Electronic – Reasons behind placing power supply parallel capacitors in the correct order

What you want is a low-pass filter. A low-pass filter will let slow changes in your signal pass, but block faster changing signals. The basic low-pass filter is this RC-filter:

\$V_{IN}\$ is the signal from the sensor. \$V_C\$ is the capacitor's voltage, which you could use to go to the ADC (Analog-to-Digital Converter) of a microcontroller, for example.
The capacitor acts as a voltage reservoir, which is filled or emptied via the resistor. The higher the capacitor's value the longer this will take. A higher resistor's value will cause less current to flow to and from the capacitor and also will deny fast changes.
Now you may think that the higher the values, the better smoothing you get, and that's true, but if you go too high you'll also prevent the normal changes due to the change in distance from the sensor. And those changes are allowed.
The key to finding the right values is the cutoff frequency, that is the frequency starting from which the filter will block a signal. For the RC filter:

\$f_C = \dfrac{1}{2 \pi \times R C}\$

Say you want to allow changes of 1Hz, that means changes which take place in 1 second. Then according to the formula

\$R C = \dfrac{1}{2 \pi \times f_C} = \dfrac{1}{2 \pi \times 1} = 0.16\$

If you choose a value of 1\$\mu\$F for the capacitor you'll need 160k\$\Omega\$ for the resistor.

I doubt that there would be much benefit to adding a capacitor across your battery. Consider the following:

• The charging of the flash can be seen as a process of moving energy from the battery to the high-voltage capacitor in the flash.
• Based on some quick searches, the capacitor in the Vivitar 285HV is 1000 uF. Assuming it's charged to 200V, it stores 20 J of energy
• A capacitor to hold 20 J at 6V would be 1.1 Farads, and that's just enough energy for a single charge of the HV cap. Assuming 50% conversion efficiency, and that we want 3 shots that's a 6 F capacitor. The least expensive capacitor I found with some margin over the 6 V working voltage and with that kind of capacitance was over $100, and 8 inches long.
• Just to supply 20 J directly from a 6V battery in 1 second with 50% efficiency would mean a current draw of about 6.7 A. That's a lot of current, but it's something that a large battery can do.
• The charging circuit in the flash will limit how quickly we can charge no matter how much power is available at the input. An easy way to see if the charging circuit is the limiting factor rather than the input power is to measure the voltage across the battery while the flash is charging. If the voltage remains close to the open circuit voltage of the battery, then it's the charging circuit that's limiting, not the battery.