I have one OSB4XNE3E1E IR LED datasheet

I use a battery type Li-ion 18650 and it's 1200 mAh - 3.7 V. The maximum voltage of the battery is 4.2 V maybe "when full charged" and when I made the circuit it was not fully charged so it read 3.99 V.

The datasheet above says the maximum voltage of the LED is 2 V at 700 mA and the typical is 1.7 at 700 mA.


Now first: every time I measure the ampere of the circuit the ampere is always under 100 mA.
Before I said the second point I can't get the current of the LED that he can take it because on the datasheet its written 700 mA and when I link it its say something under 100 mA so I was confused and all I did is get some resistors and put it from the highest to the lowest to get the resistance I need.
Second. I put a resistance about 1 kΩ and it glows nice but the voltage is under the typical voltage of the LED:
So I lower the resistance and its glow more and the voltage go more BUT! The heat is going up in the resistor. Finally I get 10 Ω resistance and the voltage on the LED read 1.5 maybe or 1.6 but the 10 Ω the resistor gets too hot and its becoming hot very fast.. any way
I link the LED directly to the battery to figure out the problem and yeah the LED is dead.
After its dead it's now reading the voltage of the battery itself which is 3.99 (with very very very very ultra low glowing)

I don't understand even with low resistance the voltage of the LED that its taken from the battery is not able to go higher than 1.6 or 1.5 (I need it 1.7 - or 1.8 - 1.9)

And I have a second question which is the company of the LED how can they use 700 mA with 1.7 V?
I'm talking about the ampere because every time I measure it its read under 100 mA so how?

And also I have a third question which is linked to the second one
Does the ampere of the battery that the battery provide it is changeable proportional to the load?


8 Answers 8


The infrared LED died from too much current. Excess current overheated its internal structure. Be aware of a caution on the data sheet:

Note: Don’t drive at rated current more than 5s without heat sink for Xeon 3 emitter series

I get the impression that too much emphasis is put on LED voltage as if its voltage of 1.7V determines light intensity. While it is true that too little voltage will result in no light, too much voltage causes excess current (and heat). It is LED current that generates heat and light intensity. Since these LEDs are often run hot, temperature affects its voltage drop too. Use 1.7V only as a rough guide to determine LED operating point.

When you use a multimeter as an Ammeter to measure LED current, you encounter a problem with its internal resistance, especially in low-voltage circuits like this. Let us assume that when measuring on a 200 mA scale, Ammeter internal resistance is \$1\Omega\$. If you use this Ammeter to measure LED current, a small voltage (a little less than 0.2V) drops across its terminals due to its internal resistance (circuit on left).
Then you remove the Ammeter and replace it with a jumper wire having close-to-zero ohm resistance (circuit on right). Current flow jumps higher than you expect:


simulate this circuit – Schematic created using CircuitLab

The QED123 infra-red LED is similar to OP's, one of few that CircuitLab simulates. When the current-limiting resistor (10 ohms in this example) is made smaller in an effort to get LED current larger, this Ammeter loading problem becomes worse.
It is probably better to monitor voltage across this current-limiting resistor, and then use Ohm's law to calculate current: \$I={V\over R} \$. It helps to use a current-limiting resistor having tight tolerance...using an ohmmeter to measure such a small resistor is usually not accurate, since the contact resistance plus ohmmeter wire resistance is significant.

  • 3
    \$\begingroup\$ So basically "Ammeters have 0 resistance" is not true in the real world, and on low voltage/resistance circuits this can have a significant impact on how your circuit functions. Which is why it is better to use a Voltmeter because even cheap ones have really high internal resistance (1 M Ohm on cheap voltmeters that I have used), so real world voltmeters come closer to an ideal infinite resistance voltmeter. \$\endgroup\$
    – Questor
    Jun 26, 2023 at 17:38
  • 3
    \$\begingroup\$ @Questor Yes! True zero-ohm ammeters are very rare (high-end) instruments. You might expect normal multimeters switch to more-ideal (close to zero ohms) on the higher amp scales, but this is not always true. Their 10A scale often suffers more from its probe resistance and its jack contact resistances. \$\endgroup\$
    – glen_geek
    Jun 26, 2023 at 17:54
  • 2
    \$\begingroup\$ yeah high precision ammeters are expensive. and a voltmeter + $I=V/R$ works pretty well. \$\endgroup\$
    – Questor
    Jun 26, 2023 at 18:04

LED is dead because you connected about 1.6V 700mA LED directly to a 3.99V battery. There was nothing except resistance of the wires and other conductors to limit the current so likely multiple amps flowed and LED bond wires acted as fuses and melted.

