By Muhammad Ibrahim · Published June 15, 2026 · Updated October 29, 2026 · 7 min read
How to Simulate a 555 Timer in Astable Mode (with the Maths Worked Out)
The 555 in astable mode free-runs as a square-wave oscillator. It is a good first analog-meets-digital circuit because every number can be predicted on paper: pick R1, R2 and C, calculate the frequency, build it, and compare. The circuit on this page uses R1 = 10 kΩ, R2 = 47 kΩ and C = 10 µF on a 5 V supply, slow enough that you can watch the LED blink and see the capacitor voltage move on the chip pins.
- CHIP
- 555 timer
- TIMING PARTS
- R1 10 kΩ, R2 47 kΩ, C 10 µF
- FREQUENCY
- about 1.39 Hz
- DUTY CYCLE
- about 55% high
Components needed for the 555 astable circuit
- A 555 Timer and a half-size breadboard
- A battery part set to 5 V
- R1 = 10 kΩ and R2 = 47 kΩ (set the values in the Inspector)
- C = 10 µF, a polarised capacitor
- A red LED with a 330 Ω series resistor on the output
Open the circuit
Open the exact 555 astable circuit. Switch to Simulate and the LED starts to blink. Select the 555 and watch the Live pin states in the Inspector: pin 3 flips between HIGH and LOW while pins 2 and 6 show the capacitor voltage.
How to wire a 555 timer in astable mode
Place the 555 and identify its pins
Place the 555 across the centre gap of the breadboard. Pin 1 is ground, 2 is trigger, 3 is output, 4 is reset, 5 is control voltage, 6 is threshold, 7 is discharge and 8 is V+. LogicBench prints the pin numbers on the package.
Power the chip and tie reset high
Connect the battery + to the top + rail and the battery − to the top − rail. Pin 8 goes to +5 V and pin 1 to ground. Pin 4 (reset) also goes to +5 V, because reset is active low and a grounded reset keeps the output off. The circuit also joins the top and bottom ground rails with one wire, so the capacitor and the LED can return to the lower rail.
Build the timing network
R1 goes from +5 V to pin 7. R2 goes from pin 7 to pin 6. Join pins 6 and 2 with a short wire. The capacitor goes from that joined node to ground, with its + lead on the node. The capacitor charges through R1 + R2 and discharges through R2 only, which is what makes the high time longer than the low time.
Add the output LED
Connect pin 3 through the 330 Ω resistor to the LED anode and the LED cathode to ground. With 5 V and a red LED the current is roughly (5 − 2) / 330 ≈ 9 mA, comfortably inside what the 555 output can drive.
Leave pin 5 open
Pin 5 is the control voltage pin. This circuit leaves it unconnected, and LogicBench shows it at 3.33 V, which is the 2/3 supply reference inside the chip. The simulator does not need a capacitor here. On real hardware you would add 10 nF from pin 5 to ground.
Run the simulation and measure
Switch to Simulate. The LED blinks a little under 1.4 times per second. Select the 555 and read the Live pin states: pin 3 alternates HIGH and LOW, while pins 2 and 6 show the same analog voltage moving between about 1.67 V and 3.33 V. You can also click pin 3 and record it with the Logic Analyzer to read the period.
Change C and compare
Set C to 100 µF and run again. The period grows from about 0.7 s to about 7 s, ten times longer, and the LED visibly takes its time. Set it back to 10 µF afterwards.
555 astable pin connections
| 555 pin | Name | Connects to | Why |
|---|---|---|---|
| 1 | GND | Ground rail | Chip ground |
| 2 | Trigger | Pin 6 and the capacitor | Starts a new cycle when the capacitor falls to 1/3 of the supply |
| 3 | Output | 330 Ω resistor, then the LED | The square wave |
| 4 | Reset | +5 V | Active low, so it must be held high |
| 5 | Control voltage | Left open | Shows the internal 2/3 reference, 3.33 V |
| 6 | Threshold | R2, pin 2 and the capacitor | Ends the charge when the capacitor reaches 2/3 of the supply |
| 7 | Discharge | Between R1 and R2 | Lets the capacitor discharge through R2 |
| 8 | V+ | +5 V | Chip supply |
555 astable timing: calculated and simulated
R1 = 10 kΩ, R2 = 47 kΩ, C = 10 µF, supply 5 V. The simulated values are the average of many cycles of this circuit file in the LogicBench engine.
| Quantity | Formula | Calculated | Simulated |
|---|---|---|---|
| High time | 0.693 × (R1 + R2) × C | 395 ms | about 0.40 s |
| Low time | 0.693 × R2 × C | 326 ms | about 0.33 s |
| Period | high + low | 721 ms | about 0.73 s |
| Frequency | 1 / period | 1.39 Hz | about 1.36 Hz |
| Duty cycle | (R1 + R2) / (R1 + 2·R2) | 54.8% | about 54.5% |
The simulated times were taken at 60 steps per second, so each one is rounded to a whole step of about 17 ms. Your own readings on the Logic Analyzer should land within a few percent of the formula.
