Electrical – buck converter, no load start

The basic answer to your misunderstanding is that the load and the input voltage are NOT connected directly when the switch is closed. The inductor is between the two.

Voltage accross a inductor can change quickly, but current only changes proportional to the applied voltage. In other words, the current will ramp up slowly. The load only sees the current coming out of the inductor. If the switch were to remain closed, that current would eventually build up too much voltage on the capacitor. However, the switch is only closed for short periods and with a average duty cycle so that the average current coming thru the inductor is exactly what is needed to maintain the desired voltage on the output capacitor.

A synchronous buck converter has no problem because it has two low impedance states in the push-pull output - it is either switch hard to the incoming supply voltage or switched hard to 0V. In other words it's a voltage waveform generator and, a simple LC low pass filter then behaves as an averager: -

However, it's different for the non-synchronous buck regulator because it only uses a single pass transistor and a flyback diode. Under these circumstances, and assuming a perfectly efficient scenario, the duty cycle MUST drop when the load current decreases. If the duty cycle doesn't drop, the output voltage will rise beyond acceptable limits.

Under near-no-load conditions a non-synchronous buck regulator must go into a different mode of operation where instead of operating at a fixed frequency of (say) 100 kHz, it pulses at a much lower rate. This is quite common to happen in many buck circuits.