Coleman 5310 lantern

Brock

Flashaholic
Joined
Aug 6, 2000
Messages
6,346
Location
Green Bay, WI USA
As I usually do when in stores I was wandering through the flashlight isle looking for new things. I came across another Coleman 4 D lantern, #5310, and it has a "dimmer" for an on switch. So for $9.95 I picked it up. I metered it and it pulled 185mA and the lamp was barely even making light. I then ran it up to hi and it was pulling about 600mA and was pretty bright.

Now for the LED part I put in a 4 cell LEDcorp PR replacement. It then ran from 8mA to 30 mA. It was pretty nice, I then put in my dummy D cell so it was a 3 cell light and popped in the 3 cell LED version. On the lowest setting it pulled only 4mA and on high it was 30mA again, same brightness as the 4 cell. So I stuck the 3 cell version.

Now a D cell is 18,000 mA so on it's lowest setting it should run 187 days straight! That's over 1/2 year! On high, 30mA, it should run 25 days, about a month. I am set. At it's dimmest it is more then enough to use as a marker or night light, and on hi it is enough to get around at night. Nice conversion, and I really like the dimmer.

I only wish the light had a white plastic top like the sides instead of the dark green plastic. I ended up painting the underside of the gree cover white and it made the light noticably brighter. If I were using it in a dry situation for a long time I would remove the top to let the LED light up the ceiling for the extra light.
 
Yup, Edwood that is the one. I didn't have to take it apart at all really. I just replaced the PR lamp with the PR LED and put in 3 alakline cells and a blank D cell. I have now hard wired across the 4th cell and it works fine. The other three cells are held in place by spacers already in the light. So it was extremely easy to do.

If you use a 4 cell PR LED, all you need to do is replace the lamp. I was suprised the dimmer worked with such a small load from the LED, maybe I just got lucky. I got mine at Fleet Farm, I haven't looked at Walmart or Home Depot yet. Fleet Farm is a local home supply chain (Midwest USA).
 
Brock - Your mention of the 3 and 4 cell versions of the LEDCorp PR bulbs reminded me of a question. Since these bulbs only use a resistor to step down the voltage, isn't the 4 cell version simply wasting more energy in order to get down to the correct voltage? I haven't played with the math yet, but my thinking is that the 4 cell version is essentially "wasting" the energy of the 4th cell or more accurately wasting 1/4 the energy of each cell. If this is the case, a 4 cell light and a 3 cell light would produce identical output and runtimes.

I have the 4 cell bulb and consider it a good product since many flashlights use 4 batts. But energy wise, it probably makes more sense to only buy the 3 cell bulb and use a dummy battery. I'd like to hear what you all think.
 
Yup, Ron I also think you are correct. That is why I beleive on the lowest setting it was pulling 8mA with the 4 cell and 4mA with the three cell. Both topped out at about 30mA so I am assuming the extra 4mA was the resistor in the 4 cell version. So I figured I might as well make it a 3 cell lantern and save the 4mA.
 
RonM: Yep. I think you are correct also. The other night I measured the amperage draw on my 4 cell LEDCORP bulb when using 4 new Lithium 1.6v+ batteries and discovered that the LED was pulling something like 98ma! I thought at first that my meter must be busted since it is an old cheap meter anyway, but then I started pondering on it and it just came to me that I was dealing with a LED with a resistor on it and the extra amperage draw must be coming from the resistor. Now I wish that I had gotten the 3 cell LEDCORP bulb and used a dummy battery. I believe, like you said, that the extra battery is serving no better purpose than to "feed" the resistor.
 
the extra Amps the led is pulling is from the extra voltage the lithium batts is giving ,, i think that is right


,,, i know how to calculate the right ressistor for a given voltage and amps

( Batt V - led V ) / Amps = Ohm

,,,, if you change the volts and keep the ressistor the same what is the formular for that ??

Ant
European Illuminati
 
Ant: If you raise the volts, but leave the resistor the same, you end up raising the voltage above the rated voltage of the LED. When you raise the voltage on a LED above what it is rated for then it pulls an unpredictable (at least by using Ohm's Law) amount of current because of thermal runaway. The reason there isn't a formula to cover this situation is because the formula would also have to take into account the external temperature that the LED is subjected to. Another reason it isn't predictable by Ohm's law is because the LED doesn't have a linear resistance (this little caveat makes for some major headaches in calculating LED's power draw). As the voltage goes up, the temperature goes up and the internal resistance of the LED decreases which leads to the thermal runaway phenomenon that we keep going on and on about. I hope I explained all of this correctly. It seems to be more and more of a complicated issue the more I delve into it.
 
What Xcal said.
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That's why I know of no way, besides trial and error, to find the dropping resistor. I don't know the formula to correspond to an LED's internal resistance. Don't even know if there is one.
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My best "tool" is a big variable resistor. It is actually a 75W stereo L-pad for volume control. I can vary it from 0 to about 30 ohms.
Once I find the current I want (to prevent runaway), I get a resistor equal to the L-pad.
You can do the same thing with individual resistors and a small breadboard.
 
Brock, what would the amp draw be if you used the 4 cell led corp bulb...but ran it on 3 cells (4.5 volts)? My goal being a night light that might last a year. I removed the top and put on a clear tupaware top (doesn`t really screw on right , sort of fudged it a little) but seems to give off more light bouncing off the ceiling. I have done this and it still is pretty bright, but I have no way to tell the amp draw.

