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LED Strip from coin cells

Discussion in 'Electronics Homework Help' started by fenner, Aug 20, 2018.

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  1. fenner

    fenner

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    Aug 20, 2018
    I have a couple of led strip lights working off 2x CR2032 cell 3V batteries and wish to run them both from one transformer I have a 12v transformer would it be possible or would I have to obtain a 6v. I ave posted this in the wrong forum sorry but I wanted to delete and post it in another but did not know how to go about it,
     
    Last edited: Aug 20, 2018
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Welcome to EP.
    I assume the CR2032 are in series, thus powering a single strip with 6 V, right?

    In theory you should be able to run 2 strips each rated at 6 V from a 12 V source by connecting the strips in series, thus creating a 12 V strip.
    A transformer alone, however, will be no good. A transformer delivers AC whereas the LED strips require DC. A 12 V DC wall wart could be used, or you'll have to rig up a small DC power supply e.g. like this one (just leave out IC1 (the 7805 voltage regulator) and the associated components.
    Other similarly constructed 12 V supplies are shown here.
     
  3. Audioguru

    Audioguru

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    Sep 24, 2016
    Most LED strips are 12V, not 6V. They have 3 LEDs in series with a resistor and if each LED is 3.5V (blue, bright green and white) then they do not light with less than 10.5V. Some red LEDs are 1.8V and a strip will light with 6V.
     
  4. fenner

    fenner

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    Aug 20, 2018
    Thank you Harald and yes the batteries are in series.I can buy a 12v DC on ebay what power should I be looking for the twin copper 0,41mm dia wire is approx 2 meters long with 20 leds
     
    Last edited: Aug 20, 2018
  5. fenner

    fenner

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    Aug 20, 2018
    Thank you Audio these are not like the normal led's and there are 20 on twin copper wire.
    IMG_7132[1].JPG
     
    Last edited: Aug 20, 2018
  6. Audioguru

    Audioguru

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    698
    Sep 24, 2016
    You said you have LED strips which are usually 12V. 20 LEDs on twin wire are not an LED strip and probably have them in parallel. They are probably powered from your two weak overloaded battery cells.

    Please post a link to the LED product or see if there is a resistor or something that limits the current.
    Also we must know the color produced by the LEDs to determine their voltage.
     
  7. fenner

    fenner

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    Aug 20, 2018
    They say they are fairy lights (these what are shown are 100 lights but mine is only 20 the colour is yellowish
    https://www.amazon.co.uk/s?url=sear...prefix=fairy+lights,aps,183&crid=BI5TCWFXGB05
     
  8. WHONOES

    WHONOES

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    May 20, 2017
    You can run LED's from ac. You just put a reverse connected diode in parallel with the LED plus the usual series resistor. You must also take into account the peak value of the ac voltage, which is going to be 1.4 time whatever the transformer says, when calculating your resistor. You will get a lot of flicker though.
     
  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Insufficient info. You need to know the power consumption of the strips. If you do not have technical data of the strips, you could measure the current consumption from the original batteries with an ammeter (or multimeter in DC A range). Say you measure x A.
    As you will have to put 2 strings in series to get to 12 V, you will need a 12 V, x A power supply.

    If you're going to buy DC power supply anyway, you could also opt for a 6 V power supply and connect the LED strinps in parallel to the power supply. You will then need a 6 V 2 * x A power supply (better get one with 4 * x A to be on the safe side).
    This scheme has the advantage that even if one LED strip fails the otjer strip can still light up (in a series scheme a failure in one strip can interrupt the current flow through both strips and you will have no light at all).
     
  10. Audioguru

    Audioguru

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    Sep 24, 2016
    The fairy lights have all the LEDs in parallel. They are normally powered by 3 AA battery cells and the ad says 4.5V so they must have a resistor in there to limit the current. The two CR2032 cells have much less capacity than the AA cells so their overload will drop the voltage so the LEDs survive.

    The ad says 4.5V so use a power supply that produces 4.5VDC. The current rating of the power supply should be 400mA for 20 LEDs or 800mA or more for 40 LEDs.
     
  11. fenner

    fenner

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    Aug 20, 2018
    Thanks Audio I am not into this sort of thing just learning and I have searched for these fairy lights online to gather some data but nothing that helps me I also have an ammeter and set it to get Amps and put the probes on the copper wire that is attached to the led's (I assumed that these copper wires are not insulated) but I did not get a reading. Going off what you say above and that the CR2032 batteries is 6V would they work from a double 5V usb hub that works from 240V supply.
     
  12. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    It is worth a try.

