As I understood it, a charge-pump increases the input voltage by stages Vout=Vin(N+1). In my example, the two stages of charge-pumps, I have Vout=3Vin. So in case of Vin=5.5V this leads to Vout=5.5*(3)=16.5V which is OK for driving the switch(if the gate voltage requires more than 10V).
However, when the input voltage hits 20V, the Vout=20*(2+1)=60V which is too high for gate voltage of 10V. So, my question is: how can the same charge-pumps circuit is used for all the varying input voltage? See figure below (Allegro A4926 MOSFET Driver)
Best Answer
The block diagrams are often, at the same time, difficult to interpret, and oversimplifying things. It is often worth reading datasheets text entirely.
In this specific case, see at the bottom of page 13:
There, you have your answer: the charge pump multiplies the input voltage by a fixed integer number, but it is followed by a regulator that clamps this voltage so the gate voltage is not excessive.