitatively the same as the waveforms we drew in the last lecture for the butt converter with reverse recovery. The only real difference is that the values are a little different so the voltages across the, the semi conductors, when they're off, is equal to the output voltage V rather than the input voltage. Just as in the buck converter, we need to correctly define the duty cycle to obtain the effective duty cycle that the power converter operates with, and we define this using the voltage wave form. So the voltage at this node, the switching node, or the voltage across the MOSFET is this blue line shown here, and the effective duty cycle is judged from the MOSFET voltage is the time when the MOSFET voltage is low. So we will define this, this is the interval DTs. And, D prime Ts is the remainder of the interval, including the diode switching time. So, we have these wave forms with the reverse recovery peak, peak in current and the and the reverse recovery time T sub r. So we the first step, then, given these transistor and diode waveforms, is to relate the inductor voltage in capacitor current waveforms to these waveforms. So, I'll define the inductor voltage here and you can see from writing the equation in this loop the inductor voltage, V sub L is equal to the input voltage Vg minus the Transistor Voltage V sub t. likewise for capacitor current, we can write the node equation here. And the capacitor current is equal to, for this converter, the diode current, i sub d, minus the load current, which is V over R, okay. So, we can find the average values of these quantities and apply volt second balance and charge balance, accounting for the effect of the reverse recovery of the the diode on these wave forms. one other small point we in general need to write the equation of the input current Ig. But for the boost converter IG is the same as IL. So, if we find the average IL or DC component of inductor current, that's the same as Ig and we don't have to do any further work. Okay, so let's do in, inductor volt's second balance first. So, VL is Vg minus Vt, and Vt is the blue waveform, right here. So, on the next slide, that waveform is drawn again. And in fact, here the waveform looks the same as in the ideal case. the reverse recovery time doesn't change this waveform, assuming we correctly define the duty cycle, as I previously mentioned. So, we can apply volts second balance. The average inductor voltage is the duty cycle times Vg for the first interval plus D prime times this Vg minus V. And we get the same equation as usual for the inductor volts second balance. And we can even construct an equivalent circuit model from this. what would we get? Vg would be the input voltage source. It has a current capital Ig, and this is, this voltage is equal to minus D prime V, so we have a dependent source that depends on the output voltage V, multiplied by D prime. Okay, so here's our loop equation corresponding to the inductor equation. And again, Ig is the same as the inductor current capital IL, which is the current in this loop. Okay, for the capacitor we apply charge balance to the capacitor and basic, basically the easy way to do this is to write the node equation at the node where the capacitor is connected Which says that the diode current is equal to the capacitor current, plus the load current. So one way to apply charge balance is to average these currents. So, we can say that the DC component of diode current equals the DC component of capacitor current plus the DC component of load current. Now, DC component of capacitor current is zero in steady state by charge balance, which says that the DC component of the diode current is equal to the DC load current. So, we've drawn the diode current waveform already for the diode. We simply find its average and equate that to the load current, and that effectively gives us the equation from capacitor charge balance. So, here is that equation repeated. And to find the average diode current we need to average this current. So the average diode current is 1 over Ts times the integral of Id of t over the period, so we get 1 over Ts times the area under the curve. Okay, so what is the area under the curve? Well first we need the area of this rectangle and this rectangle has a height of IL. And it has a length or base that is this tongue, this time, and that time is in fact what we normally call D prime, but then minus the reverse recovery time. So this length here is D prime Ts minus the reverse recovery time t sub r. And then to that we also have to add the, recovered charge QR, and in this case it's a negative quantity. We usually have data sheets express QR as a positive number so we should subtract QR. Okay, we can now divide through by the TS and what we get is D prime IL minus ILtr over TS minus QR over TS. Which is the value the expression shown here. Okay, let's construct an equivalent circuit, then, to go with that. So, this is the node equation at the output node of the capacitor, and it says that the load current, which is the current coming out of the node. And is the well we have a voltage, V across the resistor R, that is, that is our load resistor. That is equal to the sum of these three terms. the first term is D prime IL. which is the usual dependent source. That will become part of a DC transformer. The second term, is IL times tr over TS and it has a minus sign. So it's drawn out of the node, and I'll draw it as a independent source. It is a loss element. A value IL times tr over Ts. And then the last term is minus Qr over Ts which is also a loss term. So we for the boost converter we get two extra loss terms from the diode reverse recovery. And in the case of this converter. These currents are actually at the output node where the capacitor is, in fact the reverse recovery causes current to be drawn out of the output capacitor as if it's loading the output, so switching loss effectively is another load on the output, but it's a loss element. If we take the, the two models together, for the inductor volt second balance and the capacitor charge balance, and combine them, including combining the dependent sources into the usual transformer, we get this model now. So we have the ideal boost transformer that is D prime to one and then we have our two independent current sources or sinks actually that model the loss caused by the reverse recovery. And these are actual power losses. The voltage across these sources is the output voltage V. And they cause a power loss equal to that voltage V times the sum of the two currents like this. here's a sample prediction of this model this sample here actually includes an inductor resistance and so inductor resistance, if we included it, would become an additional resistor in the loop part of the inductor is. And so, with that resistor included, here's a, a solution V over Vg for this boost converter, with and without the reverse recovery. And so, the effect of the diode reverse recovery is to actually appear to load down the output. which reduces the output voltage especially at high duty cycle. Okay so to summarize we can model or incorporate switching loss into our averages model, what we have to do is draw the actual transistor and diode voltage and current wave forms. And then relate those waveforms to the inductor voltage and capacitor current waveforms, and perhaps also to Ig. We then apply volt second balance and charge balance as usual, and proceed to compute the equations of the DC conditions in the circuit including the effects of this diode reverse recovery. And we can then reconstruct the equivalent circuit by a simple extension of, of how we did it in chapter three to get equivalent circuit models that incorporate switching loss [BLANK_AUDIO]