555 timer calculator

How this is calculated

Everything a 555 does follows from one arrangement inside the chip: a resistive divider fixes two thresholds at ⅓ and ⅔ of the supply, two comparators watch the timing capacitor against them, and a flip-flop between them drives the output and a transistor that can short the capacitor to ground.

In the astable circuit the capacitor charges through R1 and R2 in series, and discharges through R2 alone into the discharge pin. Each leg is an exponential crossing between the same two thresholds, so each takes one time constant times ln 2:

t_high = ln2 · (R1 + R2) · C ≈ 0.693 (R1 + R2) C

t_low = ln2 · R2 · C ≈ 0.693 R2 C

f = 1 / (t_high + t_low) = 1.44 / ((R1 + 2R2) · C)

In the monostable circuit the capacitor starts at zero and is charged through one resistor until it reaches ⅔ V+, which is a longer journey — from 0 rather than from ⅓ V+ — and takes ln 3 time constants:

t = ln3 · R · C ≈ 1.0986 R C

Notice what is absent from all four: the supply voltage. The thresholds are fractions of V+ and the capacitor charges toward V+, so the supply cancels exactly. A 555 running from 5 V and the same 555 running from 12 V keep the same frequency. That is the single most useful property of the part, and it is why it survived fifty years of better-specified competition.

Why the duty cycle can never go below 50 %

Look at the two expressions again. The high time uses R1 + R2 and the low time uses R2 alone, so the high time is always the longer of the two, no matter what values you choose:

D = (R1 + R2) / (R1 + 2R2)

With R1 tiny that ratio approaches 50 % but never reaches it; with R1 large it approaches 100 %. Equal resistors give exactly 66.7 %, which is the default this page opens with. Every beginner meets this wall while trying to build a 50 % square wave, and the calculator refuses the request rather than printing a negative resistor the way several online calculators do.

There are three honest ways past it:

A diode across R2, anode at the discharge pin, so the charging current bypasses R2 and only R1 sets the high time. Duty then becomes R1/(R1+R2) and any value is reachable. Select it above — but note the caveat in the next section, because that diode is not free.

Feed the timing resistor from the output pin instead of from V+: one resistor from pin 3 to the capacitor, with trigger and threshold tied together. The capacitor then charges and discharges through the same resistor between the same two thresholds, so the two halves are identical by construction and the duty is genuinely 50 %. It depends on the output swinging symmetrically to both rails, so use a CMOS 555 for it; a bipolar part's high level sits about 1.7 V below V+ and skews the result.

Run at twice the frequency and divide by two with a single flip-flop. The division is exact, the duty is exactly 50 % whatever the 555 was doing, and it costs one more package. This is what you do when the duty has to be right rather than approximately right.

The diode modification, honestly

Textbooks give the diode circuit's high time as 0.693 · R1 · C. That is only true for an ideal diode. A real one drops roughly 0.6 V, and that drop is inside the charging path, so the capacitor is no longer charging toward V+ — it is charging toward V+ − V_d, while the threshold it must reach is still ⅔ V+. The exponential is chasing a target that is now much closer to the finish line, which takes longer:

t_high = R1 · C · ln( (⅔V+ − V_d) / (⅓V+ − V_d) )

At 5 V with a silicon diode that logarithm is 0.941 rather than 0.693 — the high time is 36 % longer than the textbook figure, and the duty cycle you actually measure is nowhere near the one you designed. At 12 V the same diode costs 12 %. A Schottky at 0.3 V halves the damage but does not remove it — 15 % at 5 V, 6 % at 12 V. The error shrinks as the supply rises because the drop is fixed while the thresholds scale.

This calculator uses the exact expression and lets you set the drop; enter 0 to see the idealised textbook number. Two limits are worth knowing. The supply must exceed three times the diode drop, or the capacitor can never reach ⅔ V+ at all and the circuit simply stops — a 3 V supply with a silicon diode is already marginal. And R2 is still physically in parallel with the diode during charging, so a little current takes the resistive path; the model above ignores it, which is accurate when R2 is much larger than the diode's dynamic resistance, and slightly optimistic when it is not.

Choosing R and C

A given frequency fixes the product of resistance and capacitance, nothing more, so you get one free choice — and it matters more than people expect. The table above sweeps it for you.

