Change LED resistor for higher supply voltage ?

[scottb721]

Bought a couple of 'model spotlights' to light my lego Saturn V.
Internally they're a 5mm white LED with a 51ohm resistor. They come setup to run off three 1.5v batteries but I'd like to use a 5v PSU.
Should I change the resistor for the .5v increase ?

Thanks

[Mazo]

If you would like them to have a longer life/you want to dim them yes sure add a resistor.If you don't care about them that much/will use them only short periods of time,I say don't bother.

[kjr18]

Well, even with 4.5V it's a bit high. Too keep around 20mA @5V for one white led, you would need about 100ohm resistor to keep them for long time. Anyway there is no much more brightness difference between 15 and 20mA.

[Siwastaja]

Some calculations:

Vin = 4.5V (actually a bit more with fresh set of batteries, but they drop quickly, and voltage sags under a load a bit, so let's just say 4.5V)

We don't know the Vf of the white LED. It varies from about 3.0V to nearly 4.0V, at the rated current, depending on the LED.

Assumption 1: Vf_led = 3.0V -> I_led = (4.5V-3.0V)/51ohms = 29mA
Assumption 2: Vf_led = 3.8V -> I_led = (4.5V-3.8V)/51ohms = 14mA

So they originally run the LED possibly at near 30mA, but it could be less. One way to know is to measure the current.

Then, what would be the maximum rated current for the LED? We don't know, so we would need to guess. Typically, cheap basic 5mm white LEDs are rated (absolute maximum continuous rating) between 30mA and 50mA, and recommended maximum varies between about 20-40mA.

So it's possible that they are already running the LED close to limits, but not necessarily so.

Now, assuming the "5V supply" is a regulated USB supply, they typically output around 5.1-5.2V, so let's calculate the current again:

Assumption 1: Vf_led = 3.0V -> I_led = (5.2V-3.0V)/51ohms = 43mA
Assumption 2: Vf_led = 3.8V -> I_led = (5.2V-3.8V)/51ohms = 27mA

This is highly likely to go too close to the maximum limits for comfort. It's probably going to be OK, especially in low-duty use, but I would change the resistor, or use an additional external resistor.

If you actually measure the current, you are able to adjust the resistor so that you measure the same current using the 5V supply.

[Ian.M]

The batteries' internal resistances are probably a significant contribution to the total resistance in the circuit,   Start by putting in  fresh batteries unless the originals have seen very little use, and measure the voltage across the  LED (Vf)  and across the resistor.  Then you can calculate the current (If) they run at, and, if its under 30mA, have the data to recalculate the  minimum resistor for a 5V supply.   If  its over 30mA, they are probably over-driving the  LED (a guess as its very difficult if not impossible to find the specs of an unmarked LED), and assuming Vf isn't going to drop significantly, recalculate for 5V and lets say 25mA If.

[scottb721]

Thanks all.
I did do some measurements last night and played around with some calculators and thought the standard current flow was a little high so the 2nd opinion here has given me some comfort.

Edit- I think I was seeing 2.8v across the LED and current when running off 5v was a touch over 40mA. I might try doubling R.

[Siwastaja]

The batteries' internal resistances are probably a significant contribution to the total resistance in the circuit

Only if it's using small alkaline coin cells. For example, this particular LR44 https://www.celltech.fi/fileadmin/user_upload/Celltech/Prod.sheets/11186.pdf is specified for 20 ohm max, so guestimating 10 ohms typical, and three in series for 30 ohms, you are right.

Larger cells, such as AA cells, on the other hand, are roughly around 0.5ohm when fresh and at room temperature.

[scottb721]

It takes three LR41s.

[Siwastaja]

It takes three LR41s.

That's even smaller than LR44, so expect over 50 ohms extra resistance just from the three cells. With a power supply, you'd need to add that much resistance to keep the current same.