IMPORTANT: (1) Shorting one winding of a transformer that is hard coupled magnetically will place a short on that other winding as well. You will get magic smoke as shown.
(2) Arduino as shown floats wrt TRIAC with only 1 lead connected. you MUST NOT rely on people assuming what connection you meant - assumotions will vary.
(3) Touch the Arduino and die.
It makes little sense to provide an isolated Arduiono supply and then to throw away the isolation, even if isolation was not your main aim.
(4) Having a mains load that has a designed small current flow through it when off is lethally dangerous and also illegal in all sensible administrations.
(5) Your desire -
And since I want my design working exactly as a commercial device this is the ONLY way I want to connect.
is doable but you have to do it properly. The mechanical thermostat is effectively an amplifier with an external energy source - it uses thermal energy to move a bimetal strip . This is incredibly efficint anergy wise - the fact that you are competing with an in many ways superb mechanical solution in the form of a mechanical differential to switched mains comparator is something you have to realise and deal with. Doing so by getting your powering energy via the serial leads in unacceptable.
Potential (pun noticed) solutions include a local battery charged by voltage drop when on (about 1.4 Volts needed) or a FET (so about zero drive power and cap or supercap storage or an energy harvesting solution.
The fact that this is "easy" mechanically and "hard" electronically is your problem, not nature's. Mother nature makes the rules and we are obliged to obey them.
(6) Using a ruler and a bit of preplanning when drawing your diagram (so eg the diode bridge is not jammed into a tiny space) can get you an almost Olin proof drawing. Almost. As Olin would point out, you are asking people to assist you. Giving them a modicum of respect by providing a diagram which is clear and easily readable is liable to be at least a good idea. Hand drawn can be fine (I say, some may disagree) but make it tidy.
There are ways of doing what you want - either with two wires and external power of some sort, or by running a "neutral wire, as you note. We can discuss these, but first address the above so we know what path to take.
160 mA is high. What's that used for?
There are a number of things wrong with this. The transformer voltage must be chosen to be sufficient to provide 12V at the output. If we allow 2V for the regulator, 2V for the rectifier and, say, 2V for ripple in the filter capacitor, then we need 12.7VAC on each half of the winding. Better add 15% for line voltage brownout/dips, so 15VAC is about as low as I'd want to go. So we have 30VAC CT for the winding, not 18VAC CT. Big difference.
Now, let's calculate the minimum filter capacitors. The capacitors are charged only at the peaks of the waveform so they need to hold up for 1/2 cycle. Let's use 50Hz so it will work in most places. So each capacitor is:
\$ C_{min} = (1/100) \frac{I_{out}}{\Delta V}\$ or 0.005F (5,000uF) for 2V p-p ripple. Better to use the next size up so maybe 5600uF/35V (we have to account that the transformer output voltage will rise with a light load and mains voltage might be on the high side). At this point you should check ripple current rating of the capacitor, but I'll omit that step for the sake of brevity.
Now we're in a position to calculate the transformer RMS secondary current which will be 1.61 * the output current, so 1.6A (ignoring the regulator Iq, which is pretty much negligible). Now the transformer VA can be calculated as 48VA.
The heatsinks for those regulators much each dissipate (worst case with line voltage 10% high) about 9.5W, so about 20W for the pair (at this point we can see there's not a lot of extra there for VA vs. average current- the diodes will dissipate 4W, the output 12W and the regulators 19W, so only 1W for extra heating of the transformer winding- but that's at line 10% high).
Now to the diodes- you've made a 2A/50V bridge rectifier, barely enough for the output current. Each diode sees about 55V reverse voltage every cycle worst case with a light load (assuming line 10% high and 20% regulation in the transformer) which is somewhat more than the rating even without transients on the line- fortunately 1N4001 diodes are usually made the same as 1N4004 diodes so they will not fail.. usually. I would use 4x 1N5404 diodes for this application, or a packaged 3A or 6A bridge.
Best Answer
Are you really saying that you have a 220 V supply that intermittently gives 440 V? How does anything electrical survive in your location?
You can't series connect two primaries on different cores like that unless you parallel the secondaries. Let's think about why:
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) Due to different secondary currents the primary impedances will not match. (b) and (c) will work because the secondary currents will be the same, therefore the primary impedances and currents will be the same.
Connecting the secondaries in series or parallel ensures that both secondaries pass the same current and the primary impedances will match.
This only gives you enough theory to understand why your proposed solution is poor. You need to fix your incoming power supply.
Impedance
Given that \$ Z_P = N^2 Z_S \$ where \$ N \$ is the turns ration we can see that if a seconday is open circuit then Z is infinity on both sides of the transformer.
It would if you could force current through the primary. You can't however. You only have your mains voltage and an infinite impedance. The primary of XFMR2 will appear to be an open-circuit to an AC supply. No current will flow. Full mains voltage will be across XFMR1 primary and XFMR1 secondary will have a voltage \$ \frac {1}{N} \$ times that.
Remember that we're dealing with ideal transformers for this discussion. Real ones will have some losses and leakage so the impedance will not be infinite.