0
\$\begingroup\$

I suspect that a power circuit in my home has an intermittent connection somewhere, and I'd like to test this to be sure (because if I'm wrong, then the equipment running on that circuit needs to be sent back for warranty repair).

I have the idea that if I can take either an isolating step-down transformer, or perhaps just a voltage divider, from the power rail, then feed it into an ADC on a micro-controller (probably ESP8266). I should be able to monitor continuously and trigger an alert, or record some data in a file in an SD card, if it detects that the rail has not reached a proper peak voltage within the last 1/60th of a second.

Obviously, I'm dealing with high enough voltage that it'll bite me if I mess up, and for that matter, if I divide carelessly, I could easily let the smoke escape from the ESP, but in principle I can get those things right. But, are there other concerns (or specifics about the above concerns) that might have eluded me?

It's also possible that there's a device cheaply available on the market that would do this, but if so, I've failed to find it. I'm happy enough to buy something if it's not too expensive, so if such a thing has a name, I'd be pleased to hear what that name is.

\$\endgroup\$

3 Answers 3

3
\$\begingroup\$

Easiest (and very safe) approach is to find an old non-switching (linear) wall wart and divide the DC output down to some acceptable voltage for the ADC with a simple voltage divider.

The divider resistance can help determine how fast the voltage drops when there is an interruption.

Response will be fractions of a second not cycle-by-cycle. If you need the latter, you can use a (more rare) AC output wall wart and a bridge rectifier/voltage divider.

\$\endgroup\$
2
  • \$\begingroup\$ I do need relatively fast response (the effect I'm noticing is an occasional flickering on a monitor screen). So, low-voltage side smoothing capacitors would probably be unusable. But I'm comfortable enough with my skill that connecting an transformer directly to the mains line is not a source of concern. \$\endgroup\$ Aug 6 at 17:57
  • 1
    \$\begingroup\$ In any case, I suggest using a relatively high voltage output transformer (12 or 24VAC), a bridge rectifier and voltage divider, to keep things simple. You could build an absolute value circuit but I don’t think it would add value. \$\endgroup\$ Aug 7 at 9:00
3
\$\begingroup\$

A really cheap and simple way to determine if there are glitches or brownouts on an AC line would be an AC relay that latches on by means of a NO contact. When the disturbance occurs, the relay will drop out until manually reset. You could also connect an old style electric clock across the coil, and it will stop at the instant power drops out.

If you want to know how many times a disturbance occurs, you could connect an electromechanical counter to the relay contacts.

\$\endgroup\$
1
  • \$\begingroup\$ Interesting idea. Certainly has simplicity on its side! \$\endgroup\$ Aug 6 at 17:55
2
\$\begingroup\$

Using DC gets tricky because of the capacitor(s) and the associated time constant. You can transformer the AC down then use an optocoupler (resistor needed for led value dependent on transformer output voltage) to look for missing cycles and leve. This gives you double isolation from the mains and low voltages to work with. The ADC will give you amplitude for brownout and you can use the digital input on your microcontroller to do the monitoring for you and trigger the ADC. You can if you want use a count totalizer and a RTC and log the readings and time every maybe 15 seconds (or whatever you want) minutes to see if there is a pattern. For a single instance you can use a retriggerable one shot with the optocoupler or implement it in software.

\$\endgroup\$
1
  • \$\begingroup\$ That makes sense. I guess perhaps I don't even need to use the ADC if I configure a digital signal that triggers at a suitable threshold and just ensure that triggers every 1/120th of a second. The condition that I think I'm noticing is more on the lines of a brief absence of power, rather than a true "brown"-out. \$\endgroup\$ Aug 6 at 17:53

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.