Voltage range for low-voltage LED lights

[Jack Olsen]

I'm pretty clueless when it comes to electronics.

I've got about 100 tiny LED bulbs wired inside fake candles on a deck I just built. One particular 'candelabra' thing I made has 84 of them in it. Here is what it looks like:

The problem is, there's about a 30' run for the wiring to it from the power supplies that feed it, and I'm losing half a volt along that run (skinny wire).

I believe the bulbs are supposed to run on 3 volts. But I'm probably not a guy who should even speculate about that sort of thing. The specs on them say "DC forward current: 30 mA" and "Forward voltage (typical): 2.7 V." Right now, this set that is only getting 2.48 volts is dimmer than I'd like. The power supplies I'm using switch between 3V, 4.5V and some higher settings.

My question is whether it would be safe to try feeding 4 volts (instead of 2.48) to this set of bulbs.

I don't want to burn them all out since it took a lot of work to put this thing together.

 

[66dave]

Jack, can we assume that you have these all in parallel? Or are they in a series parallel circuit? Please let us know your wiring configuration.


If they are all in parallel use the higher voltage power supply and put a simple resistor in line (in series). The math is like this ohms=[power supply-led voltage rating]/current...[4-2.7]/.03=~43ohm resistor. A 5% 43ohm resistor is a common value that you could pick up at radio shack. There are different wattage ratings, to calculate wattage multiple voltage and current.

FYI that patio you made is SWEET!

 

[66dave]

I should mention that when calculating wattage current is cumulative in a parallel curcuit. So if you had 84 LEDs at 30ma each, the total draw is ~2.5A. 2.5A*4v (Or 4.5v depending on where you measure due to the line loss that you mentioned) will equal 10watts. So if your curcuit is all in parallel, assuming 4v based on your line loss, 43ohm 10 watt resistor.

I did a quick search, did not see a43 ohm unit, but they have a 47ohm 10 watt unit for $1.21

http://www.radioshack.com/product/index.jsp?productId=12566080

 

[Jack Olsen]

Thanks!

They are wired in parallel.

If I'm understanding my results from a quick Google search, a resistor would reduce the voltage going to the bulbs?

And even though you've put the math right in front of me so that even a child could work it out... let me ask one other question. One alternate power supply I have for this is a low-voltage timer for 12v lighting that I've got a 5v step down converter for. What resistor would be correct for a 5V supply?

 

[MoonRise]

Jack,

LEDs can be a bit fussy about input power.

Too low on the input voltage and they either don't light or are too dim.

Too high of an input voltage and they either instantly burn out or if the input voltage is just a 'little' bit too high they can go into thermal runaway (the hotter they get, the less voltage they need or the less current they need, so in use if they get hot the voltage/current demand drops but the power supply may continue putting out the 'full' amount and then the LEDs keep getting hotter until they let the magic smoke out).

And sometimes the gap between too low and too high is not all that far apart.

Do NOT feed those LEDs straight 4V or 5V (or higher) power. They -will- fry.

Although you can (sometimes) successfully drive LEDs by using a higher-than-spec-voltage power source and then dropping the 'excess' voltage via an in-line (series circuit) resistor, that approach can be iffy. You have to get jsut the right resistor value and hope that the resistor is wihtin spec as well as the power supply maintaining its output voltage consistently and the LEDs also maintaining their voltage-amperage characteristics consistently with life/use and temperature.

As mentioned/calculated by 66dave, 84 LEDs at 30 ma each gives a current draw (at 'spec'), of 2.52 amps. Can your power supply do that output power?

If your power supply is not fully able to supply the voltage at the amperage draw, you may have another issue besides voltage drop through the ~30 ft (one-way distance or there-and-back distance?) of 'skinny' wire.

Also be aware that a 10W dropping resistor will get HOT. Don't start a fire or burn yourself.

As you've now experienced, LEDs have some gotcha's on their input power (especially with how many you've rigged up in your light fixture, which is rather nice looking btw ).

I would probably go more for a voltage-regulated power supply circuit to drive the LEDs.

http://www.circuitstoday.com/adjustable-voltage-regulators

Something like this, using an LM338 IC

http://www.circuitstoday.com/13v-5a-adjustable-regulator-using-lm338

Feed it your 12V DC, then pick/adjust the output 'control' resistors to get your desired output voltage (measure at the LEDs to get the desired 'spec' 2.7V at the LEDs).

Just MHO.

