Sorry if it's been asked before, I'm asking to satisfy a curiosity. I'm new to electronics, and I'm messing around with the perennial 555 IC. I've been introduced to the following formula which allows you to determine the frequency of the output when put in an astable mode:
f = 1.44 / (r1 + 2 * r2)C
I noticed how voltage is not part of the equation. However, wouldn't the VCC of the IC affect the frequency (generally speaking)? Internally, the IC creates a voltage divider and uses 1/3 and 2/3 of the VCC as references for the comparators. If VCC is 5V, and it takes x amount of time for the external capacitor's charge to build up and for the voltage to rise from 1.667V above 3.333V, would it not take longer if VCC were 9V? The references would now be 3V and 6V. If no components have been swapped (external capacitor, resistors), would it not take longer for the voltage to rise from 3V to above 6V? Or, does the increase in voltage charge the capacitor faster, and the effects cancel out?