After carefully studying the relevant patterns, I devised a simple and effective way of solving the above issues, especially when the power supply used is a transformerless one or incorporates PPC capacitors or reactance for controlling current.
Typically, a transformerless power supply will produce an output with very low values but with voltages equal to the applied AC mains (until it’s loaded).
For example, a 1 µF, 400 V (breakdown voltage) when connected to a 220 V mains supply will produce a maximum of 70 mA of current and an initial voltage reading of 220 Volts.
However this voltage will show a very linear drop as the output gets loaded and current is drawn from the “70 mA” reservoir.
In case the load consumes the whole 70 mA would mean the voltage dropping to almost zero.
Now since this drop is linear, we can simply divide the initial output voltage with the max current to find the voltage drops that would occur for different magnitudes of load currents.
Therefore dividing 220 volts by 70 mA gives 3.14. This is the rate at which the voltage will drop for every 1 mA of current added with the load.
That means if the load consumes 20 mA of current, the drop in voltage will be 20 × 3.14 = 62.8 volts, so the output now will show a voltage of 220 – 62.8 = 157.2 volts (see figure).
However the voltage across the LED terminals will show a voltage equal to forward voltage drop of the particular LED, because the above concept again becomes applicable for the section which comes after the resistor and across the LED.
Here, the resistor further controls the current and by dividing 157 by 4700 we get 0.033 or a max of 33 mA of current and a voltage of 157, which drops linearly, so dividing 157 by 33 gives 4.75, which is the rate of voltage drop for each mA rise in the load current. The connected LED draws probably the whole 33 mA, which deducts around 33 × 4.75 = 156.75 Volts from the resistor’s output 157 Volts.
That leaves the volts across the LED to the relevant values of around a couple volts or to be precise, the forward voltage drop of the particular type of LED.
Conclusion: From the above discussion and analysis, it becomes clear that voltage in any power supply unit is immaterial if the current delivering capability of the power supply is "relatively" low.
For example if we consider an LED, it can withstand 30 to 40 mA current at voltages close to its "forward voltage drop", however at higher voltages this current can become dangerous for the LED, so its all about keeping the maximum current equal to the maximum safe tolerable limit of the load.
While calculating series resistor values with LEDs, instead of using the standard LED formula directly, we can use the above rule first.
That means either we choose a capacitor whose reactance value only allows the maximum tolerable current to the LED, in which case a resistor can be totally avoided.
If the capacitor value is large with higher current outputs, then probably as discussed above we can incorporate a resistor to reduce the current to tolerable limits.
Example: In the shown diagram, the value of the capacitor produces 70 mA of max. current which is quite high for any LED to withstand. Using the standard LED/resistor formula:
R = (supply voltage VS – LED forward voltage VF) / LED current IL,
= (220 - 1.5)/0.02 = 11K,
Therefore the value of the resistor for controlling one red LED safely would be 11K.
The above theory has been assumed and deduced by me, I am not very sure about its feasibility, though