The 10 ohm resistor got hot because you had 3.99V going in and about 1.6V going out to the LED. About 239mA current passed and about 2.39V was left over the resistor, and it needed to dissipate about 0.57W of power.

The LED just has a typical voltage of 1.7V at 700mA. It could be higher or lower. The point is don't worry about what it is, but do worry about not exceeding 700mA.

  • 1
    \$\begingroup\$ Its just why? Every time i need the voltage going up to the typical wich is 1.7 and. Its didn't go its stuck on 1.6 why? \$\endgroup\$
    – Pcp115
    Jun 25, 2023 at 13:31
  • 4
    \$\begingroup\$ What why you mean? It might be 2V at 700mA. It might be 1.5V at 700mA. It depends on your LED. The datasheet does not say that your single example or all millions of LEDs will have voltage of exactly 1.7V. So, your LED may blow up at 1.7V. It certainly did at 3.99V. \$\endgroup\$
    – Justme
    Jun 25, 2023 at 13:37
  • 1
    \$\begingroup\$ Maybe i dont understand these things well... All i know is the datasheet say it work on 1.7v and 2v with 700ma can you explain more... \$\endgroup\$
    – Pcp115
    Jun 25, 2023 at 13:40
  • 15
    \$\begingroup\$ No the data sheet does not say that. It says when you forcibly put exactly 700mA through the LED, the voltage over the LED is typically 1.7V, but it might be up to 2.0V, but it might also be lower than 1.7V but there is no guarantee what the value will be. If your LED has 1.6V at 700mA, then it will surely blow up due to overcurrent if you try to increase current until you see 1.7V. \$\endgroup\$
    – Justme
    Jun 25, 2023 at 13:42
  • 8
    \$\begingroup\$ You do know ohm's law? Even if you short out the 4V battery with 10 ohm resistor, you get 400mA. With about 1.7V the LED takes away, that's 240mA. You can't have a 10 ohm resistor and expect to get more than 240mA. You need below 4 ohms to go near 0.7A. \$\endgroup\$
    – Justme
    Jun 25, 2023 at 13:53

Your LED is dead because you misread the spec:

enter image description here

This means the DC forward voltage at 700mA is:

  • Maximum 2V

This information is important to design the driver. Say you use a constant current buck LED driver, your battery has a minimum voltage of 3V, so the driver has to still be able to work with 1V difference between battery and LED voltage. Suppose this buck driver needs 0.1V on the current sense resistor, and it's not synchronous, so you'll get 0.5V on the Schottky diode, this leaves 0.4V to work with. Then you can check the voltage drop due to the ESR of your inductor and other factors, like the maximum duty cycle of the buck chip, and you can be sure there will be enough voltage for it to work down to 3V on the battery. It's the same if you use a linear current source driver, you still need to make sure its dropout voltage isn't too high.

  • Typical 1.7V

You'd use this information to optimize your driver to have best efficiency for this voltage, for example.

  • No minimum is specified

This means, at 700mA and 25°C, most LEDs will have Vf around "typical" 1.7V, some will be higher (but not more than 2V) and some will be lower (down to an unspecified minimum).

So if your LED had a Vf of 1.5V at 700mA, it was in spec. A low Vf simply means the LED is more efficient, it needs less power to operate at the same light output, so it's a good thing.

Note Vf has a strong dependence on temperature (like all diodes) and it decreases as the LED gets hot. So even if it was 1.7V, once it's hot it'll be much lower.

If you attempt to put a constant voltage on it, as it gets hot, its Vf will decrease, so it will draw more current, so it will get hotter, etc. Driving LEDs with constant voltage can end in thermal runaway and a burned LED.

