555 Timer Calculator

The 555 timer IC is one of the most widely used integrated circuits in electronics history, introduced by Signetics in 1972. In astable mode (free-running oscillator), it produces a continuous square wave whose frequency and duty cycle are set by two resistors and a capacitor.

The key formulas: High time (seconds): t_high = ln(2) × (R1 + R2) × C Low time (seconds): t_low = ln(2) × R2 × C Period (seconds): T = t_high + t_low = ln(2) × (R1 + 2×R2) × C Frequency (Hz): f = 1 / T Duty cycle (%): D = (R1 + R2) / (R1 + 2×R2) × 100

Where:

• R1, R2 = resistance in Ohms (Ω)
• C = capacitance in Farads (F)
• ln(2) = 0.693147, the natural log of 2, which is where the timing comes from

About that 1.44 you see everywhere. Datasheets write the frequency as f = 1.44 / ((R1 + 2×R2) × C), and 1.44 is just 1/ln(2) rounded down from 1.4427.
It is a fine number to do in your head and it is 0.2% low. This calculator uses ln(2) throughout so that the frequency it prints is genuinely the reciprocal of the period it prints, which the mixed 1.44-and-0.693 version is not. If your datasheet gives 138.5 Hz where this gives 138.7, that is the whole of the difference, and it is a twentieth of the error in a 5% resistor.

Where the timing actually comes from. The capacitor charges toward the supply rail but the comparators trip at 1/3 and 2/3 of it, and the time to cross that middle third of an exponential is ln(2) × RC no matter what the supply voltage is. That is the useful property of the whole chip: the frequency does not move when your battery sags.

Worked example: R1 = 10,000 Ω (10 kΩ), R2 = 47,000 Ω (47 kΩ), C = 0.1 μF (0.0000001 F)

t_high = 0.693147 × 57,000 × 0.0000001 = 3.951 ms t_low = 0.693147 × 47,000 × 0.0000001 = 3.258 ms

T = 3.951 + 3.258 = 7.209 ms, so f = 1 / 0.007209 = 138.7 Hz

Duty cycle = (10,000 + 47,000) / (10,000 + 2×47,000) × 100 = 57,000 / 104,000 × 100 = 54.8%

Component value guidelines:

• R1: 1 kΩ to 1 MΩ, and never zero, which would put the discharge transistor straight across the supply
• R2: 1 kΩ to 10 MΩ
• C: 100 pF to 1000 μF, where larger means lower frequency

Frequency ranges by application:

• LED blinker: 0.5 to 2 Hz (C = 10 to 100 μF)
• Tone generator: 200 to 4,000 Hz (C = 10 to 100 nF)
• PWM motor control: 1 to 20 kHz (C = 1 to 10 nF)
• Above about 500 kHz a bipolar 555 stops being honest about its output, so reach for a dedicated oscillator IC

Why you can never quite reach 50%

Look at the duty cycle formula and you will see the problem: (R1 + R2) is always larger than R2, so the high time is always longer than the low time. The output is asymmetric by construction, and duty cycle can approach 50% only as R1 becomes negligible next to R2, which you cannot do because R1 must stay above about 1 kΩ.

The diode fix. Put a diode in parallel with R2, anode at pin 7 and cathode toward the capacitor. Getting that orientation right is the whole trick. During charging the diode conducts and shorts out R2, so the capacitor charges through R1 alone; during discharge the diode is reverse biased and the current goes through R2 as usual. That gives:

t_high = ln(2) × R1 × C and t_low = ln(2) × R2 × C

Set R1 = R2 and you get a genuine 50%, minus a fraction of a percent for the diode’s forward drop. Put the diode in backwards and you achieve the opposite of what you wanted, shortening the low time instead.