I read that you should "always" have a resistor in series between an LED and its power source. But what if the power source already has its voltage set to a suitable max for that particular LED? Why would you need a resistor then?

So here's what I want to do:

Power a 3 watt LED with a single cell LiFePO4 battery. The only other things in the circuit would be a switch and maybe a fuse (mostly to protect the battery). The LED charts I'm looking at show the LED comfortable and productive in the voltage range between 3.3 and 3.5 volts. This is perfect for a LiFePO4 battery with 1 cell. My charger is a Turnigy Accucell 6, which can be set to always charge the battery to 3.5 volts, or any other voltage I tell it to that is reasonable for LiFePo4. I can get a good chunk of the total energy capacity of the battery just running it from 3.5 down to 3.3 volts, which is the range where the LED works ok, without over-current or over-voltage (according to the mfr charts for the LED).

For a straw-horse LED, we could look at this one, which has nice charts available here: http://www.rapidonline.com/pdf/170197_da_en_01.pdf

In this pdf, page 15 has the chart for Vf vs If, with 1 amp being the max If allowed, this happening at 3.5 volts.
Will this work ok - as long as I am careful to never let the battery voltage go higher than what produces max allowable current in the LED?

  • \$\begingroup\$ LEDs need a current drive, not a voltage one. The type of thing you are trying to do is really only workable (and even then only marginally) when the internal impedance of a small cell provides a reasonable limit to the current. \$\endgroup\$ – Chris Stratton Nov 29 '15 at 23:05
  • \$\begingroup\$ dont forget to check if your battery can handle 1Amp \$\endgroup\$ – Wesley Lee Nov 29 '15 at 23:05

This will work only if you are able to maintain the junction temperature at 25C. Normally, this is not the case and the forward voltage of the LED will drop as it warms up. This will lead to a increase in the current, the current increase will accelerate the rise of the temperature and this "snow ball effect" may destroy your LEDs. Check page 7 of the document you added as reference, it is written that the thermal coefficient is around -2 to -4mV/degC. Another interesting aspect you have to watch is the range of the forward voltage, page 7 as well, it goes from 3.19V up to 3.99V. That means, you have to find one chip with exactly the 3.5V. If the chip you receive is in one of the extremes, you may have a overcurrent or no current at all according to your proposal.

  • 2
    \$\begingroup\$ Another way to say it is that on LED's, small changes in voltage cause large changes in current. This is why LED's are virtually always driven with a circuit that limits or regulates current. \$\endgroup\$ – mkeith Nov 29 '15 at 23:18

I went ahead and tried this to see what would happen with some actual parts.
It turned out better than I expected. No resistor, just a switch, battery, and LED.
But, instead of a single LED and a 1-cell LiFePO4 battery, I used an LED module that has 3 white LED's connected in series and mounted to a .065" thick aluminum plate, round, about 1" in diameter, with holes in it that you can put small screws through to bolt it to a heat sink. And I used a 3 cell LiFePO4 battery of the type used by transmitters in RC model airplanes. The battery is 1.5 amp-hour rated and is rated to charge at up to 2C and discharge at up to 1C.

It works out like this, with these parts. The charger takes the battery up to 3.6 volts per cell, then ends the charge automatically. That's 10.8 volts. After charging, the battery absorbs down to about 3.4 volts per cell in only 30 minutes or so, just sitting there not connected to anything. So typically the "fresh charge" voltage in my light is about 10.2 volts. But even at 10.8 volts, the LED would be OK as long as it has some heat sink behind it (I have my LED module screwed to a bracket made from 1/8" aluminum plate).

At 10 volts, the current is about 200 milliamps. This is well below the 350 mA that the LED is rated "Typical". So I'm not getting the full rated light output from the LEDs, but it works well.
During use, the battery voltage comes down from 10.2 volts to 10.0 volts pretty fast, then stays in the range from 10.0 to 9.7 (measured while under load) for a long time, and that is pretty much the useful range for my purpose. Below 9.7 voltage under load, the light output is getting a little too weak, and that is about when I will put it back on the charger. (The 9.7 under load is about 9.8 after about 3 hours of "rest" after light is turned off). I could run it down to 9.6 volts where the battery is at 3.2 volts per cell which is pretty much the bottom of the well for a LiFePO4 battery and no point running it lower. Although minimum discharge voltage is usually considered to be about 2.9 volts per cell, for this type of battery. This is not the kind of battery that you want to ever let run all the way to 0. Like if you accidentally let it sit there turned on for a week and it ran down all the way, that would be a no-no.

My LED module is a Tru-Opto 3 watt OSPR3X6 series from Rapid Electronics. My battery is a Turnigy nano-tech 1500mAH LiFe 3S 9.9v Transmitter Pack from Hobby King.

I get somewhere around 5 hours of good light output from this rig per charge - somewhere around 60% to 70% of the battery capacity, with the rest of the capacity there for emergencies at lower light output. So, not optimized to the highest level, but it's a pretty nice rig.


This is a common misunderstanding of how LEDs are specced.

The spec is not saying "supply a voltage within this range and it'll live happily ever after". LEDs are not voltage devices like most things. They are current devices.

The spec is saying "drive the LED at a current within the spec, keep it cool and it'll live happily ever after. And by the way, if you are driving spec current, that will typically happen within this voltage range, a fact you may find helpful when designing your current-limiting driver circuit."

They are most definitely not saying to supply constant-voltage (and unbounded current) in this voltage range - if you do, anything could happen including overcurrent and magic smoke.

  • \$\begingroup\$ Actually the only things that could happen are determined by the I/V curve of your LED. People generally assume they are brick walls (especially on here) but you may be surprised what some of them actually look like. (However, I didn't check the one linked in the OP) \$\endgroup\$ – user253751 Jan 23 '18 at 1:27
  • \$\begingroup\$ Which is steep, and changes with temperature and age, so a small change in voltage or conditions results in a big change in current. You could stay in a "safe zone" but then you're not getting anywhere near the light that you could. (Well efficiency will be good, there's that.) \$\endgroup\$ – Harper - Reinstate Monica Jan 23 '18 at 1:28

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