That's not what happened, in this case it was extreme overcurrent (at 4V) so the chip probably got vaporized instead of slowly cooking.


If you connect an LED directly across a voltage source higher than the LED forward voltage a large current will flow, typically far larger than the LED can survive.

For example, with 10 ohms and 1.6v across the LED, the current was already (4-1.6v)/10 ohm = 240 mA. Raising it much further quickly puts you over the damage limit for the diode.

  • 1
    \$\begingroup\$ But can you answer the second quastion about the ampere (and thanks for the method about calculating the current by the resistance and voltage) \$\endgroup\$
    – Pcp115
    Jun 25, 2023 at 13:18
  • 2
    \$\begingroup\$ @Pcp115 I am not sure I understand what your second question was asking. \$\endgroup\$ Jun 25, 2023 at 13:24
  • \$\begingroup\$ Ok forget it. Can you explain this instead some answer it but i didn't understand it will... Why increase the current further maybe will damage the diode? Its says it can Handle 700ma... \$\endgroup\$
    – Pcp115
    Jun 28, 2023 at 13:28
  • 1
    \$\begingroup\$ @Pcp115 Use a 3.3 ohm resistor and you'll get ~720mA. \$\endgroup\$ Jun 28, 2023 at 13:36
  • \$\begingroup\$ 4-1.7=2.3v and than /A. So that its 2.3/0.7 =3.28571429 ohm im i wrong or something like that? \$\endgroup\$
    – Pcp115
    Jun 29, 2023 at 23:55

Let's say the battery is 4.0V, the LED drops 2V and the current through the LED is 700mA.

You have this:


simulate this circuit – Schematic created using CircuitLab

Given everything you've said about the LED and battery, these are the conditions that must be present. For the current to be 700mA, and the voltage across the LED to be 2V, then naturally the remaining 2V must be across the resistor. You could have predicted the necessary resistance R1, using Ohm's law:

$$ R_1 = \frac{V_{BAT1}-V_{D1}}{I} = \frac{2V}{0.7A} = 2.85\Omega $$

There is no other way for this equilibrium to occur, R1 must be 2.85Ω.

Don't forget, that resistance must include any additional resistance around the loop, such as wire resistance, contact resistance introduced by crocodile clips, or plugs, or breadboard connections.

Look at the power being dissipated in R1:

$$ P_{R1} = {{I_{R1}}^2R_1} = (700mA)^2 \times 2.85\Omega = 1.4W $$

R1 is going to get hot, 1.4W is easily enough to burn when you touch it, unless you take measures to cool it somehow.

The LED is dead now because you placed 4V across an LED rated for 2V, with no resistor present to limit current as R1 does in the above system. Current skyrocketed well beyond 700mA (probably many amps), and melted the poor thing.

I suspect that your ammeter's resistance (the so-called "burden" resistance, a resistance that current flows through to develop a voltage, that a voltmeter can measure and convert to and display as current, using Ohm,'s law) is significant compared to the required total resistance of 2.85Ω. With the ammeter in place, along with all the other unintended connection resistances around the loop, total resistance rises to well above 2.85Ω, and that's why you couldn't reach the desired 700mA during current measurement.

I wanted to add that we almost never use resistors in such high-current LED applications. With such a low overhead of \$V_{BAT1}-V_{LED} = 2V\$, very small resistances are required, comparable with other unintentional resistances around the loop. It's difficult to obtain the exact resistance required when the value is so small. For this reason, we usually employ active current regulation, to obtain an exact current flow, which can be achieved with a transistor or DC-DC converter.

