What do 0.693, and 1.1, and 1.44 mean in IC 555 calculations?

For calculating pulse width in a standard IC 555 astable and monostable configurations we use the following formulas respectively:

$$T(\text{high}) \approx 0.693(R_1 + R_2)C_1$$

$$T = 1.1RC$$

And for the frequency we use the formula:

$$F = 1.44/((R_1+2R_2)C_1$$

How were the constants 0.693, 1.1, and 1.44 derived in the respective calculations?

• have schema , easier to compute, or read datasheet or App Note!! Aug 24, 2019 at 7:14
• First, see what you can work out yourself. Look at the internal circuit for the 555, you'll see two comparators with trip points of 1/3rd supply voltage and 2/3rds supply voltage. Then take the equations for RC charging and RC discharging and try to produce equations that fit the 555 circuit for charging and discharging. Note that 1.44 is the approx. reciprocal of 0.693. Work at it rather than wait for answers, you'll learn more this way. Aug 24, 2019 at 7:22
• electronics-tutorials.ws/waveforms/555_timer.html you may start here Aug 24, 2019 at 7:46
• @Tony, Thanks, Yes I can surely investigate it, however I would rather appreciate a quick solution because I have some other related problems which I want to solve after knowing this answer. Aug 24, 2019 at 10:53
• Yes, you surely can investigate it, as a passionate researcher and developer. After all, you said this was interest :-) The purpose of the site is to educate and not to be an online technical encyclopedia, copied out to you on demand. There is plenty about this old, old chip on the interweb. Handing over answers is not what this site's about. Aug 24, 2019 at 11:21

The precise numbers are:

• ln(2) = 0.693147
• ln(3) = 1.098612
• 1 / ln(2) = 1.442695

It's a good exercise for you to figure out how these numbers arise from the analysis of the circuit operation.

The key facts about the 555 are:

• In monostable operation, the timing capacitor charges through a resistance from 0V to 2/3 Vcc.

• In astable operation, the timing capacitor voltage cycles between 1/3 Vcc and 2/3 Vcc.

• Great! Thanks very much for the clues, I'll check them out! Aug 25, 2019 at 12:20

Output voltage of an RC circuit at time $$\t\$$ is $$V=V_{\text{max}}e^{-t/(RC)}$$ (negative based on charging or discharging) thus $$t=RC\ln\left(\frac{V}{V_{\text{max}}}\right)$$

For an astable multivibrator, charging and discharging happens between 2/3 and 1/3 Vcc, respectively. As initially it starts from zero charge, the on time (time taken to reach 2/3 Vcc - Time taken to reach 1/3 Vcc) i.e. is $$t = RC(\ln(2/3)-\ln(1/3)) = 0.693 RC$$