If the inductor saturates it becomes little more than a wire. di/dt will be massive and the overcurrent protection may not be fast enough to prevent damage. Not very many DC-DC regs handle saturation gracefully.
By the way, Silenus asked for the peak switch current, not a rule of thumb to estimate the same in an assumed properly designed converter. This is somewhat varying. One way you can think of a buck converter is as an LC low-pass filter, which averages a square wave of varying duty cycle.
- Approximating the output voltage as fixed, the inductor will act as an integrator, where di/dt is proportional to the voltage across the inductor and inversely proportional to the inductance:
V = L di/dt
- Working from the rising slope, V across the inductor will be equal to Vin-Vout; from this we get di/dt.
V = Vin - Vout
di/dt = (Vin - Vout) / L
- We also have the duty cycle (if the square wave's top and bottom are Vin and 0, and it averages to Vout, the duty cycle must be Vout/Vin), and the switching frequency (by design).
dc = Vout/Vin
- The positive width of the waveform is equal to the duty cycle divided by the switching frequency.
Wpos = dc/fs = Vout/(Vin * fs)
- The height of the waveform is equal to the positive slope (di/dt) multiplied by the positive width.
Delta I = di/dt Wpos = (Vin - Vout)/L * Vout/(Vin * fs)
This is the ripple current. The peak will be roughly half the ripple current plus the average current, as the inductor's current waveform is equal to the ripple with an offset of the average.
Example:
Vin = 12V
Vout = 5V
L = 33uH
fs = 100kHz
Delta I = (12V - 5V)/33uH * 5V/(12V * 100kHz) = 883mA
This assumes, of course, continuous mode (the inductor's current does not reach zero).
You'll note no dependence on the average output current! This just comes from the rule of thumb - assuming L will be chosen depending on the output current.
Now, given an output current of 2A, and our approximation of the peak as the average output plus half the ripple:
Peak I = Mean I + (Delta I)/2
Peak I = 2A + (883mA)/2 = 2.44A
This is all obvious to someone who actually understands how a buck converter works - and rules of thumb like that are rarely of much use to people who don't first understand the actual operation.
This assumes ideal
everything; adding in inductor loss, FET resistance and diode V
F is an exercise left to the reader.
Where the rules of thumb
do come in is in the less ideal cases, like (as johansen mentioned) startup. This is quite variable, and I'd highly recommend
testing to find this - overspecify an inductor at first with significant excess capacity, and then
measure the peak startup current.
Soft start is nice.