# Is the use a very small value resistor negligible?

I made a simple LED circuit for a hobby project. It uses a cellphone charger rated at 5V.
There are 2 LED's rated at 700mA at 2.4 V in series. The calculated resistance for my current limiting resistor is 0.5 Ohm. This is a very small value, but is this negligible ?

I did some research on this problem, but it retuned no results..

• How did you calculate the limiting resistor? How stable is your voltage source? Are you taking into account potential tolerance issues with the LED's forward voltage? – user2813274 Mar 5 '16 at 16:53
• And the Vf v If (voltage drop v current) curve of your LED would be useful to help you. – Steve G Mar 5 '16 at 16:55
• I used two of these LED in red. My source is a cellphone charger at 5v. The LED's are rated 700 mA at 2.3 V, so the 0.4 V goes on the resitor – Reyske Mar 5 '16 at 17:01
• You should instead put the two LEDs in parallel, each with its own proper dropping resistor. If you want to put them in series, get a higher voltage source. – DoxyLover Mar 5 '16 at 17:08

It means that you're using a voltage source which is too close to the forward-drop voltage of the LED. The problem is that the values are never exact. You may have a really fresh battery which gives out a bit more voltage or the actual forward-drop of that particular LED may be a bit lower. In this case, you will burn out the LED.

Conversely, the battery may be a bit low or the FD a bit high and the LED may not light.

You can't just use a 1.5 V battery to run a 1.45 V LED. The only case where you can really get away without a significant dropping resistor is where your voltage source has a high internal resistance which acts as a dropping resistor. This is used in the cheap keychain flashlights that connect an LED directly to a button battery. That battery has a high enough internal resistance that the LED never gets too much current.

• I forgot to mention I use a cellphone charger rated at 5V, so voltage drops would normally not be a problem. Thanks for the information though ! – Reyske Mar 5 '16 at 17:05
• There is still the issue of the LED voltage not being exactly correct. Also, the voltage will change with temperature. Still not a good idea. – DoxyLover Mar 5 '16 at 17:07
• Not a good idea to make a design that requires such a low resistor value. – The Photon Mar 5 '16 at 17:09
• @Reyske - That's a much better solution. Go back to your original setup, and assume (just for grins) that your charger is actually putting out 5.1 volts instead of 5.0, and see what current you get. – WhatRoughBeast Mar 5 '16 at 17:21
• @Reyske just in case it wasn't clear, use two resistors, one for each LED, not just one. If you put the LEDs directly in parallel, it probably won't work well. – DoxyLover Mar 5 '16 at 23:51

It looks OK, because the voltge dropping will be increasing while the LED's getting to be hot (and this not stabilized current will go lower). Don't forget about heat sinker! But in this case the accuracy of current value is not perfect, but the effeciency is good :) On the other hand U must be shure, that your phone charger can handle that current for a long time. And, of coure, if you will try to connect the LEDs in the parrallel circuits - the current will be inreased twice and efficiency dropps twice

• I will use an 2 Amps charger, so no problem there. I think the parallel circuit will be the best solution here. Thanks for the reply ! – Reyske Mar 5 '16 at 17:40
• In this case, don't forget about resisters in the way that they will be very hot (because the very high power will dissipate on them) - more than a half of all consume power or near – Pinus Mar 5 '16 at 17:43
• It will dissipate 1.6 W, so a 2 W resistor will suffice in this case – Reyske Mar 5 '16 at 17:48
• Yeah, 2 big resisters :) Have a good day! – Pinus Mar 5 '16 at 17:50