Voltage deviation on the bootstrap capacitor

Question

In the application note for ILD8150 at chapter 2.2.7 I read:

The high-side MOSFET gate driver is supplied by the bootstrap circuit. Capacitor CBOOT is charged when the high-side MOSFET is switched off and the diode D conducts current from VCC. CBOOT can be calculated as:

𝐶𝐵𝑂𝑂𝑇 > 𝑄𝐺 / ∆𝑉𝐶𝑏𝑜𝑜𝑡

Where QG is the internal MOSFET gate charge, 2.5nC and ∆VCBOOT is voltage deviation on the bootstrap capacitor.The internal supply circuit provides voltage to the bootstrap capacitor VCboot ≈ 8.6 V,which is a little higher than VCC. This circuit helps to maintain the voltage in dim-to-off and standby conditions. Higher voltage improves the RON of the internal MOSFET.

I understand that VCboot is 8.6 V, but what do they mean with "delta" in this context? Perhaps the voltage across Cboot? VCBoot - VCC = 8.6 V - 7.3 V = 1.3 V

Both values are constant, so why do they provide the formula only and not the recommended value for the capacitor? I mean, I cannot change the QG neither ∆𝑉𝐶𝑏𝑜𝑜𝑡, can I?

Answer 1

'delta' means the change in the BOOT voltage from the initial charged value to the value after the gate charge has been (re)supplied to the high-side FET. The gate charge is 2.5 nC; so if you have a 25 nF capacitor, the voltage will discharge by V=Q/C = 2.5n/25n = 0.1 V. So, even if initially charged to 8.6 V, you will only actually get 8.5 V across VGS of the FET.

This isn't a significant amount, but there is little point in minimizing CBOOT.