555 Timer Calculator
Frequency and duty cycle from parts, or parts for your frequency — with E-series values.
Resistors
| Part | Exact | Standard value |
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Waveform
The capacitor charges from ⅓ to ⅔ of Vcc while the output is high, then discharges through RB while it is low.
How it was calculated
Standard values for R
| Choice | R | Pulse | Error |
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How it was calculated
About the 555 Timer Calculator
The 555 timer makes square waves (astable mode) or single timed pulses (monostable mode) from one or two resistors and a capacitor. Enter RA, RB and C to get the frequency, period, high and low times and duty cycle of an astable — with a waveform drawing — or enter the frequency and duty cycle you want with a capacitor and get RA and RB, rounded to standard E-series values together with the frequency and duty cycle those values really give.
Duty cycles of 50% or less are not possible with the standard circuit, so the calculator switches to the diode-bypass version (a diode across RB) and includes the diode’s voltage drop. Monostable mode gives the pulse width t = 1.1 × R × C, or R for a pulse you need. The formulas and limits follow the TI datasheets of the bipolar NE555 and the CMOS TLC555.
How to use it
- Choose Astable (oscillator) or Monostable (one pulse per trigger), and the NE555 or TLC555.
- Astable: enter RA, RB and C (and the supply voltage) to get the frequency and duty cycle — or switch to Find RA and RB, enter the frequency, duty cycle and the capacitor you want to use, and pick an E-series.
- For a duty cycle below 50%, tick or keep Diode across RB and enter the diode drop (about 0.6–0.7 V for a silicon diode such as the 1N4148).
- Monostable: enter R and C for the pulse width, or the pulse width and C to get R. Read the notes on the limits of the chip you chose.
Examples
RA 5 kΩ, RB 3 kΩ, C 0.15 µF
tH 0.832 ms, tL 0.312 ms, f 874.5 Hz, duty 72.7%
C 100 nF, E24
exact RA 2.886 kΩ, RB 5.772 kΩ → 3 kΩ and 5.6 kΩ: 1,016 Hz, 60.6%
1 kHz, C 100 nF, ideal diode
diode across RB: RA 3.607 kΩ, RB 10.82 kΩ
R 9.1 kΩ, C 0.01 µF
pulse 100.1 µs
t 1 s, C 10 µF, E24
R = 90.91 kΩ → 91 kΩ, 1.001 s
Astable formulas (TI datasheets)
- High time, while C charges through RA and RB: tH = 0.693 × (RA + RB) × C
- Low time, while C discharges through RB: tL = 0.693 × RB × C
- Period T = tH + tL = 0.693 × (RA + 2RB) × C, frequency f ≈ 1.44 / ((RA + 2RB) × C)
- Duty cycle (output high) D = (RA + RB) / (RA + 2RB) — always above 50% in this circuit.
The capacitor swings between one third and two thirds of the supply, so the timing does not depend on the supply voltage. 0.693 is ln 2.
Duty cycle below 50%: the diode across RB
With a diode across RB (anode at the discharge pin, pin 7), C charges through RA and the diode only and discharges through RB, so the high and low times are set separately: tH = RA × C × ln((⅔Vcc − Vd) / (⅓Vcc − Vd)) and tL = 0.693 × RB × C. With an ideal diode (Vd = 0) the high time is 0.693 × RA × C; the real drop of a silicon diode makes it longer, which the calculator includes. It ignores the small current through RB during charging.
Monostable
A trigger pulse (pin 2 taken below a third of the supply) sets the output high for tw = 1.1 × R × C, after which C is discharged. The trigger must go high again before the end of the pulse — at least 10 µs before for the NE555, 1 µs for the TLC555 — or the output stays high; the same figures are the shortest pulses each chip makes reliably.
NE555 or TLC555?
Both use the same timing formulas and the same pinout. The bipolar NE555 runs from 4.5 V to 16 V, can sink or source up to 200 mA, and TI recommends keeping it at or below 100 kHz. Its threshold current limits RA + RB to about 3.4 MΩ at 5 V and 10 MΩ at 15 V. The CMOS TLC555 runs from 2 V to 15 V, draws far less supply current (1 mW typical at 5 V), allows smaller timing capacitors and works up to about 2 MHz — above 100 kHz its propagation delays lengthen the period. Decouple the supply right next to the chip: TI calls a 0.1 µF ceramic capacitor sufficient for the NE555 and recommends at least 0.1 µF ceramic in parallel with 1 µF electrolytic for the TLC555. TI also notes that a capacitor from the CONT pin to ground (0.01 µF in its circuits) can improve operation.
Sources
- Texas Instruments, xx555 Precision Timers (NA555, NE555, SA555, SE555), datasheet SLFS022K
- Texas Instruments, TLC555 CMOS Timer, datasheet SLFS043K
Limitations
- Timing is only as accurate as the parts: a ±5% resistor and a ±10% (or worse, for electrolytics) capacitor can move the frequency by 15% or more. Use film capacitors and 1% resistors for accuracy, or trim with a potentiometer.
- Electrolytic capacitors leak, which lengthens long delays and makes them less repeatable; very long times are better made with a counter.
- The formulas ignore propagation delays and the discharge transistor’s resistance, which matter above about 100 kHz or with very small RB.
- The diode-bypass result assumes a constant diode drop; measure the circuit when the duty cycle must be exact.
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Frequently asked questions
How do I calculate the frequency of a 555 astable?
f ≈ 1.44 / ((RA + 2RB) × C). With RA = 5 kΩ, RB = 3 kΩ and C = 0.15 µF: 1.44 / (11,000 × 0.00000015) ≈ 873 Hz (874.5 Hz with the exact 0.693 constant).
How do I get a 50% duty cycle?
In the standard circuit the duty cycle is (RA + RB) / (RA + 2RB), which approaches 50% only when RA is much smaller than RB. For 50% or less, put a diode across RB so that RA alone sets the high time and RB the low time; with RA ≈ RB (and allowing for the diode drop) you get close to 50%.
What is the 555 monostable formula?
t = 1.1 × R × C. A 100 kΩ resistor and 10 µF capacitor give 1.1 seconds. To get a time, R = t ÷ (1.1 × C).
Does the supply voltage change the 555 frequency?
Hardly: the capacitor charges between one third and two thirds of the supply, so the timing cancels out. Only the diode-bypass circuit depends on the supply, because the diode drop is a fixed voltage.
What is the maximum frequency of a 555?
TI recommends 100 kHz or below for the bipolar NE555 and suggests the CMOS TLC555 above that; the TLC555 runs up to about 2 MHz.