I'm making an active rectifier, to make it work I need precise information about actual phase of main voltage.

What I need:

1) exact moment when the positive section of sine wave starts and ends. Basically ac sine -> binary.

2) solution should provide galvanic separation

My ideas so far:

1) small transformer, diode and comparator/logic gate,

If there's a diode on lower trafo side I'm losing precision, as there is drop voltage on it, so I will need to compensate it one way or another. Comparator/logic gate's purpose is to convert half-sine wave into binary.

2) diode/transoptor on upper side of trafo

Higher precision, as diode is on higher side of transformer, but not sure if it will be enough.

3) transformer to step down voltage to let's say 2.5/-2.5V, add 2,5V DC to make it 0-5V and do the math magic in uC

Not sure about it, if anything goes wrong I'd applying negative voltage to uC. Not feeling good about it. But still it sounds as the most precise solution.

What's your method? Do you know any chips that'd do the job?

  • 1
    \$\begingroup\$ Best way to sync is to use zero cross detector - ZCD, not a half wave to square. There are plenty of ZCDs. \$\endgroup\$ – Marko Buršič Oct 17 at 9:49
  • 1
    \$\begingroup\$ You say you want the "exact moment" of zero crossing, but that is not possible in the real world. You need to specify how much error you can tolerate. Is it 1ms? 100ns? 10ns? \$\endgroup\$ – Elliot Alderson Oct 17 at 11:46
  • \$\begingroup\$ Yup, I know it is not possible to get the 'exact moment', should have been more preciese. The more accurate measurment the better. Basically anything below 100us should do the job. \$\endgroup\$ – Theresnoneed Oct 17 at 14:18

Have a look here: https://cdn.hackaday.io/files/1597066832861504/SimpleIsolatedZeroCrossDetector.pdf

It is an improved version of the former EDN circuit: https://www.edn.com/design/analog/4368740/Mains-driven-zero-crossing-detector-uses-only-a-few-high-voltage-parts

You may also read many related papers:





Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.