What the calculator does
The 555 is a comparator pair, a flip-flop and a discharge transistor in an eight-pin package, and every circuit built on it times a capacitor charging through a resistor between one third and two thirds of the supply. The calculator does that arithmetic in the three ways it usually comes up. In astable mode it takes R1, R2 and C and returns the frequency, the period, the separate high and low times and the duty cycle. In design mode it takes the frequency and duty cycle you want and a capacitor you have, finds the exact R1 and R2, rounds each to the nearest E12, E24 or E96 value, and reports the frequency and duty you will actually get. In monostable mode it takes R and C and returns the width of the single output pulse each trigger produces.
Alongside the timing it reports the trigger and threshold voltages for your supply, the current the discharge pin sinks through R1, and a set of checks drawn from the datasheets: resistors below 1 kΩ or above 10 MΩ, capacitors below 100 pF, frequencies above what a bipolar or CMOS part can hold, and a supply outside the range of either. It is the arithmetic for a flasher, a delay-on relay, a small PWM source for a fan or a motor, a missing-pulse detector on a machine, or a clock for a logic counter, and for finding out why an existing 555 circuit runs 10 % off.
Formula
The timing capacitor charges from Vcc through R1 + R2 until pin 6 reaches ⅔ Vcc, when the flip-flop switches the output low and turns on the discharge transistor at pin 7; the capacitor then discharges through R2 alone until pin 2 falls to ⅓ Vcc, the output goes high again and the cycle repeats. Charging from ⅓ to ⅔ of the way to Vcc, or discharging from ⅔ to ⅓, takes exactly RC ln 2 = 0.693 RC.
T = 0.693 × (R1 + 2R2) × C f = 1 ÷ T = 1 ÷ (0.693 × (R1 + 2R2) × C) duty = (R1 + R2) ÷ (R1 + 2R2)
Design: R1 + 2R2 = 1 ÷ (0.693 × f × C) R1 = (2D − 1) × (R1 + 2R2) R2 = (1 − D) × (R1 + 2R2)
Monostable: T = R × C × ln 3 = 1.0986 × R × C ≈ 1.1 × R × C
where D is the duty cycle as a fraction (0.6 for 60 %), R in ohms, C in farads and times in seconds. The datasheets write the astable frequency as 1.44 ÷ ((R1 + 2R2) × C); 1.44 is 1 ÷ ln 2 = 1.4427 rounded, and the calculator uses 0.693 throughout so that its high time, low time and period add up exactly. The two versions differ by 0.2 %, far less than the tolerance of any capacitor. The monostable charges from zero rather than from ⅓ Vcc, which is why its factor is ln 3 rather than ln 2. Because R1 + R2 is always larger than R2, the astable output is high for longer than it is low: the duty cycle of this circuit is always above 50 %.
Worked example
An astable with R1 = 10 kΩ, R2 = 47 kΩ and C = 100 nF on a 5 V supply, the defaults above:
- R1 + 2R2 = 10 + 94 = 104 kΩ.
- f = 1 ÷ (0.693 × 104 000 × 100 × 10−9) = 1 ÷ 0.007207 = 138.8 Hz. With the datasheet's 1.44 the answer is 138.5 Hz; the difference is the rounding of ln 2.
- thigh = 0.693 × 57 000 × 100 × 10−9 = 3.95 ms; tlow = 0.693 × 47 000 × 100 × 10−9 = 3.26 ms; period 7.21 ms.
- Duty cycle = 57 ÷ 104 = 54.8 %.
- The capacitor swings between ⅓ and ⅔ of 5 V, 1.67 V and 3.33 V. While the output is low, pin 7 holds the bottom of R1 at ground and sinks 5 V ÷ 10 kΩ = 0.5 mA.
Design mode, for 1 kHz at 60 % duty with the same 100 nF: R1 + 2R2 = 1 ÷ (0.693 × 1 000 × 100 × 10−9) = 14.43 kΩ; R1 = 0.2 × 14.43 = 2.89 kΩ and R2 = 0.4 × 14.43 = 5.77 kΩ. The nearest E24 values are 3 kΩ and 5.6 kΩ, so R1 + 2R2 = 14.2 kΩ and the circuit runs at 1 ÷ (0.693 × 14 200 × 100 × 10−9) = 1 016 Hz (+1.6 %) at a duty of 8.6 ÷ 14.2 = 60.6 %. Select the design mode with the defaults to see it.
