Calculators
555 timer calculator
By Bulan Sarkar
In astable mode a 555 runs at f ≈ 1.44 / ((RA + 2RB) × C). The output stays high for 0.693 × (RA + RB) × C and low for 0.693 × RB × C. In monostable mode one trigger gives a single pulse of about 1.1 × R × C. All of these come from TI's NE555 datasheet.
Put in R and C and get the frequency, duty cycle and waveform. Or give a frequency and duty cycle and get standard RA and RB values, with the datasheet limits checked.
4k7, 4.7k and 4700 all work.
Read as 10 µF. Type 10u for 10 µF, 100n for 0.1 µF.
Only used for the datasheet checks. Timing does not depend on it.
- Period
- 1.01 s
- 0.986 Hz by the 1.44 formula
- Frequency
- 0.986 Hz
- 1/T gives 0.988 Hz
- Output duty cycle
- 53.4 %
- time high ÷ period (equation 6)
- High time, tH
- 541 ms
- 0.693 × (RA + RB) × C
- Low time, tL
- 471 ms
- 0.693 × RB × C
- Output driver duty
- 46.6 %
- time low ÷ period (equation 5)
Formulas and limits from TI's NE555 datasheet (SLFS022K, pp. 4, 6, 11–12) and TLC555 datasheet (SLFS043K, pp. 1, 4). Checked 6 October 2026. The NE555 is rated 4.5–16 V; the TLC555C from 2 V.
How a 555 keeps time
A 555 compares the voltage on its timing capacitor with two levels set by the supply: about two thirds of VCC (the threshold, pin 6) and about one third (the trigger, pin 2). Both levels move with the supply, and so does the charging current, so the timing stays the same whether you run it from 5 V or 12 V. The datasheet's one condition is that the supply holds steady during each timing interval.
When the capacitor climbs to the threshold, the output goes low and the discharge transistor on pin 7 switches on. When it falls to the trigger level, the output goes high and the transistor lets go. The resistors decide how fast each half happens, and that is all the calculator works out.
Astable: the free-running oscillator
C charges through RA and RB in series and discharges through RB alone. Because charging always goes through more resistance than discharging, the high time is longer than the low time. With this standard circuit the output duty cycle can't go down to 50 %; it gets close only when RB is much larger than RA. Equation 6 in the datasheet shows why: (RA + RB) / (RA + 2RB) is above one half for any RA greater than zero.
If you need 50 % or less, the usual fix is a diode across RB so the capacitor charges through RA alone. The diode's forward drop then shifts the timing, so the formulas here no longer hold and you should measure the result. Another option is to run the 555 at twice the frequency and divide by two with a flip-flop, which gives an exact 50 %.
Monostable: one pulse per trigger
A trigger pulls pin 2 below one third of VCC, the output goes high, and it stays high for about 1.1 × R × C. TI puts the shortest usable pulse at about 10 µs. Below that, the comparator's own delay takes over. Keep the trigger shorter than the pulse you want: held low for longer, the output just follows the trigger.
Why 1/T and the 1.44 formula differ slightly
Work out 1/T from the high and low times and you get a slightly different frequency from the 1.44 formula. That's because 1/0.693 is 1.443, which the datasheet rounds to 1.44. For TI's own example (5 kΩ, 3 kΩ, 0.15 µF) the two give 874.5 Hz and 872.7 Hz. That 0.2 % is far smaller than the tolerance of a typical capacitor, so either figure is fine. The calculator shows both, so if you check it by hand you'll see where each number comes from.
Worked examples
1. TI's own astable example
RA = 5 kΩ, RB = 3 kΩ, C = 0.15 µF. The high time is 0.693 × 8,000 × 0.15 µF = 832 µs and the low time 0.693 × 3,000 × 0.15 µF = 312 µs, so one period is 1.14 ms. By the 1.44 formula, f = 1.44 / (11,000 × 0.15 µF) = 872.7 Hz (1/T gives 874.5 Hz). The output is high for 8/11 of the time, 72.7 %.
2. An LED blinker at about 1 Hz
RA = 10 kΩ, RB = 68 kΩ, C = 10 µF. High for 541 ms, low for 471 ms, a period of 1.01 s, so f ≈ 0.986 Hz at 53.4 % duty. It's close to an even blink because RB is almost 7 times RA. These are the calculator's starting values.
3. A 10-second one-shot
C = 100 µF. R = 10 s / (1.1 × 100 µF) = 90.9 kΩ. The nearest E24 value is 91 kΩ, which gives 1.1 × 91,000 × 100 µF = 10.01 s. An aluminium electrolytic at ±20 % moves that anywhere from 8.01 s to 12 s, so for a timer that has to be right, trim R or pick a better capacitor.
