12VDC to 6VDC

[Lost_in_the_HTTP]

Is it as simple as a resistor?

I got some solar powered yard lights. They do fairly well and light up a decent area, but they don't last all night. Not sure what size battery since the ratings can't really be trusted. But no matter, in Winter when we only get 8-10 hours of daylight that is often cloudy, the cells don't gather enough light to charge the batteries enough to last the night. They often drop out before midnight, sometimes lasting until 2AM or so.

The lamp units are 6V and I can probably find a way to eliminate the solar cell and batteries and use a separate 12V battery.

I know I can use a simple resistor for a single LED and have done it many times, but these are something like 150 LEDs per unit and I'd like to do at least two from a single battery. I'm kind of wondering if 20-25 feet of 14AWG would create enough voltage drop.

..

 

[skyking]

no.
2 per battery solves your problem. Wire in series and you have 6v at each light.

 

[BoomerD]

6 volt golf cart battery.

 

[Lost_in_the_HTTP]

^^ I thought about that, but I already have a usable spare 12V battery and 10watt 12V solar panel and controller.

I'll have to look at the series wiring to see if it's doable.

 

[skyking]

it is absolutely doable only if you have them come on at the same time. It won't work for the motion lighting option unless you hack them ( my favorite pastime ) to use one motion sensor to come on at the same time.

 

[herm0016]

you would need 1100 feet of 14 gauge to drop 50% voltage with 1 amp of current.

why hack something like this that is nothing like what you want when you can just buy what you want?

https://www.amazon.com/FLOODOOR-Security-Waterproof-Daylight-Equivalent/dp/B07MV4V2GJ/ref=sr_1_1_sspa?keywords=12+volt+motion+light&qid=1674428843&sr=8-1-spons&psc=1&smid=A2BD5EBFGGS5MI&spLa=ZW5jcnlwdGVkUXVhbGlmaWVyPUEzMDBJUTkxT1lUOEVNJmVuY3J5cHRlZElkPUEwMTU4Nzc5M0E3WUJJN09FOUFKUiZlbmNyeXB0ZWRBZElkPUEwNjk2MDY4MUEyOUNGS0Y2TlA1UyZ3aWRnZXROYW1lPXNwX2F0ZiZhY3Rpb249Y2xpY2tSZWRpcmVjdCZkb05vdExvZ0NsaWNrPXRydWU=

......

I know, no fun!

 

[Lost_in_the_HTTP]

I already have those units, so no real need to buy something else, if I can make these work. But I haven't even tried to open one yet, so it make not even be possible.


I said 14, but it may be smaller. I've got some of that landscape wire laying around, whatever it is.

 

[Paperdoc]

ANY resistor, whether it is a long run of wire of an actual resistor, NEEDS some careful measurement and planning.The challenge is: HOW MUCH resistance?

Let's look at a simple case of powering only ONE of your light fixtures that needs 6 VDC from a 12 VDC battery. You want a resistor added into thee circuit that drops exactly 6 VDC across it. That Voltage drop is determined by Ohm's Law: V = I x R.That is the Voltage drop across a Resistor is the product of the Resistance in Ohms times the Current flowing through the circuit, in Amps. The trick is to recognize that the same rule applies to the total circuit, and to the half of the circuit that is your resistor. So the CURRENT running though the whole circuit is determined by the resistance of the light fixture, as well as by the resistance of the resistor in series with the fixture. You need to know that light fixture resistance to start!

Your best way to find that information will be to get into the fixture and access its components to MEASURE the right things when it is operating normally - that is, lit up fully and at full brightness from a well--changed battery it already has. In order to do that you need to identify the connection point where you can measure the electrical feed for the whole set if LED lamps inside. Then you need to OPEN the circuit at that point and insert into the circuit an ammeter. Using that you can actually MEASURE the CURRENT that flows normally through the lamps. NOW using the Ohm's Law equation you can calculate the Resistance of the fixture's lamps as R = V / I, where V should be 6 VDC (that's what you believe its batteries are supplying), and I is the measured current in amps. To VERIFY your assumptions, though, you should ALSO measure the actual VOLTAGE being supplied to the lamps during this test.

Now you have the value for I (Current) that should be running though the lamps in normal operation. If you plan to connect a battery at 12 VDC plus a dropping Resistor in place of the "normal" power supply so that the reduced Voltage at the lamps is 6 VDC, you need to create a drop of 6 VDC across the added Resistor. Again, the value of that added Resistor must be R = 6 / I, where I is the current you have measured above. NEXT you need to check and meet the requirement that the added Resistor can handle the POWER it is "wasting" to create the Voltage drop you want. The power in WATTS is V x I, or 6 x I. Your Resistor must be rated for that many Watts or slightly more, and it should be mounted so that there is some air movement around it to remove the heat it generates. NOW you have what is needed to find the correct Resistor.

