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Breadboard Wiring

Discussion in 'General Electronics Discussion' started by Ihulen, Jan 2, 2018.

  1. Ihulen

    Ihulen

    3
    0
    Jan 2, 2018
    I'm new to electronics, and would like to know why one way works, and the other doesn't(The top image doesn't work). 20180102_155657.jpg 20180102_155605.jpg
     
    Last edited: Jan 2, 2018
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,174
    2,690
    Jan 21, 2010
    Maybe one LED is the wrong way around.

    The LED is polarised and must be in the circuit the right way around.

    One lead is longer than the other, and there will be a side of the LED which is slightly flattened, both of these indicate the anode. The anode needs to be positive.

    Depending on your battery voltage, connecting them there wrong way around could kill them.
     
  3. Ihulen

    Ihulen

    3
    0
    Jan 2, 2018
    This is whats confusing me. The LED is oriented properly just as you suggested. Can you not have both the positive and negative leads on the LED in the same row on the breadboard? (The top image wont work for me).
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,174
    2,690
    Jan 21, 2010
    Well, all the holes in each row (as you have the breadboard oriented) are connected together.

    In that case you have wired the battery to the resistor, not the LED and the resistor (both LEDs of the LED and one lead of the resistor are all connected to one lead from the battery.

    It was hard to see this from your picture until you mentioned it. Also that was the one where I thought the LED was glowing so I paid less attention to it.

    p.s. It's a lot easier to see on a real screen than on my phone.
     
    Last edited: Jan 2, 2018
  5. kellys_eye

    kellys_eye

    4,284
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    Jun 25, 2010
    In the top picture you have all your connections made in row #42. All five holes are physically joined together - shorted. Your LED leads are therefore SHORTED so the green wire is connected to the resistor, bypassing the LED completely.
     
    davenn, Cannonball and Ihulen like this.
  6. Ihulen

    Ihulen

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    0
    Jan 2, 2018
    Ok thank you! So pretty much the current bypasses the LED to the green wire, because the green wire has less resistance?
     
  7. kpatz

    kpatz

    247
    52
    Feb 24, 2014
    In a breadboard such as yours, each horizontal row (of 5 holes) are connected together. So, row 1, a-e are connected together, and f-j are connected together. Same for row 2, and all the remaining rows.

    So, when placing components, you have to connect the leads to different rows, or across the middle, otherwise their leads will be shorted together. In your top pic, both leads of the LED are in the same row so the LED is shorted out. The resistor isn't shorted because it bridges the middle. So, essentially you have the battery connected to the resistor but the LED is bypassed by the lower resistance track in the breadboard (it has nothing to do with the resistance of the green wire).

    In the bottom pic, you put the LED leads into different rows so they aren't shorted anymore, and the LED works.
     
  8. davenn

    davenn Moderator

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    1,746
    Sep 5, 2009
    no, the green wire is irrelevant
    you have shorted out the pins of the LED with the row connection trip of the breadboard
     
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