555 Timer Calculator

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555 led flasher 1 hz resistor capacitor

555 LED flasher at 1 Hz: resistors and capacitor

To blink an LED at 1 Hz, the 555 period has to sit close to one second. Open the 1 Hz LED blinker example to start from RA 1 kΩ, RB 68 kΩ and C 10 µF, or use Timing to parts mode and ask for 1 Hz in the E12 series.

How to get close to 1 Hz

The astable frequency is about 1.44 / ((RA + 2RB) x C). For 1 Hz, the product (RA + 2RB) x C must be close to 1.44 seconds. With C = 10 µF, that calls for RA + 2RB near 144 kΩ. Picking RA = 1 kΩ and RB = 68 kΩ gives 137 kΩ, and the period lands at 949.6 ms, or 1.053 Hz. The LED stays on 478 ms and off 471 ms each turn, a nearly symmetric blink.
Standard values rarely land exactly on target. In the E12 series, the best combination for 1 Hz and a 70% high duty cycle that the tool finds uses RA 820 kΩ, RB 680 kΩ and C 680 nF, giving 0.973 Hz, a 2.7% difference. Moving the capacitor up a decade and dividing the resistors by 10 gives the same frequency, which is a good way to avoid very high resistors, since they suffer more from capacitor leakage.
If you want a short flash, with the LED lit only a fraction of the time, the standard circuit cannot do it, because the high duty cycle stays above 50%. In Options, turn on the diode across RB: the capacitor then charges through RA only and the high time drops to 0.693 x RA x C. It is an ideal-diode estimate that the datasheet does not give. With RA 47 kΩ, RB 390 kΩ and C 3.3 µF, the math gives 1.0004 Hz with 10.8% of the time high.

Combinations for 1 Hz

Nearly symmetric blink

Input: RA=1k, RB=68k, C=10u
Expected output: 1,053 Hz

High 478 ms, low 471 ms, high duty cycle 50.4%.

Same idea with larger RB and smaller C

Input: RA=1k, RB=100k, C=6u8
Expected output: 1,056 Hz

Period of 947 ms.

E12 series for 1 Hz and 70% high

Input: RA=820k, RB=680k, C=680n
Expected output: 0,973 Hz

High duty cycle of 68.8%, 2.7% error in frequency.

Short flash with an ideal diode

Input: RA=47k, RB=390k, C=3u3, diodo
Expected output: 1,0004 Hz

LED lit 10.8% of the time.

A little faster

Input: RA=4k7, RB=47k, C=10u
Expected output: 1,46 Hz

Period of 684 ms.

Full tool FAQ

What is the 555 frequency formula in astable mode?

Per the TI datasheet, f = 1 / T and T = 0.693 x (RA + 2RB) x C, giving f ≈ 1.44 / ((RA + 2RB) x C). With RA = 1 kΩ, RB = 10 kΩ and C = 1 µF, T is 14.56 ms and f is 68.7 Hz.

Why can a 555 duty cycle not reach 50%?

What is the monostable formula and why does 1.1 appear?

What is the maximum frequency of a 555?

Which R and C values can I use?

How do I build a 1 Hz LED blinker with a 555?

What is the difference between high duty and low duty?

Does the result account for component tolerance?

Do the formulas apply to other 555 models?

Does my data leave the browser?

Frequently asked questions

Do I need a series resistor for the LED?

Yes. The 555 output does not limit current, so the LED needs a series resistor. Use the LED resistor calculator with your supply voltage and the current you want.

Why is my blinker not at exactly 1 Hz?

5% resistors and wide-tolerance electrolytic capacitors add up error. The tool shows the possible range in Options. To fine-tune, replace one resistor with a fixed one plus a trimmer.

Can I use another capacitor, such as 100 µF?

You can, as long as you scale the resistors down in the same proportion: with 100 µF, RA + 2RB should sit near 14.4 kΩ. Very large capacitors leak more and take longer to settle at start-up.

Next steps

How to get close to 1 Hz

The astable frequency is about 1.44 / ((RA + 2RB) x C). For 1 Hz, the product (RA + 2RB) x C must be close to 1.44 seconds. With C = 10 µF, that calls for RA + 2RB near 144 kΩ. Picking RA = 1 kΩ and RB = 68 kΩ gives 137 kΩ, and the period lands at 949.6 ms, or 1.053 Hz. The LED stays on 478 ms and off 471 ms each turn, a nearly symmetric blink.

Standard values rarely land exactly on target. In the E12 series, the best combination for 1 Hz and a 70% high duty cycle that the tool finds uses RA 820 kΩ, RB 680 kΩ and C 680 nF, giving 0.973 Hz, a 2.7% difference. Moving the capacitor up a decade and dividing the resistors by 10 gives the same frequency, which is a good way to avoid very high resistors, since they suffer more from capacitor leakage.

If you want a short flash, with the LED lit only a fraction of the time, the standard circuit cannot do it, because the high duty cycle stays above 50%. In Options, turn on the diode across RB: the capacitor then charges through RA only and the high time drops to 0.693 x RA x C. It is an ideal-diode estimate that the datasheet does not give. With RA 47 kΩ, RB 390 kΩ and C 3.3 µF, the math gives 1.0004 Hz with 10.8% of the time high.

Combinations for 1 Hz

Nearly symmetric blink

Input: RA=1k, RB=68k, C=10u
Expected output: 1,053 Hz

High 478 ms, low 471 ms, high duty cycle 50.4%.

Same idea with larger RB and smaller C

Input: RA=1k, RB=100k, C=6u8
Expected output: 1,056 Hz

Period of 947 ms.

E12 series for 1 Hz and 70% high

Input: RA=820k, RB=680k, C=680n
Expected output: 0,973 Hz

High duty cycle of 68.8%, 2.7% error in frequency.

Short flash with an ideal diode

Input: RA=47k, RB=390k, C=3u3, diodo
Expected output: 1,0004 Hz

LED lit 10.8% of the time.

A little faster

Input: RA=4k7, RB=47k, C=10u
Expected output: 1,46 Hz

Period of 684 ms.

Full tool FAQ

What is the 555 frequency formula in astable mode?

Per the TI datasheet, f = 1 / T and T = 0.693 x (RA + 2RB) x C, giving f ≈ 1.44 / ((RA + 2RB) x C). With RA = 1 kΩ, RB = 10 kΩ and C = 1 µF, T is 14.56 ms and f is 68.7 Hz.

Why can a 555 duty cycle not reach 50%?

What is the monostable formula and why does 1.1 appear?

What is the maximum frequency of a 555?

Which R and C values can I use?

How do I build a 1 Hz LED blinker with a 555?

What is the difference between high duty and low duty?

Does the result account for component tolerance?

Do the formulas apply to other 555 models?

Does my data leave the browser?

Frequently asked questions

Do I need a series resistor for the LED?

Yes. The 555 output does not limit current, so the LED needs a series resistor. Use the LED resistor calculator with your supply voltage and the current you want.

Why is my blinker not at exactly 1 Hz?

5% resistors and wide-tolerance electrolytic capacitors add up error. The tool shows the possible range in Options. To fine-tune, replace one resistor with a fixed one plus a trimmer.

Can I use another capacitor, such as 100 µF?

You can, as long as you scale the resistors down in the same proportion: with 100 µF, RA + 2RB should sit near 14.4 kΩ. Very large capacitors leak more and take longer to settle at start-up.