Re: E2e\'s with underdriven 5 watts - Twins: SE & HD (beam Comparisons)
Originally posted by Jonathan:
Neglecting losses, the switching element for a boost converter is on for 1-(Vin/Vout).
-Jon
<font size="2" face="Verdana, Arial">Correct. So ideally, assuming the converter is operating in continuous mode (i.e. inductor current is always forward and never drops to zero), the duty cycle for driving a 5W LS is around 60%. High, but not unreasonable. Of course, that calculation is assuming a resistive load. An LED behaves very differently. With an LED, no matter what the duty cycle is, the voltage will rise up to Vf.
The reason a boost converter works is because inductors resist change to the current passing through them. When the transistor switches on, the current through the inductor ramps upward. Ideally, it will do this forever, but in reality the resistance of the power supply and of the transistor will stop this from happening. But when the circuit is opened (transistor switches off), the current wants to continue to flow. Ideally, opening a circuit with an inductor carrying current through it will result in an infinite voltage. (In reality, this never happens., due to imperfect inductors or a transistor that can't stand the voltage drop across it that results from such abuse.)
In the case of an LED, the voltage will rise until the current through the inductor can go somewhere. (i.e. Vf of the LED).
A BadBoy circuit can drive an output to the 7v or so required to light up a 5W LS. Whether it can do this AND supply full rated current is a different story. (From the looks of it, it CAN, but the switching transistor in the IC can't handle that much current for very long before forcing the regulator into thermal shutdown, see dat2zip's post at
http://www.candlepowerforums.com/cgi-bin/ultimatebb.cgi?ubb=get_topic;f=14;t=000444;p=16 )
At a lower output current, the BB might be able to handle the task.