555 IC astable mode - Does the voltage generally affect the frequency?

Question

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 5 V, and it takes an x amount of time for the external capacitor's charge to build up and for the voltage to rise from 1.667 V to above 3.333 V, would it not take longer if VCC were 9 V? The references would now be 3 V and 6 V. If no components have been swapped (external capacitor, resistors), would it not take longer for the voltage to rise from 3 V to above 6 V? Or does the increase in voltage charge the capacitor faster, and the effects cancel out?

Answer 1

Voltage is not part of the equation, and that is because the voltage does not matter.

I confirm that the default voltage references are 1/3 and 2/3 of the VCC. The RC time is also constant. In electronics we can use RC time simply because it does not depend on voltage when using a constant voltage accros the R+C components in series.

When the voltage scales, all other properties scale accordingly, including the compare levels.

When VCC increases from 5V to 9V, the capacitor will charge faster from 1.667V to 3.333V, but it will require the same time to charge from \$\frac{VCC}{3}\$ to \$\frac{{2}{VCC}}{3}\$.

The RC time can be used as follows: if your capacitor is charged at V0, and you apply Vtot accross the RC series circuit, then your final voltage will be V0+0.63*(Vtot-V0) after a delay of RC. The 63% percentage is constant. You can replace Vtot with VCC and V0 with VCC/3. After one RC, the voltage change is \$\frac{VCC}{3}+\frac{{0.63}{2}}{3}{VCC}= (\frac{1}{3}+\frac{{0.63}{2}}{3}){VCC}\$ = about 75% of VCC. So, expressed as a % of VCC, the voltage change is constant and it will always reach 2/3 of VCC in the same delay.

Answer 2

You're correct to think that the capacitor charges faster and cancels out the different reference voltages. Remember that ohm's law is \$I = \frac{V}{R}\$, so higher voltage with the same resistance means higher current charging that capacitor.

Answer 3

The power supply voltage DO HAVE EFFECT on the output frequency, althoug this is small. In the CMOS version of 555 this effect is much smaller.

IE. If you want to have a perfect square signal, using only one resistor and a capacitor, 10k (5%) for the resistor, 1n for the capacitor, if you use 5V as a power supply, you will have an aprox frecuency near to 65kHz, BUT.... if you raise the power supply to 12V, you will have a rise in frequency of 1 or 2kHz. This is small compared to 65kHz, but it has an effect.