Lecture 33: Soft Switching, Part 1

[@ritviksriram9918]

31:05 For anyone curious, here's an explanation for why Vds "rings down" to this level. To achieve "maximum ZVS", our goal is for the slope of Vds to be 0 (local minima) at t1 (or, if you hit a zero crossing of Vds, just switch anyway since you're already at 0 voltage for ZVS!). The purpose of this calculation is to find the lowest Vds achievable for your switch to turn on with if you're unable to reach true ZVS turn-on. As Dr. Perrault stated, after the inductor current zero crossing, the diode and FET are both "off" meaning the resonance is purely between the L and parasitic capacitance of the FET. The inductor current becomes negative since the voltage across the inductor is governed proportional to an integral of Vds - (Vin - Vout), where the voltage across the parasitic capacitance of the FET is initially ~= Vin (higher than Vout since it's a buck). Why do I care? The utility of knowing this is that the following zero crossing of the inductor current defines the local minima of my Vds! This is because if inductor current is positive, my parasitic capacitor voltage will rise (assuming both switches are still closed), so we know the minima is at this zero crossing (after all negative volt-seconds are applied). Now, all you need to do is calculate the time at which this happens: Let's consider t0 to be the first zero crossing of the inductor current and t1 to be the second zero crossing of the inductor current (as we established, this is the local minima of Vds). To calculate t1, I'll use volt-second balance across the inductor (I know the derivative of inductor current is determined by the voltage across the inductor); since my circuit is instantaneously linear, I can assume symmetry around the local minima of the inductor current. This means that the derivative at equidistant points (in this case, the first and second zero-crossings) are negatives of each other. To make the implication of this clear, to calculate my final Vds, the second zero crossing of the inductor current, I simply need to find the point at which my derivative of inductor current at t0 is equal to the negative of my derivative of inductor current at t1. Now, since the derivative of my inductor current is proportionally related by 1/L to my inductor voltage, the math becomes extremely easy. State 0 Inductor Current Derivative: 1/L * (Vds - (Vin - Vout)) = 1/L * (Vin - (Vin - Vout)) = 1/L * Vout State 1 Inductor Current Derivative: -1/L * Vout Now, all I need to do is find Vds such that the differential voltage across the inductor is equal to -Vout, as I've established for my necessary State 1 Inductor Current Derivative. Finding this is simple algebra. -Vout = Vds - (Vin - Vout) Vds = -Vout + (Vin - Vout) = Vin - 2*Vout

[@tidin_tss]

very nice