The equations for the mark and space times (Tm and Ts) of a 555 astable are:
Tm = 0.7 (R1 + R2) C
Ts= 0.7 R2 C
where R1 is the resistor from pin 7 (discharge) to V and R2 is between pins 6 and 7 (threshold and discharge).
By inspection, the mark time can only exactly equal the space if R1 is zero, but I've read that R1 should never be less than 1k.... is that right? An approximation to get Tm ~ Ts is for R2>>R1, say 10k and 1k where they would be 0.77 and 0.70 seconds with a 100uF cap for instance.
If it was critical to have Tm and Ts very close, I guess the approach would be to increase R2 >>> R1, and bring C down by the right order of magnitude?
If the output pin isn't being heavily loaded I think you can just use that instead
of the discharge pin for feedback, making 50% duty cycle easy, just use one resistor.
The output pin is push-pull (whereas the discharge is open collector).
TI's datasheet for the 555 describes how to get a 50% duty cycle. See this link, page 12: http://www.ti.com/lit/ds/symlink/lm555.pdf. If a precise 50% duty cycle matters, you'll be better off running the 555 at twice the frequency you need, and clocking an external flip-flop with the output. Otherwise, the formula in the datasheet will get you close.
MarkT:
If the output pin isn't being heavily loaded I think you can just use that instead
of the discharge pin for feedback, making 50% duty cycle easy, just use one resistor.
The output pin is push-pull (whereas the discharge is open collector).
Caveat - I've not checked this works.
Yeah I was reluctant to try a 0R resistor (aka a piece of wire) from p7 to V just in case it popped something.
It's no biggy, mostly curiosity. R2 >>> R1 is close enough.
michinyon:
In Reply #1 above, how does the first schematic have duty cycle adjustment ?
It charges through the 47k resistor, and discharges through the 68k resistor. The diode bypasses the 68k resistor while the capacitor is charging.
It isn't perfect because the diode voltage subtracts from the charging voltage, but not from the discharge voltage. You can make a slight correction to this with a diode in series with the 68k resistor, pointing up, so a diode drop is subtracted from the discharge voltage.
But if it is that critical, you wouldn't use that method.
