We’ve extended the GaN quiz to Dec. 21, 2014!
And to up the ante, here are a couple of answer hints:
Question 6 and 7 ask for the reduction in power loss in the loop inductance for the buck converter circuit by reducing the value of L.
To find the impact of L, start with the basic equation for the voltage across it: V = L di/dt. Now figure out di and dt. You know Vout and Vin and frequency. So in question 6, for example, dt(on)=1.2 V/12 V × 1 MHz = 0.1 μsec (from the definition of the output on and off times for a buck converter); (12-1.2) V = 0.33 μH × (di/0.1 μsec). Solving, di=3.27 A. Now you need the peak current…………….
Questions 8 and 9 ask for the sync FET losses using two different devices and when the converter operates at 4 MHz and puts out 25 A.
The cycle time at 4 MHz = (1/4 MHz) T = 250 nsec; then TDeadtime = 2 x 25 nsec = 50 nsec; Remember that from the duty cycle, Ton = 21 nsec. The sync FET conduction time = (250 nsec – 21 nsec – 50 nsec) = Tsync = 179 nsec. You’ll need TDeadtime and Tsync to calculate diode conduction losses during dead time and sync FET conduction losses…………..
Good luck! Look for more hints on powerelectronictips.com
Filed Under: Power Electronic Tips