Electronics calculators

555 timer calculator

Work out a 555’s astable frequency and duty cycle, a monostable pulse width, or the R1 and R2 for a target frequency, with standard values and every step shown.

Calculated with electrician.wiki · https://electrician.wiki/calculators/555-timer/

Result

f = 6.87 Hz · duty 52.4 %

R1 = 1 kΩ, R2 = 10 kΩ and C = 10 µF oscillate at 6.87 Hz (period 146 ms): high for 76.2 ms, low for 69.3 ms, a 52.4 % duty cycle.

Frequency: 6.87 Hz
Period: 145.6 ms
High time (tH): 76.25 ms
Low time (tL): 69.31 ms
Duty cycle (high): 52.38 %
1.44 / ((R1 + 2R2)C): 6.857 Hz

Show the work

1. High time: C charges through R1 + R2
$t_{H} = \ln (2) \cdot (R_{1} + R_{2}) \cdot C = 0.6931 \cdot (1 kΩ + 10 kΩ) \cdot 10 µF = 76.25 ms$
2. Low time: C discharges through R2 into pin 7
$t_{L} = \ln (2) \cdot R_{2} \cdot C = 0.6931 \cdot 10 kΩ \cdot 10 µF = 69.31 ms$
3. Frequency
$f = \frac{1}{t_{H} + t_{L}} = \frac{1}{76.25 ms + 69.31 ms} = 6.87 Hz$
Datasheet shortcut: f ≈ 1.44 / ((R1 + 2R2)·C) = 6.857 Hz. The 1.44 is 1/ln 2 = 1.4427 rounded, so the shortcut reads 0.19 % low.
4. Duty cycle: share of each period the output is high
$D = \frac{R_{1} + R_{2}}{R_{1} + 2 \cdot R_{2}} = \frac{1 kΩ + 10 kΩ}{1 kΩ + 2 \cdot 10 kΩ} = 52.38 %$

References: TI NE555 datasheet (SLFS022K) (rev. March 2026): astable tH, tL and frequency (eq. 1–4), monostable tw ≈ 1.1RC, RA + RB limits, 100 kHz guidance

Duty cycle is the fraction of each period the output (pin 3) is high. The very first high period after power-up is longer, because C starts from 0 V instead of ⅓ of the supply.
Thresholds at ⅓ and ⅔ of the supply: the control pin (5) is not driven; a 10 nF capacitor from it to ground is fine.
Timing parts at their nominal values. Electrolytic capacitors are often ±20 %, and that usually dominates the real error.
Propagation delays and the discharge transistor’s on-resistance are ignored; they start to matter above about 100 kHz.

New to this? How the 555 timer works

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How to use the 555 timer calculator

Choose a mode. Astable takes R1, R2 and C and returns the frequency, period, high and low times and duty cycle of a free-running 555 oscillator. Monostable takes R and C and returns the length of the output pulse after each trigger. Design astable works the other way: give a target frequency, a duty cycle and the capacitor you have, and it returns the exact R1 and R2, the nearest E12 or E24 preferred values, and the frequency and duty cycle those standard parts really give. Values take the shorthand printed on parts: 10u is 10 µF, 100n is 100 nF, 4n7 is 4.7 nF and 1M is 1 MΩ.

Tick Diode across R2 for duty cycles of 50 % or less. The calculator then asks for the supply voltage and the diode’s forward voltage, because a real diode lengthens the high time noticeably at low supply voltages. Enter 0 V for the textbook ideal-diode result.

555 timer formulas

In astable mode the capacitor charges through R1 and R2 until it reaches ⅔ of the supply, then discharges through R2 alone into pin 7 until it falls to ⅓ of the supply. The high and low times, frequency and duty cycle are:

t_{H} = \ln (2) \cdot (R_{1} + R_{2}) \cdot C

t_{L} = \ln (2) \cdot R_{2} \cdot C

f = \frac{1}{\ln (2) \cdot (R_{1} + 2 \cdot R_{2}) \cdot C} \approx \frac{1.44}{(R_{1} + 2 \cdot R_{2}) \cdot C}

D = \frac{R_{1} + R_{2}}{R_{1} + 2 \cdot R_{2}}

The 1.44 in datasheets and textbooks is 1/ln 2 = 1.443 rounded, so the shortcut reads 0.19 % low; this calculator uses ln 2 exactly. Because the capacitor always charges through more resistance than it discharges through, the duty cycle without a diode is always above 50 %. In monostable mode the capacitor starts from 0 V and the pulse ends at ⅔ of the supply:

t = \ln (3) \cdot R \cdot C \approx 1.1 \cdot R \cdot C

Why ln 2 and ln 3? An RC circuit closes the same fraction of its remaining gap in each time constant. Charging from ⅓ to ⅔ of the supply halves the gap to the supply, and discharging from ⅔ to ⅓ halves the gap to 0 V: both take ln 2 = 0.6931 time constants. The monostable charge from 0 V to ⅔ shrinks the gap from the full supply to a third of it, a factor of three: ln 3 = 1.099 time constants, the familiar 1.1RC. Both thresholds come from an internal divider across the supply, so the supply voltage cancels out of every result.

