Disclaimer: I am not an electrical engineer, but an environmental science student. I'm trying my best here.

astable 555 timing circuit

I have a basic astable 555 timing circuit with the following components:

\$R_1 = 10k\Omega,~~R_2 = 1k\Omega,~~C = 0.1 \mu F\$

\$T = 0.7 \cdot (R_1 + 2R_2) \cdot C = 0.7 \cdot 12k\Omega \cdot 0.1 \mu F = 0.84 ms \\ f = \frac{1}{0.84 ms} = 1.19 kHz \\ t_{up} = 0.7 \cdot (R_1 + R_2) \cdot C = 0.7 \cdot 11k\Omega \cdot 0.1 \mu F = 0.77 ms \\ t_{down} = 0.7 \cdot R_2 \cdot C = 0.7 \cdot 1k\Omega \cdot 0.1 \mu F = 0.07 ms\$

I am being asked to halve its frequency but maintain the same "up" pulse width.

I know that doubling the reactance of the capacitor from \$0.1 \mu F\$ to \$0.2 \mu F\$ will halve the frequency, as it will take roughly twice as long for it to charge. However, my "up" pulse width \$t_{up}\$ is also being doubled.

\$T = 0.7 \cdot 12k\Omega \cdot 0.2 \mu F = 1.68 ms \\ f = \frac{1}{1.68ms} = 595Hz \\ t_{up} = 0.7 \cdot 11k\Omega \cdot 0.2 \mu F = 1.54 ms \\ t_{down} = 0.7 \cdot 1k\Omega \cdot 0.2 \mu F = 0.14 ms\$

The desired output would have a frequency of 595Hz, but an "up" pulse width of 0.77 milliseconds.


1 Answer 1


The charge period is determined by R1 and R2 charging the capacitor. The discharge period determined by just R2 and the capacitor. Given that you can halve the frequency by doubling the capacitor you now have the relationships to recalculate the resistors remembering that if the capacitor doubles, R1 plus R2 has to halve to get the same charge period.

Alternatively go to one of the many 555 website calculators and try things out online.

  • \$\begingroup\$ Thanks for guiding me through the problem. I was making this more difficult than it needed to be. If I reduce the total resistance to 6k, but then increase the capacitance to 0.4uF, I get the same pulse width as the original circuit, but with the frequency being twice as long! t_up = 0.77 milliseconds and f = 595Hz. Thank you! \$\endgroup\$ Commented Jun 14, 2017 at 20:48

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