Frequency and duty cycle of an astable, pulse width of a monostable, time constant and cut-off of an RC network. Next to the result: the newer parts that do the same job with a fraction of the current or far better accuracy.
The 555 dates from 1972. These parts do the same jobs with a fraction of the current, much better accuracy, or both. The ones that suit the values above are marked.
Same pinout, so they drop into logic circuits; their output sources far less current than the NE555’s 200 mA, so check it before driving a relay, LEDs or a speaker. At 5 V they draw 60–180 µA instead of about 3 mA, and they run from 2 V (the LMC555 from 1.5 V), so they work on 3.3 V and on one Li-ion cell, where an NE555 needs 4.5 V.
Draws 35 nA and times from 100 ms to 7200 s (2 hours), set by one resistor, 1 % typical. It switches an ESP32 or a sensor on, waits for its DONE signal and cuts the power again: the node sleeps at nanoamps.
One resistor sets the frequency, with no capacitor to drift. LTC6906: 10 kHz–1 MHz, 12 µA at 100 kHz, within 0.65 % from 0 to 70 °C. LTC6900: 1 kHz–20 MHz, within 1.5 % at 25 °C.
If the project already has a microcontroller, a hardware timer or PWM output clocked by a crystal is accurate to tens of ppm instead of percent, and you change the frequency in code. ESP32 (LEDC) and RP2040 boards run from a crystal.
| R1 | R2 | Frequency | Duty | |
|---|---|---|---|---|
| 6.8 kΩ | 11 kΩ | 5.01 Hz | 61.8 % |
C charges through R1 and R2 until it reaches two thirds of the supply, then discharges through R2 alone into pin 7 until it falls to one third. The output is high while C charges and low while it discharges.
tH = 0.693 × (R1 + R2) × C tL = 0.693 × R2 × C f = 1.44 ÷ ((R1 + 2 × R2) × C) duty = (R1 + R2) ÷ (R1 + 2 × R2)
Example: R1 = 1 kΩ, R2 = 10 kΩ, C = 10 µF. tH = 0.693 × 11 kΩ × 10 µF = 76 ms, tL = 69 ms, f = 6.9 Hz, duty cycle 52 %: a steady blink.
A falling edge on TRIGGER (pin 2) sets the output high; C charges through R until it reaches two thirds of the supply, and the output goes low again.
t = 1.1 × R × C
Example: R = 100 kΩ and C = 10 µF give 1.1 s.
τ = R × C fc = 1 ÷ (2π × R × C) rise time 10–90 % ≈ 2.2 × τ
After one τ the capacitor has covered 63 % of the step, after 2.3 τ 90 %, after 4.6 τ 99 %. Example: 10 kΩ and 100 nF give τ = 1 ms and a cut-off frequency of 159 Hz.
| Part | Supply | Current | Range | Accuracy |
|---|---|---|---|---|
| NE555 | 4.5–16 V | ≈ 3 mA (5 V) | — | 1 % mono, 2.25 % astable |
| TLC555 | 2–15 V | 180 µA (5 V) | up to 2 MHz | set by R and C |
| LMC555 | 1.5–15 V | 180 µA (5 V) | up to 3 MHz | 75 ppm/°C drift |
| ICM7555 | 2–18 V | 60 µA | up to 1 MHz | 50 ppm/°C drift |
| TPL5110 | 1.8–5.5 V | 35 nA | 100 ms – 7200 s | 1 % |
| LTC6906 | 2.25–5.5 V | 12 µA (100 kHz) | 10 kHz – 1 MHz | < 0.65 % (0–70 °C) |
| LTC6900 | 2.7–5.5 V | 500 µA (3 MHz) | 1 kHz – 20 MHz | ≤ 1.5 % (25 °C) |
| MCU + crystal | board supply | depends on the chip | any, in code | tens of ppm |
Typical figures from the makers’ datasheets (Texas Instruments, Renesas, Analog Devices). The 555 errors are for the chip alone: the tolerance of R and C comes on top.
The formula is exact, the parts are not. The timing capacitor’s tolerance goes straight into the result: ±20 % for an electrolytic, ±5 % for film or C0G ceramic. Electrolytics also leak and lose capacitance with age and temperature. The NE555 itself adds about 1 % (monostable) to 2 % (astable) and drifts with temperature and supply voltage.
f = 1.44 ÷ ((R1 + 2 × R2) × C). With R1 = 1 kΩ, R2 = 10 kΩ and C = 10 µF that is 6.9 Hz. The high time is 0.693 × (R1 + R2) × C, the low time 0.693 × R2 × C.
With the standard circuit the high time is always the longer one. Make R1 much smaller than R2 to get close, or put a diode across R2 and make R1 equal to R2. A CMOS 555 can also charge and discharge C through one resistor from its output pin, which gives close to 50 % directly.
In practice a few minutes. Longer delays need a large resistor and a large electrolytic, and the capacitor’s leakage current starts to compete with the charging current. For delays up to 2 hours the TPL5110 does it with one resistor and 35 nA; beyond that, use a microcontroller.
Usually because of the capacitor: electrolytics are ±20 % and drift with temperature and age. Use a film or C0G capacitor and 1 % resistors, and add 10 nF on pin 5 and 100 nF across the supply. If you need better than about 1 %, use a crystal-clocked microcontroller or a silicon oscillator.
A CMOS 555 (TLC555, LMC555, ICM7555) draws 60–180 µA at 5 V instead of about 3 mA, and runs down to 2 V or less. To wake an ESP32 every few minutes, a TPL5110 cuts its power completely between wake-ups and draws 35 nA itself.
fc = 1 ÷ (2π × R × C). 10 kΩ and 100 nF give 159 Hz. At fc the signal is 3 dB down (70.7 %), and above it the level falls by 20 dB per decade.