Before the end of last year, dad asked me for some small red light for his cabin, to pretend there is an alarm activated. He wanted it battery powered, so it can glow when electricity is turned off. I had this piece of LED strip at my disposal. 5050 smd RGB LED I think.

Not programmable one, but with terminals 5V, G, R and B.

There are 3 resistors labeled as 101, 151 and 101.

The labels are wrong that B and G are switched and one day I accidentally plugged it to too much power and G and B stopped working, only the red one works.

I bought a 4.5V flat battery GP Greencell 312G 3R12 extra heavy duty.

The connection was made by simply wrapping wires around the battery terminals and crocodile clamps onto the LED strip. + on 5V and - onto R.

Now from what I managed to find, that the battery is supposed to have 2500 mAh capacity and that RGB LED consumes 20 mA per color on full brightness, so I was expecting it to glow for about 5 days? 2500 mAh / 20 mA = 125 hours, correct?

So how come it is still glowing when we connected it before Christmas? That's 3 months ago. It is not as bright, but shines plenty enough during night.

Is it because it is connected to 4.5 V so the current consumption is lower? Is it because something went wrong the right way when I burnt the other colors by accident? Or is it the poor connection that would basically make it blink so fast you don't see it, but enough to lower power consumption? When I fiddle with the wires it does dim considerably.

So uh... what is going on? In photo the shine doesn't look like much, but that is the poor phone camera.


  • 4
    \$\begingroup\$ The most likely explanation is that the LED is drawing a lot less than 20 ma especially since you say it;s brightness has decreased. Can you measure the battery voltage and the LED current? \$\endgroup\$
    – Barry
    Commented Mar 26, 2022 at 12:49
  • 1
    \$\begingroup\$ When battery depletes it drops in voltage so current becomes less, so if it did consume 20mA with a brand new battery, it now consumes significantly less, and it won't need 20mA to be bright enough to act as an fake alarm LED. Battery is considered empty by the manufacturer at 2.7V, but a red LED only needs about 1.6V to operate. So the battery won't suddenly stop providing power to LED. \$\endgroup\$
    – Justme
    Commented Mar 26, 2022 at 12:51
  • \$\begingroup\$ The LED strop is probably designed for a fairly low current when powered from 12V so at 4.5V and less, its current is very low. Maybe the red LED is actually two LEDs in series reducing the current even more? \$\endgroup\$
    – Audioguru
    Commented Mar 26, 2022 at 14:03
  • \$\begingroup\$ @Barry Sadly now for at least week. I don't have multimeter at home and would need to get one from work, which is different town, the LED rests in my dad's kitchen. Plus I am a bit afraid to tamper with the wires more :D and to measure current I would have to disconnect it and place the measure tool in series, right? \$\endgroup\$
    – Noir_CZ
    Commented Mar 26, 2022 at 15:44
  • \$\begingroup\$ @Justme That would make sense. I kinda assumed that the 2500mAh is what I can get out of it and divided with the 20mA I found randomly on internet that these kind of LEDs have. \$\endgroup\$
    – Noir_CZ
    Commented Mar 26, 2022 at 15:44

2 Answers 2


Because LEDs are highly non-linear.

They have a very steep voltage-current curve. A little difference in voltage causes a steep difference in current. Here's an example for a different-color LED with a different voltage (it's an illustration I happen to have lying around). The published curve is this: (pick your voltage across the bottom, then go up to find the current).

enter image description here

Looks flat enough, but look at the actual scale. The vertical scale is linear but the horizontal scale is highly compressed. So let's actually do a little Photoshop so that the bottom axis also starts at zero.

enter image description here

Wow! That's a really steep voltage/current curve! Not linear at all. A little change in voltage causes a big change in current. All LEDs are like that.

You'd have to look at the data sheet for your LED. (if there is any; lol Chinese data sheets). However, I can tell you that while blue LEDs are around 3.0 volts, red LEDs are around 2.0 volts - so most likely your red emitter there is actually two LEDs in series, plus that resistor (which is there to flatten that curve somewhat).

So by dropping the nominal voltage from 5 to 4.5, you're causing a dramatic decrease in current through the LED, and thus a reduction in total power.

But LEDs are more efficient at lower total power.

There's no substitute for just measuring it with a voltmeter and ammeter, but I bet you are nowhere near 20mA.

Also, batteries do not actually have a fixed capacity. The higher the draw the lower the total capacity. Battery capacity is specced at a particular current draw such as C/20. If your current draw is lower, your functional battery capacity will be higher.


For the red, at 5V and a 151 code or 150 ohm resistor, let's assume a 2V forward voltage drop, you get a 20 mA draw.

But adjust to 4.5V input voltage and not adjusting the forward voltage drop you get 16.6 mA. But the forward voltage drop will drop so it will be less, like 14 or 15 mA.

And as the battery voltage drops, the current through this circuit drops too. Eventually the straight rules for battery life estimation just keeps curving because of this. Leds will stay visibly lit until fractions of a mA. You could still see a pin prick of light at 0.1mA.

Unless you measure the voltage under load you can't really see where on the battery capacity vs voltage curve or led Forward Current vs Forward Voltage curve you are at.


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