Lantern MOD

Bearcat

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I have an old Coleman lantern that has a incandescent bulb powered by 4D batteries with a brightness control. I tried to putting one of those PR based drop-in LEDs in, but that did not work, because the lantern has reverse polarity and there is no practical way to reverse the polarity. So, I ordered some of these http://cgi.ebay.com/50pc-5mm-Super-Bright-White-LED-Lamp-55-000mcd-FSHIP_W0QQitemZ250070725753QQihZ015QQcategoryZ66952QQtcZphotoQQcmdZViewItem

The LEDs will be behind a large white globe, so I thought that I would arrange maybe 3+ LEDs in series. I am completely lost, because I do not know how many LEDs in series would not require a resistor, how bright, run-time, or what size of resistor I should use. Keep in mind this is my very first MOD.

Please, I need lots of help.
 

Skibane

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Bearcat said:
I thought that I would arrange maybe 3+ LEDs in series.

According to the specs on your ebay page, these LEDs require 3.2 to 3.6 volts apiece. Thus, if you put three of them in series, you would need batteries capable of supplying 9.6 to 10.8 volts - much too high to be supplied by your 4 D-cells.

However, you could instead connect as many LEDs as you wanted in parallel to the batteries (i.e., with each LED receiving current from the batteries via its own connection). Note that since your battery voltage is considerably higher than the recommended LED voltage, you would need to connect a dropping resistor in series with each LED to reduce the batteries' 6 volts down to the approximately 3.4 volts required by each LED.

Here's how to figure the proper resistor value to use:

The resistor will have to drop the difference between the battery and LED voltages:

6.0 - 3.4 = 2.6 volts

This is the voltage that will appear across the resistor.

Next, note that the specs on the ebay page also mention that the recommended current for these LEDs is 20 mA (or 0.02 amps) apiece.

Since each LED will be connected in series with a resistor (and all parts of a series-connected circuit carry the same current), you now know that the current flowing through each resistor should also be 20 mA (or 0.02 amps).

Since you now know both the current flowing through each resistor and the voltage it needs to drop, you can use Ohm's law to calculate the proper resistor value:

V = I x R, or R = V / I

Plugging in your voltage and current to this equation:

R = 2.6 / 0.02

or R = 130 ohms.

This is the value of resistor you would connect in series with each LED (using one resistor for each LED).

In practise, you could probably safely drive the LEDs a little harder than 20 mA, which would be accomplished by actually using a lower value for each resistor (perhaps 100 ohms or so).

Good luck, and happy modding! :D
 
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flash_bang

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wow, this is scary, we just finished the electricity unit with ohm's law and stuff in it last week. anywho, fwiw, what about using some A23's hooked up to a 36W LED array?
 

Bearcat

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Thanks Skibane, It is past my bedtime and I need to digest what you are saying with a clear mind before I start asking a lot of questions. I know that one my questions is going to be, will 2 leds work without a resistor or is better to have the leds in parallel with a resistor?

Thanks again for all your help.
 

Skibane

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Bearcat said:
will 2 leds work without a resistor or is better to have the leds in parallel with a resistor?

I don't see any way you're going to get around using resistors. Connecting one LED directly to the batteries would quickly fry the LED, since you'd be attempting to operate a 3.4 volt LED on 6 volts.

Connecting two series-connected LEDs to the batteries wouldn't work, either, since you need at least (3.4 + 3.4 = 6.8 volts) to run the LEDs at full brightness, and your battery bank is only capable of supplying 6 volts.

Again, as previously mentioned, if you connect each LED in series with its own resistor, and then connect that pair to your battery bank, you should have no problems. Note that you can add any desired number of LED/resistor pairs in this manner, since each one will have its own connection to the batteries.
 

Bearcat

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Baby steps for the novice here. OK, 6 volt power supply, one LED needs 3.4 volts to operate properly, 6 volts less 3.4 volts = 2.6 volts, recommend current 20 mA, need 130 ohms resistor, but can drive LED even harder with a 100 ohms resistor, thus making it brighter. Still a little fuzzy about volts vs. current, but I do not need to know every thing about an engine to drive a car either. I can always ask which resistor to use or look at one of those handy charts without completely understanding everything and get by.

If I take two or more LEDs and connect the + to + and the - to -, I would have a parallel circuit. Should I have a separate resistor for each Led or would one resistor work for all the LEDs and should the resistor go on the neg. or the pos. side of the LEDs, since it is DC current?

Since I do not know how bright these LEDs are, until I experiment with them a little, I am not sure how many I will use in the lantern. Can someone give me an estimate on run-times if I use 1,2,3 etc say using a 100 or 130 ohms resistor?

Actually, the main reason I am getting these 55,000 LEDs is to try them out in some of my Photon clones, because another member recommend them for that. Also, since they come with the resistors to use with 12 volt circuits, I am going to use them for indirect lighting in my fishing boat.

