What is the formula for 555 astable frequency?

The output charges C through R1 and R2 and discharges it through R2 alone, so the high time is 0.693(R1+R2)C, the low time is 0.693 R2 C, and the frequency is 1 divided by their sum, often quoted as f = 1.44 / ((R1 + 2R2) x C). With R1 = 10 kOhm, R2 = 100 kOhm and C = 0.1 uF the output is about 68.7 Hz.

How do I get close to a 50% duty cycle from a 555?

In the basic astable circuit the duty cycle is (R1+R2)/(R1+2R2), which is always above 50%. Making R2 much larger than R1 pushes it close to 50%, for example 10 kOhm and 100 kOhm gives 52.4%. For a true 50% square wave, add a diode across R2 so charge and discharge each use one resistor, or divide a faster output with a flip-flop.

How long is the pulse in 555 monostable mode?

A monostable 555 produces a single pulse of t = 1.1 x R x C each time it is triggered, the time the capacitor takes to charge from 0 to two thirds of the supply voltage. For example, 100 kOhm and 10 uF gives a 1.1 second pulse. Timing is independent of the supply voltage.

555 Timer Calculator

This 555 timer calculator gives you the output timing of the classic 555 IC in both of its standard modes. In astable mode the chip free-runs as an oscillator: the capacitor charges through R1 and R2 and discharges through R2 alone, so the output sits high for 0.693(R1+R2)C seconds and low for 0.693(R2)C seconds. Enter your two resistors in kilohms and the timing capacitor in microfarads, and the calculator returns the output frequency, the high and low times, the full period and the duty cycle, formatted in sensible units from nanoseconds to seconds and from hertz to megahertz. Switch to monostable mode and it becomes a one-shot pulse calculator instead, using the familiar t = 1.1RC formula with R1 as the timing resistor. Hobbyists use these numbers to set LED flash rates, tone generators, PWM experiments and time delays before committing anything to a breadboard, and students use them to check homework against the datasheet equations. Pick standard E12 or E24 component values near the calculated targets, and remember that real circuits drift a little with capacitor tolerance and temperature. For timing above a few minutes, low leakage capacitors matter more than the exact resistor value.

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Mode

Resistor R1

kΩ

Resistor R2

kΩ

Capacitor C

µF

68.71 Hz

astable output frequency

High time7.62 ms

Low time6.93 ms

Duty cycle52.4%

Formulas use the standard datasheet constants 0.693 (ln 2) and 1.1. Real-world timing varies with component tolerance, temperature and supply bypassing; ceramic and electrolytic capacitors can be 10 to 20% off their marked value. In monostable mode, R1 is the timing resistor and R2 is not used.

How it works

In astable mode the timing capacitor charges from one third to two thirds of the supply through R1 and R2 in series, taking 0.693(R1+R2)C seconds, then discharges back through R2 alone into pin 7, taking 0.693(R2)C seconds. The output frequency is the reciprocal of the period, the sum of those two times, which is the exact form of the familiar approximation f = 1.44 / ((R1 + 2R2) x C). The duty cycle is (R1+R2)/(R1+2R2), always above 50% in this basic circuit because the charge path always contains more resistance than the discharge path. In monostable mode a trigger on pin 2 starts a single pulse that lasts t = 1.1(R1)C seconds, the time to charge the capacitor from zero to two thirds of the supply. The 0.693 constant is the natural log of 2, and 1.1 is the natural log of 3, which is why timing does not depend on the supply voltage.

Worked example

With R1 = 10 kΩ, R2 = 100 kΩ and C = 0.1 µF, the high time is 0.693 x 110,000 x 0.0000001 = 0.0076 s, shown as 7.62 ms, and the low time is 0.693 x 100,000 x 0.0000001 = 0.0069 s, shown as 6.93 ms. The period is the sum, 14.55 ms, so the frequency is 1 / 0.014553, which is 68.71 Hz. The duty cycle is 110 / 210, or 52.4%. Switching the same R1 and C to monostable mode gives a one-shot pulse of 1.1 x 10,000 x 0.0000001 = 1.1 ms each time the timer is triggered.

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