I have a 10 W 308-385 nm UV LED which is rated for 3.4 V to 3.8 V at 2800 mA. I want to power it using a Li-po/Li-ion battery so that would be 4.2 V at the peak when the battery is full.

Do I need to add protection here, and if so what kind of protection, just a simple cement resistor of a very low value or some electric circuit?

When the voltage is in the range of 3.4 V to 3.8 V (or lower,) do I still need to add something to limit the current then or will at that point the LED automatically do that?

Most people talk about using LEDs in cases where the supply voltage is a great deal above what the LED can handle, but in this case it isn't far above that.

I have seen commercial high power LED floodlights also directly connect to Li-pos without current limiting (8.4 V 2S Li-po, so it might be that they use 12 V power LEDs.)

I have looked into the LM338 but that one seems to only regulate voltage.

I have also thought about using two transistors and two resistors or such to make a current limiter.

About just directly using a 1.5 ohm cement/high power resistor.

I also considered the possibility that it is so close in range that it either might not be needed or that something else much simpler or better might be possible or that I might choose the wrong parts for it.

EDIT_ I did some testing and compared it to the data in the datasheets. turns out the datasheets do in some cases show a relative correlation between voltage and current, during my tests this also showed to be pretty much right.

For many LEDs, there is actually a pretty big range of voltage offset where it will limit the current by itself thus not needing extra hardware (as long as you are okay with the LED brightnes changing with voltage.)

When using it with a powersuply with a roughly stable voltage this will make it pretty much not needed to add extra components if you look good at the datasheet or test the voltage-current relation yourself for example using a component like a Xl4015 version with voltage and current meter (for rapid simple testing.)

In my case, the speciffic 10W UV LED had a range of around 1V which is pretty much for a sub 4V LED. Actually, the 2800mA was somewhere around 4V, but it worked just drawing less current and being less bright down to around 2.6V.

This means that yes, you can power an LED directly without a current limiter if you make sure it is in the right rated range or slightly below.

In my case, however, since I had set up a screenless XL4015 board already to limit to just below 2800ma I used that anyway so I could hook up a much broader voltage range.

As my original reason for asking this question was to see if it was safe or unsafe to use an LED without a current limiter, as in if a LED will actually also limit current based on voltage, then YES LEDs do have an analog property of limiting the current based on input voltage.

Just make sure to either test it or measure the current at certain voltages or look in the datasheet.

This also explains why many commercial LED lights actually do not use current limiters.

  • \$\begingroup\$ How do you want to supply a Led with 3.4-3.8V Vf when li-ion is drained to 3.2V? Find one with smaller Vf if possible and add a current limiter with 50mV shunt. Or use a buck-boost converter with current limiting feature, then the original Led is ok. \$\endgroup\$ Jul 19 at 16:41
  • \$\begingroup\$ You must limit the current at all times. If your battery voltage is below the max Vf of the LED, expect zero brightness unless you have a buck-boost converter. \$\endgroup\$
    – winny
    Jul 19 at 19:12
  • \$\begingroup\$ I do not mind the voltage of the battery eventually going under the 3.4 minimum, since I preffer lithium batteries to be used in the more safe and durable range of around 3.4v to 4v since it lets them last many more cycles. also it directly helps as a indication if it is almost empty, so the lesser or none brightness at the end isn't really a problem for me. \$\endgroup\$ Jul 20 at 10:35

5 Answers 5


Consider the following circuit, where I use a simple resistor R1 to limit current through the LEDs:


simulate this circuit – Schematic created using CircuitLab

The LED develops 3.6V, as expected, and the resulting current is determined by the remaining voltage across R1 (as measured by VM1), accoirding to Ohm's law:

$$ I = \frac{V_{R1}}{R_1} = \frac{4.2V - 3.6V}{0.214\Omega} = \frac{0.6V}{0.214\Omega} = 2.8A $$

Already you see a problem: the resistance R1 is very, very low. Even the wiring and battery resistance in that loop might be comparable to R1, and it will be very difficult for you produce such a precisely specified loop resistance.

Secondly, consider what happens as the battery discharges to, say, 4.0V. Now (assuming LED voltage hasn't changed) the voltage \$V_{R1}\$ across R1 will have reduced to only 0.4V, a 33% reduction which will invoke a 33% reduction in current. Also, LED voltage changes as it warms up, having a similar effect on \$V_{R1}\$ and its corresponding current. That's going to produce a noticeable change in brightness.

If battery and LED voltage were less similar, this percentage change in voltage across R1 (and current through it) as the battery discharges, would be less significant, and R1 could be much larger in value and easier to acquire.

