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Getting rid of batteries... simple question but what is the best way ?

redgreenblue

Hi Guys !

 

I have 4 door sensors with an arduino to have a simple sound trigger but also calculated how much peoples are getting by (helping a small museum in town).  I'll skip the details but the thing is I need to make theses connected to DC to avoid batteries changes (staff too busy). I figure the simple way to do it is using the arduino voltage out to the sensors.

 

I find this simple question complicated becuz I wonder if the current drain from the arduino will be enough for 4 sensors, so does capacitor might do the trick or if this is not the way to do it, I wonder if the AMP from the power supply should be higher ?

 

My calculation are that each sensor need 2x 1.5 volt batteries so it's approx. 3 volt @ 500mA

But I've tried the arduino output and 3.3volt does the trick, very good with one device.

 

Here a quick diagram of the setup.

If there is an expert in the room, stand up !

 

thank for your support

Element 14 is GREAT !!

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  • 0

    Hi,

    Can you give us a little more information on your sensors?  The 500 mA seems awful high for a simple sensor. Do these devices light, ring or buzz? Are they being used as inputs to the Arduino. The 3.3 Volt regulator on an Arduino UNO (LP2985 is only rated at 150 mA so it will probably be damaged if your door units actual try to load it down to one 500 mA load let alone 4. I am sure we can come up with some suggestions but we will need more information if the advice is going to be of much use.

    John

  • 0

    Hi John,

     

    let's says the sensor are stand alone. They got sound, IR sensor and a reed switch (not 100% sure) what I do is simply catch the signal when someone is breaking the circuit by opening the door, the arduino is a 'calculator' in this setup. Since I already got wire going to them and becuz the museum has no wall plate anywhere to add a 'powersupply' we are back to the same setup, using wire to feed the 4 sensor to remplace the 8x AAA batteries.

     

    What do you think can be done ?

  • 0 in reply to redgreenblue

    Hi redgreenblue,

     

    Without knowing more details I would bring power from a separate 3 volt power supply and not try to use the 3.3 volts available on the Arduino. You could do this by having a separate wall wart and then using the wires you already have run or if the power supply for the Arduino is powerful enough you could use a regulator to bring its voltage to 3.3 volts and use that. If you are going to use a regulator I would recommend one of the Recon R78C3.3-1 which can efficiently convert the 9-12 volt power to the Arduino to 3.3 volts and can provide up to 1 amp of current. I am still assuming that the door sensors do not in fact draw a full 500 mA.

     

    http://www.newark.com/recom-power/r-78c3-3-1-0/dc-dc-converter-3-3w-3-3v-1a-sip/dp/62X9495?ost=R-78C3.3-1.0&scope=partnumberlookahead&exaMfpn=true&searchref=searchlookahead&searchView=table&iscrfnonsku=false&ddkey=http%3Aen-US%2FElement14_US%2Fw%2Fsearch

     

    John

  • 0

    Thank John again !

     

    how much current would you consider like being 'enough' to remplace batteries ?

    I seen some peoples using 1A cuz they says 2xAA or 2xAAA is doing 3 volt and 1 Amp, cuz each battery is doing 0.5 A so total make 1A.

     

    I'm not sure if over 0.5A this is too much...I'm just trying to get a frameset of voltage and amp to successfully remplace batteries, this is always a tricky question with AMP vs. Amp/hour.

     

    Thanks for the link, DC to DC converter are ALWAYS the best choice !

    I'll order fews of theses if you think 1 A is good and draw power from the powersupply.

     

    c ya

  • 0 in reply to redgreenblue

    Hi,

    Units that have 2 AA batteries for a power source will likely be designed to operate on much lower current draw than 500 mA. If not you would be replacing the batteries daily. AA batteries can deliver several amps of current but only for a short period of time before their chemical energy is exhausted. The internal resistance of the circuit is what determines how much current in mA is used not the battery. Your units may draw a very low current or none at all until they sense the opening of the door and then they will consume much more current to make the light, buzzer, or bell sound. Without putting an ammeter on the unit and measuring the actual draw we are only guessing. Incidentally if you have two batteries here is how the current and voltage add up. If you put the batteries in series you will get twice the voltage but the current limit is still the same as a single battery. If you put the batteries in parallel you will get twice the current limit of a single battery but the voltage will still be the same as a single battery. If you take the voltage of the battery or combination of batteries  and divide it by the resistance of your circuit and load you will derive the idea current that you would expect if you were to actually measure the current with a meter.

    John

  • 0

    Alright John,

     

    I like that idea...so you mean I can find out how much the circuit need amp and/or voltage ?

     

    I don't want to ask to much but can you draw me a simple schematic of that ?

    I'm not 100% sure I understand with text, drawing is always better.

     

    This is worth learning, so I want to be sure I have fully understand how you do it, simply becuz remplacing batteries is always a pratical update for home electronics. And I feel many peoples will find this post usefull.

     

    thank for this amazing help !

  • 0 in reply to redgreenblue

    Hi Redgreenblue,

     

    I think you said that the sensor units use 2 batteries and that at least one sensor will work from the Arduino 3.3 volt supply. This establishes that they are 3 volt units and will work on a supply that is limited to 150 mA of less. I am not sure where you got the 500 mA information that you provided but I am going to ignore it and assume that the sensors actually draw a lower current level. Here is a drawing of how you would measure the actual current draw of the sensors. If this isn't the drawing you were looking for let me know as I wasn't clear what you wanted a drawing of.

     

    image

     

    John

  • 0

    perfect !!!!! image

    All details are on the drawing I can't ask for more !

    thank John for help and advices.

  • 0

    Hey John, it worked, I'm happy to have discover something very usefull, the only thing is that I don't have mA option on my multimeter (not sure on drawing if this is what you meant),

     

    so I put it to 10A which was anyway the only one I got a reading.

    It gave me 0.01, so does this mean I need to X10 times becuz it was on 10A selection...

    in other words it might be 0.100 mA if this is the right way to do this... ?

  • 0 in reply to redgreenblue

    .01 would be .01 A or 10mA on the 10A scale.

    You also need to measure when the door is open and alarm is active.