Formula seems 'illogical' Captain

Status
Not open for further replies.

angie1199

New Member
Hi,

I'm new to this lectronics and now I know why I failed to understand physics.

I want to build an astable 555 timer circuit. I read a lot of info on the web in various forms and found two formulae for working out the capacitor and resistor values. But it doesn't make sense!

Tspace = 0.7R2C

which I assume is Tspace = 0.7 x R2 x C

The next formula to get R2 (as I know Tspace is 1s) and the C(apacitor) is guessed at 10uF:

R2 = 1/(0.7 x 10)

The result they give is 143K resistor. But 1/(0.7 x 10) is 0.14285 (0.143)

So I got my 143 but it's a fraction not in the thousands.

What did I miss PLEASE?
 
Cancel that... See it's me,

Just found the answer. 1uF is 1/1000000th of a farad.

I though uF was the symbol for farad
 
angie1199 said:
Cancel that... See it's me,

Just found the answer. 1uF is 1/1000000th of a farad.

I though uF was the symbol for farad

If it's of any help?, the 'u' is supposed to be the greek letter 'mu', used as a prefix for 10 to the power of minus six.
 
_nox_ said:
Actually it's written with a tiny stroke, like this: µ

I didn't attempt to show it because I've no confidence that the character will appear correctly on all computers - your's was fine on mine here though!.
 
Nigel Goodwin said:
I didn't attempt to show it because I've no confidence that the character will appear correctly on all computers - your's was fine on mine here though!.
yep i thought so ... the "u" is just to simplify the character rendering ( thus making it appear correctly on all compis )
 
Back to the original problem of finding the R & C for a given frequency, the time that the output is high is given by: T1 = 0.693(Ra + Rb)*C and the time that the output is low is given by: T2 = 0.693*Rb*C. The frequency is given by: 1/0.693/(Ra + 2*Rb)/C.
 
To restate the formula for the 555 in astable operation:
T1+T2=0.693(Ra+2Rb)C
F=1/T or F=1/((0.693(Ra+2Rb)C))
 
Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…