I have seen many questions along these lines but I don't think any is an exact match (if that's incorrect, please just point me to the right match, that would be great and save everyone some time). Apologies in advance for missing or incomplete information, I'll be happy to update with anything that is needed for a complete answer.

I've bought automotive "side rear clearance" lights to use in a Halloween costume. According to the Amazon description (https://www.amazon.com/gp/product/B01MEG40NI/ref=oh_aui_detailpage_o02_s00?ie=UTF8&psc=1) they're "3-LED" but I was not able to find any other data sheet information. I know nothing beyond 20-years-ago high school physics about electricity.

I want to run two of these red 3-LED lights from battery power, for as long as possible on Halloween night, without catching my battery/battery pack on fire, or constantly changing batteries.

Answers to similar questions seem to presuppose that I know or can find out the amount of current that these lights "want" to draw. Do I in fact need to know that, and absent information from the manufacturer, how could I find it out or estimate it?

Without knowing the current these lights want, I want to figure out how to power these lights with alkaline batteries - I have 9V and AA handy. I'd like a run time of say 5+ hours.

I notice that I can't seem to power off of 1 or 2 AA batteries in series, but can off of 3 or more in series. Is that of any help (other than knowing I will need say 4AA worth (=~6V) (or more) voltage)?

I expect the wires connecting the lights to the battery/pack will be longish - about 4 feet. Is that length of wire going to bring its own considerations like extra resistance due to the length of wire? How can I judge ahead of time whether the wiring or battery/pack will get dangerously hot?

I have a lot to learn, and appreciate your patience - like I said I'm happy to clarify as needed or seek out resources on my own. Would greatly appreciate any guidance, information, answers, etc. Thanks!

Edit: Here's the opened unit. The black box in the lower right is labeled R1 and written on it is "221". I don't see what I think of as LEDs, but these square dealies across the center are labled LD2 and LD3 (middle and right, the left I might have obliterated the label). Opened 3-Led Light Unit

  • \$\begingroup\$ Are they 12V? What color is the actual LED? Is the LED red or the lens with a white LED? All 10 are wired together with one power connector? \$\endgroup\$ Commented Oct 19, 2018 at 21:54
  • \$\begingroup\$ Thank you for your question. Yes, the product description at least says they are 12V. I haven't cracked open the lenses to see the LED color, so I can't tell you that right now. Each 3/4" light contains 3-LEDs inside the lens/reflective housing, and each set of 3 has two wires. I'd like to wire two of these together to the same battery/pack. \$\endgroup\$
    – A Barron
    Commented Oct 19, 2018 at 22:04
  • \$\begingroup\$ You've got 10 correct? As in 8 extra? Crack one open. You are going to have to do that anyhow. I am writing an answer that will explain. \$\endgroup\$ Commented Oct 19, 2018 at 22:05
  • \$\begingroup\$ Looks like you have enough answers to figure out your project, but just a general piece of advice for you about batteries: 9V batteries should be used only when strictly necessary to have a simple 7-9V in a small form factor at low currents. They have low current carrying capacity, low charge storage and cost a lot more than AA, AAA, D, Li-Ion. \$\endgroup\$
    – K H
    Commented Oct 19, 2018 at 23:41
  • \$\begingroup\$ For light weight, get a usb power bank, and a 5v to 12v step up converter. These lights are supposedly 30mA each, so any small power bank and converter would power these for a few hours minimum. Get two power banks just in case, as a backup. \$\endgroup\$
    – Passerby
    Commented Oct 20, 2018 at 0:54

2 Answers 2



A single 9V alkaline battery will easily power both LEDs at full (9V) brightness for 10 hours.

This is more complex than the typical LED circuit where you select the LED, resistor, and battery. This is a little more complex. Here we have three unknown red LEDs with a fixed resistor.

With time the battery voltage will continuously drop and the current will decrease as the voltage drops. so we cannot just say we have a 20 mA load and a 1000 mAh battery.

The amp hour rating of a battery is from full power discharging down to a cutoff voltage. 1.5V alkaline batteries have a cutoff of 0.8V. A 9V battery has a cutoff of 4.8 (six times 0.8V).