Worked examples

An LED flasher

The classic first 555 circuit: R1 = 1 kΩ, R2 = 10 kΩ and a 10 µF electrolytic, driving an LED from pin 3.

1. High time: C charges through R1 + R2
$t_{H} = \ln (2) \cdot (R_{1} + R_{2}) \cdot C = 0.6931 \cdot (1 kΩ + 10 kΩ) \cdot 10 µF = 76.25 ms$
2. Low time: C discharges through R2 into pin 7
$t_{L} = \ln (2) \cdot R_{2} \cdot C = 0.6931 \cdot 10 kΩ \cdot 10 µF = 69.31 ms$
3. Frequency
$f = \frac{1}{t_{H} + t_{L}} = \frac{1}{76.25 ms + 69.31 ms} = 6.87 Hz$
4. Duty cycle: share of each period the output is high
$D = \frac{R_{1} + R_{2}}{R_{1} + 2 \cdot R_{2}} = \frac{1 kΩ + 10 kΩ}{1 kΩ + 2 \cdot 10 kΩ} = 52.38 %$

R1 = 1 kΩ, R2 = 10 kΩ and C = 10 µF oscillate at 6.87 Hz (period 146 ms): high for 76.2 ms, low for 69.3 ms, a 52.4 % duty cycle.

A one-second-plus delay

A push button pulls pin 2 low; the output should stay high for about a second with R = 100 kΩ and C = 10 µF.

1. Pulse width: C charges from 0 V to ⅔ of the supply through R
$t = \ln (3) \cdot R \cdot C = 1.099 \cdot 100 kΩ \cdot 10 µF = 1.099 s$
2. Check against the 1.1RC rule of thumb
$t \approx 1.1 \cdot R \cdot C = 1.1 \cdot 100 kΩ \cdot 10 µF = 1.1 s$

R = 100 kΩ and C = 10 µF give a 1.1 s output pulse (t = ln 3 · RC ≈ 1.1RC) each time the trigger pin (2) is pulled below ⅓ of the supply.

Designing a 1 kHz, 60 % oscillator

A test tone at 1 kHz with a 60 % duty cycle, built around a 10 nF film capacitor and E24 resistors.

1. High and low times from the target
$t_{H} = \frac{D}{f} = \frac{0.6}{1 kHz} = 600 µs t_{L} = \frac{1 - D}{f} = \frac{1 - 0.6}{1 kHz} = 400 µs$
2. R2 from the low time
$R_{2} = \frac{t_{L}}{\ln (2) \cdot C} = \frac{400 µs}{0.6931 \cdot 10 nF} = 57.71 kΩ$
3. R1 from the high time (C charges through R1 + R2)
$R_{1} = \frac{t_{H}}{\ln (2) \cdot C} - R_{2} = \frac{600 µs}{0.6931 \cdot 10 nF} - 57.71 kΩ = 28.85 kΩ$
4. Nearest E24 values and what they give
$f = \frac{1}{\ln (2) \cdot (R_{1} + 2 \cdot R_{2}) \cdot C} = \frac{1}{0.6931 \cdot (30 kΩ + 2 \cdot 56 kΩ) \cdot 10 nF} = 1.016 kHz$

For 1 kHz at 60 % duty with C = 10 nF, R1 = 28.85 kΩ and R2 = 57.71 kΩ exactly; the nearest E24 values, 30 kΩ and 56 kΩ, give 1.016 kHz at 60.6 % duty.

A 38 kHz infrared carrier on 5 V

IR remote receivers expect a carrier near 38 kHz at about 50 % duty. On 5 V with a 1N4148 across R2, the diode drop makes R1 much smaller than R2.

