Hi again antknee,
How accurate do you need the 50 percent duty cycle to be? Would it be a problem if it was 51 percent, or 49 percent?
Yes, the data sheet shows a 555 with connections for a true 50 percent duty cycle, but the design information is lacking.
We need to get t1=t2 to get a 50 percent duty cycle, so that means:
ln((R2-2*R1)/(2*R2-R1))/(R2+R1)-ln2/R2=0
where
R1 is the upper resistor,
R2 is the resistor in series with pin 7,
ln2=ln(2)
and
Frequency=1/(2*ln2*R1*C)
or alternately:
R1=1/(2*Frequency*ln2*C)
So say we want 10kHz, that makes R1=7213.475 ohms, and using the equation above to get R2 we end up with R2=3053.726 ohms.
To make it simpler, the following may be true (i'll have to test this idea out better later though):
R2=R1*0.4233363273
so we can select R1 with the formula for R1 above and then calculate R2 like that (test R2 with the big equation above first though just to make sure).
This design info is only for the circuit on the data sheet of course and is only for a theoretical 50 percent duty cycle output.