Next comes the complexity. That was for powering ONE light fixture from your battery-plus-resistor combo. But you have more than one fixture. IF you have only TWO of them and want to connect both in PARALLEL to that power source, now the AMPS total load on that source is twice as much, and that means a DIFFERENT calculation for the values (Ohms and Power) of the Resistor needed, given that the current though that resistor is twice what it was in the first case. NOTE, too, that this means you MUST have BOTH fixtures connected and working for this to work - if one fixture quits, it changes the Current and the Voltage drop and it's all wrong again!

What about having MORE fixtures connected to that one power source? Needs re-calculation and building again.

IF you are right and the actual Voltage being supplied to the LED lamps in one fixture IS 6 VDC, then the SERIES connection proposed by others above is MUCH easier. In that situation, connecting the lamp assemblies of TWO fixtures in SERIES to a 12 VDC battery without any Resistor DOES guarantee 6 VDC supply to EACH fixture's lamps. IF one fixture fails then BOTH fixtures will not work and there is no danger that one will be over-Volted. IF you have more than two fixtures, you can arrange to have PAIRS of two fixtures in SERIES so that each pair requires 12 VDC.THEN you can connect each PAIR to the battery in PARALLEL.

A further hint. If you do go ahead, that 12 VDC battery you will use will require some means of keeping it charged up. A small automotive "trickle changer" able to supply from 2 to 5 Amps charging current at 12 VDC would keep it fully charged.

By the way, how do you plan to have these lights turn on and off?

 

[TheELF]

Get a 6V voltage regulator that can handle 12V input and the amount of Amps the lights need, this one is just an example, the first result that showed up on google.
The datasheet shows that it can handle up to 18V input and it handles 1.5A output, it also shows you the application circuit, as in what capacitors to put on the in and output.

Voltage regulator L7805CV

Basic voltage regulator in the TO-220 package. A must have for basic 6V electronics.

grobotronics.com

http://grobotronics.com/images/companies/1/L78xx.pdf

 

[Paperdoc]

Yes, a Voltage Regulator built using such devices is relatively simple to build and MUCH easier to use because it can handle some range of load. Resistors need careful design and can NOT adjust to load changes.

 

[mindless1]

You really need to open one up and reverse engineer it. How do you know "the lamp units are 6V"? What does that even mean when it has dozens of parallel arrays?

Random guess based on the 4000mAH spec, is it is using either one, fraudulently rated Li-Ion cell for 3.7V, or two in series for nominal 7.4V.

Either way, your best bet is remove the batteries from the circuit (or use two schotttky diodes in an OR-ing configuration so the external power source isn't charging the batteries, -OR- reverse engineer the solar charging circuit to see if you can tap into that and have it control the internal battery recharge.

Regardless of all that, you'd just want a switching buck regulator in each lamp, will be much more efficient and create much less heat than a linear regulator like a 7806, which loses HALF the power as heat if the target were dropping to 6V from 12V but as mentioned above, it's possibly some other voltage you want, should open it and measure working voltage if nothing else.

Current consumption would be good to know too, as it lets you know how small a buck regulator you can use. There are some tiny ones costing under a buck a piece in bulk, and others a little higher powered but still pretty small. I will link to the latter as these are some I have used before, nothing fancy but sometimes you don't need continual pushbutton changes, nor an LCD display at extra size and cost:

2pcs XL4015 5A DC Buck Step Down Voltage Converter Constant Current Power Module | eBay

This module combines the high efficiency XMLSEMI XL4015 DC-DC buck converter IC with a TI LM358 Op-Amp for stable current regulation. Current limiting function requires at least 5.6V input. 5x 5V Fixed Output Buck.

www.ebay.com

You'd take the output from those, as the input where the batteries connected to the light, not trying to direct drive the LEDs arrays with them. That "might" work too if it they just use a series resistor built-in for current limiting but this is unknown without some reverse engineering of the light. Above linked regulators are small enough that they "might" fit in the area that the removed batteries leave behind. Again, need to take the light apart to see what you're dealing with.

No, 25ft of 14awg would not have created enough voltage drop, on the contrary it seems like a fine gauge to use from the battery, to the linked buck regulators above, to do this. It is assumed you realize that you need to adjust the buck regulators to the target voltage (measure what the charged battery in the light reaches, not just 6V since IDK where you got 6V from), measured with a multimeter, before powering the light, then if there's a little extra voltage drop, can fine tune it from there.

 

[mindless1]

Also, what is the point of a separate 12V battery? You would have to periodically recharge that too, unless you have a larger solar panel doing so? If you are just using mains AC/DC adapter/charger to recharge the 12V battery, then why not just use a mains AC/DC power supply without the battery?

If you measure the voltage and current across the LED array then you know the range of AC/DC adapter to look for, and can determine if you want an LED-driver specific, current regulating power supply, or just a voltage regulated to emulate the power source from the existing batteries, and if you can't find a voltage regulated at the right voltage, get one at least a couple volts (3V or more higher, better still) above that and use a step down buck regulator as I linked in my prior reply.

Option # 3, measure how charged the solar panel gets the battery, and add additional solar panels to recharge to a higher level. This assumes the lamp has overcharge protection, which it certainly *should* if charging Li-Ion cell(s).

 

[aigomorla]

Stepdown converter like mindless1 links.

You just put it between the battery and light.