# using led stripes with external battery: any side effects?

i have to power some led stripe from ikea with a drill battery:

On led stripe power is written: 12V 0,25A. and on drill power: 12V 2A.

Now i want to power n led strip (for example: n=4) and i want to understand if there are some side effects or some safety way to do it: do i have to add some resistor? and how to have the best result (=more light) for example mayebe if i use more then 10 stripes the light is reduced or whatever..

and if i want to add more kind of light? (ex: 4V lamp like this black one below)

i had some trials and it seems that there are no problem with 4 stripes in parallel, and i think that there is no need for resistors because both the parts works with 12 volts, but my theory knowledge is very poor (i have only some trial and error experience) and i have no idea with 4V lamps (i think that here i need some resistor) or if i change the power and if i use a different power supply (or even some normal battery like 2 or 3 9V battery in series)

mayebe here there are someone that can explain me how things works and how to design the circuit when i have decided how many light i want.

Yes, you can use you 12V battery to power your led stripes.

The Ikea stripes are internally connected that way :

12v
---------------------------------------------------------  -  -
|              |               |
---            ---             ---
| |            | |             | |
|R|            |R|             |R|
| |            | |             | |
---            ---             ---
|              |               |
|              |               |
|              |               |
---            ---             ---
\ /            \ /             \ /
---            ---             ---
GND        |              |               |
--------------------------------------------------------  -  -


This means that the same voltage ( 12V ) is applied to every resistor/led branches. If you connect other led stripes the way Ikea recommends to. It's ok. You will still have 12V applied to every resistor/led branches.

The problem will be the current drawn by the system. If I assume that Ikea power adapter (0.25A) is design to power up to 4 stripes. Then, every stripe consumes : $$I_{stripe} = 0.25A/4 = 0.0625A = 62.5mA$$

Thus your battery, which is able to deliver 2A can power : $$N_{stripe} = 2A/62.5mA = 32 units_{max}$$ 32 units max.

But what you really cares is for how long the battery will last. You have a 2000mAh battery. Thus, roughly, the time required to deplete the battery will be:

$$t_{battery} = 2000mAh / (Nb_{stripe} \centerdot 62.5mA)$$

This is a rough approximation, because as the battery voltage drops, the drawn current drops also and the battery will last longer. But this is balanced a little bit by the fact that I have not taken into account that, after some time, the voltage will not be sufficient to light up the lights.

Another thing to notice : Here you don't have any voltage regulator if you use a battery alone. This means that the 12V at the battery output will decrease as the battery get depleted. And the emitted light will reduce over time.

Concerning the 4V lamps, you can connect 3 of them in series. If I combine this with the stripes:

12v
---------------------------------------------------------  -  -
|              |               |           |
---            ---             ---         ---
| |            | |             | |         |X| )4V
|R|            |R|             |R|         ---
| |            | |             | |          |
---            ---             ---         ---
|              |               |          |X| )4V
|              |               |          ---
|              |               |           |
---            ---             ---         ---
\ /            \ /             \ /         |X| )4V
---            ---             ---         ---
GND        |              |               |           |
--------------------------------------------------------  -  -


You will have 4 V at each lamp.

• quite clear thanks! it is 2Ah! but what with 4V black lamp? – nkint Sep 24 '12 at 15:50

Your assumptions and the answer by mehdi on the LEDs are correct, but I should add that LEDs are current devices, not voltage devices. So they will go slightly dim as the voltage drops as the battery drains. A DCDC converter with current feedback would normally be used to control the brightness. But as the LEDs are designed to work from 12V this is not essential (it's just most effecient).

On the 4V lamp, you have more of a problem. Those sort of lamps often use a halogen bulb, these take a lot of current. A resistor to drop the voltage to that bulb would be physically massive and get very hot. E.g. if the bulb requires 0.5A @ 4V, this is 2W (0.5 * 4 = 2) But the resistor will have to drop 12V to 4V (12-4=8V) 4V@0.5A is 4W. The resistor will have to dissipate 4W of heat continuously (so you'd choose a much larger wattage part or it will break). It's value would be 8/0.5 = 16 Ohms.

If you are running from a battery, then more than 2/3 of your battery energy would go into wasted heat.

Similar to the LEDs, a much more effecient way of driving the bulb would be a DCDC converter. This would only dissipate a few 10ths of a watt and convert the 12V into 4V effeciently, so the bulb would last longer.

When you use your led strip in parallel, currents are added to each other. So, as you write in parallel manner 4*0,25A=2.0 A is needed and more kind of light could not be used. If you are interested to use another light, add 12v power in parallel with your first one, this power could increase current in circuit.