# Total newbie question - resistor size for led

Discussion in 'General Electronics Chat' started by Traveller, Jan 19, 2013.

1. ### Traveller Thread Starter New Member

Jan 19, 2013
3
0
Totally new to electronics but my son wants to learn so we endeavoring to do so together. I hope someone can validate or correct my question regarding sizing or resistor for a single led, as well as the impact of changing the resistor.

Power supply = 5.15V / 2.5A
LED = 3.0Vf / 20mA

As I am trying to fully grasp Ohms law, I came up with the following:

V=IR
I=0.02A ( from the LED spec)
V=2.15V (5.15 from the supply - 3.0 drop from LED spec)

So R=0.02/2.15
R=107.5 Ohms

Common sizes of resistors near this seem to be 110 and 120 Ohm, and I'd probably go with the 120 to give a little safety margin given its an unregulated supply and may fluctuate.

Question 1 - is my thought process correct so far?

Now, I don't have any resistors of that size on hand but I do have a 210 Ohm on hand. I confused myself trying to figure out what this would do. If the resistance is now fixed, but I am not sure which would change - voltage or current. My gut says voltage would stay at 2.15 and it is current that would change. So that means the I would now solve for I:

I=V/R
I=2.15/210
I=10.24mA

Running almost half the current would result in a much dimmer led is my assumption and the voltage would not be changed at all due to the different resistor size.

Question 2 - do I have this part right?

Thanks for putting up with some very basic questions.

2. ### nerdegutta Moderator

Dec 15, 2009
2,565
805
It sounds right to me, after all, it is a current limiting resistor.

3. ### Traveller Thread Starter New Member

Jan 19, 2013
3
0
Nerdegutta - yeah, I guess I see the obviousness....

Thanks!

4. ### kubeek AAC Fanatic!

Sep 20, 2005
4,687
805
Yes you are right about the voltage remaining almost the same. But I have a thing or two about the brightness, LEDs are non-linear and the more you add current the less the brightness increases, so the brightness at 10mA would be more than half at 20mA I would guess.
Second, tody´s LEDs are much more efficient than they used to be, so I would say that the brightness will drop from "annoyingly bright" to "normal". I usually run indicating LEDs at 5mA, unless I am making something where the LED really needs to be seen, like a warning red light.

But anyway, electronics is a lot about experimenting, so feel free to try using the resistor in series or parallel combination, so you get 105, 210 and 420Ω and see what suits you best. The slight overcurrent with the 105Ω combination should not be problem, but I wouldn´t recommend running it like this for years especially with the unregulated supply.

5. ### Traveller Thread Starter New Member

Jan 19, 2013
3
0
Thank you kubeek - that makes sense.

6. ### djsfantasi AAC Fanatic!

Apr 11, 2010
2,900
876
or a combination of parallel-series, which could get you about 122 ohms.

7. ### Audioguru New Member

Dec 20, 2007
9,411
896
Your math is correct but the forward voltage of the LED is A RANGE of voltage with a minimum and a maximum, not a fixed number like an incandescent light bulb.

8. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
With how bright LEDs are currently, for 5V, I usually use a 220Ω or even a 330Ω (since I have a ton of 330 resistor networks in DIP Pakages).

That is a safe value for an indicator type LED, for an app where maximum brightness is needed, I use next R24 standard value above whatever the calculation shows.