  • \$\begingroup\$ As the other answers explain, the LED is not at all "rated for 2V". \$\endgroup\$
    – Ben Voigt
    Jun 27, 2023 at 21:40
  • \$\begingroup\$ @BenVoigt, that's true, but as I explained here, that has almost nothing to do with why everything went wrong for OP. \$\endgroup\$ Jun 28, 2023 at 2:35
  • \$\begingroup\$ Literally, it doesn't. Conceptually, it has everything to do with it. OP's entire problem is designing for voltage control of a device which requires current control. glen's answer correctly diagnoses "too much emphasis is put on LED voltage", a mistake that your answer perpetuates (although it makes a bunch of other good points, such as parasitic resistances and the need to cool the series resistor). \$\endgroup\$
    – Ben Voigt
    Jun 28, 2023 at 15:03

With constant current passing through a LED, the voltage drop can vary a lot due to temperature and manufactoring. When LED is heated up, the voltage across it will drop. Therefore, you don't have to care about the voltage at all because few hundreds of mV is acceptable. Notice that the datasheet doesn't tell you the minimum voltage.

Also, if the LED's brightness is enough for your application, you don't need 700mA. When voltage or current is larger than some values, LED tends to dissipate more heat than light.

Talking about my experience. I have some small blue LEDs. They almost reach their peak brightness at about 3V 20mA. Surprisingly, when I accidentally connected one of them to 5V supply without any protection, it survived, passing through 50mA. The brightness didn't increase much, but it became very hot.

Sadly your LED is not as touch as mine. Yours blow up.


How are you checking that IR Led is glowing? By pointing a smartphone camera at it?

IR means Infrared, and infrared light is not visible to human eye.
But, typical digital cameras will react to it (you can try pointing a tv remote at smartphone camera and pressing some buttons - it will flash white on the screen)

If you can see the IR led glowing with bare eyes, you are overdriving it (and destroying).

For your test circuit you want a digital camera, to verify whether the diode is glowing, and measure the current across the diode.
Then its a matter of picking correct resistor, but at 4V battery, it needs to dissipate 700Ma@2V which is 2Watts of power - the same power the led uses.
You can put couple rectifier diodes in series to reduce the voltage, each of those eats around 0,5V so resistor has less to do.

  • \$\begingroup\$ Im using camera without ir filter so i can see it clearly. And can you explain more? Did you mean to get high power resistance or something like that? \$\endgroup\$
    – Pcp115
    Jun 28, 2023 at 12:45
  • \$\begingroup\$ @Pcp115 A rectifier diode (or several) can be used to achieve small voltage drop. This way the resistor will have less voltage thus less power to dissipate. A voltage regulator would be better ofc. \$\endgroup\$
    – Thomas
    Jun 29, 2023 at 7:23

First, let's tackle the primary issue of understanding how LEDs work, and then we can consider special circumstances of this particular LED.

An LED, or Light-Emitting Diode, is a diode. These are non-linear devices, which means that they do not follow Ohm's Law. If the current flowing through them doubles, the voltage drop will not double. Instead (to a first approximation), it will increase by a quite small, constant amount.

Correspondingly, forcing the LED to carry a significantly larger voltage drop than its typical "operating voltage" will mean that a lot of current flows through it, far more than it can handle.

Never connect an LED directly across the terminals of a (voltage) power source (such as a battery or electrical mains).

  • If the voltage is less than the typical operating range of the LED, it will simply do nothing.

  • If it's in range, then you will at best have a very unpredictable circuit for a short period, because the "expected" voltage across the LED will depend on the individual device, temperature, and possibly other factors. Even if you find the exact right voltage, if you leave this circuit on, the heat generated by the LED itself will lower its expected operating voltage, i.e. increase the current that flows through. This is a positive feedback loop that will burn out the LED - called thermal runaway.

  • If the voltage is more than what the LED wants (for example, a lithium-ion battery sourcing about 4V to an LED rated for typically 1.7 to a maximum of 2.0V), this is effectively a short circuit that will test the internal resistance of the power supply. The best case here is that the LED shorts out and becomes useless. (While the LED will itself also have some internal resistance, it will definitely not be enough to protect the device - if it were, it would never be able to turn on in properly designed circuits.) To my knowledge, it is unlikely to damage the power supply, start a fire etc. this way, but there is no reason to risk it.

