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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)

Block diagram from Allegro A4926 MOSFET Driver datasheet

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    \$\begingroup\$ Your final question does not match the title. Which question are you asking - high side or low side? Pick one, please. \$\endgroup\$ – WhatRoughBeast Sep 4 '18 at 11:27
  • \$\begingroup\$ actually, I want to understand both . nut the first question is more important to me. should I ask the second question in a different post? \$\endgroup\$ – Yaakov Sep 4 '18 at 11:53
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    \$\begingroup\$ can you edit and formulate your question better. \$\endgroup\$ – Navaro Sep 4 '18 at 20:20
  • \$\begingroup\$ I have made the changes \$\endgroup\$ – Yaakov Sep 11 '18 at 7:13
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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:

Pump Regulator

The gate drivers are powered by a programmable voltage internal regulator which limits the supply to the drivers and therefore the maximum gate voltage.

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.

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