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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?

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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.

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    \$\begingroup\$ Nicely explained you should certainly have been the selected answer. Extremely good first answer post. \$\endgroup\$ – Jack Creasey Jul 13 at 20:08
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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.

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  • \$\begingroup\$ Not sure what you are trying to say here, you certainly didn't answer the question directly ….the 555 timer is supply voltage insensitive. The calculation for frequency does NOT include the supply voltage. Calculate the values for a given frequency and double then supply voltage ….there will be no significant change in timing. The basic frequency calculation is f= 1/tau (where tau is RC), no reference to supply voltage since the threshold point define the percentage of the charge/discharge curve used …..and that is fixed. \$\endgroup\$ – Jack Creasey Jul 13 at 20:06
  • \$\begingroup\$ @JackCreasey I think Heart correctly confirmed OP's last question: Or, does the increase in voltage charge the capacitor faster, and the effects cancel out? \$\endgroup\$ – Huisman Jul 13 at 20:22
  • \$\begingroup\$ @Huisman I'd suggest the effects do not 'cancel out'. No matter the VCC, the trigger points are at a fixed ratio of VCC. The Frequency is always 1/tau.The 555 is a brilliant design, the only problem is the first cycle when the capacitor charges from 0V to 0.66VCC ...that time period is longer and many tames causes unexpected difficulty in its use. It's semantics, but may confuse some. \$\endgroup\$ – Jack Creasey Jul 13 at 22:43
  • \$\begingroup\$ @JackCreasey I'm not sure how the effects don't cancel out. This cancellation is exactly why the supply voltage doesn't matter, as long as you have the same supply for the capacitor charging as for the 555. \$\endgroup\$ – Hearth Jul 14 at 0:06

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