Isolating a secondary backup power source?

Discussion in 'The Projects Forum' started by Conductanc3, Mar 1, 2013.

  1. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    I have a remote controlled AC switch, which plugs into mains power and allows an appliance load to be plugged into it. Its wireless remote then allows this switch to be turned on and off from a distance.

    Most of these kinds of switches available today will, if the mains power is cut (as in a power outage) result in the attached appliance defaulting to be OFF once the power returns to the remote switch. I have managed to find one that will, for a short time, retain "knowing" that the remote switch had been in the ON position, and when mains power is re-applied it will default back to having the load outlet switched ON. This condition will last only for a minute or so of mains power loss. After that, the outlet reverts to defaulting as OFF when power returns. So if you pulled the switch's cord from the wall outlet and then plugged it back in, the powered appliance will resume being on. If you wait a few minutes and then plug it back in, the powered appliance will be off.

    I would like to make this switch remember to return as ON after a power outage for a longer period of time. A LOT longer. 12 hours, to be precise. I want to power this remote switch with a typical programmable wall timer, set to run 12 hours, then be off 12 hours, then repeat.

    Experimenting, I have found a small electrolytic capacitor in the circuit which, while I monitored its voltage, starts out at 5.5VDC when the mains power is applied, and eventually bleeds off when mains power is removed. The 'memory' to default back to ON when mains power is reapplied is tied directly to the level of charge in this capacitor. As long as it exceeds 0.5V, the switch will remember to turn its relay ON when mains power is reapplied. This reminds me of some battery powered devices that have little internal "backup batteries", which will retain information while changing their main battery due to the temporary power provided by the little backup battery.

    As such, I would like to find a safe way to keep that electrolytic capacitor charged, when mains power is removed, somewhere above 0.5V, so the remote switch's "memory" to default back to ON is preserved. Perhaps through the use of an additional battery or regulated DC supply.

    I have some hobby electronics experience, but am far from well-versed in any particular aspect of it. I know that I can't just add a battery or secondary power source in parallel to the capacitor, because when the mains power were restored there would likely be a current passage back into whatever I had used as the auxiliary supply, and cause problems. So I am looking for a way to supply a few volts DC to this application but ideally have it either cut off when the mains power energizes the circuit itself, or else implement some sort of current limiting or blocking to prevent current feeding backwards into my auxiliary source of power. I'd also like to keep whatever circuit I have to make as simple as possible in order to do this.

    I appreciate any suggestions on how to accomplish my goal. :)
     
  2. #12

    Expert

    Nov 30, 2010
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    You seem smart enough to figure this out. Maybe a 3 volt button cell battery and a diode. I forgot which chemistry it is that you can not get a 1.5 volt cell, but from the description you gave, even a 1.5 V cell would work.
     
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  3. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    Hey! I found the Thanks button! :) Maybe they are not displayed in the off topic section?

    OK, thanks. I was hoping it would be something this simple. I'm not sure how much actual current would be involved to feed the capacitor I want to keep alive when the device is powered down, but I can measure it once I have the backup circuit set up for testing. If it kills a button cell too quickly I may opt for a regulated wall wart to take its place. 3V should do it, because I have found I can retain the relay "ON" status in the device if I switch the mains power back on briefly before the capacitor gets below 0.5V, and then watch it fall from its 5.5V full charge back down toward 0.5V again. So 3V should be plenty.

    Any particular recommendation on the diode I should use? I have some 300V IN5394 rated for 1.5A that I think would work OK.
     
  4. #12

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    Bit of an overkill, but this requirement is so small that anything with a pn junction will work.

    Off Topic has, "Thanks" buttons. Maybe you're just too new here.
     
  5. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    I just happen to have those on hand for sure. I may have smaller ones if I look through things.

    Thanks! :)
     
  6. #12

    Expert

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    I was gonna say, "That's what the button is for" but quickly remembered...you don't have buttons yet:D

    As for the pn junction...ANYTHING (except an LED) will work, including 2 legs of a transistor. Cheap and convenient is the clue.
     
  7. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    I like how you think, #12 :)

    I'm getting ready to hook up the test circuit, so I'll let you know what happens.
     
  8. #12

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    Some people like my style, some don't.;)
     
  9. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    For a while I had a retired Bell Labs EE mentoring my efforts. I'll just say that your approach so far, by comparison, is.........refreshing........:) We think more alike, you and I. Keep it sweet and simple, and not necessarily "by the book", so long as the solution works, and is safe.

    The circuit is working apparently well. When the mains power is cut to the device (it is a remote controlled switch for AC appliances) my aux supply (currently using a pair of AAA because they were easy to set up) is feeding 2.75 mA into the capacitor, holding it between 2.5 and 3V (when I added a DVM in the line to watch the current it dropped from 3V to 2.5, I am guessing due to the meter's resistance for such a low current). When I apply mains power to the device, the capacitor's voltage rises to 5.5, the current from the AAAs drops to 0 mA, as it should. And the switch preserves the previously "ON" status. Hopefully indefinitely, as long as the batteries hold out.

