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I was asked by my uncle to convert 22 units of gx53 LED Bulbs that are powered by 230VAC 5W to DC voltage power.

The LED Bulb itself consists of 5 LEDs(string) in series, in parallel to another 5 LEDs(string) in series.

I have disconnected the LEDs inside the Bulb from its inner driver and through a resistor i connected it to a DC power supply.

I learned using a measuring device(voltmeter) that each LED lights up with 8VDC and the current flowing through it is 8ma so the Bulb itself can be powered by 40VDC with current of 16~20ma.(5 * 8V=40V, 2 branches of LEDs * each branch is 8ma=16ma)

Now what i need is the right led driver.

I learned that LED drivers supply constant current while the output voltage is within a range of voltages.

So if i have 22 LED Bulbs that each need 20ma with 40VDC then the driver, to my understanding, should supply current of about 450ma with output voltage of at least 40VDC.

I ask for advice whether i'm right or wrong.

enter image description here

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Your calculation is incorrect. When you connect LEDs in series, the voltage required increases, but the current through the string is the same as for 1 LED. (The same electrons are flowing through all LEDs).

So if 1 led requires 8mA at ~8V, then 5 LEDs will require 8mA at ~40V.

Two of these strings in parallel would require 16mA at 40V (if your math for 1 led is correct).

An easy check you could have done is to calculate the wattage. 450mA at 40V would be 18W, where it should be around 5.

Your initial assumption of 8V and 8mA for 1 led would give a wattage per led of 64mW. So for all 10 leds we're looking at 0.64W, which is not correct. Again it should be around 5W.

You should hook up the driver again, and adjust it until 1 led is drawing 0.5W. Then redo the subsequent calculations with the current and voltage you measure.

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  • \$\begingroup\$ Inside the Bulb there are two strings of 5 LEDs in parallel to each other. So if one string draws 8ma then two strings draw 16ma. I measured these values and not calculated them. The total wattage is for all 22 Bulbs and not one. The LED Bulb consist of its own driver that i disconnect and use only the LEDs so i guess the 5W written on the Bulb is more then what the LEDs themselves need. \$\endgroup\$
    – Alon Lavi
    Jun 11, 2020 at 23:49
  • \$\begingroup\$ @AlonLavi Info added \$\endgroup\$
    – Drew
    Jun 11, 2020 at 23:54
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No, I think your 8V 20mA for one LED is wrong. Usually these type of LED diode (a more exact term than "bulb") are 3V 150mA. They will still turn on at 8V at overcapacity. Don't try this for a long time. To measure the voltage of the LED, you have to measure the voltage drop.

  1. Apply 5V (or 8V if you don't have a 5V supply).

  2. Measure the exact voltage at the supply output or at the LED anode (its plus side). It may be slightly different than indicated. Call it Vsupply.

  3. Measure the voltage at the LED cathode (it's minus side). Call it Vdrop

  4. Calculate: Vled = Vsupply - Vdrop. If by applying 8V you measure 5V, then the LED is 3V. And you have 5V too much.

A better way would be to make this test with two LEDs in series after the 8V supply. measure the voltage drop after each LED. If the LEDs are off when you use two LEDs in series but are on when only one LED is connected, then you already know they are more than 3V. Still measure to know how much exactly.

About the amperes, it's a wild guess from my part, looking at the picture. IMO, it must use between 100 and 150 mA. If you have 5x3V = 15V, ideally, you will have to find two constant current power supplies of 120 or 150 mA with a voltage range around 15V, or 3W. If you can't find this or if you want to do it cheaper, you can use simply a classical 15V power supply and add one resistor of +- 30 ohms before each series of LED. These resistors must be rated 1/2W or more. Should be OK too.

Make sure the LED are ventilated to avoid overheating.

