555 Timer Calculator
Work out a 555 timer circuit. Astable frequency and duty cycle, or the RA and RB for a target frequency; monostable pulse width, R or C; bistable latch levels; Schmitt-trigger thresholds; and R or C for a frequency divider, a missing-pulse detector or a pulse stretcher.
How to Use
- Pick the circuit: Astable (oscillator), Monostable (one-shot), Bistable (latch), Schmitt trigger, Divider, Missing pulse or Stretcher.
- Enter the supply voltage; it sets the ⅓ and ⅔ Vcc switching levels and the largest timing resistor the 555 can use.
- Enter the parts. Type a value and choose its unit, or type it with a suffix such as 10k, 4.7uF or 100ms.
- Where a part is unknown, choose it in the Solve for list: its box turns read-only and shows the answer.
- Read the results, the circuit and its waveform, then check Show Work for each step and any datasheet warnings.
Worked Example
An astable LED flasher. RA = 10 kΩ, RB = 100 kΩ, C = 10 µF. The capacitor charges through RA + RB: tH = 0.693 × 110 kΩ × 10 µF = 0.762 s. It discharges through RB only: tL = 0.693 × 100 kΩ × 10 µF = 0.693 s. The period is 1.455 s, so f = 0.687 Hz and the output is high 0.762 ÷ 1.455 = 52.4 % of the time.
A monostable one-shot. TI’s datasheet example uses RA = 9.1 kΩ and C = 0.01 µF: tw = 1.1 × 9.1 kΩ × 0.01 µF = 100.1 µs. To hold a relay for 1 s with a 10 µF capacitor, R = 1 s ÷ (1.1 × 10 µF) = 90.9 kΩ (use 91 kΩ).
The common mistake: RA + RB instead of RA + 2RB. Plugging the flasher into 1.44 ÷ ((RA + RB) × C) gives 1.44 ÷ (110 kΩ × 10 µF) = 1.31 Hz, almost twice too fast. RB is in the circuit twice, once while charging and once while discharging, so the right sum is RA + 2RB = 210 kΩ and the answer is 0.687 Hz.
Show Work
Formulas
Background: Inside the 555
Hans Camenzind designed the 555 at Signetics; it went on sale in 1972, and TI’s own datasheet for it (SLFS022) dates from September 1973. Inside are a resistor divider across the supply that makes reference points at ⅓ and ⅔ Vcc, two comparators, a flip-flop, an output stage and a discharge transistor on pin 7. When the trigger (pin 2) drops below ⅓ Vcc the flip-flop sets, the output goes high and the discharge transistor lets go of the capacitor. When the threshold (pin 6) rises above ⅔ Vcc it resets, the output goes low and pin 7 shorts the capacitor. Pin 4 resets everything, and pin 5 exposes the ⅔ Vcc point so it can be moved.
Every circuit on this page is a way of wiring that one block. Pins 6 and 2 tied to the capacitor make it oscillate (astable). The trigger used alone, with the capacitor emptied by pin 7, makes a one-shot (monostable); because a monostable ignores triggers while it is timing, the same circuit divides a frequency, and with a transistor emptying the capacitor on every input pulse it becomes a missing-pulse detector. With no capacitor at all and pin 6 grounded, the trigger and reset pins make it a latch (bistable). Tie pins 2 and 6 together as an input and it is a Schmitt trigger with ⅓ Vcc of hysteresis.
The timing numbers come from the capacitor charging toward a supply: v = V(1 − e−t/RC). Going from ⅓ to ⅔ of the way takes ln 2 × RC = 0.693 RC, and going from 0 to ⅔ takes ln 3 × RC = 1.0986 RC, which TI rounds to 1.1 RC. Both the charging and the thresholds scale with Vcc, so the timing does not depend on the supply, and TI rates the drift at 0.1 %/V (monostable, typical).
About This Calculator
One page for seven 555 circuits. It works out astable frequency, duty and high and low times, or designs RA and RB for a target; monostable pulse width or the R or C that gives one; bistable switching levels and output voltage, with SET and RESET buttons to drive the latch; Schmitt-trigger thresholds with pin 5 open or driven; and the R or C for a frequency divider, a missing-pulse detector or a pulse stretcher. Each mode draws its circuit and waveform.