Go too far toward small resistance and the discharge transistor has to sink V+/R1 every cycle, which heats the chip and softens the low level; below about 1 kΩ that starts to be a real effect at 12 V and above. Go too far toward large resistance and the threshold and trigger pins' own bias currents become comparable to the current charging the capacitor. On a bipolar NE555 those inputs draw a fraction of a microamp, which puts a practical ceiling near 1 MΩ; a CMOS 7555 or TLC555 draws picoamps and works happily up to 10 MΩ, which is exactly why long timings belong to the CMOS parts.

That leaves the capacitor to take the strain, and it is the weakest link in the circuit. Anything above about 1 µF means an electrolytic, whose tolerance is commonly −20/+80 %, whose leakage current fights the charging current, and whose value walks with temperature and age. A 555 asked to time ten minutes with a 100 µF electrolytic and a 10 MΩ resistor will be wrong by minutes, and the chip is not at fault. Below 1 µF you can use film or C0G/NP0 ceramic at 1–5 % and the circuit becomes as accurate as the 555 itself; below about 1 nF, stray capacitance and the chip's own input capacitance start to show up as an unwanted addition to C.

The practical sweet spot for the capacitor is 1 nF to 1 µF, with the resistors carrying the rest of the range. The calculator marks the rows that satisfy that. One catch when you go shopping: resistors are stocked in E12 and E24 steps, but capacitors mostly are not — outside precision film and C0G parts you will usually find only the six E6 values (1, 1.5, 2.2, 3.3, 4.7, 6.8), so treat an E24 capacitor suggestion as something to check availability on.

What really sets the accuracy

The three error sources are not equal, and they are almost never the ones people blame.

The capacitor dominates. A 10 % capacitor gives you a 10 % frequency error, full stop, and its temperature coefficient adds more. Unlike the resonant circuits on the LC page, where a square root halves component error on its way to the answer, here the relationship is linear and nothing softens it.

The resistors are next and are easy to fix: 1 % metal film costs the same as 5 % carbon film and removes the term from the budget.

The chip is a distant third. The internal divider is made of matched resistors on the same die, so the ⅓ and ⅔ ratios hold to well under a percent, and because the answer depends on those ratios, supply variation contributes very little — datasheets typically quote around 0.01 %/V for the astable frequency. Temperature drift of the timing is on the order of 50 ppm/°C. In other words the 555 is more accurate than the parts you are likely to hang on it.

One thing that is not in the budget but will ruin your measurement anyway: supply noise. The 555 draws a current spike of tens of milliamps for a moment at every output transition, and if that spike moves the supply rail it moves the thresholds along with it and can retrigger the comparators. A 100 nF decoupling capacitor across pins 8 and 1, right at the chip, is not optional. Add 10 nF from pin 5 to ground when you are not using the control input for anything, for the same reason.

Bipolar or CMOS

The original bipolar NE555 sources and sinks up to about 200 mA, which is enough to drive a small relay or a lamp directly, and needs 4.5 V or more to run. It also draws a few milliamps just idling, and its output high level sits roughly 1.7 V below the positive rail.

The CMOS derivatives — 7555, ICM7555, TLC555, LMC555 — swing very close to both rails, run from as little as 2 V, draw a few hundred microamps, generate a far smaller supply spike, and work at much higher frequencies (the TLC555 is specified past 2 MHz against a few hundred kilohertz of usable range for the bipolar part). They give that up on drive current, typically tens of milliamps rather than hundreds.

For battery work, long timings, high frequencies, or anything sharing a supply with an analogue circuit, use the CMOS version. Use the bipolar one when you need it to drive a load on its own. The pinout and every formula on this page are identical for both: 1 ground, 2 trigger, 3 output, 4 reset, 5 control, 6 threshold, 7 discharge, 8 V+.

FAQ

My measured frequency is off by 10–20 %. What is wrong?

Almost certainly the capacitor. Measure it, or substitute a known-good film part, before touching anything else. If the error is stable and proportional it is component tolerance; if it drifts as the circuit warms up it is the capacitor's temperature coefficient or an electrolytic's leakage; if it changes when you touch the board it is stray capacitance and your C is too small. A 555 whose timing changes when you change the supply voltage is telling you something different — that the supply is sagging under the output current, or that decoupling is missing.

