When looking at example schematics of a 555 astable circuit, they all show the two resistors in series with the C. I'm wondering how this differs from the schematic I've created below;
Here R6 is used to load C3 and R5 to discharge, which should allow for more control of the duty cycle, e.g. having 50% duty cycle. What would be the reason not to use this schematic, am I missing something?
Imagine what happens when Discharge is pulled low by the internal transistor, R5 and R6 will form a voltage divider with Trigger and Threshold connected to the center of the divider. With it like that the voltage on those pins will be never get below Vcc/2 and it needs to get down to Vcc/3 for the timer to be triggered.
simulate this circuit – Schematic created using CircuitLab
You may be able to use resistors with a different ratio than 1:1, something just under 2:1 seems to work in simulation and may give you a 50% duty cycle but you'd have to try it with real components to see if it will work reliably. There are other circuits to get a 50% duty cycle.
The biggest drawback is that the math gets a lot more complicated. The discharge time is no longer a linear function of one of the resistor values.
When the discharge pin is grounded, you now have both R5 and R6 determining the voltage on the capacitor. You can treat them as a Thevenin source with voltage
$$V_{TH} = V_{CC} \left(\frac{1}{1 + \frac{R6}{R5}}\right)$$
and resistance
$$R_{TH} = \frac{1}{\frac{1}{R6} + \frac{1}{R5}}$$
So the discharge waveform becomes an exponential curve dropping from \$\frac{2}{3}V_{CC}\$ down toward \$V_{TH}\$. If we define \$V_0 = \frac{2}{3}V_{CC} - V_{TH}\$ and the time constant \$\tau = R_{TH} C\$, then this curve is
$$V(t) = V_{TH} + V_0 e^{-\frac{t}{\tau}}$$
The discharge ends when \$V(t) = \frac{1}{3} V_{CC}\$, so we can make that substitution on the left hand side and then solve for \$t\$ to find out what the discharge time is. If we say \$k = \frac{V_{TH}}{V_{CC}} =\frac{1}{1 + \frac{R6}{R5}}\$, we can write it as:
$$t = -\tau \ln\left(\frac{1 - 3k}{2 - 3k}\right)$$
From this, we can make a couple of observations. First, there is no solution when k is 1/3 or greater, since this makes the argument of the logarithm zero or negative. This means that R6/R5 must be less than 2 — R5 must be less than 1/2 of R6.
Second, solving this to get a value of R5 for a particular time value (e.g., the same value as the 0.693×R6×C charging time) is a similarly complicated derivation. I'll leave that for you to work out if you're interested in pursuing it.
In more practical terms, the closer R5 gets to 0.5×R6, the more sensitive the circuit becomes to component tolerances, including things like the exact voltage on the discharge pin during the discharge cycle.