 

[gatchel]

Here is a calculator if you are running some numbers.

http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator


I would use a higher voltage, series few of the LED's together and use a resistor per LED series string if it were me. Less possibility of any failure causing other failures because of voltage and current changes caused by the original failure...

I made some LED rear turn signals for my R6 in 2004 /2005. At that time there were no LED options for aftermarket turn signals and I really wanted them. They are sill working today.

 

[Kevin C]

LED's produce light as current flows through a diode junction. The issue is that the amount of current that flows VS the voltage is not linear.

There is a knee in the curve where a small increase in voltage results in a much larger increase in current.

There are two methods of dealing with this. One as others noted is to put a resistor in series. The resistor has a linear voltage VS current curve and serves to limit how much the current changes as the voltage changes.

The other option is to use a constant current drive. For high output LED's this is preferred.

For a normal lighting system that's not trying to run the LED's close to there max performance a dropping resistor is a reasonable choice. When your setting the system up, I would measure the current.

http://www.amperor.com/products/led/constant_voltage_constant_current_led_driver.html

I linked a discussion on LED drivers... Worth noting is the current VS voltage curve.

My experience is that for good life you want to be a bit back from the manufactures max current rating.

If you can supply the number of LED's and a spec sheet we might be able to get something a bit closer. Even the wire gauge and the length of the run can be factored.

 

[ishiboo]

I would use a PWM LED driver. They are not expensive.

 

[gatchel]

As long as the resistor values used are conservative, there is nothing wrong with using an appropriately sized resistor.

If you are using larger LED's like the Luxeon stuff or equivalent, then a constant current driver makes more sense due to the cost of the LED's alone.

 

[ForceFed70]

He's already got the LED's and the power supply. Upping the voltage and installing a resistor is the easiest and cheapest way to do this.

Aiming for 30mA each and with 84 LED's in parallel the total current is approx 2.5A

As a starting point I would aim to scrub off 1V.

Currently you have a line loss of approx 0.5V which you would think would give you 3.0V at the LED's (which is probably in range of what your LEDs will handle - check the Vmax rating) BUT, your LED's won't see that much. As you increase the voltage you are going to increase the current. More current = more line loss. So It'll be less than 3V at the LED's. How much less - you won't know until you try it. Although we could make a more accurate guess if you are able to measure the output current as your circuit sits now.

OK, so we want to scrub off 1V at 2.5A
V=IR
R=V/I
R=1/2.5
R=0.4

So you will want to try a 0.4ohm resistor in series. Which may be a little difficult to find, but I'm sure you can put some common sized resistors in parallel to find the net resistance you desire. Or buy a rheostat (adjustable resistor).

Also make sure you use resistors of a high enough wattage. You're at approximately 1W of power that you need to remove. So use resistors rated at 1W or higher.

Oh.. also consider just putting in a larger wire. That will drop your line loss and negate the need for resistors, etc all together.

 

[Kevin C]

I missed some of the original info reading the post on my phone.

This is what I came up with.

100 LED's in parallel that each draw 30 ma.

At 2.7 volts across the junctions they should draw 30 ma each 100 * .03A = 3 amps of draw for the whole system.

If your source voltage is 4 volts and you have a .5 volt drop on your wires ( assuming that's the total drop and not each side). The actual voltage that is available is 3.5 volts.

3.5-2.7 = .8 V

V=IR Solving for R

V/I = R .8/3= .266 ohms.

The power dissipated is I squared * R 3*3*.266= 2.5 W

If you have a higher voltage supply ~ 12 V It would look like this.

Wire loss 12-.5 = 11.5V

11.5- 2.7 = 8.8 - this is what you need to drop across the resistor.

V= IR Solving for R
V/I = R 8.8 / 3 = 2.93 ohms
Power dissipated in the dropping resistor ~ 3 squared * 2.93 = 26 watts.

The voltage drop on the wire will increase as the current increases. As you get closer to 2.7 volts the current will increase pretty quickly , as will the voltage drop across the feed wire.

I attached a link to several resistors that would be a decent match with a 12 V supply. Things to double check would be can the gauge of the wire you used handle the current?

For long term life, your better off running at 70% or less of the chips rating.

http://www.digikey.com/scripts/dksearch/dksus.dll?pv1=3708&pv1=849&pv1=850&pv1=3713&pv1=382&pv2=17&pv2=544&pv2=239&pv2=81&pv2=13&pv2=1199&FV=fff40001%2Cfff80482&k=resistor&mnonly=0&newproducts=0&ColumnSort=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25


That's the best I've got on a Thursday night....