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good afternoon i bulid a circuit to convert 220v ac to 12v dc so i can power a 12v dc relay. once i attach the output to the relay the voltage drop to 2v once i disconnect it and attach the positive 12v to the relay it work but then i have the disconnect it again if i want to powr it again.

this is the transformation circuit

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    \$\begingroup\$ Measure the load. Simulate it it LTspice and it will become apparent where your dimensioning failed. Also, are you aware of the dangers of lack of isolation? \$\endgroup\$ – winny Jan 13 '17 at 19:32
  • \$\begingroup\$ what you mean! ? \$\endgroup\$ – Ahmad Jan 13 '17 at 19:34
  • \$\begingroup\$ Measure the resistance of the load. Download and install LTspice/simulator of your choice. Draw your circuit. Add the load as per your measurements. Poke around and learn how to dimension it. Also, how will you isolate yourself/anyone else from coming in contact with the output? \$\endgroup\$ – winny Jan 13 '17 at 19:38
  • \$\begingroup\$ i make this circuit upon a copper sheet so i can solder it \$\endgroup\$ – Ahmad Jan 13 '17 at 19:42
  • \$\begingroup\$ so once i connect parts my hands will not connect with the input and output \$\endgroup\$ – Ahmad Jan 13 '17 at 19:46
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Although your circuit may produce 12 V for your relay in principle, I suspect that the impedance of your R-C network is too high at your load current. I have no calculations to back that up, though.

The issue that I see is danger, in the case of a component failure. In your circuit, a diode failing to short-circuit could put mains onto your low-voltage relay and risk a fire. Unless it is critical to your application to save space or parts cost, I would strongly recommend using a step-down mains transformer and bridge rectifier instead.

As a component, the reliability of a transformer will be extremely high and it will ensure that your load circuit cannot receive a dangerously high voltage. Otherwise you are avoiding a fire or similar damage only if every component works well for the lifetime of the circuit. The cost savings will look very unimportant in that light.

As I imagine you know, electronics and circuit design are only a part of engineering. Electrical safety and reliability are just as important, as are cost, environmental safety, suitability for test, suitability for servicing and a good few more.

I have designed equipment for approval to worldwide electrical standards, and before CE marking made it much easier. We used to export my designs of office equipment into 45 countries. To meet many of these approvals, our mains circuitry (PSUs, cabling) had to stay safe and protected under Single Point Of Failure (SPOF) conditions. This means that an engineer from an approving body would visit and open-circuit any single PSU/mains component or solder a short-circuit across it. The equipment would be switched on and operated and was required to not suffer a fire or present any safety hazard to the user.

In that environment, designing mains circuitry with SPOF protection becomes a lifelong discipline that you always follow. Unfortunately, your circuit is far from safe under such testing. I hope this information helps.

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    \$\begingroup\$ Well spoken about the danger. The failure to deliver 12 V stems from too low capacitance. \$\endgroup\$ – winny Jan 13 '17 at 20:43
  • \$\begingroup\$ so what you suppose! to add a higher capacitor \$\endgroup\$ – Ahmad Jan 13 '17 at 21:21
  • \$\begingroup\$ @Ahmad, no, I recommend you (a) forget your circuit and use a small transformer (you calculate but maybe 230V-to-9V transformer), bridge rectifier and smoothing capacitor or (b) tell us why this is not possible. My answer above explains why :-) \$\endgroup\$ – TonyM Jan 13 '17 at 21:45
  • \$\begingroup\$ @TonyM i have small capabilities to buy \$\endgroup\$ – Ahmad Jan 13 '17 at 21:49
  • \$\begingroup\$ @Ahmad, other here might advise you more on your circuit but I cannot - I do not recommend you use your circuit, as I have said. Good luck to you and I wish you well :-) \$\endgroup\$ – TonyM Jan 13 '17 at 22:05
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I think the coupling C is too small, or the resistance is too high. simple calculation. R1 + R2. If R1 is too big, most voltage will be on R1.

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