Need some basic Electronics and circuitry help

[JohnWPB]

Well, this is the wiring closet... and as it states "Post anything" so here goes

Ok, I have started a small project, that I hope I can pull off with out to many problems <_<

Firstly, I have built a pond and waterfall in my front yard. What I want to do is power the lighting with solar power. All of the crappy "over the counter" solar lights from Home Depot and the likes, are just not bright enough.

So far I have a 12v motorcycle battery, and have ordered a $25 solar panel designed to keep a 12v battery charged.

I have ordered THIS LIGHT, made to be a bike headlamp actually. I chose this one for the inexpensive cost, as well as it is rated to be pretty bright with the 36 High powered LED's, and should work well to illuminate the water fall.

I have a light housing, that you can see in the photo below. It is currently halogen, but am converting it. I am going to take the bike lamp apart, and mount just the LED cluster into that housing, which is waterproof. (I may have to add a diffuser to disperse the light, but will not know that till I get the light, and see just how tight the beam is)

So far all is well...

Now, this is where I need the help..... I need to step the voltage down from the 12v battery to 4.5 volts to power the LED's. I am sure the amperage will need reduced as well.... I think..... I am not sure if I need a potentiometer, or some sort of resistors, voltage regulator or what. I know just about zero about electrical circuits. So, any tips and suggestions are greatly appreciated!

If all goes well, I intend on putting a couple of the solar panels in parallel, using a larger battery, and adding more landscape lighting, as well as possibly wiring up the front porch light to be solar as well, and it will just come on at sunset, and remain on. I will make something similar to THIS, as I have the perfect beam across the the front door that will hide it from view. This will also leave the standard porch light in place, for when more light may be needed ect. (Can you say Pizza delivery!)


Here is a snapshot I took a couple days ago of the pond:

 

[mdesmarais]

You need to know the current draw of the lamp before you can design the circuit.

The problem is that the cheap ways to drop the voltage (regulator or resistor) are going to throw away most of your power as heat.
4.5/12 = .375, so you are using 37.5% as light, and 62.5% as heat.

Looking at the item, I noticed that there is some kind of an IC controller (or so they say ;-) in the light already. This also has an unknown impact on the design. . .

If you want to just use a power resistor, you can, but your light will dim as the battery drops. . . probably not desirable. You will be at that 37% eff number.

A regulator will get you a stable voltage- but still at the 37% eff.

It may well be that the lamp would run OK at 5v. Then you could use a fixed voltage regulator like a 7805.

If you want to get to exactly 4.5v, then an adjustable regulator is the way to go. A couple of resistors will let you set the voltage point.

BTW, I'm not endorsing Fairchild, just the first link to pop up. Both these components are commodities, you can buy them or their equivalents all over.

Now, if you don't mind spending a little more to get better efficiency, you could use a small DC/DC converter and get up into the 80-90% range.
I've used these types of TI parts quite a bit. Again, there are other options, and knowing the current draw is key.

A third option would be to buy two more of these lamps and hook them up in series- 4.5*3 = 13.5v. Now you are converting all your power to light. You will still have the issue of having the lights dim as the voltage drops when you battery discharges.

Markd

 

[JohnWPB]

Thank you very much for all the info, I can see you spent some time on it searching and such! Well I got the gist of most of what you were saying. Yea, operating at 1/3 the efficiency is definitely not desirable.

A DC/DC converter seems to be the best route. However, I am not quite sure how to get the current draw of the lights.... I am sure my multimeter is not going to help with that in any way, and a way to measure what is being used would be needed. Kind of like the KillAwatt device for determining the draw of an appliance. Being they are intended to run off of 3 AA batteries, I can not imagine that the draw is that much.... How accurate does this figure have to be?

Also, is there anything to worry about as far as amperage? I mean going from a .00001 MA AA battery to 5 Amps or so is a pretty big jump <_<

Lastly, I again am clueless when it comes to electronics and the lingo.... The page you linked to went on about Vin Min, Vin Max, Vout Min, Iout Max..... reading that....not a clue what I was looking at... sorry!

What would the DC/DC converter roughly cost, and what one do you think would be the one to work for this project?

On a last note, are there a couple of other options? Such as a 3-way switch or something like a standard fan uses for low, med and high? They do not seem to convert the spare energy into heat. Possibly even a DC Potentiometer that I could adjust to down to 4.5 Volts using a multimeter, and just lock in place?

EDIT: on a last note, possibly use an existing DC/DC converter, like a 12v car charger used to power a cell phone? Or, again, would there just be too much power loss using something like that? These things can be picked up at the dollar store these days, and I am sure I could find one around the 4.5 volt range......

 

[electron]

LOVE the pond. I just installed one of those large Lowes pond shells, with plans to build it up like you did, but I just can't find that kind of stone locally <_<

 

[JohnWPB]

Thanks, was a lot of work! Started with the pond shell, then when digging, decided to go trade it in for the liner instead...

