# Connecting two 10m fairy lights

#### James90s

Joined Mar 3, 2021
31
Hello.
This is my first time posting. I am very new to any form of working with basic electrics and lights. I am doing a home project and need to connect two 10m led fairy lights also called dew drop lights together to make 20m. They are battery operated, and a single battery holder has three AA batteries producing 4.5V. The lights run in a parallel and the LED lights are diodes. I connected the two 10m together by cutting and soldering. The lights are very dim, so i added a 9v battery. The lights are bright for the first 5 meters, but as the string continues the lights get dimmer and dimmer, and after 20 minutes the 9v drops down to 2.4 volts and eventually all the lights go off. I don't understand why they start bright and get dimmer, and also why the battery drops the voltage so dramatically at such a fast rate. why is it doing it and is there a way to extend the 10m with an additional 10m to create a 20m while running off batteries that produce no higher than 9 volts.
Also the lights seem to run best when each light receives between 3 to 4 volts.

#### Ya’akov

Joined Jan 27, 2019
8,505
Hello.

...
I don't understand why they start bright and get dimmer, and also why the battery drops the voltage so dramatically at such a fast rate. why is it doing it and is there a way to extend the 10m with an additional 10m to create a 20m while running off batteries that produce no higher than 9 volts.
Also the lights seem to run best when each light receives between 3 to 4 volts.
Hello, and welcome to AAC.

While there may be other things going on here, some very basic things may explain it.

First, voltage and current (amperage) are entirely interrelated and in powering a practical device they can't be considered separately. So, the battery supplying the lights may have an "open terminal voltage" of 4.5 or 9 volts but under load—that is, once it is connected—that will drop in proportion to the amount of current it is required to supply.

The three AA cells are much larger then the six AAAA cells that are hidden in your 9V battery. They can both provide more current and have more capacity to deliver current over time usually measured in mAh (milliamp-hours) and vary with the discharge rate. The AA cell has a capacity of ~2500mAh while the AAAA has only ~600mAh in the best case.

The heavy draw you are imposing certainly reduces that further.

Try putting the two battery boxes in series and in parallel and observe the results.

#### LesJones

Joined Jan 8, 2017
4,106
A 10 meter length of LEDs would put quite a high current load on three AA cells. There are three ways the LEDs could be connected. 1/ Each LED could have its own current limiting resistor. This is the best way but most expensive . 2/ The LEDs could be in groups of LEDs directly in parallel with a limiting resistor for each group. The LEDs in each group SHOULD be selected so the forward voltage drop of each LED in the group was very similar. 3/ All the LEDs in the string are connected directly in parallel with a single current limiting resistor. Again the LEDs SHOULD be matched for forward voltage drop. It is also possible that no current limiting resistor is used and they rely on the internal resistance of the battery to limit the current. For us to fully explain what is happening you will need to inspect the construction of the LED strings to see which wiring method has been used. When you connected two strings to the three AA cells there would have been a larger voltage drop across the internal resistance of the battery due to the higher current and if you measured the terminal voltage of the battery you would have found that it had dropped. You do not say what kind of 9 volt battery you used so I will assume it was the small PP3 layer type battery. This will have a higher internal resistance and a lower mAH rating than the three AA cells in series. So I think if you measured its terminal voltage when you connected it to the LED string it would have dropped to much less than 9 volts.

Les.

#### BobTPH

Joined Jun 5, 2013
8,092
Putting these in series is not recommended.

For two strings, you need twice the battery capacity as 1 string (there is no free lunch.)

Why not just run each string off its own battery?

if you MUST have 1 battery, use 3 C cells, or, better yet, D cells.

Bob

#### James90s

Joined Mar 3, 2021
31
I would like to thank everyone for their replies and inputs. I put 2 battery boxes in series and parallel, and now that the mAh between each of the individual types of cell has been pointed out to me, I could see some differences. the main difference was time and when in series I had the best results. However, after 20 minutes the lights began to dim. After 40 minuets they all had the exact same brightness, and after a full hour all the lights were dead and the batteries were heavily depleted. I inspected the led's and could not find or see any form of a resistor, so I am going to assume that the only resistance is from the led its self and the internal resistance of the battery. The lights are setup in a parallel circuit. the lights are still very bright for the first 5m and then get dimmer and dimmer but when I reach the last 10m the brightness is the same. I wanted to use two boxes to power them individual, but I can't find a way to put them in the circumference that has been designed to house the batteries, and I would like only one on and off switch. Is their any type of a battery box that would power two individual parallel circuits while sharing a single on off switch?

#### LesJones

Joined Jan 8, 2017
4,106
I found this battery box on ebay that takes 3 x D cells. It would be better to have the battery box between the two strings so that you don't have the resistance of the wires of the first string in series with the second string. If this is not possible then run a 10 meter twin cable from the battery box to the start of the second string. (Probably 0.5 sq mm cable would be large enough to give a low enough voltage drop.) Another possible option to get more capacity than D cells is to use a type 4R25 - 996 6 volt lantern battery and add a suitable value series resistance. (Calculated to give 1.5 volt drop with the current taken by the LED strings.) Have you considered using a suitably rated USB power supply to supply the lights ?

Les.