Monostable mode, with R = 100 kΩ and C = 10 µF: T = 1.0986 × 100 000 × 10 × 10−6 = 1.10 s, the "1.1 RC" of the datasheet. A 10 µF electrolytic with a −20/+80 % tolerance would put that anywhere between 0.9 and 2 s, which is the point made under the limitations below.
Pin functions of the 555
| Pin | Name | What it does |
|---|---|---|
| 1 | GND | Ground, 0 V. |
| 2 | TRIG | Trigger. Taking it below ⅓ Vcc sets the flip-flop: output high, discharge off. In the astable it is tied to pin 6. |
| 3 | OUT | Output. Push-pull; a bipolar NE555 sources or sinks up to 200 mA, a CMOS TLC555 sinks 100 mA and sources about 10 mA. High is about 1.7 V below Vcc on the bipolar part. |
| 4 | RESET | Active-low reset. Below about 0.7 V it forces the output low and discharges the capacitor. Tie it to Vcc when unused; never leave it floating. |
| 5 | CONT | Control voltage: the ⅔ Vcc node of the internal three-resistor divider. Decouple it to ground with 10 nF; drive it to modulate the frequency or pulse width. |
| 6 | THR | Threshold. When it rises above ⅔ Vcc the flip-flop resets: output low, discharge on. |
| 7 | DIS | Discharge. An open-collector (open-drain) transistor to ground, on while the output is low. It discharges the capacitor through R2 in the astable and directly in the monostable. |
| 8 | VCC | Supply: 4.5–16 V for the bipolar NE555, 2–15 V for the CMOS TLC555 (2–18 V for the 7555). Decouple it with 100 nF at the pin plus 10–100 µF nearby. |
Assumptions and limitations
- Component tolerance dominates the accuracy. The formula is exact for ideal parts; the parts are not. An aluminium electrolytic is typically −20/+80 % or ±20 %; a class 2 ceramic (X7R) is ±10 % and loses a further 10–30 % of its value under DC bias and with age, and Y5V is far worse; a 5 % resistor is 5 %. A 555 timed with a film or C0G/NP0 capacitor and 1 % resistors holds a few per cent; timed with an electrolytic it holds tens of per cent, and the datasheet's own timing accuracy of about 1 % is the chip alone.
- Temperature moves it. The chip's own drift is small (about 50 ppm/°C for the bipolar part). The capacitor is the problem: X7R varies ±15 % from −55 to 125 °C, and an electrolytic's value and leakage both rise with temperature. Inside a panel at 60 °C the frequency is not what it was on the bench.
- The bipolar NE555 is hard on its supply. It draws 3–6 mA quiescent at 5 V (up to 15 mA at 15 V), and on every output transition its push-pull stage conducts briefly from Vcc to ground, a current spike of the order of 100 mA lasting tens of nanoseconds. Without a 100 nF capacitor at pin 8 and a 10–100 µF electrolytic nearby that spike resets microcontrollers and shows up in audio. The CMOS TLC555 and 7555 have neither the quiescent current (tens to hundreds of microamps) nor the spike, and they hold their thresholds at exactly ⅓ and ⅔ Vcc, so their timing is a little closer to the formula.
- Frequency limits. Above about 100 kHz the bipolar part's output rise and fall times (about 100 ns) and its internal delays make the formula optimistic and the waveform ragged; the calculator warns there. A CMOS part is usable to about 1 MHz (the TLC555 datasheet quotes 2 MHz). Above that, use a different oscillator.
- Resistor range. Keep R1 above 1 kΩ, because pin 7 shorts its lower end to ground for the whole low time and the discharge transistor's saturation voltage then adds to the low time. Keep R1 + R2 below about 10 MΩ on a bipolar part, whose threshold pin draws up to 0.25 µA; a CMOS part tolerates much more. Between 1 kΩ and 1 MΩ, with C from 1 nF to 100 µF, the formula holds.
- Duty cycle below 50 %. The basic astable cannot do it. A diode from pin 7 to pins 6 and 2 (anode at pin 7) bypasses R2 during charging, so that thigh ≈ 0.693 × R1 × C and tlow = 0.693 × R2 × C and any duty is possible, at the cost of a diode drop that makes thigh a little shorter than the formula and dependent on Vcc. The calculator does not model the diode.