4. Design backwards: 1 kHz at 60 % from 0.1 µF
Equation 4 sets RA + 2RB = 1.44 / (1,000 × 0.1 µF) = 14.4 kΩ. The duty cycle fixes the split: RB = (1 − 0.6) × 14.4 kΩ = 5.76 kΩ and RA = (2 × 0.6 − 1) × 14.4 kΩ = 2.88 kΩ. Neither is a stock value. The calculator tries the E24 values either side of each and keeps the pair with the smallest combined error: RA = 3 kΩ, RB = 5.6 kΩ, giving 1.014 kHz at 60.6 %, +1.4 % on frequency.
Datasheet limits the calculator checks
- Supply: 4.5 V to 16 V for the NE555 (18 V absolute maximum). The CMOS TLC555C runs from 2 V to 15 V.
- Frequency: up to 100 kHz on the bipolar NE555 for a clean waveform. Above that, TI points to the TLC555, rated up to 2 MHz.
- Timing resistance: RA + RB at most about 3.4 MΩ at 5 V and about 10 MΩ at 15 V, because the threshold input draws a small current.
- Output current: ±200 mA recommended (±225 mA absolute maximum).
- Decoupling: a 0.1 µF ceramic capacitor from VCC to ground, close to the chip.
These show as warnings, not errors. A circuit outside them may still run on the bench, but the datasheet no longer promises the timing.
Tips and tricks
- Choose C first, then the resistors. Capacitors come in fewer values and wider tolerances, so fix C and let the calculator fit RA and RB around it.
- Don't make RA tiny. When pin 7 switches on, RA sits straight across the supply: at 9 V a 100 Ω RA pulls 90 mA through the discharge transistor, and none of it does anything for the timing.
- Use film or C0G capacitors for steady timing. A high-k ceramic such as X7R or Y5V can lose a large part of its value with DC bias and temperature, and an electrolytic's value drifts with age. Check the code on yours with the capacitor code calculator.
- Tie RESET (pin 4) to VCC if you don't use it. Left floating, it can pick up noise and stop the timer at random.
- Decouple pin 5. A 10 nF capacitor from CONT to ground keeps supply noise off the comparator levels.
- Expect a longer first cycle. At power-up the capacitor starts from 0 V, not from one third of VCC, so the first high time is about 1.6 times the rest (ln 3 / ln 2).
What we would do
For a blinker, buzzer or slow clock from 5 to 15 V, an NE555 with values from this calculator is fine. Start from a 10 µF or 0.1 µF capacitor, take the E24 pair it suggests and expect the real frequency within the capacitor's tolerance. From 3.3 V, above 100 kHz, or where battery current matters, buy the TLC555 instead; the formulas and pinout stay the same. If the timing has to be accurate to better than a few per cent, a 555 is the wrong part, and a microcontroller timer or a crystal oscillator will do better.
Questions people ask
Why can't my 555 give a 50 % or lower duty cycle?
In the standard astable circuit the capacitor charges through RA + RB and discharges through RB alone, so the high time is always the longer one. Making RB much larger than RA brings it close to 50 %, but the duty cycle never reaches 50 % or goes below it.
Does the supply voltage change the 555 frequency?
No, as long as the supply holds steady through each cycle. The threshold and trigger levels are fixed fractions of VCC, and the charging current scales with VCC too, so the two cancel. TI states that the timing is independent of the supply.
What is the maximum frequency of a 555 timer?
TI recommends keeping the bipolar NE555 at or below 100 kHz. For anything faster it points to the CMOS TLC555, which is rated up to 2 MHz.
Can I run a 555 from 3.3 V?
Not the NE555: its recommended minimum is 4.5 V. The CMOS TLC555C is rated from 2 V to 15 V and uses the same timing formulas.
Why does the datasheet use 1.1 RC for the one-shot?
The capacitor charges from 0 V until it reaches two thirds of VCC, which takes ln 3 × RC, or 1.0986 RC. TI rounds that to 1.1, which is high by about 0.13 %. A capacitor tolerance of 10 or 20 % swamps that difference.
Related calculators
Read the bands on RA and RB with the resistor colour code calculator, size the LED's series resistor for a blinker with the LED resistor calculator, and build a steady 5 to 15 V supply for the timer with the LM317 calculator. To build a two-LED flasher from start to finish, follow the 555 LED flasher system.
Sources: Texas Instruments, xx555 Precision Timers datasheet, SLFS022K (revised March 2026), pp. 4, 6, 10–12 and 18; Texas Instruments, TLC555 LinCMOS Timer datasheet, SLFS043K (revised January 2026), pp. 1, 4, 14–15. Checked 6 October 2026.