My monostable output stays high and never falls.

The trigger input is level sensitive, not edge sensitive, once it is below ⅓ V+. If the trigger is still held low when the timing period ends, the output stays high until it is released. Fix it with a differentiator on the trigger — a small capacitor in series and a pull-up to V+ — which converts your long low into a short pulse. The rule is simply that the trigger pulse must be shorter than the output pulse you are trying to produce.

Can I retrigger a monostable to extend the pulse?

Not in the standard circuit. Once the flip-flop is set, further triggers are ignored and the pulse ends when the capacitor reaches ⅔ V+ regardless. A retriggerable one-shot needs the timing capacitor to be discharged on each new trigger — usually a transistor across it driven from the trigger — or a part designed for it such as a 74HC123. The dedicated part is the better answer when the requirement is real.

What is the highest frequency I can get?

A bipolar NE555 is usually specified for astable operation up to a few hundred kilohertz, but the duty cycle and the edges degrade well before that as the internal propagation delays become comparable to the timing intervals. Treat 100 kHz as comfortable and 300 kHz as the point where you should be measuring rather than calculating. A CMOS 555 roughly extends that by an order of magnitude. Above a megahertz or so, a 555 is the wrong part — use a crystal oscillator, a proper function-generator chip, or a microcontroller timer.

And the lowest?

Set by leakage rather than by any limit in the chip. With a CMOS 555, a 10 MΩ resistor and a 10 µF low-leakage film or tantalum capacitor you can reach hours, but the capacitor's own leakage is then a significant fraction of the charging current and the accuracy is poor. Beyond a few minutes, the standard approach is a 555 running at a convenient rate followed by a ripple counter such as a 4020 or 4060, which divides exactly and costs nothing in accuracy.

What does pin 5 do, and can I leave it open?

Pin 5 brings out the ⅔ V+ node of the internal divider, so anything you impose on it moves both thresholds together and therefore changes the timing. That is how a 555 is turned into a voltage-controlled oscillator or a PWM modulator — drive pin 5 and the pulse width follows. When you are not using it, tie it to ground through 10 nF; leaving it bare lets supply and radiated noise modulate your thresholds directly.

Do I need to connect pin 4 (reset)?

Yes. Reset is active low and it overrides everything, forcing the output low and holding the discharge transistor on. Left floating it will pick up noise and stop your oscillator at random intervals. Tie it to V+ if you do not need it, or drive it from a gate signal to switch the oscillator on and off cleanly.

The output drives an LED strip or a relay and now the timing is erratic.

The load current flows through the same ground and supply pins as the timing comparators. Give the chip its own 100 nF decoupling capacitor at the pins, keep the load's ground return away from the capacitor's ground, and consider buffering the output with a transistor or MOSFET so the 555 sees a light, constant load. An inductive load also needs a flyback diode, without which the switching spike will reach the chip.

Which resistor do I change to adjust just the frequency, or just the duty?

Neither one is independent in the standard circuit — both times contain R2. To sweep frequency at roughly constant duty, scale R1 and R2 together, or vary C. To adjust duty at a given frequency you have to move both, which is what solve-for-resistors above does for you. If you need independent control, the diode variant separates them cleanly: R1 owns the high time and R2 owns the low time.

Is a 555 a good choice for a PWM motor or LED dimmer?

It works and it is the classic hobby circuit, using two diodes so that a single potentiometer shifts the charge and discharge paths in opposite directions at constant total resistance. Expect the achievable range to stop short of 0 % and 100 %, expect the frequency to wander a little across the sweep, and expect the two diode drops to skew the curve exactly as described above. For anything where the duty needs to be accurate or the range needs to reach the ends, a microcontroller PWM output is a better tool and usually a cheaper one.

How does this relate to the RC filter calculator?

The same exponential, read differently. A first-order RC network has the time constant τ = R·C, and every number on this page is that constant multiplied by a logarithm: ln 2 for a threshold-to-threshold leg, ln 3 for the monostable's run from zero. If you want to know how the same R and C behave against a continuous signal rather than a comparator, that is the page for it.