    How did you connect the Ammeter? You need to put it in series between the power supply (aka batteries) and the LED strip:
    upload_2018-8-21_10-5-12.png
    You also need to set the meter to a suitable measuring range, e.g. 20 mA. When set to 2 A the resolution of the meter may not be good enough to show a few mA flowing.
     
  13. Audioguru

    Audioguru

    3,136
    698
    Sep 24, 2016
    Each LED draws about 20mA to be very bright but the fairy lights probably draw 5mA or 10mA so that they are not too bright. You have 40 of them so their total current is about 200mA to 400mA.

    They are designed for 4.5V from powerful three AA cells. CR2032 cells are small and weak so their 6V drops a lot with such a high current.

    If you use USB 5V to power them then their current will be higher than designed and a few of the LEDs and maybe the current-limiting resistor might burn out soon. To reduce the voltage and current a resistor can be added in series.
    0.5V/200mA= 2.5 ohms (a 1/4W resistor) or 0.5V/400mA= 1.25 ohms (a 1/2W resistor).
     
  14. fenner

    fenner

    42
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    Aug 20, 2018
    Thanks Harald.
     
  15. fenner

    fenner

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    Aug 20, 2018
    Thank you Audio, there does not appear to be a resistor in the circuit, I have enclosed a photo of the battery compartment and switch and there does not appear to be one where the plastic covered flex meets the copper wires. So could this mean (if there is no resistor) I could use a 5v supply. Perhaps I should tell you my intentions my Grand-daughter has an hairdressing salon and has these lights scattered around but she is forever replacing the batteries so I want to power the two sets to run from the mains, whether parallel to run two sets or series to run as one. What would be the decision you would adopt.
    IMG_7133[1].JPG IMG_7134[1].JPG
     
  16. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    It is a common practice to use CR2032 batteries to power LEDs directly. The technique relies on the comparatively high inner resistance of the batteries to limit the current. It cannot be recommended to operate the LEDs from a more powerful source without current limiting series resistor.

    To get an idea which resistor to use you need:
    1) current consumption of the LED strip (see post #12). Write down current ILED
    2) real operating voltage of the LED strip. This will be less than 6 V due to the voltage dripping at the output of the CR2032s when current is drawn. Use your multimeter and measure the output voltage of the battery box with the LEDs turned on (note to connect the multimeter in parallel to the battery box for this measurement). Write down VLED

    Calculate the required resistor from R = (Vsupply - VLED)/ILED

    More about operating LEDs in our ressource.
     
  17. fenner

    fenner

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    Aug 20, 2018
    Thanks Harald, I don't know If I have done this correctly (as I said I am a learner) Calculating the resistor,
    Current consumption;- I placed the meter probes in series on the wire + battery side and the neg probe on led side with the meter set on 20m the reading was 1. Real operating voltage . Reading 5.6v. Reading with the lights on:=2.9v So R = (Vsupply - VLED)/ILED 5.6v-2.9v - 2.7ohms so is this my resistor
     
  18. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    No. Your reading of 1 in the 20 mA range means 1 mA.
    Vsupply is the output voltage of your power source, the 5 V wall wart you want to use.
    (5 V - 2.9 V)/1 mA = 2.1 V / 1 mA = 2.1 kΩ
    The power rating of the resistor is P = 2.1 V * 1 mA = 2.1 mW. A standard 1/4 W resistor is more than good enough here.

    With a 5 V wall wart (e.g. a typical off the shelf USB supply) you can only run the strips in parallel with one resistor for each strip (preferably). You may connect the strips in parallel and use a single resistor, but you'll have to calculate the resistor anew for the number of strips in parallel.
    I prefer the one resistor per strip solution. That way a defect in one strip doesn't affect the other strips.
    A 5 V power source is not suitable for series connecting these LED strips. In series connection teh min. voltage is 2*2.9 V = 5.8 V, more than the power supply delivers. You may use another type of wall wart with e.g. 6 V output voltage, but the 5 V USB types are ubiquitous and therefore the least expensive ones.

    If you don't want to mess with the separate power supplies and resistors, you can get mains powered fairy lights.
     
  19. fenner

    fenner

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    0
    Aug 20, 2018
    Thanks once again Harald, She has two sets of these so I will try to use them. I have found a resistor it is colour coded Red/Red/violet/gold have I read it that it is 2.2kohm
     
  20. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    R/R/V/G is 220 MΩ
    2.2 kΩ is red/red/red/gold or silver - the last color band indicates accuracy (gold = 5%, silver = 10%) which is pretty irrelevant in this application.

    See e.g. this calculator.
     
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