0
\$\begingroup\$

I have a 7020 white light LED strip, which includes a resistor on each module as you can see: enter image description here

I would like to power a little strip of five of these modules with a 9 V battery, but I wonder :

1- Do I need to put a resistor beetween the battery and the strip (so the leds do not catch fire or early die, or anything inappropriate)?

2- If I do, how many ohms do it needs to be?

EDIT :
The question is not about the intensity of the light that the LEDs can emit, nor about how long the battery will operate.

As the strip comes prewired and includes a resistor on each module, I suppose that this is not required, but I read everywhere that lighting a single LED requires a resistor upfront, and I can't find anything about how this applies or not to LED strips.

\$\endgroup\$
  • \$\begingroup\$ If it's the type with builtin resistors, then no, you don't. Also notice that it says 12V and they might not be very bright, or maybe won't work at all, with 9V. \$\endgroup\$ – user253751 Sep 6 '18 at 3:01
  • 2
    \$\begingroup\$ No, and in addition the LEDs will be very dim, if they light at all on 9V. \$\endgroup\$ – Jack Creasey Sep 6 '18 at 3:02
  • 2
    \$\begingroup\$ I have no idea why you want to use a low capacity 9V battery with a 12V LEDstrip, but it won't work very bright or very long You have 2 choices, short out 1 LED , short out the resistor. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 6 '18 at 3:16
  • \$\begingroup\$ @immibis,Jack Creasey,Tony EE rocketscientist With a 9V battery without added resistor, they light up, and they are really bright. Not as much than with a 12V power, but they are very bright. With a resistor, of course, they are less bright, reason why I ask if I need to put it. \$\endgroup\$ – Oliver Sep 7 '18 at 1:35
  • \$\begingroup\$ @Tony EE rocketscientist I want to do this because 9V battery light them up quite well, and it is small, light and autonomous (not like a huge 12V battery, or a wall plug). \$\endgroup\$ – Oliver Sep 7 '18 at 1:36
3
\$\begingroup\$

The typical 12V strip has a voltage range of 10 to 14 volts. Less than 10 and they barely light. The leds alone have a forward voltage of 3.6V at their typical nominal forward current, so you add in the voltage drop across the resistor, and 9V won't cut it.

You could short out the resistor as suggested and connect the 9V battery. It's internal equivalent series resistance and the lower voltage will light the leds at less than normal brightness. Unfortunately a 9V battery has low capacity and won't last long. Slightly better, you can short 1 led, and add a resistor in series. The resistor should be (Vs - Vf) / If = R1 + R2, where r1 is the existing resistor, vs is voltage source 9V, and Vf is 3.6 times 2 for 2 leds. If may be 0.06 A or 60mA for those leds. This would last a little longer.

If you want 5 of these in parallel, your 9V battery ain't going to last half an hour...

Instead, get 5V led strips, and a usb power bank. Rechargeable, no resistors needed. Just an old usb cable.

\$\endgroup\$
  • \$\begingroup\$ When you suggest to short one led, do you mean one on each of the 5 modules ? Also, on your formula, what is If ? Is [(9-(3.6*2)) / If] -39ohms is the good calculation ? \$\endgroup\$ – Oliver Sep 7 '18 at 1:54
  • \$\begingroup\$ Why are you making a MIckey Mouse flashlight? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 7 '18 at 4:00
  • \$\begingroup\$ @Oliver yes one per module. And If is the forward current at the forward voltage. I'm assuming it's going to be like 60 mA. \$\endgroup\$ – Passerby Sep 7 '18 at 4:20
  • \$\begingroup\$ Current white lighting LEDs typically have a Vf of 2.8V and exceed 3V only at currents approaching 1 Amp. 9V will cut it just fine. Less than 9V, which is where a 9V battery quickly goes, will not work well. Also the brightness will quickly diminish significantly. Neither LEDs nor batteries have an equivalent series resistance. Batteries have IR and LEDs have a Vf. The Vf is a function of the energy (number of electrons) in the band gap and the number of electrons converted to photons, not a resistance. LEDs need either a current source or a resistor to allow the LED to float to its Vf \$\endgroup\$ – Misunderstood Sep 7 '18 at 19:16
  • \$\begingroup\$ An LED can, but should never, be connected to a battery without a current limiting resistor. It is a myth that the battery's IR will do the current limiting. Look at the IR as a parallel resistance not series. The IR has no current limiting. What actually happens with a higher IR button cells is they have low current capabilities but the battery voltage forces the LED to operate at the battery voltage where good design practice dictates the LED to operate at its Vf. A resistor is needed to allow the LED to operate at its Vf rather than be stressed to operate at the battery voltage. \$\endgroup\$ – Misunderstood Sep 7 '18 at 19:32
1
\$\begingroup\$

The question is not about the intensity of the light that the LEDs can emit, nor about how long the battery will operate.
Just, do I need to put a resistance between the battery and the strip.

No, you do not need to add another resistor, this strip already has a 39Ω resistor.
enter image description here



With 3 LEDs and a 39Ω Resistor the LEDs will draw about 3-4 mA @ 9V. At this load a 9V alkaline will have a capacity of about 600 mAH as it discharges to 5V.

At 4 mA the forward voltage is likely to be below 9V. So it should be visible for a while. How long depends on how low the forward voltage can drop before it no longer emits light.

With 3 LEDs and 9V the resistor is going to be very efficient because the forward voltage will be very close to 9V. efficiency will improve as the battery voltage drops.

This strip was intended to draw about 60 mA @ 12V and would likely be fairly bright. At 3-4 mA not so bright but should be visible. I would guess brighter than moonlight in a small room.

You can always solder another resistor on top of the existing resistor to draw more current. Any value over 2Ω will work fine.
2Ω will give you about 50 mA.
5Ω will give you about 20 mA.
10Ω will give you about 10 mA.
The above numbers are for a fresh battery.

Do NOT short the resistor. A resistor is need to allow the LED to operate at its forward voltage. Without a resistor the LED is forced to operate at the battery voltage putting stress on the LED. It is a myth that it's okay to connect an LED(s) directly to a battery.

Below is 3 deep blue LEDs at 10 mA. Deep blue is what is under the white LED's yellow phosphor. Deep blue light is not very luminous. After the blue light is absorbed then reemitted by the phosphor it will appear to be about 10x brighter than this blue.

enter image description here



Keep in mind that a 9V battery is 9V for only a very short period of time. enter image description here
Source: Energizer 9V Alkaline Datasheet

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.