1. High and low times from the target
$t_{H} = \frac{D}{f} = \frac{0.5}{38 kHz} = 13.16 µs t_{L} = \frac{1 - D}{f} = \frac{1 - 0.5}{38 kHz} = 13.16 µs$
2. Charge factor with the diode’s voltage drop
$k = \ln (\frac{2 \cdot V_{CC} - 3 \cdot V_{D}}{V_{CC} - 3 \cdot V_{D}}) = \ln (\frac{2 \cdot 5 V - 3 \cdot 0.7 V}{5 V - 3 \cdot 0.7 V}) = 1.002$
3. R2 from the low time
$R_{2} = \frac{t_{L}}{\ln (2) \cdot C} = \frac{13.16 µs}{0.6931 \cdot 1 nF} = 18.98 kΩ$
4. R1 from the high time (C charges through R1 alone)
$R_{1} = \frac{t_{H}}{k \cdot C} = \frac{13.16 µs}{1.002 \cdot 1 nF} = 13.13 kΩ$
5. Nearest E24 values and what they give
$f = \frac{1}{k \cdot R_{1} \cdot C + \ln (2) \cdot R_{2} \cdot C} = \frac{1}{1.002 \cdot 13 kΩ \cdot 1 nF + 0.6931 \cdot 20 kΩ \cdot 1 nF} = 37.19 kHz$

For 38 kHz at 50 % duty with C = 1 nF and a diode across R2, R1 = 13.13 kΩ and R2 = 18.98 kΩ exactly; the nearest E24 values, 13 kΩ and 20 kΩ, give 37.19 kHz at 48.4 % duty.

555 frequency and pulse-width tables

These tables are computed by the same engine as the calculator. Frequency scales inversely with C, so each row is ten times the one below it; duty cycle depends only on the ratio of R1 to R2.

Astable frequency with R1 = 1 kΩ, no diode
C	R2 = 1 kΩ	R2 = 4.7 kΩ	R2 = 10 kΩ	R2 = 47 kΩ	R2 = 100 kΩ
1 nF	481 kHz	139 kHz	68.7 kHz	15.2 kHz	7.18 kHz
10 nF	48.1 kHz	13.9 kHz	6.87 kHz	1.52 kHz	718 Hz
100 nF	4.81 kHz	1.39 kHz	687 Hz	152 Hz	71.8 Hz
1 µF	481 Hz	139 Hz	68.7 Hz	15.2 Hz	7.18 Hz
10 µF	48.1 Hz	13.9 Hz	6.87 Hz	1.52 Hz	718 mHz
100 µF	4.81 Hz	1.39 Hz	687 mHz	152 mHz	71.8 mHz
Duty cycle	66.7 %	54.8 %	52.4 %	50.5 %	50.2 %

Monostable pulse width t = ln 3 · R · C
R	C = 10 nF	C = 100 nF	C = 1 µF	C = 10 µF	C = 100 µF
10 kΩ	110 µs	1.1 ms	11 ms	110 ms	1.1 s
100 kΩ	1.1 ms	11 ms	110 ms	1.1 s	11 s
1 MΩ	11 ms	110 ms	1.1 s	11 s	110 s

Duty cycles below 50 %: the diode and its voltage drop

A diode across R2 (anode at pin 7) lets the capacitor charge through R1 alone and discharge through R2, so the duty cycle becomes roughly R1 / (R1 + R2) and can go well below 50 %. The textbook formula, tH = ln 2 · R1 · C, assumes an ideal diode. A real one drops its forward voltage V_D, so the capacitor charges towards V_CC − V_D and takes longer to reach ⅔ of the supply:

t_{H} = k \cdot R_{1} \cdot C

k = \ln (\frac{2 \cdot V_{CC} - 3 \cdot V_{D}}{V_{CC} - 3 \cdot V_{D}})

How much longer the high time is than the ideal-diode formula ln 2 · R1 · C
Supply	Schottky, 0.3 V	Silicon (1N4148), 0.7 V
5 V	+15.0 %	+44.6 %
9 V	+7.8 %	+20.4 %
12 V	+5.7 %	+14.5 %
15 V	+4.5 %	+11.3 %

At 5 V a silicon diode stretches the high time by almost half, which is why a “50 %” 555 built with equal resistors and a 1N4148 runs noticeably asymmetric on a 5 V rail. This calculator includes the drop; use a Schottky diode, a higher supply, or a smaller R1 to compensate.

Practical pitfalls

Bipolar or CMOS. The original bipolar NE555 is rugged and drives heavy loads, but draws milliamps and tops out around 100 kHz. CMOS versions run from lower voltages, draw microamps, accept far larger timing resistors and run into the megahertz range, at the cost of output drive. The datasheet figures:

Datasheet figure	NE555 (bipolar)	TLC555 (CMOS)	LMC555 (CMOS)
Supply voltage	4.5–16 V	2–15 V	1.5–15 V
Typical consumption at 5 V	3 mA	180 µA	under 1 mW
Output drive	up to 200 mA, sink or source	100 mA sink, 10 mA source (typical)	tested at 50 mA sink, 10 mA source
Highest astable frequency	100 kHz or below recommended	2.1 MHz typical (1.2 MHz minimum)	3 MHz typical
TI datasheet	SLFS022K	SLFS043K	SNAS558N