Stone.... Locally!?!?! Hah! I live in South Florida.... there are no rocks in this whole state, as its just a huge sandbar. The 3 Large tan rocks are fake, and the ones in the falls were bought at a nursery here, and cost like 50 cents a pound! Highway robbery for when I grew up up north, you could get rocks free anywhere......

 

[BLH]

You may be able to check the current draw of the LEDs.

I have used two THIN pieces of metal with a piece of paper between them. Between one of the battery ends and the contact it normally touches. Then use the DC current part of my meter to read the current. By connecting one lead of the meter to each piece of metal. As always with current use the highest range first and work down until a reading is in range.

 

[mdesmarais]

Thank you very much for all the info, I can see you spent some time on it searching and such! Well I got the gist of most of what you were saying. Yea, operating at 1/3 the efficiency is definitely not desirable.

Not too much- its my day job, so I'm pretty fast at it. ;-) (You're still welcome though)

A DC/DC converter seems to be the best route. However, I am not quite sure how to get the current draw of the lights.... I am sure my multimeter is not going to help with that in any way, and a way to measure what is being used would be needed. Kind of like the KillAwatt device for determining the draw of an appliance. Being they are intended to run off of 3 AA batteries, I can not imagine that the draw is that much.... How accurate does this figure have to be?

If your multimeter has a current setting (mAmps or Amps) it certainly will. BLH mentions one method below. Since you are already commiting to modifying the unit, it shouldn't be too hard to put your meter in series with the batteries and find out what the draw is.
If you can, take the whole unit apart and post what you find inside- it may make sense to pull the control circuits out and drive the LEDs directly.
As far as draw, you'd be surprised with what you can pull out of batteries. I noticed in the reviews that this seems to eat them up.

Also, is there anything to worry about as far as amperage? I mean going from a .00001 MA AA battery to 5 Amps or so is a pretty big jump <_<

No worries here- the LEDs will only pull what they can use. We just have to get the voltage right.

What would the DC/DC converter roughly cost, and what one do you think would be the one to work for this project?

Get the current measurement and I will help you with component choices. I'd guess we can get you set up for ~$10-$15.

On a last note, are there a couple of other options? Such as a 3-way switch or something like a standard fan uses for low, med and high? They do not seem to convert the spare energy into heat. Possibly even a DC Potentiometer that I could adjust to down to 4.5 Volts using a multimeter, and just lock in place?

EDIT: on a last note, possibly use an existing DC/DC converter, like a 12v car charger used to power a cell phone? Or, again, would there just be too much power loss using something like that? These things can be picked up at the dollar store these days, and I am sure I could find one around the 4.5 volt range......

I'm not sure about the fan controllers. . . if you are talking AC though (like a ceiling fan) it won't work for this application, unless it IS just dumping the heat.

If you are buying it for a dollar, it is 99% likely it is just a regulator and it is dumping as heat.

Markd

 

[JohnWPB]

Also, is there anything to worry about as far as amperage?

No worries here- the LEDs will only pull what they can use. We just have to get the voltage right.

Ok, again, I am no electronics person by any means, but this is confusing me a bit further than when I started (NOT hard to do! <_< ). If there is no concern at all for the amperage, and all that is needed is the correct voltage..... Would that not simply be 4.5 volts? (1 AA battery = 1.5v x 3 = 4.5, as we know it is designed to run off of 3 AA batteries.......)

As for reducing the voltage in a 12v cell phone charger.... doesn't that just use a coil with very thin wire, to reduce the voltage? Or is that just bleeding of the extra voltage via heat exchange?

 

[BraveSirRobbin]

I have to agree with Markd on getting a second light instead of trying to regulate the voltage. For ten bucks more you will double your light output and have better use of your battery's current.

Current is the important factor here. You will want to calculate how many hours (max season) you want your light on, then multiply that by the current it draws. This will give you amp*hour rating of your power use. You then want to double this number and that should be the amp*hour rating of your battery (you never want a "common" battery to go below 50% of its capability if possible).

So now you know how much current your are "drawing" from the battery. Now you have to calculate the size of solar panel needed to charge it during the daytime. For this you need to know the current output of the solar panel or (little less accurate) the output wattage. The solar panel you choose should be for charging 12 volt batteries so you can (roughly) calculate the current output by WATTAGE OF SOLAR PANEL/12.

Now you need to know the minimum hours of sunlight (again seasonal) for your area. There are plenty of places to get this, but I'm guessing something around five to six hours is probably appropriate.

So your charge system in amp*hours will be the solar panel's current capability times the minimum hours of sunlight. I like this number to be a lot greater than the battery amp*hour draw.

One other thing to consider is the solar panel's voltage (depending on the specific brand and wattage) will vary with the amount of solar radiation it is absorbing. This voltage can go over 18 volts. A charge controller is recommended that will sense the battery's voltage, regulate the voltage of the solar panel, and charge the battery when it needs it.

