Timeline for How do I calculate the impedance needed to drop 120V AC?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 28, 2016 at 2:23 | comment | added | brhans | Aah ok - I was just looking at the diagram. No extra diode necessary then. | |
| Feb 28, 2016 at 2:12 | comment | added | Spehro 'speff' Pefhany | @brhans That particular optoisolator is 'AC input' - it has two back-to-back LEDs (not shown by the OP) so the diode is not required in this case. | |
| Feb 28, 2016 at 1:45 | answer | added | Arvind | timeline score: 0 | |
| Feb 28, 2016 at 1:24 | comment | added | brhans | I think you're over analyzing this, and why are you expecting a phase-shift with no reactive components? Just choose a resistance which ensures that Irms stays safely under 60mA. And put a regular diode in parallel with, but opposite polarity to your LED. | |
| Feb 28, 2016 at 0:17 | answer | added | Spehro 'speff' Pefhany | timeline score: 1 | |
| Feb 27, 2016 at 23:25 | comment | added | crasic | Take a safe margin 50ma. Round to 100v peak. This is 5W peak of passive dissipation, equivalent of a 2k 5W resistor (which would work). With a 120V rated small current transformer (like the kind for off line smps) you can use standard 1/2W parts to set the detector current | |
| Feb 27, 2016 at 23:17 | review | First posts | |||
| Feb 28, 2016 at 1:24 | |||||
| Feb 27, 2016 at 23:15 | history | asked | Joshua Paul Chan | CC BY-SA 3.0 |