Please look if the resistor I chose is the right one.

Discussion in 'The Projects Forum' started by Lightfire, Aug 5, 2011.

  1. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
    21
    Hello,

    Catapult is in need of your help. Please look at the image I attached.

    I use 51 Ω resistor and not 47 Ω because I guess 51 is much better.

    Please help and any help would be? Greatly appreciated.

    Catapult
     
  2. bertus

    Administrator

    Apr 5, 2008
    15,649
    2,348
    Hello,

    There is a mistake in your drawing.
    The resistor should be in series with the leds.

    Bertus
     
  3. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
    21
    Thank you for notifying me. Here is the modified and corrected one. Please take a look. Thanks.
     
  4. bertus

    Administrator

    Apr 5, 2008
    15,649
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    Hello,

    As for the value of the resistor.
    You state the leds are 2 Volts (3 in series), the powersupply is 12 Volts.
    The voltage accross the resistor will be 12 - (3 X 2) = 12 - 6 = 6 Volts.
    You state you are using a resistor of 51 Ohms.
    The current will be 6 / 51 = 0.118 A = 118 mA, wich will probably be to much for the leds.
    If you want 20 mA in this case the resistor must be 6 / 0.02 = 300 Ohms.

    Bertus
     
  5. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    Still wrong, currentfrom + to - through a diode goes only in the direcrion of the "arrow", so you have all backwards.

    Are you sure the leds have only 2V of Vf? Even then the resistor is wrong. It drops 6V at 20mA, so that is 300 ohms.
     
  6. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
    21
    Oh. I calculated the whole schematic with 12 volts. I forgot, it 9 volts. Sorry. And sorry for the backwards. :) So if it is 9 volt, it is now okay?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    If you are going to be using the LEDs while your 12v SLA battery is charging, then the battery may be as high as 13.8V.

    In that case, you need to calculate for the maximum battery voltage.
    Rlimit >= (Vsupply - (Vf_LED * Number_of_LEDs_in_series) ) / Desired_Current
    (13.8v - (2.0v * 3) ) / 20mA
    (13.8v - 6) / 0.02
    7.8 / 0.02
    390 Ohms, which is a standard value of resistance.
    For the wattage rating, you will need 7.8v * 0.02a * 1.6 = 249.6mW; 1/4 Watt is perfect.

    If you operate the LEDs until the battery is discharged down to 12v or 50%, the LEDs will have about 15.4mA current flowing through them.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    No, you still did not calculate the resistors correctly.
    Start with this formula:
    Rlimit >= (Vsupply - (Vf_LED * Number_of_LEDs_in_series) ) / Desired_Current
    And then substitute your voltages, LED count, and current in the formula, and then tell us what you calculated.
     
  9. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
    21
    Okay.

    My resistor should be 150 ohms. Isn't it?

    Thanks
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, 150 Ohms is correct.

    Now, please re-draw your schematic with the resistor values and the LEDs with their Vf and current ratings, and your battery voltage range, from full to discharged.
     
  11. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
    21
    I don't know the discharge and full battery rate. But here is the schematic I redraw. I hope it's correct.
     
  12. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    The leds are the wrong way around.
    They are back-biased now and will not give any light.

    Bertus
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    You have the battery shorted out, and the LEDs backwards.

    Re-draw the schematic with the battery on the left side, positive terminal up.

    Then add the two resistors to the right of the battery, oriented vertically.
    Then connect the top of the two resistors to the battery positive terminal.
    Then place three LEDs below each resistor, anode up, cathode down, in two vertical columns.
    Then wire the LEDs in series, and connect the bottom cathodes to the battery negative terminal.

    I thought I'd shown you that kind of schematic layout before, but perhaps you forgot.

    Anyway, some general rules for schematic drawings are:
    1) More positive voltages towards the top, more negative towards the bottom.
    2) Inputs come from the left, outputs flow towards the right.

    Since the only "input" is your battery; the power supply, it goes on the left, with the positive terminal up.
    Since the only "output" is the light from your LEDs, they go towards the right.
    The LEDs should be oriented so that the cathodes are pointing down, since that is the more negative part of the schematic.

    You need to document the part number and manufacturer of the battery. It might be a "PP3" type, or other type. Once we know what the manufacturer and part number is, we can find the documentation for the battery.
     
  14. iONic

    AAC Fanatic!

    Nov 16, 2007
    1,420
    68
    There are chargers out there that apply voltages as high as 14.4V, which would require a
    larger resister. I have more than one charger that can reach this voltage level.
     
  15. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    A little 9V battery is 9V when new. Its voltage quickly drops to 7.2V then drops slower to 6V when it is said to be dead.
    So your LEDs will dim as the battery voltage drops.
     
  16. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    [​IMG]
    When we apply positive to anode and negative to cathode we say the diode is forward biased and when we do the opposite we say the diode is reverse biased .When a diode is forward biased it conducts but when its reverse biased it does not (in some special diodes in reverse bias it also conducts).In your circuit you have reverse biased all your diodes, so it will not conduct i.e.. it will not illuminate.

    Here is how it should be...
    [​IMG]

    Good Luck
     
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