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I need to put quite a few LEDs on a PCB, and I want to reduce the number of components.

The LED I want to use is NCD0603G1, which has a forward voltage of 2.6V to 3.6V (documented.) I have a 5V power supply.

I have a few questions regarding this:

  1. Since 2XLED voltage is ~5V, can I use it without a resistor (I tested this and LEDs did light up.)
  2. I checked thevforward voltage using the diode voltage function of a multimeter, that shows a voltage of 2.3V, why is it different from the datasheet? Is it common?.
  3. If putting it in series without resistor is not a good idea, can you please suggest a recommended method?

I am quite new to electronics, please let me know if anything is wrong in the question itself.

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    \$\begingroup\$ For a one-off circuit that you've checked it's probably fine. For a production run you could run into reliability problems. See if what I've written on LED binning helps. \$\endgroup\$
    – Transistor
    Commented May 16, 2020 at 14:55
  • \$\begingroup\$ @Transistor, thanks, the variation is quite higher than I expected, is there any parameters for LED(like the one for resistor) where we can find the tolerance? \$\endgroup\$
    – Jithin Jose
    Commented May 16, 2020 at 15:14
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    \$\begingroup\$ You can get SMD or through-hole LEDs with the resistor integrated. This is a much safer, more consistent option if you need to reduce components. \$\endgroup\$
    – user1850479
    Commented May 16, 2020 at 15:14
  • \$\begingroup\$ @user1850479, can you please give me one example, but resistance value will be power supply voltage right? \$\endgroup\$
    – Jithin Jose
    Commented May 16, 2020 at 15:16
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    \$\begingroup\$ @JithinJose, the linked article shows you how to read the variation from the datasheet. It is not given as a tolerance or percentage. \$\endgroup\$
    – Transistor
    Commented May 16, 2020 at 15:22

3 Answers 3

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Since 2XLED voltage is ~5v, can I use it without resistor(i tested this and LEDs did light up).

Many LEDs have a negative temperature coefficient of forward volt drop versus temperature and, if the device gets warmer, it may disproportionately take more current and get warmer and warmer until the LED fails. You must satisfy yourself that this does not happen and I can't find any info in the data sheet to state what the tempco is so it's a bit tricky to guess. I would use as much resistance as I could afford.

I checked forward voltage using diode voltage function in multimeter, that shows a voltage of 2.3 V, why is it different from the datasheet? Is it common?.

Look at the graph for typical forward voltage: -

enter image description here

At 10 mA the forward voltage (typical) is 3.0 volts but, your multimeter will probably use something like 1.0 mA as a diode test current so what will be the typical forward volt drop then? I estimate from the graph it will be 2.625 volts or about 88% of the volt drop at 10 mA.

Now, if the device you chose had a typical forward voltage of (say) 2.7 volts at 10 mA then it might only register as 2.36 volts on your meter.

So, the value you measure on your meter depends on several things.

If putting it in series without resistor is not a good idea, can you please suggest a recommended method?

Plenty of folk use constant current feed circuits.

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In addition to the answer by Andy Aka, I would say that if you want to make safe, you should put 3 or even 4 SMD leds in serie. Not only two. Because two is just theorically ok without any margin, for example in case of heat. This is not really "reducing the number of components".

As it's a 0604 SMD LED, I imagine you want to use it as an indicator. If so, you certainly have SMD resistors between 500 ohms and 22 Kohms elsewhere in your schematic. Don't you? Use the same resistor for the LED. For example 4.7K or 10K if you don't have 1K. If you are afraid that the light output will be too small, use green leds.

Like this you won't increase the number of different components.

(The absolute number of components would still be the same with 1 Led + 1 resistor and with 2 Leds.)

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more than 2 in series won't light up enough to see in this application.

The reason to use resistors is because a) the 5 V supply will vary over time and temperature; b) the LED forward voltage required to obtain a certain brightness will vary significantly over temperature (decreases by about 2 mV/degree C). Thus to have any control over the LED brightness, you use a resistor to 'swamp' the other variations.

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  • \$\begingroup\$ Can you please recommend a resistor value to put for 2LED in series? Mt calculation says 20ohms. \$\endgroup\$
    – Jithin Jose
    Commented May 17, 2020 at 8:18
  • \$\begingroup\$ a LED doesn't have a simple 'forward' voltage'. The voltage depends weakly on the current; the current controls the brightness. As your data sheet says, VF is over 2.6 V; therefore two in series won't work. \$\endgroup\$
    – jp314
    Commented May 18, 2020 at 4:50

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