How the 555 astable works: frequency and duty cycle
Frequency, high time and low time
For the astable configuration: t_high = 0.693 × (R1 + R2) × C and t_low = 0.693 × R2 × C. With R1 = 10 kΩ, R2 = 47 kΩ and C = 10 µF: t_high = 0.693 × 57 000 × 0.00001 ≈ 395 ms, and t_low = 0.693 × 47 000 × 0.00001 ≈ 326 ms. The period is about 721 ms, so f ≈ 1.39 Hz. The shortcut f = 1.44 / ((R1 + 2·R2) × C) gives 1.44 / 1.04 ≈ 1.38 Hz, the same answer within rounding, because the period is 0.693 × (R1 + 2·R2) × C and 1 / 0.693 ≈ 1.44.
Why the capacitor swings between 1/3 and 2/3 of the supply
Inside the 555 two comparators watch the capacitor. When its voltage reaches 2/3 of the supply (3.33 V at 5 V) on pin 6, the output goes low and pin 7 starts discharging the capacitor through R2. When it falls to 1/3 of the supply (1.67 V) on pin 2, the output goes high and the capacitor charges again through R1 + R2. That is why pins 2 and 6 show a voltage between 1.67 V and 3.33 V in the Live pin states, and the 0.693 in the formulas is ln 2, the exponential charge time between those two levels.
Why the duty cycle can never drop below 50%
Duty cycle = t_high / (t_high + t_low) = (R1 + R2) / (R1 + 2·R2). Because R1 is added to the top line and R2 appears in both, the result is always above 50%, here 54.8%. If you need a duty cycle at or below 50%, the standard fix is a diode across R2, so that charging bypasses R2, or a different circuit. Making R1 small only pushes the duty cycle towards 50%.
Why component tolerance matters on a real bench
A typical electrolytic capacitor is rated ±20%, and the frequency is inversely proportional to C, so a real build can easily land 15 to 20% away from the calculated value even when it is wired correctly. A simulator uses the exact value you enter, which makes it a clean way to separate wiring mistakes from tolerance effects.
Expected 555 astable output
- Pin 3 is a continuous square wave: high for about 0.4 s and low for about 0.33 s, a period of about 0.73 s (roughly 1.4 Hz).
- The red LED blinks a little under 1.4 times per second and is on slightly longer than it is off.
- Pins 2 and 6 always show the same voltage, rising and falling between about 1.67 V and 3.33 V. Pin 5 stays at 3.33 V.
- With C = 100 µF the period grows about ten times, to roughly 7 s.
555 astable not oscillating? Common mistakes
The output never changes
Check that pin 4 (reset) is tied to +5 V and that pins 8 and 1 are really powered. A grounded reset holds the output low.
The output stays high or low
Pins 2 and 6 must be joined and connected to the capacitor. If the capacitor goes to pin 7, or only one of the two pins is connected, the capacitor cannot cycle between the two thresholds.
The frequency is far from the calculated value
Re-check which resistor is R1 and which is R2, and confirm the capacitor value (µF against nF is the usual slip). R1 sits between +5 V and pin 7, and R2 sits between pin 7 and pin 6.
The output toggles but the LED stays dark
Check the LED polarity (anode towards the 330 Ω resistor, cathode to ground) and that the resistor really sits between pin 3 and the LED.
R1 is very small
R1 should be at least 1 kΩ. In astable mode a smaller R1 lets too much current flow into the discharge transistor on pin 7 every cycle.
Frequently asked questions
How do you calculate the frequency of a 555 astable circuit?
Use f = 1.44 / ((R1 + 2·R2) × C). With R1 = 10 kΩ, R2 = 47 kΩ and C = 10 µF the bottom line is 104 000 × 0.00001 = 1.04, so f is about 1.4 Hz. Working it from the high and low times (0.693 × (R1 + R2) × C and 0.693 × R2 × C) gives a period of 721 ms, which is 1.39 Hz.
Why is the duty cycle of a 555 astable always above 50%?
The capacitor charges through R1 + R2 but discharges through R2 only, so the output is high for longer than it is low. The duty cycle is (R1 + R2) / (R1 + 2·R2), which can get close to 50% when R1 is small but never reaches it. Here it is 57 / 104 = 54.8%.
How do I make the 555 LED blink slower or faster?
The frequency is inversely proportional to C and to R1 + 2·R2. Changing C from 10 µF to 100 µF makes the period ten times longer, about 7.2 s, and 1 µF makes it ten times shorter. Changing R2 changes the speed and the duty cycle together.
Do I need a capacitor on pin 5 of the 555?
Not in the simulator. Pin 5 is left open in this circuit and LogicBench shows it at 3.33 V, the 2/3 supply reference inside the chip. On a real board a 10 nF capacitor from pin 5 to ground keeps noise off that reference, and it does not change the timing noticeably.
What happens if the 555 reset pin is grounded?
Reset (pin 4) is active low, so a grounded reset holds the output low and nothing oscillates. Tie it to +5 V for normal operation and do not leave it floating.
Try it yourself
Open LogicBench and follow along — the simulator is free and needs no account.
Launch the simulator