555 Timer Astable
f = 1.44 / ((R1 + 2×R2) × C)
Quick answer
This calculator solves f = 1.44 / ((R1 + 2×R2) × C) for Frequency. With R1 10 kΩ, R2 22 kΩ, Capacitance 100 nF, the result is 0.2667 kHz. It solves in 4 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
The 555 timer in astable mode generates a continuous rectangular wave without external triggering. The capacitor charges through R1 + R2 and discharges through R2 alone, creating an asymmetric waveform. The duty cycle is (R1 + R2) / (R1 + 2×R2), which is always greater than 50% in the standard configuration. For a near-50% duty cycle, make R1 much smaller than R2. The 555 is one of the most versatile ICs ever made, used for timing, pulse generation, clock signals, PWM, and tone generation in countless applications since its introduction in 1972.
Variables
- f — Output frequency (Hz)
- R1 — Timing resistor between Vcc and discharge pin (Ω)
- R2 — Timing resistor between discharge and threshold pins (Ω)
- C — Timing capacitor (F)
Practical Notes
For stable operation, R1 and R2 should be between 1kΩ and 10MΩ. The CMOS 555 (TLC555, LMC555) draws much less supply current than the original bipolar NE555 and works at lower voltages (2-15V). For 50% duty cycle, add a diode across R2 to bypass it during charge, or use the 555 in toggle mode with a divide-by-2 flip-flop.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| f | Frequency | Hz | Hz, kHz, MHz, GHz |
| R1 | R1 | kΩ | mΩ, Ω, kΩ, MΩ |
| R2 | R2 | kΩ | mΩ, Ω, kΩ, MΩ |
| C | Capacitance | nF | pF, nF, µF, mF, F |
Step by step
- 1. What you know: R1 10 kΩ, R2 22 kΩ, Capacitance 100 nF
- 2. The relation: f = 1.44 / ((R1 + 2×R2) × C)
- 3. Substitute: R1 = 10 kΩ, R2 = 22 kΩ, C = 100 nF
- 4. Result: Frequency = 0.2667 kHz
Worked examples
Minimum range
Given: R1 100 Ω, R2 0.22 kΩ, Capacitance 100 pF
Result: Frequency = 26.67 MHz
Low range
Given: R1 1 kΩ, R2 2.2 kΩ, Capacitance 1 nF
Result: Frequency = 0.2667 MHz
Typical
Given: R1 10 kΩ, R2 22 kΩ, Capacitance 100 nF
Result: Frequency = 0.2667 kHz
High range
Given: R1 100 kΩ, R2 0.22 MΩ, Capacitance 1 µF
Result: Frequency = 2.667 Hz
Maximum range
Given: R1 1 MΩ, R2 2.2 MΩ, Capacitance 10 µF
Result: Frequency = 0.02667 Hz
Solving for each variable
Solve for Frequency
R1 100 Ω, R2 0.22 kΩ, Capacitance 100 pF → 26.67 MHz
Solve for R1
Frequency 5 Hz, R2 0.22 kΩ, Capacitance 100 pF → 2880 MΩ
Solve for R2
Frequency 5 Hz, R1 100 Ω, Capacitance 100 pF → 1440 MΩ
Solve for Capacitance
Frequency 5 Hz, R1 100 Ω, R2 0.22 kΩ → 0.5333 mF
How Frequency changes with R1
| R1 (input) | Frequency (result) |
|---|---|
| 1 kΩ | 0.32 kHz |
| 2.5 kΩ | 0.3097 kHz |
| 5 kΩ | 0.2939 kHz |
| 7.5 kΩ | 0.2796 kHz |
| 10 kΩ | 0.2667 kHz |
| 15 kΩ | 0.2441 kHz |
| 20 kΩ | 0.225 kHz |
| 40 kΩ | 0.1714 kHz |
| 70 kΩ | 126.3 Hz |
| 100 kΩ | 100 Hz |
The same result in other units
| Unit | Value |
|---|---|
| Hz | 266.7 |
| kHz | 0.2667 |
| MHz | 0.0002667 |
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<p><a href="https://vbatpower.com/formulas/555-astable/">555 Timer Astable</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate 555 Timer Astable?
Use f = 1.44 / ((R1 + 2×R2) × C). With R1 10 kΩ, R2 22 kΩ, Capacitance 100 nF, Frequency works out to 0.2667 kHz.
What is the formula for 555 Timer Astable?
The formula is f = 1.44 / ((R1 + 2×R2) × C). The 555 timer in astable mode generates a continuous rectangular wave without external triggering.
How do you find R1 when you know the other values?
Rearrange for R1: with Frequency 5 Hz, R2 0.22 kΩ, Capacitance 100 pF, R1 is 2880 MΩ.
How do you find R2 when you know the other values?
Rearrange for R2: with Frequency 5 Hz, R1 100 Ω, Capacitance 100 pF, R2 is 1440 MΩ.
What units does 555 Timer Astable use?
Frequency is shown in Hz, kHz, MHz. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
Related Concepts
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