Diode conduction loss

What happens if Vsw < VD In fact that is the situation in most commercial Flybacks. But note that to do a proper comparison, you have to reflect the diode drop to the primary side. And for that we have to multiply the diode drop by the turns ratio (see the equivalent Buck-Boost models of a Flyback section in my book, Switching Power Supply Design Optimization). So, for example, if the turns ratio is 20 and the diode drop is 0.6V, the effective VD we need to compare with Vsw for our time-sharing analysis is 0.6 x 20 = 12V. And that is usually greater than the (average) drop across the switch. Therefore, we tend to say that in a Flyback, decreasing D (increasing input) will worsen the total conduction loss and decrease the efficiency. But of course that never happens, because as we increase the... 

We analyze this in Figure 10-10, and the steps should be fairly obvious. Basically, we are writing the loss at each point as the sum of the switch conduction loss, the diode conduction loss, and a generic switching loss (crossover) term. We thus arrive at the general solution to the equations. We then take the published efficiency curves for the 2593HV (see... 

The essential difference from a conventional buck regulator is that the low-side mosfet in a synchronous regulator is designed to present a typical forward drop of only around 0.1 V or less to the freewheeling current, as compared to a Schottky catch diode which has a typical drop of around 0.5 V. This therefore reduces the conduction loss (in the freewheeling path) and enhances efficiency. 

The diode conduction loss is the other major conduction loss term in a power supply. It is equal to Vd x Id avg, where Vd is the diode forward-drop. Idavg is the average current through the diode — equal to Io for the boost and the buck-boost, and Io x (1 — D) for the buck. It too is frequency-independent. 

We realize that the way to reduce conduction losses is by lowering the forward-drops across the diode and switch. So we look for diodes with a low drop — like the Schottky diode. Similarly, we look for mosfets with a low on-resistance Rds. However, there are compromises involved here. The leakage current in a Schottky diode can become significant as we try to choose diodes with very low drops. We can also run into significant body... 

I am fully planning to reduce these delays at the breadboard stage. Delays this long cause the diodes to conduct too long, thus causing the losses to be high. But this is operating on the safe side. 

Another way of reducing the reverse recovery current shoot-through is simply to ensure that the boost diode is carrying no forward current at the moment when the switch starts to turn ON. The diode then blocks reverse voltage instantly. In other words, running the Boost in DCM or BCM (boundary conduction mode, i.e., at the critical boundary) will produce higher peak currents, but smaller inductors (yes, if r is large, the size of any inductor typically reduces ), and perhaps much better efficiency too, because now, the turn-on crossover loss becomes zero. 

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