- What is V at the time the circuit is connected to supply? To me, it seems like it's floating without a reference voltage, since C3 is empty (i.e. basically open circuit) and D2 does not conduct since OUT is low. (I am pretty sure I am wrong here, but I don't know why/where).
V is 0 volts. And no, it isn't open-circuit. A capacitor will "hold" 0V just as well as it holds any other voltage.
- How to determine the period and the duty time? (I am looking more for a formula and some pointers to understanding its derivation)
In astable mode, V is cycling between 1/3 Vcc and 2/3 Vcc. It does this at a rate determined by both the capacitance and resistance that is applied during each phase of operation.
It can be shown that the exponential curve moves from 1/3 to 2/3 (or vice-versa) in a time that is equal to \$-\ln (0.5) \cdot R \cdot C\$, or 0.693×R×C, which is the formula used in the datasheet.
EDIT: Specifically, the voltage across the capacitor (during discharge) is
$$v(t) = V_0 e^{-\frac{t}{RC}}$$
We want to know how long it takes to fall from 2/3 Vcc (\$V_0\$) to 1/3 Vcc. Therefore, we set the expression equal to 1/3 Vcc and solve for t:
$$\frac{1}{3}V_{CC} = \frac{2}{3}V_{CC} \cdot e^{-\frac{t}{RC}}$$
Isolate the exponential:
$$\frac{1}{2} = e^{-\frac{t}{RC}}$$
Take the logarithm of both sides:
$$\ln \frac{1}{2} = -\frac{t}{RC}$$
And finally, isolate t:
$$-\ln \frac{1}{2}\cdot R \cdot C = t$$
A similar formula applies when charging from 1/3 Vcc to 2/3 Vcc using a supply voltage of Vcc.
- The examples of 555-based timers I have seen around usually charge C3 from Vcc. In this case, it is charged from the output pin. Are there any advantages/disadvantages of this approach?
This circuit topology allows the timer to operate with a duty cycle that is less than 50% — in this case, it can be MUCH less than 50%. A disadvantage is that the charging voltage is not exactly Vcc (and this gets worse if there's a significant load on the output), which throws off the timing calculation for the "on" time.
The relay is an inductive load, which means that there's a significant voltage spike when you turn it off that could damage the output of the IC. D1 "shorts out" this spike to protect the rest of the circuit.
