hi guys, I have a simple question:
which type of LED(under same condition, ie same voltage, temperature etc), SMD or DIP, draws more current and consumes more power?
SMD or THD
Good question.
From what I see these days both are about the same.
BTW, you can get many SMD LEDs that are plenty bright at 1 or 2 mA.
Sorry package does not matter. The color etc is determined by the die, not the package. The same die is used in several different packages. Also many packages have more then one die, that now becomes a factor of each of the dies. Now if you place them at different temperatures the current will vary a small amount until failure.
You might as well ask which car consumes more petrol driving the same distance on the same day. ![]()
so you say that the current only depends on the color? so which color draws more current?
also, do you think for VCC=3v is 1K ohm resistor in series good enough?
not bad really!
the answer shows how much I distinguish between different cars, which in this case, are LEDs that I do not have enough info about them ![]()
Point is, it has almost nothing to do with the package.
Why on earth would it?
Clearly, there are many different LED chips and a particular form of package may be manufactured with various alternate chips.
Your question indicates you do not understand how LEDs work.
The LED does not as such determine how much current it draws. You design a circuit, usually using a series resistor, to supply a certain current to the LED. Different colour LEDs generally have a different voltage/ current response curve - they are not strictly a resistance nor are they a fixed voltage drop - but this also depends on the manufacture.

So having a general idea of the voltage drop of a particular LED - reading the datasheet is a good start - you calculate a series resistor to provide the desired current given the voltage across the resistor being the difference between the nominal voltage drop of that particular LED and the supply voltage.
And then you test it out and measure the current, taking into account the voltage drop your multimeter may introduce to measure that current.
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