Where is that current value coming from?

Per the datasheet...should be more like 7.5ma@3v...

eT

I'm pretty sure it was just a value used as an example (notice the "IF" that starts the sentence).

The figure that you are getting your 7.5 mA figure from is clearly generic, since it gives no indication of which voltage ranking is being used. Looking at the binning specs, the V0 part specifies that the voltage is between 3.0 V and 3.1 V at a current of 60 mA.

 

The datasheet for the LED is posted in the first post.
The absolute maximum allowed current listed in the datasheet is listed at 90mA.
The forward voltage, brightness and color are listed when the current is 60mA.

 

Dear Sir/Madam,

I want to have a very simple circuit, one 5050 3V SMD LED and one CR2032 battery cell, according to the datasheet

5050 LED http://ncled.koreasme.com/eng/download/solleds/5050-3Chip-specification_WHITE(6500K).pdf
Vf: 3.0 ~ 3.4V (0.1V Sorting)
If: 60mA

CR2032 https://html.alldatasheet.com/html-pdf/95177/SONY/CR2032/80/1/CR2032.html
Nominal Volage 3V

If connecting the LED and battery in series, an experienced engineer asks me to add 1 Ohm resistor between the LED and battery to protect the LED by limiting the current pass through the LED. According to the Ohm's Law, ( Vs - Vled ) / I = R, (3 - 3) / 0.06 = 0, what is the theory behind adding the 1 Ohm resistor?

Best regards,

Kelvin.

Hello there,

Normally you would treat the LED as another voltage source (sounds crazy right, but it's not).
That means if the LED characteristic voltage is 3v and you put a 3.1v battery in parallel with it, you get an infinite current which blows the LED out.

However, the LED does have some equivalent resistance in series with it, and so does the battery.
This means there is a delicate balance between the voltage across the LED and the battery voltage and the current though both.
This means a departure from the ideas in the first paragraph above because the two work together. When you start with the LED and put the real battery across it, the LED draws current but also the battery voltage falls. So if you put a 3v LED across a 3.1v battery the current draw causes the LED voltage to rise and the battery voltage to fall, thus reaching an equilibrium point where there is some current flow but certainly not infinite.

So now the question is just what is that current. If it is too high, it could blow the LED or limit it's life. However we have wiggle room here too, because we may not be using this light very often for very long because it would drain the battery too fast anyway when the battery is new. So we end ujp with another parameter to think about, and that is the time it runs for and the absolute voltage limit of the LED.

Amazingly, 20ma LEDs can be overdriven quite a bit, as long as they are not run for too long. You can to u to 50ma but you cant run it for a long time like that. Now also interesting is that if you do draw 50ma from the battery AT FIRST, later the battery voltage drops and the current becomes more "safe" for the LED.

So you see so far we are juggling three parameters, voltage, current, and run time.
Now the next parameter is battery capacity. if we have a low battery capacity then the current does not stay high for too long.

With all this then, how do we decide what is the best way to run the LED.
Well with many practical problems like this the answer is, with a measurement or two.
Starting with a new battery, run the LED and see what the initial current is. If you use a 0.1 ohm resistor in series you can use Ohm's Law to calculate the current, and 0.1 Ohms does not add too much to the series resistance. If you want to use the 1 ohm resistor that's up to you but it is probably not needed, and the measurement will tell you for sure if it is or not, or even if a higher value is needed.

SO that's really the long and the short of it. Measure the current and go from there, and if you want to you can measure the voltage of the battery too. If you want to report back here we can figure out what is best from there. Be prepared to make the current measurement fast in case the current is really high so you can disconnect the circuit quickly. Also have spare LED's on hand unless you want to go the safer route, which is start with a 10 ohm resistor and measure, then go down to maybe 2 ohms, then down to 1, then down to 0.1 and see what happens to the current for each resistor value.

Also keep in mind that the LED half life is HIGHLY dependent on the current. If the current is 20ma the spec might be 10000 hours, but if you go up to 40ma it could be 2000 hours.

 

This is not an ordinary 3mm or 5mm 20mA LED. Instead it is fairly large to be surface mounted and has its forward voltage,brightness and color rated at 60mA. Its maximum allowed continuous current is 90mA and peak pulses can be 240mA.
The datasheet mentions Korea so maybe is translated. The datasheet says that the allowed changes of forward voltage and brightness is at an ambient temperature of 25 degrees C and at 60mA for a duration of only 1000 hours (shouldn't it be for 50 thousand hours or more?).

I wonder why the datasheet says, "Do Not Duplicate"? Maybe it is out of date (10 years old) and the Korean company began 11 years ago.

 

You don't need a resistor.

If the LED draws 20mA @ 3V, for example, and your battery puts out exactly 3V then the LED will draw 20mA.

However, if I were at my computer I would show you graphically why this is a very bad idea. Moreover, as others have pointed out, a 1Ω resistor is a terrible idea.

To put it in words, you need a realistic current source.
An ideal current source would have a very large resistance and very large voltage. Now you have to extrapolate that backwards to a realistic solution. You need a voltage source that is much higher than the LED voltage and a series resistance that will give the proper LED voltage and current.

You can learn a lot from this simple exercise. In a couple of weeks I will be back at my home computer and then I will be able to show you why this is very educational.

 

No need for a resistor , but the output voltage will vary if from a cell .... and the parameters of the led will vary depending on temp (current) ... If you take this into account and drive at well below max current (50%) all will be fine , the eye perceives light logarithmically so will not notice the loss of brightness , also leds work more efficiently at below max rating and will last much longer.

 

Here is my blog as promised.

https://forum.allaboutcircuits.com/ubs/why-you-need-a-series-resistor-when-driving-an-led.1651/