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  • \$\begingroup\$ be very careful to apply the current in the right direction. Otherwise you would destroy the LED instantly. \$\endgroup\$
    – Fredled
    Jun 11, 2020 at 23:35
  • \$\begingroup\$ I tried to operate the Bulb - 5 LEDs string in parallel to 5 LEDs string - at the beginning with only 12V cause i assumed each LED operate at about 2V and the LEDs didn't light up. Eventually only After raising the output voltage to 40V(8V for each LED) the LEDs light up. \$\endgroup\$
    – Alon Lavi
    Jun 12, 2020 at 0:02
  • \$\begingroup\$ 2V is for red and yellow led. White leds and other colours are 3V. If these ones are operating at 8V, They must be special LEDs. But it's possible. \$\endgroup\$
    – Fredled
    Jun 12, 2020 at 0:15
  • \$\begingroup\$ Well that is what i measured. But there is another way...to connect the Bulb itself to 230VAC as it was created for, and measure the voltage on each LED and also the current through it. Once i do it, I'll update you. \$\endgroup\$
    – Alon Lavi
    Jun 12, 2020 at 0:17
  • \$\begingroup\$ Did you measrured it with a voltmeter, the way I explained? \$\endgroup\$
    – Fredled
    Jun 12, 2020 at 0:18
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The best idea by far would be to measure the voltage/current from the original driver and then replicate it. However, from your measurements some guesses are possible.

If they light up at 8V, then they're probably 9V nominal (3 diodes in each SMD) 0.5W diodes. Ten of them gives the nominal 5W rating for the whole bulb, so that works out.

If there are two parallel strings of 5, then the total voltage is 45V, and the max current per string is probably about 50mA, with a little less than that for safety being a good idea. Times two gives 100mA. I would aim for a little less than that to be safe since you aren't sure what the original driver was giving.

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  • \$\begingroup\$ Thanks for the answer cause it clarifies everything. :-) I will anyway connect the Bulb again to the 230VAC and measure the actual voltage and current as it was meant to be and I'll update again. So from what i understand from you is that i should choose a driver with at least 45V output and constant current that will be be enough for 22 Bulbs.(100ma * 22 = 2.2A) \$\endgroup\$
    – Alon Lavi
    Jun 12, 2020 at 0:49
  • \$\begingroup\$ @Alon Lavi You should probably use 22 individual power supplies, or else put groups of a few bulbs in series. Putting 22*2 = 44 strings of 5 LEDs in parallel isn't the safest idea since the forward voltages (and thus current received) will vary between devices. With so many, you run the risk of blowing up one or more, which will result in more current to the survivors, and more blowing up, and so on until they're all dead. \$\endgroup\$
    – user1850479
    Jun 12, 2020 at 2:11
  • \$\begingroup\$ Although I suppose that since there are 3 diodes in each SMD, it is really 44 strings of 15 LEDs, which is a little more reasonable (more series diodes to average out variation). Still, if you choose to parallel lights, make sure that the temperatures will be very similar and that you are well below the maximum current rating. \$\endgroup\$
    – user1850479
    Jun 12, 2020 at 2:33
  • \$\begingroup\$ Thanks again for your comments. I did as you said and connected the LED Bulb to its original 230AC voltage and measured the current and voltage on the LEDs and it is the same as you said. 44VDC for each string of 5 LEDs and 110ma for two strings in parallel which means about 55ma for each string alone. But still it doesn't change the fact that the LED Bulb also lighted up with 40VDC on each 5 LEDs string, powered by a DC power supply with current of 16ma for both strings.(measured) \$\endgroup\$
    – Alon Lavi
    Jun 12, 2020 at 17:33
  • \$\begingroup\$ Since these LED Bulbs meant to serve as decorative lighting on a garden's ground, it won't change much if they will light up a little weaker so i may use a driver of 42VDC with constant current of about 500ma ?. (20ma for each Bulb's two LEDs strings * 22 Bulbs = 440ma) \$\endgroup\$
    – Alon Lavi
    Jun 12, 2020 at 17:33

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