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Advice needed Please.
I live on a boat; everything is 12v
I'm going solar and need to be as energy-conscious as poss.

I looked at LEDs in maplin and the diagram shows a 50k ohm resister either side of the 12v LED (potential devider). I want to put several LED's in parallel. What I dunno is the resistance of the LED and hence what size of resister to put either side of more than one. Any help welcome

You need to find the current used by each LED. If you have a Digital Volt Meter that shouldn't be hard to figure out. Just hook the meter up in series with one LED and measure the current (I). Now measure the voltage drop (E) across one of the resistors. Then multiply that current by the number of LEDs you are going to use. The resistance (R) you will need is calculated by using Ohm's Law E = I * R. So R = E / I. So just divide the voltage (E) by the current (I) and that is the resistance you need. Remember that the current you measure will be in milli-amperes (mA), that is one thousandth of an Ampere, so be sure that you get the decimal point in the right place.

Of course a quick way to figure it is to take the number of LEDs you are using and divide that into 50k. So if you use 5 you would need the resistors to be 10k.

Also check the wattage. You probably won't have a problem, but power (P) is equal to E * I or it is equal to I^2 / R. Resistors are rated for the amount of power they can dissipate without overheating and you don't want to overload them.

And of course there is the old fashioned brute force way. Just hook up each LED with its own 2 50k resistors.

Good luck with solar power project and keep up the good work.

Bill Gill

[Quote=Mike Kremer]

Hi Andist, Im pretty certain we have spoken before. You wrote that you live on a barge, and I am sure I told you to hotfoot it down to Camden town, Maplins...to buy a Solar 12volt battery charger.??
I do not think they are REAL substitute for a real car battery charger. Rather they just keep the car battery topped up (I think).
Your Car or Marine battery will soon run down when you place any definate load upon it...like a TV or a small Fridge cooler? I dont think the solar charger that Maplins sold at that time gave out enough Milli-amps to run a decent appliance
for more than a few hours, from your 12volt Car/Marine battery.

Re your new above qustion.. you are not giving out enough clear information.
Idividual LED's run off approx 3.1--3.3 DC Volts, not 12 Volts as you suggest.
You might get 4 0r 5 LED's wired up in series to a 12 volt battery to provide say lighting? But individually those LED's will only draw about 3m/a to maybe 30m/a ? depending upon the type of LED
I sm surprised regarding the 50K ohm resistors EITHER side of the Led? A single 5K resistor in series with your LED should be Ok....or two 5k on either side if the LED (as you state?)
..equaling 10k (much less bright, but should be Ok as well)?
Why one either side?
Otherwise not enough info.. so I cannot help further. Sorry

http://white-leds.co.uk/led-wiring-guide.htm

in the above it states that the resistors can develop heat
which is wasting the electric.

it might be best to use a step down transformer to reduce
the current and voltage that you will be using in the light circuit , it would be really nice if you could get your circuit to run without resistors wasting too much of your electric.



as you can see above the current (I) of (0.020 Amps) is extremely small compared to the possible current that your battery will have when fully charged up.

and the more current your resistors resist the more heat they will generate thus the more electric you will waste.

you should first find out your battery amperage and voltage
before you begin to figure out your light circuit/s , or you might flip on the switch and blow all the LED's.

Unfortunately transformers don't work with DC, which is what batteries produce. You can find voltage converters that will do the job, but they add to the cost and complexity.

As far as power consumed. Assuming the 20 mA (.02 A) in your example and 2 50k resistors the power dissipated would be I^2*R or

.02 * .02 * 2 *50,000 = .0004 * 100,000 = 4 * 10 = 40 W for each LED.

That does sound kind of high. The voltage converter may be a better choice. In any case a 40 Watt 50k resistor is going to be a problem. It will definitely get hot. I realize that original design called for 2 50k resistors and each one will only dissipate half the power, but you need a big safety margin. I would not use less than a 40 W resistor. If I put the 2 resistors together and used one 100k resistor I would go for an absolute minimum of 50 W.

Bill Gill

Thanks guys, I had assumed that the LEDs ran off of less than 12v , with resisters either side. thanks for the advice