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Arduino and AC powered circuit Upper left corner is a simple AC powered circuit that can supply 15mA of DC current. Lower left is an Arduino connected via USB to desktop computer. The meter shows a voltage of 1.58V between the Arduino ground and negative of the DC output of the circuit. So there must be a potential conduction path through the USB, computer power supply. Shouldn't the transformer in my computer power supply isolate the USB ports?

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    \$\begingroup\$ Stop before you burn something or get electrocuted. PC computer power supplies are referenced to ground. I repeat, they are not isolated. You must not connect anything that is mains referenced to it, or be able to touch mains referenced circuits. Stop using capacitive droppers as power supplies right away and get a laboratory power supply. Cheap life insurance to avoid accidents with mains voltage. \$\endgroup\$ – Justme May 4 at 21:08
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    \$\begingroup\$ Everything in your photo is live: the power supply, the Arduino and the mutimeter leads and meter. You can buy a USB power supply for < €5. Why take chances with your life? \$\endgroup\$ – Transistor May 4 at 21:17
  • \$\begingroup\$ ... and if you've plugged the Arduino into a laptop then that must be considered live too. \$\endgroup\$ – Transistor May 4 at 21:45
  • \$\begingroup\$ Is that a capacitive dropper supply on the top left? \$\endgroup\$ – marcelm May 5 at 11:33
  • \$\begingroup\$ Yes. 470nF in this case. In parallel with 1M ohm resistor. \$\endgroup\$ – Ernst van Zyl May 5 at 15:10
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Don't use mains power for anything!

Simply use any common-as-dirt, UL-listed DC power supply. They are made by the billions at trivial prices. 5V supplies are particularly abundant, being sold at every convenience store and gas station. This is not a wheel you need to reinvent. Certainly not worth dying for! Let alone smoking a laptop.

Where you went off the rails is by taking 120V or 230V mains (which you should be mortally afraid of), and homebrewing a power supply from that. You go "Well, it's only 1 volt!" Right, you're testing DC. Flip the meter to AC setting. I bet you see a much larger number.

DC and AC can coexist in the same place, the DC is a bias upon the AC. Normal 120VAC has 170V peaks. If you had a sinewave with -165V to +175V peaks, that's still 120VAC, but has a 5 volt DC bias. It will read 5V on DC.

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  • \$\begingroup\$ This is not realistic. Yes it's dangerous I will admit. How do you learn anything if you can't experiment? The reason for building this circuit is purely out of curiosity. It can power a few LED's. An AC spike might fry anything connected to it. I won't consider using it to power anything \$\endgroup\$ – Ernst van Zyl May 5 at 7:08
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    \$\begingroup\$ @ErnstvanZyl If you're building the power supply to build it, fine, you've done that, now take it apart. This project with the Arduino, get a listed COTS supply for that. Don't poke the dragon. \$\endgroup\$ – Harper - Reinstate Monica May 5 at 7:20
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I think Justme hit the nail on the head. The computer power supply is not isolated from AC mains. Now the moment of truth: I destroyed two Arduinos, two ports on my USB hub, a bluetooth dongle and a wireless keyboard and mouse dongle. Tripped the earth leakage a few times.

The idea was to use the Arduino as a crude oscilloscope to measure the voltage across the resistor (and hopefully visualize the ripple voltage), 3V according to multimeter.

This "circuit" was simply an experiment, not meant for anything I planned to use in everyday life. Hopefully someone will learn from my stupid mistakes and not repeat them.

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What you’re measuring is the voltage difference between your ground/earth and the floating potential of your AC to DC circuit. Since arduino is connected to usb of laptop, then ground on arduino=ground of computer=ground of your house since your laptop should have grounded power supply. Since your AC to DC circuit is only using LINE and NEUTRAL, then DC is essentially ~peak AC voltage with respect to neutral. Your negative terminal of AC to DC circuit is at neutral point, ie not perfectly grounded.

WARNING: Your circuit is dangerous since you are using LIVE MAINs (without ground) and have high voltage dc output.

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    \$\begingroup\$ OP is not using a transformer. Hence your answer does not apply. The DC output is referenced to live mains. \$\endgroup\$ – Justme May 4 at 21:22
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    \$\begingroup\$ Oh wow, so that’s live and neutral itself!!! That is dangerous! \$\endgroup\$ – Leoman12 May 4 at 21:31

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