Battery Operated 24v,17w Solenoid

Discussion in 'The Projects Forum' started by jh72i, Nov 23, 2012.

  1. jh72i

    Thread Starter New Member

    Nov 23, 2012
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    Hi,
    I feel like an intruder here because I know close to nothing about electronics and my questions may be so dumb you wonder who let me on here!

    Anyway, I need an injection of knowledge and Googling has just left me more confused than I started out being so here goes...

    I want to put together a small device that allows me to control the flow of water through some pipes (4 to be explicit).

    The device is to be located in an environment where the direct power source is untrustworthy and can be out for, at a maximum, a few days at a time. So I need the device to have backup power from batteries.

    The device will pulse the valves no more than a few times every day – on/off.

    I have some 12v, 17w latching solenoid valves to control the water flow and offer low power consumption. I don’t have any measurements on pulse duration right now but assume 200ms.

    I need the batteries to be cheap, accessible/easy to buy, and as neat/small as possible – car batteries or anything of that nature are totally out of the question.

    Now I have 2 electronics acquaintances with differing views. The not being friends is important and the reason I am on here – I cannot contradict them and still expect their help!!!.

    The first indicates that a string of 10 standard 1.2v AA batteries will be sufficient to power the unit for a very long time (given the criteria outlined – 4 valves, a few toggles every day, an LED, small screen, etc.). The second indicates that this would work only for a few minutes.

    Personally, I don’t even like the idea of 10 batteries at all. I was hoping that, for example, 4 3volt batteries might work. Or, even better, a couple of those neat rectangular 9v ones (is having 18v a problem?).

    Can you folks help me out with my severe knowledge gap?

    Would be very appreciated.
     
  2. wayneh

    Expert

    Sep 9, 2010
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    You really need a bit more information; some estimate of current draw for the entire system so yo can calculate an average mA draw. Without this, you and we can only guess.

    If your solenoids are the biggest drain on power, and they only are energized as briefly as you've described, then I'm inclined to agree with the 10 AA suggestion - not the actual 10 batteries, but something of that general size, about 2,000mAH. That should be plenty of capacity and enough instantaneous current to trip a solenoid. There are other similar solutions that wouldn't require 10 individual cells. A small 12V sealed lead acid (SLA) battery comes to mind.

    But again, I wouldn't do this by guessing. Maybe if you have part numbers for your components, we can help.
     
  3. jh72i

    Thread Starter New Member

    Nov 23, 2012
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    Thanks so much for taking the time to reply.

    I'm getting cheeky now but could you send me a link to a battery you think would be up to the task (and, sorry, I don't have part numbers, etc. yet).

    Can you also tell me what scientific way I can actually calculate how long any battery configuration would last - as in how many on/off cycles I could realistically expect?

    Thanks again
     
  4. John P

    AAC Fanatic!

    Oct 14, 2008
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    In your title, you say 24V but then the text says 12V.

    As far as capacity is concerned, the AAs should run the setup for months, if it's true that the solenoid gets operated a few times a day. But to get 17W at 12V, you'd be drawing close to 1.5Amps, and that's a lot. But then I did see somewhere on the Internet (so you know it's true) "I looked around, and the consensus seems to be 2400mA for most AA batteries. Energizer claims a peak of 2700 on theirs." So it seems it can work.

    What I'd do if I had to build this thing, would be to put the batteries in parallel with a great big capacitor. That way when the switch was first operated, there'd be a pulse of current available from a source which had very low impedance, and that would hopefully get the solenoid moving. If the current dropped after that, a good part of the work would already have been done.

    Re the lifetime, power divided by voltage gives you current. Current times activation time gives you charge used per cycle (amp-hours). Then you can divide that into the capacity of the battery to get the number of cycles you should be able to achieve. The capacity of an alkaline AA cell is about 1.8 amp-hours, so there you are. Actually the numbers are so simple that it's obvious--1.8 amps from a 1.8 amp-hour battery is 1 hour of operation. Except it'll be a lot less, because the battery won't deliver such a high current for as long. But in increments of a fraction of a second, you've got a lot of time there.
     
    Last edited: Nov 24, 2012
  5. wayneh

    Expert

    Sep 9, 2010
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    Here are examples of what I was thinking of.
    Another.
    The AAs approach would give you over 2 amp•hrs. A comparable 12V should be under $20 and would not be at risk of a bad connection, which gets multiplied every time you add another cell. I believe the SLA would also have a much higher instantaneous current capacity.
    The calculation is simple - power consumed is just current (amps or milliamps) multiplied by the time (hours) that the current flows. You add up all the "on" intervals and compare to the rated capacity of the battery, usually quoted in mAh for smaller batteries and Ah for the larger ones.

    Simple enough, but it gets complicated when you have intermittent and brief pulses that may not even be constant current. You'd really need to be able to plot current versus time and integrate the area under the curve to get power.

    I think in your shoes I'd make a series of worst case assumptions (high currents, long pulses, frequent service, etc.) to size the battery. Or, just buy the biggest one you can live with. If you do this right, the battery might be "too big" early on. But it will stay in service even as it ages and degrades.
     
