I want 12 v DC output from a Transformer less power supply.
For starters, transformerless power supplies are dangerous, potentially lethal even, especially to people with little or no experience - and your post gives a strong hint that you belong in that category!
The safe way is to use a transformer and isolate the mains carrying parts completely.
But, since I have little hope that I can dissuade you from continuing your walk on the wild side by a warning, I'll try with facts instead
I have used the attached circuit with little modifications. (Image is attached)
AC Input Voltage = 220V , 50 Hz
R1 = 2.7 ohms
C1 = 1.25 uf, 400V
D1 = 12 V
Your choice of R1 shows that you really shouldn't be doing this!
Why make such a radical change, when you clearly don't know the purpose of it?
R1 is an inrush current limiter for when the circuit is turned on.
A value of 2.7 Ohm equals a toasted circuit in a short time (if not instantly) as it will try to pass a huge current when you hit the switch at, or near, 90° from the zero cross (as the cap isn't limiting anything immediately at turn-on).
What is your reason for changing C1 to 1.25µF?
I'm guessing that you think, that the value of the cap determines the voltage (which it doesn't) and then approximated 12/5*0.47µF
I connected R1 & C1 in phase line instead of Neutral and changed diodes polarity accordingly. Does this matters?
I'd prefer C1 and R1 in the phase line as well, keeping the polarity of the zener and moving D2 to the phase line as well, but pointing towards the cap of course.
That way the 0V of the low voltage will be directly at Neutral , which matters in some circuits and it's a good practice to adhere to.DON'T be mistaken, the "low voltage side" is still lethal!
When i run this circuit.
AC Voltage after C1 drops to 2.5 V and output DC Volts are1.5 v.
I want 12 v ½DC output. How to get it?
Another clear indicator that you'd be much better off with a transformer.
Working with stuff that is both lethal AND that you don't understand, is like playing Russian Roulette with an automatic.
Is the "½DC" just a typo or??
How does a capacitor reduces AC voltage. what is the theory behind it?
A capacitor can be seen as an AC "resistor" with a value that's inverse proportional to the frequency. In this application you can see it as the top resistor in a voltage divider, where the bottom resistor is your load.
Equation 3 in the schematic holds the formula for this "resistance" (capacitive reactance).
= capacitive reactance [Ohm]
pi = 3.14
f = frequency [Hz]
C = Capacitance [F]
Given your numbers:
= 1/(2*pi*50*1.25*10^-6) = 2,546 Ohm (~2.5kOhm)
Since you want 12VDC, you want to drop approximately: 220-12 = 208V
This leaves a current of: 208V/2.5kOhm = 81.7mA
And this will be able to drive ~980mW of load.
If you thought you could drive a heavier load (your numbers point towards something like a windshield wiper motor or heavier), I have to disappoint you - you'll need around 15.5µF/A and R1 needs to be able to handle the power dissipated in it as well.
High power zener diodes are expensive and hard to find (and it needs to be able to handle the full power for at least a couple of cycles).
Your 220V (RMS) is: 220*sqr(2) = 310Vpeak
, so you should use a 630V (non-polarized) cap.
(230V is the EU harmonized mains voltage BTW).
The sensible limit for transformerless supplies is way below 1A, as both the cost and the physical size is then on the ugly side of a simple transformer solution.
I'm not trying to scare you off experimenting, but when you are walking down new territory, at least go with non-lethal voltages