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I am designing a DC-powered ceiling fan. Fan is working on AC and I convert it into DC when light is out; we attach a 12V battery so that fan runs continuously.

I want to make a circuit which senses that AC power is gone and automatically switches fan supply to the battery sources.

JYelton
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user37391
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  • http://electronics.stackexchange.com/questions/50782/ac-detection-for-microcontroller – Butzke Feb 18 '14 at 19:04
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    Agh !!!!!!!!!!!!!!!. The question is absolutely specific an clearly understandable and is about design. And yes, I read the original version, which has been tidied. As the question has now been improved so that it is tidy understandable are we going to make an exception and actually re-open it as we always say we will do but about never do? – Russell McMahon Feb 19 '14 at 03:05
  • @User37391 - I assume that the fan has a 120 VAC to ~= 12VDC supply to operate it from mains. | If so, easiest of all is to make the mains based supply produce a voltage that is slightly higher than the battery voltage, and feed both supplies to the fan via diodes (Schottky diodes perhaps, but silicon diodes probably OK. ) Whichever supply has the highest voltage will operate the fan and the diode to the lower voltage supply will be reverse biased so the lower voltage (battery) supply will supply no power. When the mains supply ceases the battery will automatically operate the fan. – Russell McMahon Feb 19 '14 at 03:10
  • As in diagram d in second row here. http://www.allaboutcircuits.com/vol_3/chpt_3/10.html – Russell McMahon Feb 19 '14 at 03:22
  • @PeterJ - (1) FYI - this is Stack Exchange - editing answers is part of what the opensource bunfight is about. My edit was apposite. I'd have placed it in a separate answer box if I could have, but the system allows modifying other answers. My apology was to Spehro as I'd much the rather not use the offered facility even though available. (2) Thanks to drawing my attention to the reopen button. Alas, it's fatally damaged. I've tried pressing it in very similar situations in the past with absolutely zero (intended) results. ... – Russell McMahon Feb 19 '14 at 10:19
  • @PeterJ - ... I think the battery may be flat, or [**The Bellman**](https://www.google.co.nz/search?q=%22the+bellman%22+the+hunting+snark&source=lnms&tbm=isch&sa=X&ei=YoUEU5b4CIyBogTwwoGAAg&ved=0CAcQ_AUoAQ&biw=1745&bih=938) may be asleep. (Or the Snark is a Boojum?) – Russell McMahon Feb 19 '14 at 10:21
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    @Russell, the question is reopened so you can post your answer. However, while editing others' posts is fine you should only do it to improve their content and not to inject your own, that can be unrelated. And if a question is closed it's to prevent it from being answered, so please follow the procedure (and possibly flag for reopening). – clabacchio Feb 19 '14 at 10:41
  • @clabacchio The way we consistently choose to abuse a significant proportion of new members here is appalling. And the lack of understanding of English language which is 'within the range of normal' (say 3SD from mean) is puzzling. New members are punished for their lack of English competence by people exhibiting often equal or greater lack of English competence. The original question was entirely understandable. If you used it as a comprehension example in a final year high school English exam I'd certainly hope the majority could understand it adequately. – Russell McMahon Feb 20 '14 at 06:40
  • @clabacchio Your input on editing and content is puzzling, not according to any official or written or otherwise-known guidelines that I'm aware of. Material gets added to answers during editing 'often enough', and it is my understanding of the ethos of open-source Wiki like editable information sources that this is exactly what is intended. A pointer to formal guidelines would be useful. – Russell McMahon Feb 20 '14 at 06:44
  • Re " ... And if a question is closed it's to prevent it from being answered ..." -> I haven't seen that stated anywhere, although it obviously makes it harder to do. If this is the case should the system not also lock existing answers to stop their original poster making them more useful and helpful? – Russell McMahon Feb 20 '14 at 06:46
  • @RussellMcMahon Some reference: [this post](http://meta.stackexchange.com/questions/120576/why-can-any-user-edit-any-other-users-question-or-answer), at the section **What should I *not* do?**. And in the [general guidelines](http://meta.stackoverflow.com/help/editing) it's explained what you *can* edit for. Re "Why can't I answer closed questions?" [link](http://meta.stackexchange.com/a/182124/177507) – clabacchio Feb 20 '14 at 07:08

3 Answers3

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I am designing a DC-powered ceiling fan.
I want to make a circuit which senses that AC power is gone
and automatically switches fan supply to the battery sources.

I assume that the fan has a 120 VAC to ~= 12VDC supply to operate it from mains.

If so, easiest of all is to make the mains based supply produce a voltage that is slightly higher than the battery voltage, and feed both supplies to the fan via diodes
(Schottky diodes perhaps, but silicon diodes probably OK.)
Whichever supply has the highest voltage will operate the fan and the diode to the lower voltage supply will be reverse biased so the lower voltage (battery) supply will supply no power.

When the mains supply ceases the battery will automatically operate the fan.

See diagram d in second row here. or in commented form below.

If "V+ ex line" is greater than "V+ bat" then the top diode will conduct and the bottom diode will be reverse biased. The fan will operate from the top supply and the battery will provide no current.

If the mains power is removed the battery will operate the fan via the bottom diode. TYhe series diodes will cause a small power loss compared to operation directly from battery or mains supply.

enter image description here

Russell McMahon
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An SPDT relay operated from the AC will do this.

However: will the fan run from 12 V dc? I wouldn't expect a fan intended to operate from 120VAC to do anything on 12V DC.

Peter Bennett
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Easiest would be to use an AC relay with form "C" (changeover) contacts (or DPDT if you need to isolate from the line for safety reasons, if isolation is not provided elsewhere).

enter image description here

clabacchio
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Spehro Pefhany
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