2

I have some led strings of the type shown:

led string

...which run off two C2032 batteries, so 6V.

I am cutting off the battery cases and connecting them to a 6V supply which is a step down converter on a 12V battery.

However, the led's are overheating and burning out.

Isn't one 6V DC supply the same as the next? How would they draw more current, just because it's from a larger battery?? Could it be something basic?

Update:

I looked up the nominal resistance of new CR2032's and introduced a resistor of equivalent size. This worked a treat. So a) I learned something, and b) it saved my project, which was part of a public performance which would have failed if my led's failed. So thanks all for sharing the knowledge.

SamGibson
  • 17,231
  • 5
  • 37
  • 58
rsk
  • 21
  • 1
  • 3
    Yes, it's basic. The CR2032 batteries have a high internal resistance which limits the current. The step down converter does not. – Dampmaskin Dec 01 '17 at 13:43
  • Great, thank you. So I would need to introduce a resistor equal to the resistance of the two cr2032's? Would that be enough or would I need to make the whole new circuit have the same resistance as the original circuit? – rsk Dec 01 '17 at 13:55
  • 1
    adding a resistor should do the trick. – Makoto Dec 01 '17 at 13:58
  • 1
    Yes, a series resistor would be sufficient. Too small and the LEDs will burn out. Too large and the LEDs will be too dim. To calculate the right resistance we need to know three things: The voltage drop of one LED, the number of LEDs in series, and how much current the LEDs can withstand. If you can't get this information, you'll have to experiment until you have a satisfying solution. – Dampmaskin Dec 01 '17 at 13:58
  • @Bort: That would work fine until the forward voltage of the LEDs change due to changing temperature or other factors. In other words, not long. – Dampmaskin Dec 01 '17 at 13:59
  • 1
    @Dampmaskin he could just use the nominal internal resistance of a CR2032 as a guide. – Makoto Dec 01 '17 at 14:00
  • 1
    *"Isn't one 6V DC supply the same as the next?"* No. Series resistance matters. Look up *Thevenin equivalent*. – Olin Lathrop Dec 01 '17 at 15:09
  • You could power this from 3 to 5V without issue. The resistor would change value accordingly. – Passerby Dec 07 '17 at 07:50

1 Answers1

4

enter image description here

Figure 1. The simple key-fob LED power circuit works on the principle of current limiting by battery internal resistance.

As mentioned in the comments, those little cells have some internal resistance which acts as the current limiter. You can confirm this by measuring the open-circuit voltage and then measure again on-load. You will see a voltage drop.

If you can measure the current as well then you can calculate the internal source resistance value from \$ R_S = \frac {V_O - V_L}{I} \$.

Your circuit seems to be using 2 x 3 V which is on the high side for a single LED but it isn't possible to figure out the wiring from your photo. The same principle of battery resistance current limiting is applicable.

Reference: Battery plus LED without resistor.
Disclaimer.

Transistor
  • 168,990
  • 12
  • 186
  • 385