When we put an appropriate resistance in series with the LED, these problems disappear. Since the resistor does follow Ohm's Law, the current across it will be stable if the voltage across it changes a little bit (whether because of a glitch in the power supply, or because the LED is heating up a little bit) or is a little different from what was planned (because the individual LED is on the high or low end of its specification). And since the resistor is in series with the LED, the current through it must be equal - so we know that the current across the LED will be stable, and in the appropriate range.

Because of this stability vs. the diode characteristics, we do not need an exact calculation in order to build such a circuit. Instead, we follow a simple process:

  1. Assume it's possible to choose an appropriate resistance.
  2. Given this assumption, we expect that the potential difference across the LED will be close to some target value (say, 1.7V).
  3. Therefore, we can subtract this from the potential difference provided by our voltage source (say, a lithium-ion battery operating at 4V) to see what the voltage drop will be across the resistor (somewhere in the neighbourhood of 2.3V). As long as this value is positive, the assumption in step 1 is validated.
  4. Since there is a specific current we want to flow through the circuit (say, 700mA), we can compute the resistance needed (about 3.3Ω).
  5. If it turns out that the actual voltage drop across the LED is slightly different, or the voltage source changes, or our 10% tolerance resistor is a bit off, all of these things could mean the actual current we get is somewhat different. But it won't be an order of magnitude off like it would be if we tried to control the voltage directly, so we won't break the LED nor fail to turn it on.

This is the part where I point out some more specific realities of the specific situation:

  1. As described, this is clearly not an ordinary consumer device. You generally need to be pretty careful when working with something that draws 700mA, for hopefully obvious reasons. Of course, the resistance of your body means that you are not realistically going to pull a 700mA electrical shock out of a 4V Li-ion battery - but it's worth keeping this kind of thing in mind, on principle. (It would definitely be a problem if you were using an industrial active current source, rather than a battery which is a voltage source.)

    To put this in perspective: an LED-based light bulb might draw on the order of 9W of power from a 120V (AC) mains source, i.e. 75mA (root-mean-square) of current will flow through the bulb (it seems there are leading-edge models that draw half this much or less to get the same brightness) And that is for multiple separate LEDs connected in parallel (perhaps 16, from some pictures I could find). That is to say, ordinary LEDs work in the range of a few mA, not hundreds of mA.

  2. A resistor with 2.3V across it and 700mA through it is therefore drawing about 1.6W of power. That might not sound like a lot if you're thinking of home appliances, but it's a ton for a breadboard-sized resistor. You need to be thinking about the power rating of the resistors you buy, and building in a factor of safety. If your 10Ω resistor was apparently struggling (getting unexpectedly hot), a 3.3Ω one of the same form factor would definitely have been a problem.

  3. As noted in other answers, multimeters are not perfect. Using a typical multimeter as an ammeter might involve the current flowing through a "test" load on the order of 1Ω. For normal circuits that you build on a breadboard, that's insignificant; but if you're building something that's supposed to have specifically a 3.3Ω resistor in it, clearly that will throw things off by a lot.

  4. Infrared light is not visible to the human eye, by definition. You should not expect to see this device glow with your own eyesight. It seems you understood this, but it's a good idea to be more careful and explicit when describing the problem.

  5. Electricity can be dangerous. This is not like computer programming. In programming, with a simple personal-scale test project (let's suppose this doesn't have to worry about anyone else's data and doesn't use an Internet connection), the worst thing that can realistically happen is that the program crashes and a bunch of data gets corrupted. In electrical engineering (let's suppose that you buy off-the-shelf components and use them with power sources you can easily find around the home), it is entirely possible to die.

    Therefore, if something in your circuit is acting unexpectedly (it's getting hot, sparks are flying...), do research first, before trying to fix the problem. Especially if you observe "I tried reducing a resistance value from 1kΩ to 10Ω and the resistor was heating up", the thought "maybe I should try reducing the value to zero (i.e. taking out the resistor completely)" is dangerous and makes no sense.

    Especially if you are still asking questions like "Does the ampere of the battery that the battery provide it is changeable proportional to the load?" (i.e., a complete lack of understanding of fundamentals), when you see "700mA" on a spec sheet, that is a good reason to put the part back on the shelf and study more. Next time, it might not just be the LED getting destroyed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.