I'm building LED driver based on current mode control buck converter. My question is quit general. If load is disconnected then output capacitor voltage shoots up. If load is reconnected it gets overcurrent spike from that charged capacitor. If output capacitor cannot be removed, what else can be done? Is there a general approach to this problem?
Thats a simplified circuit. If LED is disconnected, output capacitor charges to high voltages and can destruct the LED if it is reconnected.
My biggest worry is if I'm missing something simple and obvious? If not, thats ok, I will do my work somehow.
Best Answer
If instead of using a MOSFET and diode non-synchronous buck regime you chose a synchronous (2 MOSFET) scheme you would get what you want without the hassle of a control loop.
A synchronous buck converter power stage is this: -
Either Q1 is on or Q2 is on - this translates to a rapidly alternating changeover contact producing a square wave of varying duty cycle (duty controlled by possibly a MCU).
The average value of the output square wave is Vin x duty cycle. This output voltage is low pass filtered by the inductor and capacitor so, there is need for a closed loop control should the load become disconnected.
The usual caveat of a sync buck converter applies; constant output voltage with load is only guaranteed if the input supply voltage (Vin) remains constant. However a non-sync buck converter cannot deal with input voltage variations either without a feedback loop. It's easier for a sync buck because duty cycle can be determined by an ADC measuring the input voltage and making a simple correction to duty cycle.