I am just a flashoholic that does not know a whole lot and needs a helping hand ever once in a while, so just bare with me, because I will ask questions.

Thanks for all the help.
 

Scott Packard

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Your boat might be a 11-16V power supply, so you may get irritating changes in brightness levels with a straight resistor. Bet you didn't know that your car's dashboard lights have a voltage regulator circuit driving them. I'm guessing it was put in there because of that.
 

Bearcat

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Scott Packard said:
Your boat might be a 11-16V power supply, so you may get irritating changes in brightness levels with a straight resistor. Bet you didn't know that your car's dashboard lights have a voltage regulator circuit driving them. I'm guessing it was put in there because of that.

I have not thought about that, but the 12 volt incandescent lights, such as the running lights and anchor light seems to work OK. Been thinking about trying these LEDs for my anchor light, since the incandescent light pulls a lot more current.
 

wwglen

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If it is going to be behind the frosted globe just:

1. Buy a PR based socket.
2. Wire the PR socket into the socket on the lantern making sure the polarity changes.
3. Put the PR based LED bulb into the socket.

wwglen
 

watt4

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Bearcat said:
the lantern has reverse polarity and there is no practical way to reverse the polarity.

old, incan, ......


try reversing the batteries ?
 
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Bearcat

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watt4 said:
old, incan, ......


try reversing the batteries ?[/QUOTE]

There for a while, I felt like the guy who has been working and pulling on a lawnmower all day, who forgot the obvious; check to see if the tank had any gas in it.
ohgeez.gif


Reversing the batteries would work, but the only PR based LED drop-ins that I have are 3 volt and the lantern is 6 volts. I'm not going to spend big bucks on that old lantern. Besides, I already got the LEDs ordered and I really would like to do my first MOD.

Thanks, I am going to store "try reversing the batteries?" in my memory bank for future reference.
 

Skibane

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Bearcat said:
Should I have a separate resistor for each Led or would one resistor work for all the LEDs and should the resistor go on the neg. or the pos. side of the LEDs?

It's better to use a separate resistor for each LED. When you are using more than one LED in your project, this approach ensures that each LED will have the desired current (i.e., 20 or 30 mA) flowing through it. If you were to use one resistor for several parallel-connected LEDs, they wouldn't share the current equally (due to slight variations during their manufacturing process) - some LEDs would be brighter than others (and might possibly be damaged due to excessive current). Anyway, resistors are dirt-cheap, and available at your nearest Radio Shack - might as well use 'em!

Here's the hookup for each LED/resistor pair:

ledhookupnd3.gif


Note that the LONGER lead on the LED needs to go to the POSITIVE side of the battery - If you accidentally reverse the leads, the LED won't light, and might even be damaged.

It doesn't matter which side of the LED you put the resistor on - although this diagram shows it connected to the LED's more negative (shorter) lead, it could have just as easily been shown connected to the LED's more positive (longer) lead. As long as the LED's longer lead still goes to the battery's positive side (even if it has to travel through the resistor to get there!), you're OK.

You can build as many of these circuits as you want, stuff them all inside the same flashlight, and connect them all to the same batteries. (BTW, this is the same circuit that some PR-base LED bulb conversions use - the resistor is hidden inside the metal bulb base).

Can someone give me an estimate on run-times if I use 1,2,3 etc say using a 100 or 130 ohms resistor?

In theory, Alkaline D-cells are capable of providing approximately 20 amps for one hour, or 10 amps for 2 hours, or 5 amps for 4 hours, etc. - any combination of amps and hours that adds up to 20 Amp Hours is what they'll supply.

Since your 100 ohm resistor should make each LED draw 0.03 amps (or 30 mA), your D-cells should be capable of operating one LED for

20 Amp Hours / 0.03 amps = 666 hours (or almost 1 entire month! :D)

If you install 2 LEDs (each drawing 0.03 amps), the run-time of your batteries will be cut in half (333 hours).

If you install 3 LEDs, it will be cut in third (222 hours), and so on.
 
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Bearcat

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Thanks a million Skibane, for the lesson in basic 101 LED modding, you are a great teacher. I am watching for the mailman everyday to bring me my LEDs, because I have several lights that I can use them in. I am headed to Radio to get the resistors that I need.

Thanks so much!
Bearcat

PS: To run these LEDs at 30ma, with just 2 cells rather than 4, what size resistor would I need? 50 ohms?????
 

2xTrinity

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Instead of using 4 D-Cells and dropping the voltage down using resistors, you would be better off replacing one of the cells with a dummy cell and running this with 3 D-Cells and a much lower value resistor. The fourth battery's power will be completely wasted if you use a resistor to drop the voltage from 6V to 3.6V, so it doesn't benefit you at all. I would also recommend overdriving the LEDs slightly at the beginning -- this is because the batteries themselves have internal resistance that increases as they wear down -- so when the cells are no longer fresh, current will be closer to 20mA for a longer period of time. This is the general reason why flashlights gradually get dimmer throughout their runtime, though this won't be as noticeable running such low current, and due to the fact that D-cells have fairly low resistnace (the bigger the battery, the lower the resistance in general).

