The 555 is the part most people meet the first time they want a circuit to do something by itself. Two resistors, one capacitor, and an output that switches forever. The arithmetic fits on one line — and that one line hides two things that cost beginners an afternoon each.
The formula
f = 1.44 / ((R1 + 2·R2) · C)
duty = (R1 + R2) / (R1 + 2·R2)R1 runs from the supply to pin 7 (discharge). R2 runs from pin 7 to the timing capacitor. The capacitor goes from the junction to ground, and that junction also feeds pins 2 (trigger) and 6 (threshold) — the same node driving both is what makes the circuit oscillate.
| R1 | R2 | C | calculated | measured in simulation |
|---|---|---|---|---|
| 10 kΩ | 10 kΩ | 1 µF | 48.0 Hz | 47.95 Hz |
| 10 kΩ | 47 kΩ | 10 µF | 1.385 Hz | 1.385 Hz |
| 1 kΩ | 100 kΩ | 10 µF | 0.716 Hz | 0.709 Hz |
Those right-hand numbers are not the formula printed twice. They come from running the circuit, and the small gaps are the chip's own switching behaviour rather than rounding.
The first thing the formula hides: duty cycle
The capacitor charges through R1 + R2 and discharges through R2 alone. The high time is therefore always longer than the low time, and the duty cycle can never drop to 50% in this configuration — it approaches it from above as R1 shrinks.
With R1 = R2 = 10 kΩ, duty is 66.7%. Measured: 66.5%. If you wanted a square wave and got something visibly lopsided, this is why, and no amount of adjusting R2 fixes it. Make R1 small relative to R2 and you get close; put a diode across R2 so charge and discharge take different paths and you get all the way.
The second thing: the supply does not set the frequency
The thresholds inside the chip are ⅓ and ⅔ of VCC, produced by three equal resistors across the supply. Because both the charging target and the trip points scale together, the frequency does not change when the supply does. Run the same circuit at 5 V and at 9 V and the period is the same.
This is the property that made the 555 useful in 1972 and it is worth seeing rather than being told. It also means pin 5 is already defined — the internal divider holds it at ⅔ VCC whether or not you connect anything.
Why a correctly wired 555 sometimes does nothing
Pin 4. It is RESET, it is active low, and floating it is the single most common fault. Tie it to the positive supply. On a breadboard a floating reset pin will happily pick up enough noise to hold the output low and look like a dead chip.
The output cannot drive much. A 555 sources a couple of hundred milliamps at best, and its output sits about a volt below the supply under load. Hanging a motor directly on pin 3 sags the output badly — in simulation a 6 Ω motor pulls a 5 V 555 output down to 1.46 V, which is a useful thing to watch happen. Put a transistor in between.
Monostable: one pulse per trigger
Move R2 and the capacitor around and the same chip gives a single pulse instead of a train:
pulse width = 1.1 · R · CWith R = 10 kΩ and C = 1 µF that is 11 ms; the simulator measures 10.97 ms. The trigger on pin 2 has to go below ⅓ of the supply, briefly, and the pulse runs to completion — pulling the trigger again during the pulse does nothing.
Simulating it
An astable has no steady operating point, which is why a lot of simulators show it as a flat line: asked to find a DC bias first, the solver settles into one of the two states and stays there. CircuPilot runs the 555 in the time domain and plays it back at real speed, so a 1 Hz blinker blinks once a second on the canvas rather than being a number in a table.
Drop a 555 from the sidebar, wire the two resistors and the capacitor, press Run, and compare the measured frequency against your arithmetic. If they disagree by more than a percent or two, the wiring is telling you something the formula cannot.
Where to go next
- The two-transistor version of the same idea: simulate a blinking LED with an astable multivibrator
- Using a 555 to control a motor: Arduino DC motor speed control with PWM
- The capacitor charging curve underneath all of this: RC time constant, charge and decay
Frequently asked questions
What is the 555 astable frequency formula?
f = 1.44 / ((R1 + 2·R2) · C), where R1 is the resistor from the supply to pin 7, R2 runs from pin 7 to the capacitor, and C is the timing capacitor from pins 2 and 6 to ground. With R1 = R2 = 10 kΩ and C = 1 µF that gives 48 Hz; a real simulation of the same circuit measures 47.95 Hz.
Why can't I get a 50% duty cycle from a 555?
In the standard astable the capacitor charges through R1 + R2 and discharges through R2 alone, so the high time is always longer than the low time. The duty cycle is (R1 + R2) / (R1 + 2·R2), which approaches 50% as R1 gets small but never reaches it. Making R1 much smaller than R2 gets close; a diode across R2, or a 50% mode chip like the 555 in a different configuration, gets you the rest of the way.
What does pin 4 do on a 555?
Pin 4 is RESET and it is active low. Left floating it can pick up noise and hold the output low, which is the most common reason a correctly wired 555 does nothing at all. Tie it to the positive supply unless you actually want to gate the oscillator.
Do I need a capacitor on pin 5?
No. Pin 5 is the control voltage input and the internal divider already defines it at two thirds of the supply. A 10 nF capacitor to ground is conventional noise decoupling, not a requirement, and the circuit works without it.
How do I calculate a 555 monostable pulse?
The output pulse width is 1.1 · R · C. With R = 10 kΩ and C = 1 µF that is 11 ms; simulating the same circuit measures 10.97 ms. The trigger input must be pulled below one third of the supply briefly, and the pulse does not retrigger until it finishes.