  6. jh72i

    Thread Starter New Member

    Nov 23, 2012
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    Fantastic information. Thank you soooo much. I'm going to read and reread and see if my brain cannot click into understanding and applying this info.
     
  7. jh72i

    Thread Starter New Member

    Nov 23, 2012
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    This is embarassing - I'm really struggling with this I have to admit :( I have an Excel full of calculations and just cannot get my head around it.


    I found this link too which confuses me: http://oregonembedded.com/batterycalc.htm


    with values in order:
    2400 (from your post John P)
    0.2 (just left this default)
    1500 (or 1.417 being 17watts/12v)
    1 (pulse per hour although it'll be a lot less - more like .17)
    200 (estimated duration of pulse)


    The output suggest a single 1.5 AA battery would stay working for 2.25 years which, of course, cannot be right.


    I wonder....ahem...if you could...cough...think of me as a 4 year old and actually spell out the calculation. I know you probably think you already have but I'm lost :(


    Valve: 12v, 17watts, 200ms pulse required 4 times a day as an example - or just take 1 full second of operating time (5 times a day).


    I'm assuming 10 1.5AA batteries because surely they have to add up to at least 12v do be able to do anything?!


    How long will the little fellows last and what is the smallest configuration of 'ordinary', available-at-your-corner-shop batteries I can get away with?



    I know asking like this after all the info you've given is not in the spirit of forums such as this but please believe me I have spent half the day trying to figure this out on my own. I knew I was electronically limited and now it seems I am also very much mathematically limited too :(
     
  8. #12

    Expert

    Nov 30, 2010
    16,298
    6,810
    From P=IE, 17 watts/12 volts = 1.416 amps
    1.416 amps X .2 seconds = .283 amp seconds or 283 milliamp seconds.
    throw a 3600 in that and get .0787 milliamp hours or .0000787 amp hours.
    That's a really small amount of power per pulse.
    If your battery contains 2 amp hours, 2/.0000787 = 25,441 pulses.
    I can see why a bulky battery is not necessary and an answer in the range of years is valid.

    Gotcha started?
     
  9. jh72i

    Thread Starter New Member

    Nov 23, 2012
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    #12 thank you so much for this. All the calculations make sense and matched what I had tortured out of myself with Excel yesterday (except the final number of operations/pulses of 25,412).

    BUT... how can this be??? I am totally missing something - like these calculations refer to a single battery??? Not 10!!?

    Mr. Utterly Confused (4 years old)
     
  10. wayneh

    Expert

    Sep 9, 2010
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    In series, the capacity of the entire pack is the same as the capacity of a single cell. More cells give you more voltage, but not more current. All the cells, in theory, will deplete at the same rate as a single cell. You can get more current and more capacity by putting cells in parallel, but then you won't have enough voltage to operate the circuit.

    BTW, a 9V cell might be a viable option for this circuit. The voltage is a bit low and will sag as the battery ages, but maybe it would last long enough. It's worth trying, IMHO. If it's not good, move up.
     
  11. jh72i

    Thread Starter New Member

    Nov 23, 2012
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    Hi Wayne. So am I to understand that it's pointless talking about amps and watts if the voltages don't match up - as in, I need all those batteries to level the voltage playing field?

    Wondering then how the 9v could work - if you're talking the small 9v batteries used on things like smoke detectors, etc. then that'd be a very attractive option.



    I have another question (of course!!). It's a little different but still very much in this area. My friend is trying to test out the valves we have. He has three 12v, 18watt solenoid valves and one 12v, 17watt valve from a different manufacturer. We have no electronics components built yet (thankfully, say you, with our lack of ability!) so he tests the valves out by applying current directly. All the valves work fine.

    Then he wants to test them out as a group so he just introduces and simple junction box - wires the valves to one side and a battery to the other.

    Now, the problem he reports is that the 17w valve does not work in this configuration. Replace it with an 18w one and all is fine.

    To rule out current issues he's tried using a car battery but still the 17 won't toggle in a 3 valve configuration. It does, however, work if one of the 18w valves is removed.

    It's confusing. Any ideas what is going in with that?
     
  12. wayneh

    Expert

    Sep 9, 2010
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    Pretty much. In the analogy of electricity to water, voltage is like pressure, or height of head. Current is the flow per time, and power is the ∆P for that flow. So water falling from a greater height gives more power with the same flow. Or conversely, a slow moving river may have plenty of current like your battery, but just not enough pressure (drop in height) for the job at hand.

    It'd be easy enough to test, first if they can even activate the solenoid at all, and then if they last an acceptable time.
    No idea without a schematic and/or picture. Are the valves in parallel? Maybe there's a short somewhere?
     
  13. jh72i

    Thread Starter New Member

    Nov 23, 2012
    7
    0
    Again, thank you so much for taking the time answer my questions and attempt to educate me. I think I've got it now. Whew!

    I'm going to try your suggestion on the 9v and perhaps also two 6v camera batteries!?
     
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