    My goal here is to be able to use this remote controlled switch to temporarily cut off the device it powers during the day. The switch will be fed by a timer, 12 hours on/ 12 hours off. Most of these kinds of switches, as I mentioned, default to OFF when the power is cut for more than a moment or two. I wanted a remote switch that could be "ON" as soon as the timer cuts in every morning, without my having to use the remote to turn it on each day.

    This project will allow me to do just that! :)

    One last question. What kind of mAh rating is the typical coin cell, such as a CR2032? I need to decide what I am going to use to supply the aux power once I set this up. I do have a bunch of wall warts, but most of them are not regulated, so at such a low current load they would likely spike over their stated voltage unless I added a regulator. I don't want to take the chance of overpowering the circuitry within the switch by supplying more than 5.5VDC.

    Thanks for the help!! :)
     
  10. #12

    Expert

    Nov 30, 2010
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    200 ma hours for a CR2032. That's only 72 hours...6 days at 12 hours per day, and you're right about not over-volting this kind of stuff. A lot of digital chips smoke at 6 volts.

    AAA alkaline batteries are not much better at 300 hours (25 days).
    You're going to end up with AA batts at 700 hours (2 months).
     
    Last edited: Mar 2, 2013
  11. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    Thanks. I found some mAh info on AAAs. Alkaline AAAs show 1150 mAh, which works out to about 34 days. Still too frequent to be swapping batteries. I have a pair of LM350 regulators in the parts cabinet, so I guess I'll build a regulator circuit for a 3.7VDC wall wart I've got, and use that.
     
  12. #12

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    I'm going to have to quit depending on battery books I bought 30 years ago:mad:
     
  13. #12

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    LM350??? Huge overkill! An LM317LZ would do this job.
     
  14. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    I agree, but the thing is that I already have "some" parts here on hand. I don't have an LM317. I had some, but when it was time to order more, I decided to go with the LM350 because it has a 3.5A current capacity (in case I needed it for some future project). If I had an LM317 I would use it. ;)

    Same story with how I got those diodes. I have learned that, to maximize shipping costs, when I need one or two of something inexpensive, I order 5 or 10 (depending on their unit cost) so I have spares. Those diodes were used in a previous project that was run on mains power and carried more current, and I have some extras left over. Might as well use what I have on hand. That way I don't need to order anything. :)
     
  15. #12

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    How about this for 1.2 volts applied?
     
  16. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    I don't understand how that circuit works. Could you clarify how it would work? I know:

    The 2 diodes in series would drop about 1.4V

    2.75mA through the resistor would produce a voltage of 2.25V

    Aside from that, I am not sure what this would do. Is that resistance enough to keep an unregulated wall wart close to its stated voltage? I've seen them go several volts over the rating on the label when not used to power their intended load.
     
  17. #12

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    If any DC supply provides from zero to 14.3 volts, the power in an 820 ohm resistor is less than 1/4 watt. You can use a 1/2 watt resistor for any voltage up to 14.3 plus the voltage used by the diodes, and that sums up to about 15.5 volts.

    3.7 volts will send 3 ma through the resistor. The thing you'r working on will use 2.75 of those milliamps and the leftover .25 milliamp will be shunted to ground. There will be 1.2 volts available at the top of the diode string.

    If the wall wart supplies 15.5 volts, 17.4 milliamps will go through the resistor. 2.75 of those milliamps will go into your thingy and the remaining 14.65 milliamps will go to ground through the diodes. There will still be 1.2 volts available at the top of the diode string.

    Two diodes in series is the same as a 1.2 volt zener diode (at low currents).
     
  18. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    After I asked for the clarification and headed off to bed I was thinking about how it might work (when I don't have all the pieces I allow for lots of possibilities) and it finally dawned on me that the voltage drop across the resistor would give me the 1.2 volts, as you explained. But the 2 diodes had me confused, because I never thought about using a Zener diode to get the results we wanted. Your mentioning it in the last sentence made the lights come on in my head. :) I did not know a series of 2 diodes could work as a Zener diode.

    Thanks for sticking with me and explaining this. I have a lot of fun with this hobby when I can make things work, but I have always had trouble wrapping my head around certain concepts, and I suck at understanding schematics, especially if they involve several pathways. I can identify the components, but the ways they will interact often eludes me. That previous helper I mentioned had a gift for looking at a complex schematic and quickly gaining full understanding of it. For me its always very difficult, and I need guidance, unfortunately. I'm better with mechanical things that I can see. Electronics has always seemed a bit magical to me, because I can't "see" the current doing things without the help of meters and a scope, and when AC is involved I can quickly get lost. Too many variables to follow with AC. :) So I try to stick to DC projects.

    I'll try this simple method of regulation later today when I get time, and let you know how it works out. Thanks!
     
  19. #12

    Expert

    Nov 30, 2010
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    "Regulating" a voltage has several levels of accuracy, power, smoothness, etc. Proper cheapitude depends on choosing the right level of quality for the circuit you are using.

    I have to go to work now. Not be back for hours.
     
  20. Conductanc3

    Thread Starter New Member

    Feb 28, 2013
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    OK. Have a great day at work! (Is that even possible? :p )
     
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