I bought one of these: https://www.ebay.ca/itm/High-Power-Warm-White-DC-32-34V-100W-LED-light-Lamp-SMD-Chip-9000-10000LM-LM/272210214196

I am powering it with a 12v 8ah battery and a voltage booster: https://www.ebay.ca/itm/150W-DC-DC-Boost-Converter-10-32V-to-12-35V-6A-Step-Up-Voltage-Charger-Power/171907240597

I adjusted the voltage booster to output 34v exactly (measured with my multimeter)

Problem?: It's not as bright as I would have hoped ... so with my multimeter I measured the current flowing from the battery and it is only drawing about 4amps from the 12v, so about 50W.

Am I missing something or is the seller just selling under wattaged chips?


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    \$\begingroup\$ We can't know if it is a 100W LED bulb. However, LEDs have quite large variation in their forward voltage, so they really should be driven with regulated current, not regulated voltage. So instead giving it regulated 34V, adjust the voltage until current through the LED matches the rated 3A current. \$\endgroup\$ – Justme Jan 8 '19 at 20:10
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    \$\begingroup\$ Is 3000 mA flowing through the LED, as specified? If not, you need to turn the voltage up a bit. But it would be better to use an LED driver that regulates the current directly. \$\endgroup\$ – Dave Tweed Jan 8 '19 at 20:11
  • \$\begingroup\$ Probably cheap versions ( surplus) with higher than average Vf. ( out of spec) Raise output to 38V or until input reaches 110W at 90% efficiency or measure output current \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '19 at 20:15
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    \$\begingroup\$ But then you need a supercooler CPU heatsink + fan or equiv 32~34V is false advertising \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '19 at 20:23
  • \$\begingroup\$ You can get thermal runaway if your V booster is stronger than your heatsink towards max current, so better get a proper current source. Meaning ESR of booster+battery is less than Thermal resistance of heatsink and effect on -ΔVf / +ΔIf with - xxx mV/'C \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '19 at 20:31

Well, there's your problem:

I adjusted the voltage booster to output 34v exactly (measured with my multimeter)

Look - on the one hand, LEDs (even the same model from the same manufacturer) do not all have the same voltage vs current curve. Assuming that 34 volts is an exact value will cause you nothing but heartache. On the other hand, attempting to drive a bare LED (without some sort of current limiter) from a voltage source is really great way to kill an LED.

The problem is that LEDs have what's called a negative temperature coefficient. That is, for the same current, as the LED gets hotter its voltage drops. Put another way, for a fixed voltage the current will increase as the LED gets hotter. So let's say you have an LED connected to a voltage source. At first, it will draw some current value, and dissipate power appropriately. In the process, this will cause the LED to heat up. This will cause it to draw more current, and therefor dissipate more power and get even hotter. Depending on your cooling system, it's entirely possible that this will form a vicious circle which eventually burns out the LED.

So don't do it.

Instead, either use a current source or put a current-limiting resistor in series with the LED. Figure on dropping something like 10 to 20% of the LED voltage in the resistor, if that's the way you go.

Now, as to your specific current. You have measured the converter output voltage - but did you do so with the LED connected? Having measured the LED voltage, why didn't you measure the current as well? If you do, you'll be able to calculate the LED power directly, rather than trying to back it out of the converter input numbers. And you should be aware that any voltage converter will have an efficiency less than one. That is, if the input power is 100 watts, the output power will be less. It depends on circuit and load, but 85 to 90% efficiency is a pretty standard number. Problem is, you don't actually know what yours is.

So. First, change your LED circuit. You want to set the converter output to several volts greater than the nominal LED forward voltage, then add a resistor to limit the current to a reasonable value. Second, measure both LED voltage and current, rather than messing around with the converter inputs.

In your particular case, you got lucky. Your LED actually needs more than 34 volts in order to pull 3 amps, so you wound up not killing your LED.

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  • \$\begingroup\$ Thank you @WhatRoughBeast. I will look into this in the next few days. \$\endgroup\$ – Daniel Rivera Jan 9 '19 at 15:31

ok just in case lets set the premises. V*A=W and the voltage booster sacrifices amps in order to boost the voltage, meaning that we gain voltage, but we lose amperage.

with that said, the voltage booster would be drawing 6amps and 12volts from your battery. 6*12=72, therefore the wattage coming out of the voltage booster would be 72watts.

if you want 100watts, I recommend getting a 4volts battery and hooking it up in parallel with the battery that you have now. that way the 4volts are going to add up to 12volts, making it 16volts and 16*6=96, so that would be 96watts

if we add more than 4volts, we might risk burning the LED

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    \$\begingroup\$ Welcome to the site. "I recommend getting a 4volts battery and hooking it up in parallel with the battery that you have now" Not a good plan. The OP's existing battery is 12V. Adding a 4V battery in parallel will not end well :-( Later in your answer, it looks like you meant to recommend connecting the batteries in series. However, connecting dissimilar batteries in series is also unlikely to end well, due to their different capacities and the liklihood of the larger capacity battery driving the lower capacity one into reverse charging, when the lower capacity battery is exhausted :-( \$\endgroup\$ – SamGibson Jan 9 '19 at 1:37

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