Some electronics help needed please

These 3xAA powered lights are usually direct driven, using a resistor to limit the current, alkaline cells also have very high internal resistance (like more than 1Ω) which added to the resistor limits further the current. NimH AA cells have much lower DCIR (~50mΩ) so even if the voltage at rest is lower the voltage under load is higher, more current, shooter runtime.

Anyhow what you would need to do is increase the resistor value since with the USB PSU you have a constant 5V. Measure the Vf of the LED with alkalines (if that’s a suitable brightness) and calculate the new resistor :

Rnew = (5-Vf)/I
With I = VRold/Rold

Check that the current is lower than the USB PSU rating, and power in the resistor lower than its rating too.

Thanks Yuval, but I wanted to get away from batteries entirely - that said I did fancy putting a dimmable maybe anduril driver in there so I could dim it to the level I like, and also have the flashing effect lightening etc, but couldn’t source a board anywhere. That would mean a lot more hassle, new switch etc
thefreeman soloution seems better for my needs, and I guessed the reisitor value would need changing, but without know how many amps it’s drawing with the nimhs, I can’t work it out lol. When my mate pops round I’ll get him to help me measure it as he’s more electronic minded than me. I have a multimeter, so it should be doable, even if we need to unsolder a lead to do so, which I think is what will need doing (in series)

Maybe a charger board like this

USB-C input and 4.2 volts output with UP to 1000 mA output. I have not tried using one to power a led light directly but have thought about it. I could connect up a trial with one to a driver and led but not until 24 to 36 hours from now.

No you can just measure the voltage across the resistor (VRold), and if you know the resistor value (code or measurement, though DMMs are bad at measuring low resistance values), you can calculate the current.
Series measurements with a DMM is not very accurate because it adds resistance to the circuit.

That looks good Don, I doubt it’s drawing much more than 1A as there is no real heatsinking inside barring a very thin aluminium led board (standard cheap one)
The board I have gives no data, so i can’t tell you anything about it (amps wise) other than it’s a power board…useless I know.
Still before deciding I have to find out for sure how many amps it is drawing, or I could end up doing it all then being disappointed lol!

Thevoltage across the resistor seems to be 0.72v
anyway, here is a pic of the resisitor if it helps. red yellow gold gold

Don’t desolder, turn on the light and measure the voltage across the resistor.
Then the current is V/2.4

0.72v if I’m doing that correct - does that / mean ‘x’? so 1.728 amps?
If so then the nimhs give 3.6v and they are supplying 1.728 amps - is that right?
sorry for sounding so dumb, but like I said I know nothing about electronics or it’s symbols lol

Or you could just add a diode between the usb board and the + terminal of the light to drop the 5V down to 4.5.

The photo is above, there is no board.
I’m thinking I’m taking on more than I can understand, my head is already hurting lol, I know for those of you who understand it , it looks simple but for me, well, I’m lost lol!
If my calculations are in fact correct, I could buy a 4.5v or 3.5v 2 amp power supply (plug in type, not usb) and wire it right in?
would this - power supply be what I need?

0.72/2.4 (the color code says 2.4Ω) = 0.3A, that’s your current.

Now measure the voltage across the LED (Vf), which should be arround 3V anyway, and you can calculate the new resistor :
Rnew = (5-Vf)/0.3
If Vf = 3V you would need 6.67Ω, 6.8Ω for an existing value.

The voltage across the led is 3.23v
so that being correct, I need to order a 0.68 ohm resistor, solder that in series off the ‘+’ of the usb power board and it should be good to go?
Finally, if that is all correct, is 0.68 ohm resistor what i need?
Thankyou for all your help by the way :+1:

Better get a 6.2 Ohm, your maths are not correct.

Or a 6.8 Ohm, 1 Watt power rating resistor to replace the 2.4 that is in the lamp. Unsolder the old one and put in this new valued resistor.

The resistors in your link are only 1/4 Watt (0.25 Watts), you need higher power rating.

I don’t know how you end up with 0.68Ω, (5-3.23)/0.3 = 5.9Ω

And with P = R x I2 , P = 5.9 x 0.32 = 0.53W, better use a 1W rated resistor.

Sorry for my ignorance guys - I have no idea what your symbols (I don’t understand any of them) or working out mean, or how they work. I was taking the 6.8 ohm bit from the answer abover when freeman said ‘Now measure the voltage across the LED (Vf), which should be arround 3V anyway, and you can calculate the new resistor :Rnew = (5-Vf)/0.3If Vf = 3V you would need 6.67Ω, 6.8Ω for an existing value.’

So, I need a 1w resistor with 5.9 or 6.2 ohm rating which is it please? or is the difference so negligable it doesn’t matter?
Better still if you could link one off ebay uk, that would be an immense help. I for the life of me can’t find one on there. I can find 1w ones, but no 5.9 ohm ones

edit -
I have found this 6.2 ohm 1 watt - is this the one I need to order? - link

Why is the USB not gonna work? Just place that resistor in series with the LEDs? Alkaline batteries are 1.6V at full charge, I think 0.2V more isn’t gonna burn the LEDs out.

Perhaps you can try running it on USB for a few seconds, if it is not abnormally bright then it should be fine.

Or perhaps, you can get a constant current driver and adjust the brightness to your preference.

https://m.aliexpress.com/item/32842135081.html

P = power (watts), R = resistance (ohms), I = intensity (current). How it works, well, it’s a math equation. Plug in the values and solve.

Yes good find, that would work.

Thankyou all very much for your help everyone, it’s ordered and I’ll show you a photo when it’s all wired in!

:+1: