Before we learn the formula for calculating and optimizing the mains capacitor current in a transformerless power supply, it would be important to first summarize a standard transformerless power supply design.
The following diagram shows a classic transformerless power supply design:
Referring to the diagram, the various components involved are assigned with the following specific functions:
C1 is the nonopolar high voltage capacitor which is introduced for dropping the lethal mains current to the desired limits as per the load specification. This component thus becomes extremely crucial due to the assigned mains current limiting function.
D1 to D4 are configured as a bridge rectifier network for rectifying the stepped down AC from C1, in order to make the output suitable to any intended DC load.
Z1 is positioned for stabilizing the output to the required safe voltage limits.
C2 is installed to filter out any ripple in the DC and to create a perfectly clean DC for the connected load.
R2 may be optional but is recommended for tackling a switch ON surge from mains, although preferably this component must be replaced with a NTC thermistor.
In the entire transformerless design discussed above, C1 is the one crucial component which must be dimensioned correctly so that the current output from it is optimized optimally as per the load specification.
Selecting a high value capacitor for a relatively smaller load may increase the risk of excessive surge current entering the load and damaging it sooner.
A properly calculated capacitor on the contrary ensures a controlled surge inrush and nominal dissipation maintaining adequate safety for the connected load.
The magnitude of current that may be optimally permissible through a transformerless power supply for a particular load may be calculated by using Ohm's law:
I = V/R
where I = current, V = Voltage, R = Resistance
However as we can see, in the above formula R is an odd parameter since we are dealing with a capacitor as the current limiting member.
In order to crack this we need to derive a method which will translate the capacitor's current limiting value in terms of Ohms or resistance unit, so that the Ohm's law formula could be solved.
To do this we first find out the reactance of the capacitor which may be considered as the resistance equivalent of a resistor.
The formula for reactance is:
Xc = 1/2(pi) fC
where Xc = reactance,
pi = 22/7
f = frequency
C = capacitor value in Farads
The result obtained from the above formula is in Ohms which can be directly substituted in our previously mentioned Ohm's law.
Let's solve an example for understanding the implementation of the above formulas:
Let's see how much current a 1uF capacitor can deliver to a particular load:
We have the following data in our hand:
pi = 22/7 = 3.14
f = 50 Hz (mains AC frequency)
and C= 1uF or 0.000001F
Solving the reactance equation using the above data gives:
Xc = 1 / (2 x 3.14 x 50 x 0.000001)
= 3184 ohms approximately
Substituting this equivalent resistance value in our Ohm's law formula, we get:
R = V/I
or I = V/R
Assuming V = 220V (since the capacitor is intended to work with the mains voltage.)
I = 220/3184
= 0.069 amps or 69 mA approximately
Similarly other capacitors can be calculated for knowing their maximum current delivering capacity or rating.
The above discussion comprehensively explains how a capacitor current may be calculated in any relevant circuit, particularly in transformerless capacitive power supplies.