What we do not know with this LED is the acceptable operating voltages before the LED gets too dim as the battery discharges. We do not know how much the current will be at the various operating voltages. The human eye (and mind) do not perceive LED brightness the same as a light meter. We do not know how much capacity is required due to the above variables.

The Dynamic Variables

  • Voltage
  • Current
  • Perceived LED brightness
  • Battery capacity


I wanted to see for myself how the advice given actually worked out.
I bought ten of these LED lamps from the link provided.
They came with 40 AAA/LR03 Phillips "Power Alkaline batteries.
I used six of them in serial for testing.

In the Lab

At 12V I measured 45 mA.
At 9V I measured 22 mA.
At 8V I measured 11 mA.
At 7V I measured 6 mA.
At 6V I measured less than 0.1 mA.

The 6-9V measurements were lower than I would expect with respect to 12V so retested and results were confirmed. I then also used another LED and at 12, 9, and 7V and got very similar results: 45, 22, 5 mA.

Current measurements were made measuring the voltage across a 0.01 Ω 1% resistor in series with the LED.

Perceived Brightness

At 8-9V the brightness is consistent.
At 7-8V the brightness is slightly dimmer.
At 7V is where the brightness begins to significantly dim, yet at 7V still very acceptable nearly as bright as at 9V.
At 6V the brightness is borderline acceptable.
At 5.75 it is very dim and begins flickering.


I tested a single LED lamp powered by 6 Phillips AAA alkaline batteries.

12 hrs, 8.5V
15 hrs, 8.3V
18 hrs, 8.23V
20 hrs 8.2V
24 hrs 8.13V
29 hrs 8.03V
30 hrs 7.97V
40 hrs 7.93V

I estimate the LED could have easily lasted more than another 40 hours.

An AAA battery has about 1100 mAh discharged down to 0.8V @ 25 mA. A 9V 550 mAh @ 25 mA.
It is fair to say if six AAA lasted more than 40 hours, a 9V will last 20 hours. So both LEDs powered by a single 9V alkaline would conservatively last more than 10 hours.

After 40 hours of viewing the LED brightness in my office, I took it back to the lab and compared the brightness to another LED on my bench supply at 9V. The difference was nearly imperceptible.

I would go with a 9V alkaline battery for each LED.
One alkaline could power both, I prefer a margin of error.
If I used one 9V alkaline for both, I'd bring and extra battery.

An AAA has a capacity of about 1000-1200 mAh.
A 9V has a capacity of almost 600 mAh.
mAh capacity is not so useful. The cutoff for the 9V capacity is 4.8V.
We need a cutoff of 7V with a load that ranges from 22 mA @ 9V to 6 mA @ 7V.

The dynamic resistance of the LED ranges from 400 Ω at 9V to about 700 Ω at 7V.
In the discharge curves below, we need to look at the number of service hours at 7V.
The service hours according to the discharge curves, is 15-35 hours.
Our 400 Ω load is better than the 35 hour point.
Our 700 Ω load is better than the 15 hour point.

so saying a 9V alkaline will last 20-40 hours for one LED lamp is very reasonable.
And very safe to say a single 9V alkaline could power both LEDs for 5+ hours.

Energizer AAA Discharge Curves
enter image description here

A 9V battery actually has six AAAA cells in them.

enter image description here

Six Fresh AAA
enter image description here

After 24 hours, very slight perceptible difference
enter image description here


9V alkaline
9V lithium

Red with cover sawed off
enter image description here

These are peculiar LEDs. There is a metallic coating with a mirror like finish and a very small hole for the light to exit. In the above photo you can see the mirror in the center LED and the reflection of the red plastic on the right side of the right LED. Below the red regions are the mirror coating.

The LED on the left, all the red light (washed out in photo) comes from the little hole, The one on the right I scratched off some of the coating and it appears brighter. Still emits all red (not white as it appears in photo).

enter image description here

End of Update


To run these from battery you would likely need a 12V battery.
My guess is that each lamp contains 3 LEDs and a resistor to set the brightness.

If you can, you should rewire the 3 LEDs in each lamp to be powered individually rather than 3 in series.

If the LEDs are red (not white) you will need a little over 2V to power them. If white you will need a bit over 3V.

