resistors in the right place?

Discussion in 'General Electronics Chat' started by scytzoh, Nov 8, 2015.

  1. scytzoh

    Thread Starter New Member

    Nov 8, 2015
    14
    0
    i have posted a really bad drawing showing how i am thinking to solder together a very simple led array (simple for a majority besides me) the color code on the resistors is not correct..lol....im pretty sure this is how i have it done but the problem im haveing is the leds only flicker when i turn on the power then remain off led.jpg
     
  2. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,804
    833
    Can you post a picture? What are the ratings for the LEDs? Forward voltage and milliamperes required? And what is the power supply? Volts? As a point of information, you only need one resistor in this layoutr. And how is your soldering skills.
     
  3. scytzoh

    Thread Starter New Member

    Nov 8, 2015
    14
    0
    the led calculator i found gave 2 options both with 2 resistors (see attachment)
    the only thing is neither looks like the pattern that the leds must be in ( its a computer fan) forward voltage is 3.3 i think 12volt from computr
     
  4. Veracohr

    Well-Known Member

    Jan 3, 2011
    551
    76
    Computer fan? How does that relate to this LED string?

    No attachment.
     
  5. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,804
    833
    The circuit that you gave only requires one resistor. And if the forward voltage is 3.3v, then a 12v supply is not enough (4x3.3v=13.2v). There are other ways to connect the LEDs
     
    Last edited: Nov 9, 2015
  6. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,804
    833
    Here is another way to connect the LEDs, that will run on 12V. Note that this circuit does require two resistors.

    LED-Circuit.JPG

    I calculated the resistor value for you. First, I assumed that the LEDs operate at 20ma. The formula below will explain the process of finding the resistor value, and help you find out the resistor value in this circuit if my assumption is wrong.

    R = \frac{(Vs-Vdiodes)}{Idiodes} "Vdiodes" is the same as "Vforward"

    R = \frac{(12V-(2*3.3V))}{.020A} two LEDs at 3.3Vf each

    R = \frac{5.4V}{.020A}

    R=270\Omega
     
  7. scytzoh

    Thread Starter New Member

    Nov 8, 2015
    14
    0
    Untitled.jpg the fan is where im installing the leds into this is the attach ment
     
  8. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,804
    833
    Note that solution 2 is almost exactly what I presented in my schematic. The difference is the location of the resistors. In this simple circuit, it does not make a difference (Note that if you're going to be working in electronics, you will need experience in reading schematics.) The major difference between solution 2 and your diagram, is where you connect the + and - of your power supply. As you drew it, the voltage is passing through all four resistors, dropping a little (3.3V exactly) each time. If you hook series in parallel, voltage drops less, but the current requirements of the entire circuit, go up. To match the solution, you needed to join the other ends of the two resistors together, turn around two of the LEDs, and finally make a connection in the middle of the four LEDs...

    Like this:
    myleds.jpg
     
  9. ian field

    Distinguished Member

    Oct 27, 2012
    4,415
    783
    The sum of the Vf of each LED in series must be less than the voltage available.

    If that sum is too close to the available voltage, current limiting in the resistor will be rather "stiff" - that is to say small variations in voltage will affect the current more.

    If that sum is small compared to that voltage; you'll need a higher resistance to set the current - voltage fluctuations will affect the current value less, but more energy is wasted in the resistor.
     
  10. scytzoh

    Thread Starter New Member

    Nov 8, 2015
    14
    0
    ...............................i notice that un my attachment all the leds are conected - to + but in yours 2 are and 2 are not and the resistore is in different place so how is that almost exactly?? just curious not trying to slander......so your correction to my awesome painting is the way you recomend that i solder them up then??
     
  11. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,804
    833
    I don't see, in both my schematic and the modified awesome painting, where the LEDs are not connected - to+. I just want to understand what you are seeing in the pictures. When I modified your painting, I reversed two of the LEDs, so that they are arranged as in your painting BUT still connect - to +. Here is another modified painting, with the LEDs arranged in two columns of two each, that I hope will illustrate this point further.
    myleds2.jpg

    As far as the location of the resistor, I thought I had explained that in this circuit, the resistor can be placed on the + side of the series LEDs or the - side of the series LEDs. It is there to maintain the proper current to the LEDs and can do so from either side.

    Pay close attention to the polarity of the LED legs. My assumption was where you painted a leg red, it is the positive lead; where you painted the leg black, it is the negative lead.

    As far as your last question, wire the LEDs as shown, paying attention to polarity, and they should work.
     
  12. scytzoh

    Thread Starter New Member

    Nov 8, 2015
    14
    0
    ok well i did as above and when i turn on the computer they go on for 1\4 or hald a second then off.. there not blown because it does it every time i turn computer off then on... wrong resistors?? im sure they are the ones i had asked for when i used the led calculater i got a big strip of them for use with 12 volt 4 led i dont have random rolls of resistors around
     
  13. markdem

    Member

    Jul 31, 2013
    73
    38
    Do you have any way of measuring the voltage on the power supply output after the LEDs go off? I had a look at the post, but can't see what power supply you are using but keep referring to a fan and a PC. Is this to say you are trying to use the fan output in your computer as the power supply? If so, is it just that the PC is not sensing a fan connected and turning the output off, or simply lowering the speed via PWM?
    I am thinking there is something else going on here.

    If you are using your computer to supply the voltage, maybe try a different power supply. A old 12v charger from something works very well.
     
    djsfantasi likes this.
  14. ian field

    Distinguished Member

    Oct 27, 2012
    4,415
    783
    Or just pinch a Molex connector off a scrap drive and get the power from one of those leads.

    Even if you still use a FDD, there's usually a connector left over.
     
  15. scytzoh

    Thread Starter New Member

    Nov 8, 2015
    14
    0
    OK i went bought new everything and im posting another awesome painting please tell me if this is right ..thank you for patience the final.jpg the final.jpg
     
  16. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,804
    833
    Other than why are there two awesome paintings, It looks correct.
     
  17. scytzoh

    Thread Starter New Member

    Nov 8, 2015
    14
    0
    ok sucsess! the only thing now the wires that went to the fan originally must be underpowerd or thermally controlled because they are really really dim but when put to another source it works perfect
     
  18. AlphaOmega

    New Member

    Apr 30, 2012
    4
    1
    If you are trying to power the LEDs from the fan supply, note that most fans driven from the motherboard will be controlled via a PWM supply to enable the speed to be managed.

    Try taking a supply from a 12V connector

    Edit:
    The apparent effect of changing the PWM modulation is to increased or decrease the voltage between about 0 and 12V and thus the fan speed.
    (note that some fans are 5V)
    Some motherboards allow you to turn off the variable speed, but your PC will be noisy!
     
Loading...