555 Timer Calculator
Frequency, period and duty for a 555 astable, pulse length for a monostable, and a straight answer about why 50% duty is out of reach.
A plain 555 astable cannot do 50% duty
The timing capacitor charges through R1 and R2 in series and discharges through R2 alone. The charging path is always the longer one, so the output high time is always longer than the low time and the duty cycle is always above 50%. It gets closer as R1 shrinks against R2 and never arrives. Any schematic promising a 50% square wave from a bare 555 is wrong — the standard fixes are a diode across R2, so charging bypasses it, or running at double the frequency and dividing by two with a flip-flop.
- The timing does not depend on the supply. The comparator thresholds sit at one third and two thirds of whatever the supply is, so a bigger supply charges the capacitor faster to a proportionally higher threshold and the two effects cancel exactly. A 555 clock runs at the same frequency on 5 V and 15 V.
- Keep R1 above about 1 kΩ. The discharge pin has to sink the capacitor's current through R1 every cycle. Below a kilohm that exceeds what the pin is rated for, which runs the chip hot and shortens its life.
- Above about 1 MΩ, use a CMOS version. A bipolar 555 has enough input leakage to shift the timing at high impedance and to drift with temperature. The 7555, TLC555 and LMC555 are built for it, draw far less current, and do not produce the supply spike a bipolar 555 does on every transition.
- Long timings are not precise. Multi-minute intervals need large electrolytics, which leak, vary 20% part to part and drift with temperature. For anything that has to be repeatable, a counter chip or a microcontroller is a better answer than a bigger capacitor.
How to use
- Pick astable for a free-running oscillator or monostable for a one-shot pulse.
- Enter R1, R2 and C in any notation — 1k, 10k and 100n works.
- Read the frequency, period and duty, with the high and low times drawn to scale.
- Check the warnings: they cover the values that damage the chip or drift.
Frequently asked questions
Why can I not get a 50% duty cycle?
Because the timing capacitor charges through R1 and R2 in series but discharges through R2 alone. The charging path is always longer, so the high time always exceeds the low time. Duty is (R1 + R2) divided by (R1 + 2 times R2), which approaches 50 percent as R1 shrinks and never reaches it. Any schematic promising a 50 percent square wave from a bare 555 is simply wrong.
So how do I get a square wave?
Two standard fixes. Put a diode across R2 so charging bypasses it, which makes the two halves independent and gives exactly 50 percent when R1 equals R2 — at the cost of the diode drop shifting the timing slightly. Or run the 555 at double the frequency you want and divide by two with a flip-flop, which gives a genuinely exact 50 percent because a divider cannot do anything else.
Where do 1.44 and 1.1 come from?
From natural logarithms, not from fudging. The 555 switches at one third and two thirds of the supply, so an astable transition moves the capacitor across that middle third, which takes ln(2) or 0.693 time constants. One over 0.693 is 1.44. A monostable charges from zero to two thirds, which is ln(3) or 1.0986 time constants, universally rounded to 1.1.
Does the supply voltage change the timing?
No, and this is one of the genuinely elegant things about the chip. The comparator thresholds come from an internal divider set at fractions of the supply, so a higher supply charges the capacitor proportionally faster toward a proportionally higher threshold. The two effects cancel exactly. A 555 clock runs at the same frequency on 5 V as on 15 V, though it draws more current.
How small can R1 be?
About a kilohm. Every cycle the discharge pin has to sink the current flowing through R1 while it pulls the capacitor down, and below a kilohm that exceeds what the pin is rated for. The chip runs hot, and over time it fails. This matters because the obvious way to chase 50 percent duty is to shrink R1, which walks you straight into the limit.
When should I use a CMOS 555?
Above about a megohm of timing resistance, where a bipolar 555 leaks enough current at its inputs to shift the timing and drift with temperature. Also whenever supply current matters, since a 7555 or TLC555 draws a fraction of what the original does, and whenever the supply is shared with analogue circuitry — the bipolar 555 produces a substantial current spike on every output transition that a CMOS version does not.
Can I use a 555 for a long delay?
You can, but not precisely. A multi-minute delay needs a very large capacitor, which means an electrolytic, and electrolytics leak, vary 20 percent part to part and drift badly with temperature. The leakage competes directly with the current through the timing resistor. For anything that has to repeat reliably, a counter chip driven from a fast oscillator or a microcontroller is far more predictable.
🔒 This tool runs entirely in your browser. Nothing you enter is uploaded, logged, or stored.