- The control pin moves the thresholds. Anything connected to pin 5 shifts the ⅔ and ⅓ Vcc points and with them the timing; so does noise on it. Decouple it with 10 nF to ground and the calculator's thresholds apply. Driving it deliberately is how the 555 becomes a voltage-controlled oscillator or a pulse-width modulator.
- Monostable trigger width. The trigger at pin 2 must return above ⅓ Vcc before the timed pulse ends; if it stays low, the output stays high until it rises. A trigger longer than the pulse needs a small RC and diode in front of pin 2 to turn it into an edge. The standard 555 monostable does not retrigger: a second trigger during the pulse is ignored.
Frequently asked questions
Why not just use a microcontroller?
Often you should: a microcontroller timer is crystal-accurate, costs the same, and can do everything a 555 does with firmware. The 555 wins where there is no firmware to write or maintain, where the circuit must keep working when a supply of anything between 5 and 15 V is all that is available, where 200 mA of output drive with no extra transistor is useful, and where a standalone, one-function board must be understood and repaired by someone who is not a programmer. A delay-on relay in a control panel, a flasher on a machine and a watchdog that resets a controller are the typical cases. It is also robust: a bipolar 555 survives abuse that kills a microcontroller pin.
Why is my frequency 10 % off what the calculator says?
Almost always the capacitor. Check its tolerance and type: a 100 nF X7R ceramic at 12 V bias can be 80 nF; a 10 µF electrolytic marked ±20 % is legally anything from 8 to 12 µF; a Y5V can be half its marked value warm. Measure it, or use a film capacitor. The next suspects are a floating or loaded control pin 5, a supply that dips on the output transition because it is not decoupled, and R1 below 1 kΩ. The chip itself is good to about 1 %.
How do I get exactly 50 % duty?
Three ways. Make R1 much smaller than R2, say 1 kΩ and 100 kΩ, which gives (R1 + R2) ÷ (R1 + 2R2) = 50.2 %: close enough for a flasher, not for a clock. Put a diode across R2 (anode to pin 7) and make R1 ≈ R2, which gives roughly 50 % but drifts with the diode drop and the supply. Or, for a true 50 %, run the 555 at twice the frequency into a D flip-flop wired to divide by two: the output is then exactly square whatever the 555 does. A CMOS 555 can also be wired with a single resistor from the output to pins 6 and 2, which is close to 50 % because the CMOS output swings almost rail to rail.
Can it drive a relay or an LED strip directly?
A bipolar NE555 output can source or sink 200 mA, which covers a small relay coil, a few LEDs or a piezo sounder. Put a flyback diode across any relay coil or motor, and remember that the high output is about 1.7 V below Vcc on the bipolar part, so a 5 V supply gives a 3.3 V high. For more than 200 mA, or for a motor, drive a logic-level MOSFET from pin 3. The CMOS TLC555 sources only about 10 mA, though it sinks 100 mA; drive the load between Vcc and pin 3 so that it is on when the output is low.
What does the CONTROL pin do?
Pin 5 is the top of the internal divider that sets the ⅔ Vcc threshold, and through it the ⅓ Vcc trigger level. Left alone with a 10 nF capacitor to ground it just holds those levels steady against noise. Fed with a voltage, it moves them: raising it lengthens the charge time and lowers the frequency, so a slow signal on pin 5 frequency-modulates an astable, and in a monostable it stretches or shortens the pulse, which is one way to make a pulse-width modulator. The internal divider is about 5 kΩ per section, so whatever drives pin 5 must be able to source and sink a milliampere or so.
References
- Texas Instruments, xx555 Precision Timers (NE555, SA555, SE555) datasheet — supply range, quiescent current, output drive, timing accuracy and the astable and monostable formulas
- Texas Instruments, TLC555 LinCMOS Timer datasheet — CMOS 555: 2–15 V supply, low supply current, frequency to 2 MHz, output sink and source ratings
- H. Camenzind, Designing Analog Chips, 2005 — the design and origin of the 555 by its designer
- P. Horowitz and W. Hill, The Art of Electronics, 3rd ed., Cambridge University Press, 2015 — §7.1.3, the 555 and its relatives
- IEC 60063:2015, Preferred number series for resistors and capacitors — E12, E24 and E96 values