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Schematic of Batteries

  1. Nov 23, 2013 #1
    Hi, I am working on a PCB design and the mock-up prototype is now done, thanks for all your help by the way!
    now its time to get it mobile. I am trying to add a battery (CR2032) to the design of the PCB,

    the thing is that I have my voltage regulator set up as so:
    http://img607.imageshack.us/img607/2811/ao8x.png [Broken]

    but I dont know how to hook up the batteries,

    the voltage regulator is supposed to take a voltage in of 7.2V (from 2 CR2032 rechargeable batteries at 3.6V each)

    and one stream is supposed to feed a RF transmitter at 3.3V (the bottom regulator)
    the top one is a 7805 that outputs 5V to drive the other circuit.


    I was confused as the battery symbol (that comes in adafruit library) has 3 leads, from what I remember soldering, 2 of them are positive? and 1 is negative, but should I link the negative to the ground? or how should I wire it up



    thank you!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 23, 2013 #2
    Your schematic symbol looks like it has two pins on the cathode. It's hard for me to tell from the image but I think the symbol has two bold vertical lines, one long one for the positive side and one shorter one for the negative side.
     
  4. Nov 23, 2013 #3
    yeah the 2 pins are positive, and the single (middle) pin is the negative, my question is, should the negative go to ground? or where should I connect it to?
     
  5. Nov 23, 2013 #4

    davenn

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    what is the bottom regulator ?
    looks like a negative regulator ?
    what's it outout voltage supposed to be ?

    Dave
     
  6. Nov 24, 2013 #5
    The top regulator is a 7805, it receives up to 13V input and outputs 5V

    the bottom regulator is a UA78M00, recieves up to 13V input and outputs 3.3V, or thats what I want it to do.

    Thing is most of my board works on arduino (5V) but the RF transmitter works on 3.3V

    I found these CR2032 , they are 3.6V and I am using 2 that I want to connect in series to give an input of 7.2V

    should I set them up in series instead of parallel then?

    VCC = 5V
    VEE = 3.3V
     
    Last edited: Nov 24, 2013
  7. Nov 24, 2013 #6

    davenn

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    OK
    thanks for your clarifications

    in this case you have indicated the lower 2 capacitors with the wrong polarity

    here's a circuit I drew using the battery supply you wanted....

    attachment.php?attachmentid=64240&stc=1&d=1385280714.gif

    Now the major problem with your thought of using only 2 batteries is that you are not really going to sustain the 2 to 2.5V headroom difference between the input and output of the 5V regulator

    I would suggest that you will find you need 3 of those 3.6V batteries in series, rather than the 2 you wanted and that I have shown

    OHHH and you can see I have chosen the BA033T regulator for the 3.3V rail

    cheers
    Dave
     

    Attached Files:

  8. Nov 24, 2013 #7
    So, given that I am limited in size, would you suggest I move to 3x CR1220 or CR1225 ? (3volts each) and get a 9V total voltage?

    Also, is ground the same as 0V ? (I know they are similar in AC or DC, but I cant remember where? or maybe im completely wrong)
     
  9. Nov 24, 2013 #8

    davenn

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    Im not too familiar with the 1220 or the 1225, you would need to consider if they have enough current supply capabilities for what your circuit requires
    NOTE tho that none of those small button batteries have much current capability
    the CR1220 for example is only 20mAH
    you are not going to power much off that for any length of time
    These styles of batteries are used in digital watches and for memory backup where the required current for the circuit is extremely low

    I suspect you are going to have to reconsider your project size and a different type of battery

    in this case, the same thing

    cheers
    Dave
     
  10. Nov 24, 2013 #9
    You put capacitor backward for 3.3V
     
  11. Nov 24, 2013 #10

    davenn

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    OK
    thanks for your clarifications
    in this case you have indicated the lower 2 capacitors with the wrong polarity
    here's a circuit I drew using the battery supply you wanted....

    https://www.physicsforums.com/attachment.php?attachmentid=64262&stc=1&d=1385332430

    Now the major problem with your thought of using only 2 batteries is that you are not really going to sustain the 2 to 2.5V headroom difference between the input and output of the 5V regulator

    I would suggest that you will find you need 3 of those 3.6V batteries in series, rather than the 2 you wanted and that I have shown

    OHHH and you can see I have chosen the BA033T regulator for the 3.3V rail

    cheers
    Dave
     
    Last edited: Nov 24, 2013
  12. Nov 24, 2013 #11

    meBigGuy

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    The lm7805 is a voltage hog, needing 2V headroom. Use an LDO regulator like the http://www.ti.com/lit/ds/symlink/lm2940-n.pdf, which is more like 0.5V @ 1A.

    Or, any LDO regulator that fits your needs. Just be sure to use modern LDO regulators, not another battery.
     
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