Mains voltage onto the opto coupler would work, if you can find an opto-coupler that is rated to the peak main voltage.

I suggest an Omron G3VM-351B, available from RS or Mouser. In fact, there are similar devices that would switch the current you want without using a triac at all.

I also have doubts whether the circuit in the example would work. When the main voltage goes negative, that would give a negative voltage on the opto isolator, which would probably cause breakdown.

Any voltage that feeds the gate of the triac has to be relative to T1 of the triac, so there has to be another connection to T1 of the triac. That means that there cannot be isolation between the whatever feeds the gate voltage and the triac. That means that you can't take the supply from the same transformer windings.

The whole reason to use a transformer is to have isolation so that the micrcontroller cannot have any dangerous voltages on it.

On your capacitor calculation, you would do better to parallel the two tranformer secondaries.

Although the nominal voltage is 6V, that is at full load and it is the RMS voltage so the peak voltage is 1.4142 times a big, and bigger still at low loads. The current is whatever the circuit take, not 2A. The transformer is only rated at 0.25A for both windings in parallel.

The resistance of the transformer is quite large on these small transformers so the output voltage will drop a lot as more current is taken.

You want to aim for about 0.5V ripple, so the capacitor has to be larger than:-

I / 0.5 / 2f so if you are on 50 Hz mains the answer is 5000µF, so you should maybe use a 4700uF 16V capacitor. If the current that the microcontroller takes is less, you could get away with more ripple so a much smaller capacitor would do, but 4700µF, 16V is much smaller and cheaper than the tranformer anyhow.

 

I'll start a new thread about this later.. understanding the AC to DC is confusing enough! I shouldn't have posted two questions in one thread here.

Thanks, I was adding 6v+6v instead of remembering it is 6v RMS when in parallel. I now understand your calculation for the capacitor value at 0.25A which is what I wanted.

I now have a circuit drawn up (to make a board from later), it is attached.

When converting DC to DC capacitors are used on the 7805 input and output. Do I still need these? I don't think I need the one on the input side as the 4700uF is doing the same job.. but I'm not sure. That should be my last question before buying the parts and blowing a few fuses!

If you have any comments on the circuit that would be great, I'll remove the capacitors later if they aren't needed.

Attached Images

Full-transformer.jpg (147.4 KB, 10 views)

 

If wires were had no impedance and capacitors had no inductance, then the 100 µF would be pointless, but in the real world, you should always have capacitors close to the input of a regulator, and the inductance should be low. That means that you should have a small (therefore low inductance) capacitor right beside the regulator. For that reason, the 100 µF could help.