Also, evne though NiMH battery is a lower nominal voltage, those also have lower internal resistance, so in higher current direct drive applications, those are actually more likely to overdrive LEDs than alkaline.

To give an extreme example of internal resistance, look at a typical coin cell flashlight. Those typically produce a current of about 60mA (major overdrive of the LEDs), from two 6V batteres. This means that about 2.5 volts are dropped, so the internal resistnace of the two 2016 coin cells is 42 ohms, or 21 ohms each.

PS: To run these LEDs at 30ma, with just 2 cells rather than 4, what size resistor would I need? 50 ohms?????
2 Cells will be 3.0 volts when fresh. Not enough to drive the LED at any more than ~8-10mA. You will need to start off with at least three as the forward voltage of a typical LED is 3.6V.
 
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Skibane

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Bearcat said:
To run these LEDs at 30ma, with just 2 cells rather than 4, what size resistor would I need? 50 ohms?????

Your LEDs require around 3.4 volts. Two D-cells would only provide around 3 volts (or maybe a tad more when they're brand-new). So, it is unlikely that two cells would provide enough voltage to operate your LEDs at full brightness, even with NO resistor present (i.e. both LED leads connected directly to the batteries).

2xTrinity makes a good point about using a dummy cell. Three cells would provide around 4.5 volts, which is still plenty to operate 3.4 volt LEDs. In this case, your resistor would only need to drop (4.5 - 3.4) = 1.1 volts, so

1.1 volts / 0.03 amps = 36 ohms.
 

Bearcat

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I must have had a bad brain cell connection thinking that 2 cells (3 volts) would run these LEDs (3.4 volts), because I remember way back reading something in a post by Skibane saying that he modded old pr based bulbs with 3 volt LEDs without using any resistors for 2 cell flashlights, but those low voltage 3 volt LEDs were hard to find.

2xTrinity, you have good point about using a dummy cell and to run the LEDs off of 3 cells instead of 4. I guess that is why there are so many cheap 3 cell LED flashlights, because it is cheaper to drop voltage than increase it.

Question, do resistors consume energy, if so, how much?

Also, I have numerous 2 cell incandescent flashlights that I would like to convert to LEDs. What is the most economical way to do it without purchasing pr based LED drop-ins? Are there any bright LEDs available that would operate off of 2 cells (3 volts)?

Thanks again, if y'all got the answers, I am loaded with questions.:)
 

Skibane

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Bearcat said:
do resistors consume energy, if so, how much?
Yes, they do - it's dissipated as heat. The amount of power a resistor dissipates is determined by the current flowing it, and the value of the resistor:

Power = Current(squared) x Resistance

For example, with a 36 ohm resistor and 0.03 amps of current,

Power = (0.03 x 0.03 x 36) = 0.0324 watts (or 32.4 milliwatts).

Are there any bright LEDs available that would operate off of 2 cells (3 volts)?
MAG-Light makes a 2-cell version of their 3 watt LED conversion bulb. However, it uses an oversized PR bulb base which doesn't fit many non-MAG flashlights, and also gets hot enough to melt plastic bulb holders. Cost is around $18 at Wal-Mart, etc.

The Super MJ LED (SMJLED) was an excellent PR bulb conversion for 2-cell flashlights that produced quite a bit of light on just 80 to 100 mA of current. Cost was around $7. Unfortunately, it is no longer manufactured, and the replacement version (SMJLED-II) has problems with a short life expectancy. The original version still occasionally shows up for sale on CPF - definitely worth keeping an eye out for.

Nite-Ize makes a low-brightness LED bulb conversion that will work with 2-cell flashlights (but is much brighter and more efficient in 3+ cell lights). It costs less than $9 at Target, Wal-Mart, etc.

Another option is to make your own bulb conversions. As mentioned in a previous thread, the Lumex SLX-LX5093UWC/C is a good LED for this purpose - In a 2-cell (3 volt) flashlight, it draws around 20 mA (direct-drive - no resistor required). It's not as bright as a PR-2 incandescent bulb, but will produce adequate light for a long, long time on one set of AA cells. It costs about 78 cents apiece in 10-piece quantities from Digi-Key. (Digi-Key's part number for this LED is 67-1691-ND).

Also, there are still a few bare (unbased) SMJLEDs for sale from some dealers. This LED will operate OK from 3.0 volts, and is quite bright. The Sandwich Shoppe still shows them in stock for $4.50 ("4-die", "uncut") apiece.

Finally, just about any of the new Cree XR-E high-power LEDs (or the similar Seoul Semiconductor P4 versions) have low enough operating voltages to produce a significant amount of light with just 2 cells (direct-drive/no resistor). They're very efficient (plenty of light for the small amount of current consumed), and don't get hot enough with just 2 cells to require any heat sinking. Look in the Dealer's Corner section for good deals.
 
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