No announcement yet.

Calculations for Making 60 watt DC LED Light

  • Filter
  • Time
  • Show
Clear All
new posts

  • Calculations for Making 60 watt DC LED Light

    I decided to make own 24 volt DC led light. I strongly believe I need help Here is my calculations

    [B]this is the leds that I am going to buy on ebay: [/B]

    [B]DC Forward voltage : 3,2 volt to 3,4 volt
    Current: 700 mA[/B]

    Because I want to make 60 watts led light I am going to use 60 of led diode (because it says each one of them is 1 watt)

    According to my research I can not exceed 24 volt constraits in paralel and I need to use resistor for current limitation.

    Amount of Led that I can use in series : 24 Volt / 3.4 Volt = 7.05 = [B]7 Led Diode in series

    *Because light is going to be 60 watt lets use 6 led diode in series so it is going to be 10 paralel strings

    KVL for determining resistor in every string= [/B]

    - 24 V + 6 (diodes) x 3.4 volt + 700 mA x R = 0

    R = 5.14 ohm is the resistor that every paralel string requires

    Is my calculation is correct ? Do I need to use LED driver or something else ?

    Thank you

  • #2
    You do know you can buy 50 watt ($9) and 100 watt LEDS ($10.50) right? LED's demand a current supply, not voltage.
    Last edited by Sunking; 02-11-2016, 04:58 PM.
    MSEE, PE


    • #3
      Your basic math is correct, but your assumptions need to be examined.

      1. What do you need this 60W LED light for? Outdoor illumination at a great height?
      60W of LED is the equivalent of about 500W of incandescent.
      2. You indicate that it will be powered from 24V and calculate a resistor value based on that. But if is is a 24V battery system rather than a regulated power supply the actual voltage will range as high a 30V while the battery is being charged, Using a resistor and a 3.6V drop at 24V input will instead give you a 9.6V drop from 30V and the triple current will burn out the LEDs and the resistors very quickly.
      3. You use 3.4V per LED. If for some reason (high temperature, for example) the voltage drop goes down to 3.2, then even at an exact 24V the drop across the resistor will now be 4.8V instead of 3.6V, again causing a damaging current.
      4. That 1W per LED figure probably assumes some sort of heat sinking for each diode. Will you incorporate that into your design?

      Points 2 and 3 argue strongly in favor of using a linear current regulator (or constant current LED driver) instead of a simple resistor. If efficiency is very important to you, then a switching mode regulator would be better still.
      SunnyBoy 3000 US, 18 BP Solar 175B panels.


      • #4
        What is providing the 24 V supply?

        What you need to consider is that the LED forward voltage is temperature sensitive, and drops as the LED heats up. As the forward voltage drops, more current is going to flow.

        For example, in your base design, solve for I:

        I = [24 - (6 * 3.4)] / 5.14 = 0.700 A

        What happens as the LED's heat and the Vf drops to 3.2 V?

        I = [24 - (6 * 3.2)] / 5.14 = 0.934 A

        Now that more current is flowing, the temperature will tend to climb even more, potentially running away.

        One way around it is to size the string + resistor in a way that the resistor is big enough to prevent the current from ever getting high enough to cause runaway, but of course, the more power that goes into resistors, the less efficient the light becomes.

        An LED driver controls the current directly, which is much more robust way to prevent runaway and can eliminate the need for the resistor entirely.
        CS6P-260P/SE3000 -


        • #5
          As a general rule of thumb for long-term success, any LED or string of LEDs that requires over 100 mA should not use a simple current limiting resistor circuit. You cannot manage the heat and current properly with such a simple circuit design. LEDs in that power range need current regulation. Placing 60 LEDs like those in a large series-parallel arrangement is a definite taboo. The design may well work for some number of hours but the LEDs will not live the long life you expect. Those LEDs also require a heat sink. What are your plans there?
          Dave W. Gilbert AZ
          6.63kW grid-tie owner


          • #6
            Originally posted by malborn View Post
            *Because light is going to be 60 watt lets use 6 led diode in series so it is going to be 10 paralel strings

            R = 5.14 ohm is the resistor that every paralel string requires

            Is my calculation is correct ? Do I need to use LED driver or something else ?
            You need to use an LED driver. They are available widely. Meanwell makes cheap LED constant current power supplies; for example their CLG-60-24 will give you 24 volts at 2.5 amps, and will accurately regulate current. Generally you are better off making each string as long as possible; that way you end up with fewer parallel strings. Since LED's care about current not voltage a single long string is most ideal, but for your case you are not going to be able to series 60 LED's since that would require over 200 volts of drive, and such supplies are harder to find. Most LED supplies max out at around 50 volts.


            • #7
              First you cannot parallel LED strings because you cannot regulated current in each string. Nor can you use a Resistor to regulate current. What is throwing off is you are stuck in Voltage Source box when you need to be in a Constant Current Source box across town. As has been mentioned many times you need a LED Driver which is a fancy name for a Constant Current Supply. With a 24 volt supply, 60 diodes with 6 diodes in each string means you need 10 LED drivers.

              With a 24 volt source means you can use from 1 to 7 LED's in series. Constant Current Source supplies are very simple devices using analog IC's. In its simplest form is a [B][U]variable resistor [/U][/B]in series with a voltage source. The value of the resistor is a range from [U]0 Ohm's up to R[/U] where R = Supply Voltage / Current. So for example if you use a 24 volt source and want a .75 amp supply is:

              24 volts / .75 = 32 Ohms.

              The series resistor has to be 0 to 32 ohms, but it cannot be fixed. You use a simple IC to act like a variable resistor. The resistor value will be whatever it needs to be to maintain the designed current. In addition current supply can be made adjustable so you can adjust light intensity.

              If you short out the supply you have .75 amps at 0 volts. If you disconnect the load or open circuit Voltage= Supply Voltage and 0 amps of Current. This gives you the ability to power 1 to 7 LED's in series. If you connect just 1 LED, the voltage out of the current supply is 3.2 volts. 2 Diodes 6.4 volts up to 7 LED's at 22.4 volts.
              Last edited by Sunking; 02-12-2016, 12:54 AM.
              MSEE, PE


              • #8
                And God Said Let There Be Light.

                How about a 90,000 Lumen, 1000 watt LED Flashlight

                MSEE, PE


                • #9
                  You need to use a current regulator for each string of leds. But even more fun, you need to heat sink EACH of the LED die. You must go to the die mfg's site to locate the heat sink requirements.
                  But why not use larger die - 10 watts, just need 6 of them, 6 drivers, 6 heatsinks
                  Powerfab top of pole PV mount (2) | Listeroid 6/1 w/st5 gen head | XW6048 inverter/chgr | Iota 48V/15A charger | Morningstar 60A MPPT | 48V, 800A NiFe Battery (in series)| 15, Evergreen 205w "12V" PV array on pole | Midnight ePanel | Grundfos 10 SO5-9 with 3 wire Franklin Electric motor (1/2hp 240V 1ph ) on a timer for 3 hr noontime run - Runs off PV ||
                  || Midnight Classic 200 | 10, Evergreen 200w in a 160VOC array ||
                  || VEC1093 12V Charger | Maha C401 aa/aaa Charger | SureSine | Sunsaver MPPT 15A