There are three kinds of technologies for charge controllers (shunt, PWM, and MPPT). For your situation you may want to consider a cheap PWM controller such as THIS. They also help with battery life.

Now for just running LED lighting you may not want to use a controller and hopefully the battery will always be in some sort of charge state so the voltage may always be under 14 volts. I don't know if the output current/voltage will damage the LED's but I would not let the light connected to the system unless the battery was always in place.

Also remember that you don't have the "optimal" battery for a solar charge situation. Most will use a glass mat (AGM), but again this is probably more advanced than you need for just running a light.

I would also fuse the output of the battery (near it's terminals) with an inline fuse and have the fuse's current rating slightly over your total lighting current output.

I'm sure others here that are actually more familiar with solar systems can chime in (and correct any mistakes I might have made) <_<.

 

[JohnWPB]

Good Lord!

4.5/12 = .375, so you are using 37.5% as light, and 62.5% as heat, Amp Hours, Algebra, Calculus....................

I may just go to Home Depot and buy those crappy dim lights with a 1" square solar panel on them! LOL

I never knew something that SEEMED so simple could turn into all of this! And it did seem Sooooo simple at the time:

buy a charger made to charge a 12v battery, hook up a light to it <_<

 

[BraveSirRobbin]

Well, I'm probably making this more complicated than it should be, was just trying to give various scenarios.

If you have the components, just connect together, get a current draw of the lighting, place a resistor in series (between the battery and light) and connect the sucker up. <_<

 

[WayneW]

depending upon how much current you need, there are 12VDC-4.5VDC assembled converters available.
http://www.powerstream.com/dc7.htm
http://cgi.ebay.com/4-5VDC-650mA-COBY-CAR-...3QQcmdZViewItem
http://cgi.ebay.com/PANASONIC-DC-DC-CONVER...808011187r22079

 

[BraveSirRobbin]

John;

Do you have a link to the solar panel you purchased?

 

[mdesmarais]

Well, I'm probably making this more complicated than it should be, was just trying to give various scenarios.

If you have the components, just connect together, get a current draw of the lighting, place a resistor in series (between the battery and light) and connect the sucker up. <_<

Definitely the simplest- although I'd recommend the extra $1 or so to use a regulator, then the brightness will be constant until the battery is drawn down below 4.5v.

A regulator will also protect your LEDs if the charger starts putting out 14v. . .

Ok, again, I am no electronics person by any means, but this is confusing me a bit further than when I started (NOT hard to do! ). If there is no concern at all for the amperage, and all that is needed is the correct voltage..... Would that not simply be 4.5 volts? (1 AA battery = 1.5v x 3 = 4.5, as we know it is designed to run off of 3 AA batteries.......)

BSR covered this in pretty good detail in his post. Just because a battery is rated at some number of amps doesn't mean that they will all come out at once- it depends on the load you put on it. The LED light will be a pretty small load, so the current will just trickle out. Yes, 4.5v is correct- all I meant was that we need to set that voltage by some means, resistor drop, regulator, DC/DC.

As for reducing the voltage in a 12v cell phone charger.... doesn't that just use a coil with very thin wire, to reduce the voltage? Or is that just bleeding of the extra voltage via heat exchange?

Thin coil of wire is a resistor- yes, it is dropping as heat. The current is probably very low, so the heat is small.

depending upon how much current you need, there are 12VDC-4.5VDC assembled converters available.

Yep, another option. Make sure that it is stating the eff. though, or you could be just getting a regulator.
Measuring the current draw is still the 1st step!

Markd

 

[JohnWPB]

John;

Do you have a link to the solar panel you purchased?

This is the solar panel I ordered. It should be enough to charge the battery from sunrise to set.... I think

Just DO NOT believe everything you read! hahaah
"This panel is even powerful enough to function as a straight power generator . It can easily power most laptop computers, DC refrigerators, television sets and even small air conditioning units "

Directly power a small Air Conditioner?!!??! LOL

If the light only lasts till midnight or so, that's fine by me. I would think that once the battery is fully charged, that the low draw of the LED's should never drain the battery over night... I really have nothing solid to base this on, other than the 3AAA batteries should last at least an hour in the light, so I would think it should be at least 10 times that considering the battery being used. It is actually an alarm system battery, 12v and not too sure on the amperage. The battery itself is around 6" x 3" by 4" tall or so.

I have the same exact high output LED that is in the "bike light" on a key chain light from the same manufacturer. It runs off of 2 CR2032 batteries, and I have been using it for 6-8 months, and and use it quite frequently. I buy em by the 10 pack, as its far cheaper than paying $3 for a single CR2032 battery. Buying them in a 10 pack is around $7, or around 70¢ each.

Last time I ordered an extra 10 of em, and ripped em all apart, and put all new batteries in all my X10 slim sitck-em switches