For these reasons, R1 is a terrible way to regulate current in this loop, where battery and LED voltage are so similar. If you used a DC-DC boost converter to obtain a greater source of voltage, say 8V, your circuit might look like this:


simulate this circuit

Now R1 is much more reasonable and we still have 2.8A through the LED. The voltage drop across R1 is 4.4V, meaning that a change of battery or LED voltage of 0.2V will produce a much smaller change in current, so you might think that the problem is solved.

Power dissipated in R1 is:

$$ P = IV = 2.8A \times 4.4V = 12W $$

That's huge, and will require an enormous heat sink and/or forced air cooling to keep it alive.

High current and similar supply/LED voltage mean that a passive resistive current limit is not feasible. You can't just connect the LED directly to the battery, as this will result in currents well beyond what the LED can tolerate.

You will require active current regulation of some kind. In this case, with such a small difference between battery and LED voltage, a linear current regulator will be difficult (but not impossible) to implement. Probably this situation calls for a switching solution, a current regulating DC-DC converter. This is usually how high power LED drivers do the job, by producing exactly the right voltage across the load that results in exactly the prescribed current through it.


The LED needs to have a current regulator. The challenge for your design is that the LED forward voltage is very close to the battery's voltage, so there isn't lot of overhead to play with. And, the current is kinda big, plus I assume your LED is expensive.

This is a common issue as it turns out. This answer has some ideas: Help please! I need to give constant current to an LED from a Lipo battery


So I looked into this and came up with a solution that uses two op-amps and an n-FET (simulate it here)

enter image description here

The circuit senses the LED current. As the battery voltage approaches the LED forward voltage, the LED current will rise rapidly until it reaches 2.8A, where it will stay due to negative feedback throttling the FET. Below that limit threshold the FET is full on.

The sense resistor is 10 milliohm, chosen to limit losses. The first sense op-amp has a gain of 43.2, so at 2.8A it makes an output of 1.2V. This is compared in the second op-amp against a 1.2V reference to make the FET drive signal.

This would work with an LMV324 type op-amp, which has an input range of GND to Vcc - 1V. The 1.2V reference could be a dedicated IC or constructed from discrete components.

That said, you may be sacrificing battery life because as the battery drains its voltage will fall below a useful LED Vf. To overcome this, a buck-boost converter could be used instead of the FET, with a similar feedback scheme. Here's a TI appnote for that.


What kind of protection should be used for an LED if input voltage is almost equal to the rated voltage?

LEDs have a rated current and, when that current passes there will be an associated terminal voltage. However, it would be wrong to call it a rated voltage and, in most situations, it would be wrong to regulate that voltage across a diode because, you won't know if the current exceeds its rated or limit value.

when the voltage is in the range of 3.4 V to 3.8 V (or lower), do I still need to add something to limit the current then or will at that point the LED automatically do that?

You regulate the current and then you don't have to worry about the voltage.

Most people talk about using LEDs in cases where the supply voltage is a great deal above what the LED can handle, but in this case it isn't far above that.

I think you may be misinterpreting something here or not understanding the bigger picture.


You need supply voltage high enough to power the LED. If not, no [linear] regulator will be able to maintain regulation, and LED output will be less than expected.

Note that LM338 requires almost 4V above the LED voltage to guarantee regulation; it would be a very wasteful choice, getting less than 50% efficiency.

Since the battery minimum voltage is less than the LED maximum voltage, a converter is a more likely solution. That is, a circuit which can boost the battery voltage as needed; but it may also need to drop voltage. The most common example is a SEPIC converter. Which can be designed to also limit current into the LED (so that its exact voltage rating is unimportant, while operating at rated intensity).

(If you just need one, beware that this site is centered more on design, and we can't help you select a commercial product.)


You should use an LED driver to regulate the LED current and keep it from exploding. You could also use a resistor but as the battery discharges the LED will rapidly become much dimmer.

  • \$\begingroup\$ With 3.2 V \$\lt U_{batt} \lt\$ 4.2 V and 3.4 V \$\lt U_F \lt\$ 3.7 V, a resistor won't do to usefully define an operation point. \$\endgroup\$
    – greybeard
    Jul 19 at 20:11
  • 1
    \$\begingroup\$ @greybeard You won't be able to calculate the current from that spec, but you can still use a resistor to set current by starting with a larger resistor and then gradually decreasing until finding the set point. Not elegant but the specs on these "10W" ebay specials are often made up anyway so trusting them isn't the best idea. \$\endgroup\$ Jul 19 at 21:49

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