555 Timer Astable Calculator
555 Timer Astable Calculator
Frequency, period, high and low times and duty cycle for a 555 in astable mode — or work backwards from the frequency and duty you want to R1, R2 and the nearest E12 or E24 parts.
555 astable oscillator
R1 = 8.2 kΩ, R2 = 68 kΩ, C = 10 nF, standard astable
555 astable timing
- ln 2
- 0.693 — the datasheet’s 0.693, because ⅓Vcc to ⅔Vcc is one halving of the remaining gap
- 1.44
- 1 ÷ ln 2 = 1.4427, the constant TI prints in the frequency formula
- R1
- from Vcc to the discharge pin; carries the charging current only
- R2
- from the discharge pin to the threshold/trigger node; carries both
- C
- the timing capacitor, from the threshold node to ground
Worked example
R1 = 8.2 kΩ, R2 = 68 kΩ, C = 10 nF, standard astable
thigh = 0.693 × (8.2 kΩ + 68 kΩ) × 10 nF = 528.2 µs
tlow = 0.693 × 68 kΩ × 10 nF = 471.3 µs
T = 528.2 µs + 471.3 µs = 999.5 µs, so f = 1 ÷ T = 1 kHz
Duty = (8.2 + 68) ÷ (8.2 + 136) = 76.2 ÷ 144.2 = 52.84%
Check with TI's constant: 1.44 ÷ (144.2 kΩ × 10 nF) = 1.44 ÷ 1.442 ms = 998.6 Hz, the same figure to the two digits the 1.44 carries.
One capacitor, five values of R2
| R2 | Frequency | Duty cycle (high) | Time high | Time low |
|---|---|---|---|---|
| 1 kΩ | 1.202 kHz | 91.7% | 762.5 µs | 69.31 µs |
| 4.7 kΩ | 743.7 Hz | 75.8% | 1.019 ms | 325.8 µs |
| 10 kΩ | 480.9 Hz | 66.7% | 1.386 ms | 693.1 µs |
| 47 kΩ | 138.7 Hz | 54.8% | 3.951 ms | 3.258 ms |
| 100 kΩ | 68.7 Hz | 52.4% | 7.625 ms | 6.931 ms |
How the 555 astable works
In astable mode the 555 is a relaxation oscillator. The timing capacitor charges from Vcc through R1 and R2 in series; when it reaches ⅔Vcc the internal comparator flips the flip-flop, the output goes low and the discharge pin pulls the junction of R1 and R2 to ground, so C now discharges through R2 alone. At ⅓Vcc the lower comparator flips it back and the cycle repeats. Each leg is an exponential that covers exactly half the gap to its destination — ⅓Vcc to ⅔Vcc is half of the remaining ⅔Vcc, and ⅔Vcc down to ⅓Vcc is half of ⅔Vcc — so each takes ln 2 = 0.693 time constants. That is where the datasheet’s 0.693 comes from, and why the supply voltage cancels out of the timing entirely.
Why the duty cycle is always above 50%. The charging path is R1 + R2 and the discharging path is R2 on its own, so the high time is longer than the low time by exactly 0.693 × R1 × C, whatever the values. The duty (R1 + R2) ÷ (R1 + 2R2) approaches 50% as R2 grows relative to R1 but can never reach it. Put a diode across R2, anode at the R1 end, and the charging current bypasses R2 altogether: the high time becomes 0.693 × R1 × C and any duty is available. Expect a few per cent of error there, because the diode’s forward drop is in series with the charging path.
Design mode runs the equations the other way. Choose C and a target frequency and duty; the page solves R2 = (1 − D)T ÷ (ln2 × C) and R1 = (2D − 1)T ÷ (ln2 × C), picks the nearest E12 or E24 values from IEC 60063, and reports what those real parts actually give. For 1 kHz at 60% with a 10 nF capacitor the exact values are 28.85 kΩ and 57.71 kΩ; the nearest E24 pair is 30 kΩ and 56 kΩ, which oscillates at 1,016.0 Hz (1.60% high) at 60.56% duty. Choose the capacitor first: capacitors come in far coarser steps than resistors, so it is the capacitor that should be a round value.
Practical limits. Keep the resistors between about 1 kΩ and 1 MΩ. Below 1 kΩ the discharge pin has to sink a lot of current the moment it turns on; TI limits R1 + R2 to 20 MΩ at the top end, and long before that the capacitor’s own leakage competes with the charging current, which is why megohm resistors want a film capacitor rather than an electrolytic. A bipolar LM555 draws 3–6 mA at 5 V just to run itself, so it is a poor choice on a battery; TI’s CMOS LMC555 dissipates under 1 mW at 5 V and is specified to 3 MHz. One more trap: TI’s own two datasheets define “duty cycle” oppositely — the LM555 sheet gives (R1 + R2)/(R1 + 2R2), the high fraction, and the LMC555 sheet gives R2/(R1 + 2R2), the low fraction. This page shows both. For a single pulse rather than a train, use the 555 monostable calculator; to drive a motor or an LED from the same square wave, the PWM duty cycle calculator; for the RC exponential behind it all, the RC time constant calculator.
Frequently asked questions
What is the formula for a 555 astable frequency?
f = 1.44 ÷ [(R1 + 2R2) × C], which is 1 ÷ [0.693 × (R1 + 2R2) × C]. With R1 = 8.2 kΩ, R2 = 68 kΩ and C = 10 nF that is 1 kHz.
Why can’t a 555 astable give a 50% duty cycle?
The capacitor charges through R1 + R2 and discharges through R2 alone, so the high time is always 0.693 × R1 × C longer than the low time. The duty gets closer to 50% as R2 grows but never reaches it. A diode across R2 lets the charging current bypass R2, which makes 50% and below possible.
Does the supply voltage change a 555’s frequency?
No. The switching thresholds are ⅓ and ⅔ of Vcc, so the capacitor always covers the same fraction of the gap and the times are the same at 5 V and at 15 V. Supply voltage does change the output drive, the supply current and the charging current shown above.
What resistor and capacitor values should I use in a 555 astable?
Pick the capacitor first from what is available (100 nF, 1 µF, 10 µF), then let the resistors fall between about 1 kΩ and 1 MΩ. Below 1 kΩ the discharge pin has to sink a large current; above about 1 MΩ the capacitor’s leakage starts to matter and TI caps R1 + R2 at 20 MΩ.
How do I get a duty cycle below 50% from a 555?
Put a diode across R2 with its anode at the R1 end. The capacitor then charges through R1 only and discharges through R2, so the high time is 0.693 × R1 × C and the duty is R1 ÷ (R1 + R2). Allow for the diode’s forward drop, which lengthens the high time slightly.
Related calculators
References
- Texas Instruments. LM555 Timer, datasheet SNAS548. Astable: t1 = 0.693(RA+RB)C, t2 = 0.693 RB C, T = 0.693(RA+2RB)C, duty cycle (RA+RB)/(RA+2RB); monostable t = 1.1 RA C; supply 4.5–16 V, supply current 3–6 mA at 5 V.
- Texas Instruments. LMC555 CMOS Timer, datasheet SNAS558N (revised March 2024). “Fast astable frequency of 3MHz”; “Less than 1mW typical power dissipation at 5V supply”; specified down to a 1.5 V supply; f = 1.44/[(RA+2RB)C].
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. §7.1.3, “The classic oscillator–timer chip: the 555”.
- IEC 60063:2015. Preferred number series for resistors and capacitors (E6, E12, E24, E48, E96 and E192). International Electrotechnical Commission.