If these are 2V red LEDs a lot (50%) of power is being wasted. That's not good for battery powered projects. If 3V white then only 25% is being wasted.

We can estimate the battery life. If each lamp draws 30 mA, for 6 hours of run time you will need a 180 mAH battery for each LED.

First Update

I can definitely verify that I can get these light units to light up with less than 12V input. Is that important/relevant? For example, they'll light when I attach the wires to both sides of a 9v battery, or to 4 AAs in series.

Four AAs, that's good, only 6V.
Are they bright enough for you with the four AAs?

The problem I see with AA alkaline is the voltage drops fairly fast.

enter image description here

But the Ultimate Lithium start at a higher voltage and do not drop much over the battery lifespan.

This is a AA, but the AAA will work just as well for you.

enter image description here

you mention that each (LED I think, not 3-LED light) draws 20mA - how do you know that?

I don't know, I said "let's say..."
20 mA is a good guess.
If you could measure the current that would be a big help in determining how long the various battery choices would last.
Otherwise you can only find pout how long by experimenting as I suggested.

Or when you take one apart we may be able to find the resistor value to find the current.

Not what I was expecting. This appears too well made for the price.

enter image description here

The bad news is, if the above is true then with four AA at 6V then there would only be about 1 mA flowing.

The good news is any voltage over 6V will work. The higher the voltage, up to 14V, the brighter they will be.

If you have the capability to add a resistor you could make them bright with as little as 7V, maybe 6V. At 9V I estimate they will be half as bright at 9V as at 12V.

The big question is how bright are the LEDs with four AA compared to the when they are powered with 9 or 12V?

  • \$\begingroup\$ Thanks @Misunderstood. I can definitely verify that I can get these light units to light up with less than 12V input. Is that important/relevant? For example, they'll light when I attach the wires to both sides of a 9v battery, or to 4 AAs in series. Also, you mention that each (LED I think, not 3-LED light) draws 20mA - how do you know that? Note, I still haven't pulled one apart but will try to do so today and see where we're at. \$\endgroup\$
    – A Barron
    Commented Oct 20, 2018 at 16:07
  • \$\begingroup\$ How bright are the LEDs with four AA compared to the when they are powered with 9 or 12V? See latest update. \$\endgroup\$ Commented Oct 21, 2018 at 4:12
  • \$\begingroup\$ With 4 AAs they are significantly less bright than with a 9V. I couldn't say with any technical accuracy. You can see them even in a lit room with the AAs but the difference is notable vs a 9 volt. \$\endgroup\$
    – A Barron
    Commented Oct 22, 2018 at 21:03
  • \$\begingroup\$ I agree, very noticeable difference. 6V is unacceptable. To be sure I bought 10 of them myself. See the update top my answer. \$\endgroup\$ Commented Oct 23, 2018 at 23:26
  • \$\begingroup\$ FINAL UPDATE POSTED \$\endgroup\$ Commented Oct 24, 2018 at 16:58

This can be done easily with three 9V batteries in series. You will get 27 volts when the batteries are fresh.

As others already suggested, i believe each lamp unit contains 3 LEDs in series connected to a current limiting resistor possibly a 330 ohm to be powered by automotive voltage 12V.

My suggestion will be like this. You connect the two lamp units in series so that they will be happy to be powered by 24V.

Now you can power them safely from 27V. You don't have to worry about extra 3 volt, that won't cause any harm because automotive voltage ratings can go as high as 14V which means the LEDs guaranteed to work just fine even at 28V.

Ps; Advantage of this scheme will be the running time of the lamps will greatly increased because three 9V batteries have more energy storage. You can definitely use this longer than 5 hours.

  • \$\begingroup\$ Good idea. That 27V is going to drop below 24V fairly fast as 9V batteries do not have much energy capacity. 9V batteries are considered dead at 5V. It's going to be close as to whether 3 batteries will last above 24V for 5 hours. A 9V has 400-500 mAH when drained down to 4.8V. I do not use 9V of LEDs because of the steep discharge and the LED's intensity range to too big. I think this will work. Check out a 9V datasheet: data.energizer.com/pdfs/522.pdf \$\endgroup\$ Commented Oct 20, 2018 at 0:15

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