Decoupling and the control pin. A bipolar 555 pulls a sharp current spike from the supply every time its output switches, enough to retrigger itself or upset nearby logic. TI recommends a 0.1 µF ceramic capacitor from V_CC to ground right at the chip (for the TLC555, 0.1 µF ceramic plus 1 µF electrolytic). A 10 nF ceramic from the control pin (5) to ground keeps noise off the thresholds; avoid a leaky electrolytic there, because its leakage shifts them.
Choosing the timing capacitor. Electrolytics are typically ±20 % and leak, so long delays wander. Film capacitors (polyester, polypropylene) hold their value; C0G/NP0 ceramics suit small values. High-capacitance ceramics such as X7R and Y5V lose capacitance with applied voltage and temperature: fine for decoupling, poor for timing. To read a capacitor’s markings, decode its value code.
Resistor limits. Keep R1 at 1 kΩ or more: whenever pin 7 discharges, R1 connects the supply straight to it. At the top end, TI limits RA + RB on the bipolar part to about 3.4 MΩ at 5 V and 10 MΩ at 15 V, because the threshold input draws up to 250 nA. For long delays use a bigger capacitor, a CMOS 555, or a 555 driving a counter.
The first cycle is longer. At power-up the capacitor starts from 0 V rather than ⅓ of the supply, so the first high period of the LED flasher above is 121 ms instead of 76.2 ms. It matters for tone bursts and anything that counts pulses from power-on.
Reset and trigger. Tie RESET (pin 4) to V_CC when unused. In monostable mode the trigger pulse must end before the output pulse does; if the input can stay low, couple it through a small capacitor with a pull-up so only the falling edge triggers.
Driving loads. An NE555 output can drive a small relay directly; put a flyback diode across the coil. For LEDs on the output, size the LED’s series resistor from the output high voltage, which sits a volt or two below V_CC on the bipolar part.

To check the parts themselves, read the resistors’ colour bands; to bias the control pin from a fixed voltage, calculate the voltage divider that feeds it.

Method and verification

The calculator uses the timing equations in TI’s NE555 datasheet (SLFS022K, March 2026), with ln 2 and ln 3 evaluated exactly rather than as 0.693 and 1.1, and double precision throughout. With the diode option, the high time is solved from the capacitor’s exponential charge towards V_CC − V_D; with V_D = 0 it reduces to the textbook ln 2 · R1 · C. In design mode the exact R1 and R2 come straight from the target high and low times, and each is then rounded on its own to the log-nearest E12 or E24 value before the real frequency and duty cycle are recomputed. Propagation delays, the discharge transistor’s resistance and capacitor tolerance are not modelled.

Test cases include the default flasher (6.87 Hz, 52.38 %), the component values of TI’s typical astable waveform (RA = 5 kΩ, RB = 3 kΩ, C = 0.15 µF: 874.4 Hz), checks of the diode factor against the raw charging equation at 5 V, 9 V and 15 V, design-to-analysis round trips with and without a diode, and rejection of duty cycles of 50 % or less without one.

Results assume nominal part values; real timing is usually limited by the capacitor’s tolerance. Updated 2026-10-14. Found a problem? Report it and we log every fix publicly.

Questions people ask

What is the formula for 555 astable frequency?

f = 1 / (ln 2 × (R1 + 2R2) × C), usually written f ≈ 1.44 / ((R1 + 2R2) × C). With R1 = 1 kΩ, R2 = 10 kΩ and C = 10 µF the 555 runs at 6.87 Hz with a 52.4 % duty cycle.

How do I get a 50 % duty cycle from a 555?

Put a diode across R2 so the capacitor charges through R1 alone, then make R1 equal to R2 for an ideal diode. A real silicon diode at 9 V lengthens the charge time, so R1 has to be about 83 % of R2; the design mode works this out for your supply.

What is the 555 monostable formula?

t = ln 3 × R × C ≈ 1.1 × R × C. With R = 100 kΩ and C = 10 µF the output pulse lasts 1.1 s.

What is the maximum frequency of a 555 timer?

TI recommends the bipolar NE555 for 100 kHz or below. CMOS versions go much higher: the TLC555 typically reaches 2.1 MHz (1.2 MHz minimum) and the LMC555 about 3 MHz, though timing accuracy falls off well before those limits.

Why does my 555 run at a different frequency than calculated?

Usually the timing capacitor: electrolytics are often ±20 % and leak. Other causes are the drop of a diode across R2, stray capacitance with very small capacitors, comparator delays above about 100 kHz, and the first cycle after power-up, which is longer than the rest.

Does the supply voltage change the 555 frequency?

Hardly. Both thresholds are fixed fractions of the supply, so the supply cancels out and the datasheet sensitivity is a fraction of a percent per volt. The exception is the diode-across-R2 circuit, whose high time depends on the supply through the diode’s voltage drop.