Every formula and limit is from Texas Instruments’ NE555 datasheet (SLFS022K, revised March 2026): the 0.693 and 1.1 constants, the 4.5–16 V supply range, the ±200 mA output, the 10 µs minimum pulse, the 100 kHz recommended ceiling, the 3.4 MΩ / 10 MΩ resistor limits, the 0.3–1 V reset level, and the output-high voltages (13.3 V at 15 V and 100 mA, 12.5 V at 200 mA). The page warns when a value falls outside them. These are for the bipolar NE555; CMOS versions such as the TLC555 have different limits. Everything runs in your browser and nothing you type is sent anywhere.
Related calculators: RC Time Constant Calculator, Clock Duty Cycle, and Capacitor Code Decoder.
Frequently Asked Questions
How do I calculate a 555 astable frequency?
TI’s NE555 datasheet gives f ≈ 1.44 / ((RA + 2RB) × C). The capacitor charges through RA + RB for tH = 0.693 (RA + RB) C and discharges through RB alone for tL = 0.693 RB C. With RA = 10 kΩ, RB = 100 kΩ and C = 10 µF that is 0.762 s high, 0.693 s low, a 1.455 s period, 0.687 Hz and a 52.4 % duty cycle.
Why is the 555 monostable pulse 1.1 × R × C?
The capacitor charges from 0 V toward Vcc and the pulse ends when it reaches ⅔ Vcc. That takes ln 3 × RC = 1.0986 RC, which TI rounds to 1.1 RC. Because both the charging and the threshold scale with Vcc, the width does not depend on the supply. 10 kΩ and 4.7 µF give 51.7 ms; TI’s own example, 9.1 kΩ and 0.01 µF, gives 100.1 µs.
Why can’t a basic 555 astable reach 50 % duty?
It charges through RA + RB but discharges through RB only, so the high time is always longer. Duty = (RA + RB) / (RA + 2RB): 10 kΩ and 100 kΩ give 52.4 %, and TI’s 5 kΩ and 3 kΩ give 72.7 %. A diode across RB lets the capacitor charge through RA alone; with RA = RB = 10 kΩ and 100 nF both halves are 0.693 ms, 50 % at 721.5 Hz.
What are the largest and smallest resistor values for a 555?
TI limits RA + RB to about 3.4 MΩ at a 5 V supply and 10 MΩ at 15 V, because the threshold input draws up to 250 nA that must come through the timing resistor. TI’s timing curves start at 1 kΩ; below that the discharge transistor sinks a large current. TI also recommends staying at or below 100 kHz.
How does a 555 divide a frequency?
Wired as a monostable, it ignores triggers while it is timing. Make the pulse longer than N − 1 input periods but shorter than N, and it fires on every Nth edge. Aiming at the middle, T = (N − 0.5) / fin: dividing 10 kHz by 4 needs 350 µs, which is R = 31.8 kΩ with 10 nF. The timing then has ±14.3 % of room before the count slips.
How do I use the 555 Timer Calculator?
Simply type your numbers and read the result, which refreshes the instant you change something. There is nothing to submit and nothing to wait for.
Is it free? Does it work without internet?
Yes to both. It is free with no sign-up, and once the page has loaded it keeps working even with no internet.
Where does my data go?
Nowhere — every calculation runs on your own device. Nothing you enter is uploaded, logged, or stored.
Common Use Cases
LED flasher
RA = 10 kΩ, RB = 100 kΩ and C = 10 µF blink an LED at 0.687 Hz: on for 0.762 s, off for 0.693 s, from any supply between 4.5 V and 16 V.
Piezo buzzer tone
1 kΩ, 10 kΩ and 10 nF oscillate at 6.87 kHz. For a 1 kHz tone at 60 % duty with 100 nF, design mode gives RA = 2.886 kΩ and RB = 5.772 kΩ.
Switch debounce and timed lights
A monostable with 10 kΩ and 4.7 µF turns every press into one clean 51.7 ms pulse; 100 kΩ and 10 µF hold a light or relay on for 1.1 s.
Motor stall and watchdog alarm
A missing-pulse detector watching a 10 Hz tachometer with a 1.5× margin times out after 150 ms; with 10 µF that needs R = 13.64 kΩ.
Stretching short sensor pulses
A 10 µs pulse from a sensor is stretched to 100 ms (10,000×) so an MCU or an LED can see it: R = 9.09 kΩ with 10 µF.
Latch and level-sensing
As a latch on 12 V the 555 drives a 100 mA relay at about 10.3 V; as a Schmitt trigger on 9 V it switches at 6 V and 3 V, 3 V of hysteresis.
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