Using an LED string of 12 LEDs, with a forward voltage of 3.15V (Vout = 38.1V), at a certain point the LEDs don't conduct and there is a large amount of charge in the output capacitor. Is there a way to use a bleed off resistor in parallel with the output capacitor or from Vout to GND to allow for the capacitor to bleed off its charge after the IC is powered off WITHOUT ruining the LED ripple current or regulation capability?