Solving for logarithms

Hi MrAl,

Nope no logs solved at all, I used to be able to do maths like that, i am a physicist by training, but its too difficult! lol. I discovered the duty cycle is determined by the ratio of R1 to R2, as long as its big you get pretty close to 50%. The biggest determinant of frequency is the cap. So I have R1 at 1K and R2 at 33K in series with a 10K pot. So I can vary the frequency between 90 and 110KHz and the duty cycle stays around 50-51%.

I made the circuit up on a breadboard originally then put it on a stripboard but that isn't space saving and was more of a tester really so I am currently making up another which will be tidier, space saving and adding in a few other bits I've had floating around.

I have a copy of Matlab which I'm fairly sure will solve equations like this, I just haven't learnt how to use the program.

Hi again,


Oh ok, sounds good, so i guess you are on your way now.

If you want to try solving that in Matlab, by all means do that and post the results (if any) here so we can take a look and compare.

In the mean time for that first 555 circuit (with 50 percent duty cycle and three resistors) you posted, for q such that:
0.001<=q<=0.495

we get a pretty darn good approximation with this function:
q=(p*0.462253+0.182796)/(p*p+1.43675*p+0.431798)

I tried it and it comes out close enough to be practical for almost every circuit.
Check it out and see how you like it

Hi MrAl,

I have used both of my 555 timers now, I did have a look to see if I could find another, because I know I have an extra one somewhere. But that one has been missing in action for a while now, it'll turn up when I'm least expecting it

I quite like the utility value of being able to tune the frequency with a potentiometer and probably using that would necessitate R3 is changed too. I will need a fixed value so I will use your equation on the next and hopefully last iteration. Thanks.

I have just copied it into my notes. Thanks again.