Theoretically, any correct ratio works but of importance is not loading the 5V source too much that it sags a little i.e. don't choose resistor values that are so low that what you measure changes due to the introduction of the two resistors.
Next is consideration of what is measuring the output from the potential divider - you don't want the resistors to be so high in value that the impedance of the instrument or circuit changes the ratio. For instance an oscilloscope might have an input resistance of 1Mohm - using a 100k top resistor and a 900k bottom resistor will not be good. Try to aim for 0.1% (just a figure that is reasonable to me) so that the bottom resistor would be about 900 ohms and the top resistor 100 ohms.
Some DVMs are 10Mohm input resistance so a divider formed from 1k and 9k would be OK.
- "which will eventually become an SMPS"
Unless you are thinking of low voltage and low current, DONT TRY TO BUILD YOUR OWN SMPS.
- "If I only supplied one voltage divider, every time a transistor turned on all the current would be sucked to ground."
I can't imagine what you are talking about, which even more convinces me that you should not be building an SMPS.
- The only difference I can see here is the bottom resistors forming the voltage dividers.
A difference I see is that you use 1nF instead of 100 nF capacitors, so you frequency would be 100 * higher, which might be a problem.
A big difference is that you seem to want to replace the pull-up resistors by equivalent voltage dividers, presumably because you have a higher power supply than 9V? (You probably don't need to bother, those 2N3904's will wrok well up to much higher voltages.). But you screwed up your calculations: take the leftmost resistors (R1, R2): 500/700 in parallel will act a s a resistor to 9V if you power is 1200/500*9v = 22V, but the equivalent series resistance is ~ 390 Ohm. That probably does not matter much for those resistors, but for R3/R4 it is a big problem.
IME the circuit you show works well without R2/R5 and the diodes. In most cases larger values are choosen for R3/R5, with correspondingly lower capacitor values.
PS I think you approach of taking a know-working circuit and repurposing it for your own projects is fine and can be educational, but you should select a project that is reasonably free feasible and free of danger, and you your theoreticalk knowledge (in this case, calculating the Thevenin equivalent of a voltage divider) should match the level of thinkering you want to do.
Best Answer
You can make a voltage divider with any number of resistors, but you must remember that any current drawn from an output of a voltage divider must come through the resistors in that divider, so will affect the output voltage.
simulate this circuit – Schematic created using CircuitLab
In Fig. 1, we have a simple voltage divider. Since the resistors are equal value, the output voltage will be half the input voltage (10V in gives 5V out).
However, in Fig 2 we've added a load (R5) intended to draw 5 mA from the voltage divider. However, that load is in parallel with R4, so we effectively have a 1K resistor (R3) in series with a 500 ohm resistor (R4 and R5 in parallel), so we only get 3.33 volts for our load, rather than the desired 5 volts.
When designing a voltage divider you must always make allowance for any currrent drawn from a tap on the divider, usually by ensuring that the current drawn by the load is much less than the current through the divider chain.