I would like to repurpose eight, 8' LED strip lights that were previously individually powered by 1600ma constant current drivers. Is it possibble to power these LED fixtures with a more readily available Constant Voltage driver? I only have one of the previously used drivers (dimmable) which had an output voltage between 10 and 55 so my plan would be to dial down a 60 V power supply to 55 V. As far as limiting the current, my plan would be to put enough fixtures parallel to the power supply that it couldn't exceed the 1600ma per fixture rating. For example if I was to use a 600W power supply my logic is this: 600w / 55v = 10.91A / 7 = 1558ma per fixture at 55V

Is this logic at all correct or will I just smoke the LED's attempting this?


  • \$\begingroup\$ for clarity I would power 7 of the fixtures in parallel \$\endgroup\$
    – Jake H
    Commented Oct 12, 2020 at 22:25
  • 1
    \$\begingroup\$ Sounds like a bad plan to drive strips with constant voltage without current limiting if they are meant to be driven with constant current. The current would not divide equally, as strip with lowest voltage would light up first, and draw all the current through it, most likely breaking and then the next strip would light up, drawing all the current, etc. \$\endgroup\$
    – Justme
    Commented Oct 12, 2020 at 22:32
  • \$\begingroup\$ even if Justme stated the worst case scenario with the led-strips vaporizing,... if not you will for sure have different currents in each stripe, resulting in different brightness. But you might be able to use one central DC voltage controler - adjusted to say 58-60V and then build a simple current source (e.g. 2 transistors, 2 resistors - one being adjustable) for each stripe, just for balancing the parallel stripes... that should be feasable \$\endgroup\$
    – schnedan
    Commented Oct 12, 2020 at 23:36

1 Answer 1


If you dial down the voltage to 50 V then monitor the current rise with temperature as Vf reduces -0.1%/‘C or ~-0.5V then you should be able to adjust the current and temperature so that they do not overheat with a 700W supply to have margin, as it is a bad idea to load a supply to >=100% for reliability reasons.

You might then be able to Make a dimmer with a Nch Fet rated for 10x the current or a really good heat sink.

Use a a smallish power shunt switch resistor to have a minimum load current of 5% max. if the supply is unhappy then add an LC LPF with a 2 Ohm impedance at the switching rate for good dampening.

You may adjust my estimates.

  • \$\begingroup\$ Thank you for all of the advice. I am just an electrician so I am not familiar with some of the technical recommendations but I think I was able to get the general concepts by doing some Google searches. I purchased a CC CV Step Up Regulator off of Ebay to test on a single fixture. If it works out I intend on using one per fixture to dial the voltage and amperage to the appropriate levels. Ill only load the Power Supply to 75 or 80 percent as suggested. Thanks again!! \$\endgroup\$
    – Jake H
    Commented Oct 14, 2020 at 16:25
  • \$\begingroup\$ The LEDs behave exactly like Zeners so the CC supply had a wide range, yet your string will draw current at exponential levels above threshold. The effective series resistanceESR above 1/3 power is almost linear rather than exponential. ESR approx = S/P /2Pmax for S number in series and P in parallel. So if say 12S4P , S/P=3 and if each LED Pmax=1W, ESR=3/2 Ohms. Your numbers may be different, Each LED has a threshold voltage ~ 2.85 and rated Vf~3V roughly the current is thus proportional to VI and ESR is the ratio of the slope above 1/3 power \$\endgroup\$ Commented Oct 14, 2020 at 16:48
  • \$\begingroup\$ Using Vth and Vf and number of LEDs in series, S you can now estimate your required Voltage for the array, current depends on P \$\endgroup\$ Commented Oct 14, 2020 at 16:54
  • \$\begingroup\$ So imagine each LEd as a 3V Zener \$\endgroup\$ Commented Oct 14, 2020 at 16:55
  • \$\begingroup\$ |------------| | Mind | | Blown!! | |------------| (__/) || (•ㅅ•) || /   づ \$\endgroup\$
    – Jake H
    Commented Oct 14, 2020 at 22:35

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