Timeline for How efficient is a capacitive power supply?
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:32 | history | edited | CommunityBot |
replaced http://electronics.stackexchange.com/ with https://electronics.stackexchange.com/
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| Jun 29, 2016 at 17:08 | comment | added | Dave Tweed | That's good, but it would be even better if you followed up on your speculations at the end and found answers for them. | |
| Jun 29, 2016 at 16:51 | comment | added | davidcary | @DaveTweed: Better now? | |
| Jun 29, 2016 at 16:49 | history | edited | davidcary | CC BY-SA 3.0 |
fix mis-guided "improvements", in response to comments.
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| Jun 3, 2016 at 23:30 | comment | added | davidcary | @DaveTweed: Good points. As you point out, at half load a linear regulator would be less efficient than the zener regulator, so this answer needs some changes. | |
| Jun 3, 2016 at 14:30 | comment | added | Dave Tweed | As an example, suppose the capacitor is selected to provide 30 mA @ 3.3V (100 mW total). If the load goes to half power, the zener will be dissipating 50 mW (50% efficiency), but a linear regulator would be dissipating 15 mA * 60 V = 900 mW, which is an overall efficiency of only 5%. | |
| Jun 3, 2016 at 14:28 | comment | added | Dave Tweed | Also, the linear regulator would NOT be more efficent than the zener. At the full load allowed by the capacitor, both would be equally efficient, with neither regulator dissipating much power. At lower loads, the zener would simply dissipate whatever power the load isn't consuming. However, the linear regulator would be dissipating increased power because its input voltage will rise, and this excess voltage will have to be dissipated as heat in the regulator. | |
| Jun 3, 2016 at 14:16 | comment | added | Dave Tweed | This is a fairly good answer overall, but the part regarding replacing the zener with a linear regulator is misguided at best. The zener functions as a shunt regulator, and this is important to the operation of the circuit overall, because it keeps the current flowing in the series dropping capacitor constant despite any variations in the load current. A linear regulator is a pass regulator that does not have this property -- the source current is the same as the load current, which means that the regulator's input voltage would vary widely as the load current changes. | |
| Apr 25, 2016 at 16:03 | comment | added | Jules | @supercat according to the wikipedia article on the standard quoted in this answer, "the following equipment are excluded: Equipment with rated power less than 75W, except class C equipment." I'm not sure what "class C equipment" is, exactly, but other than that, such a device would appear to be compliant. | |
| Jun 16, 2015 at 12:25 | history | edited | Ricardo | CC BY-SA 3.0 |
Minor edit, admitedly, but I couldn't resist. Formatted expression using MathJaX. Otherwise flawless post (even links are dressed). Great answer!! +1
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| Oct 14, 2011 at 16:41 | comment | added | supercat | Do the power-factor rules restrict the absolute amount of power one can absorb which doesn't fit the profile, or do they restrict the relative amount? For example, would a device that consumed 10uA at a power cycle of zero be compliant or non-compliant? | |
| Feb 7, 2011 at 19:10 | history | answered | davidcary | CC BY-SA 2.5 |