Would I be better off going with 2 dummy cells (3 volts) and a yellow or red led ?
 
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by KenB:
Brock, what would the amp draw be if you used the 4 cell led corp bulb...but ran it on 3 cells (4.5 volts)? My goal being a night light that might last a year. I removed the top and put on a clear tupaware top (doesn`t really screw on right , sort of fudged it a little) but seems to give off more light bouncing off the ceiling. I have done this and it still is pretty bright, but I have no way to tell the amp draw.

Would I be better off going with 2 dummy cells (3 volts) and a yellow or red led ?
<HR></BLOCKQUOTE>

Ken, you still don't have a DMM?
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Did you not find one, or just way short on cash??

Have you heard of Harbor Freight Tools? They sell tons of cheap tools, both online and in retail stores. I buy stuff from them when quality isn't my main concern.
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How far are you from Orlando? They will be opening a retail store there soon.
The reason I mention them, they have a small DMM for $5 this week in the retail stores. It measures 0-1000vDC, 0-750vAC, 0-10A dc, 0-2Mohms, transistor test, and diode test. Not bad at all and should do everything you need.
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I checked it's accuracy against my $80 full function Craftsman meter. Right on the money. I bought 4 for myself.
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They also offer it online, but it is a few bucks higher, plus shipping.
 
I don't know the formula to correspond to an LED's internal resistance

A dynamic resistance of a device is by definition dV/dI, or the inverse of the slope of the I-V characteristic of the device at a certain V. (Inverse, because if we take a derivative, we get dI/dV, but we need dV/dI).

For a resistor, the I-V charateristic is a straight line, so the inverse of the derivative at any point will be constant: R.

For a diode, the I-V charateristic is an exponential function. So the derivative will be another exonential. The diode dynamic resistance is therefore decreasing exponentially with V.

I also see people use the formula
R = (V_batt - V_led)/I.

That formula is not technically correct, but it does estimate an approximate value, because V_led is changing with I and the formula assumes that it is constant.

There is a "more exact" way to derive R, but there is a bit of algebra involved, and probably no one cares anyway
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.
 
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by mad_scientist:
.....I also see people use the formula
R = (V_batt - V_led)/I.

That formula is not technically correct, but it does estimate an approximate value, because V_led is changing with I and the formula assumes that it is constant.

There is a "more exact" way to derive R, but there is a bit of algebra involved, and probably no one cares anyway
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<HR></BLOCKQUOTE>

Thanks for the explanation.

So, if we had an accurate graph of LED voltage vs current, the simple formula would be accurate?
R = (V_batt - V_led)/I_led

I'm not big on calculus, but I do see where you're going.
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Some of us "care".
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Thanks!
 
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Gadget:
Ken, you still don't have a DMM?
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Did you not find one, or just way short on cash??


How far are you from Orlando? They will be opening a retail store there soon.
The reason I mention them, they have a small DMM for $5 this week in the retail stores. It measures 0-1000vDC, 0-750vAC, 0-10A dc, 0-2Mohms, transistor test, and diode test. Not bad at all and should do everything you need.
smile.gif
I checked it's accuracy against my $80 full function Craftsman meter. Right on the money. I bought 4 for myself.
wink.gif


They also offer it online, but it is a few bucks higher, plus shipping.
<HR></BLOCKQUOTE>

Hi Gadget, I was saving up my nickels and dimes for either a 4000 or LS...but you have wore me down...I`ll have to drag myself out and go shopping for one.

I live in Tampa...Orlando is about one hour away...but...I will be going to Disney World for the 4 th of July (fireworks) and if I can convince my wife to let me, I`ll try to hunt down this new store
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Thanks for the tip
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Ken
 
You don't need the whole graph, actually.
You would just need too points: (V1,I1), and (V2, I2).

The realionship between I and V for any diode is:

V = V1 + k*ln(I/I1), where (V1,I1) is a known "data point", V is the voltage accross the diode at a current I. The constant k is calculated from the knowledge of the second data point: (V2,I2).

This formula is also not ideal, as it does not account for the effect of temperature, that would affect k.
 
LOL Doug, you remember my post about how to get my Radio Shack one to do this eh eh.

Well Gadget, give me the weekend to check out some local pawn shops. I tried the link and can get to the first page...but the links all give me a server error message, I will try again later...if the shippings not much I may spend the extra 2 bucks and order it sraight from them.

Thanks

Ken
 
KenB, I don't know how you survive without a DMM.

Make sure to get one that has protection (fuses) that can be replaced when you accidently blast it with more current than it can handle in the current setting. It's easy to do that (at least on my meter) when you think you are measuring volts, but forgot and left it set to measuring current.
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Just so you are fully informed...
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I bought 4 of these for myself. The meter in question looks like so:

3u10981.jpg


Low amperes are fused (250ma).
9v battery is included.
10 amp range is unfused (like many meters).
Leads are 32" long. (not 23" like I put before)

The $5 price is at stores only. Online price is $10. But, there is another identical meter on sale online that adds a backlight for $8.
However, online orders incur a $9 S&H charge.
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Just trying to help.
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I am in no way associated with this place, I just like their deals. I don't order online 'cause their S&H charges suck.
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So, after all that, my offer still stands. If you want one, I'll get one for ya. $5.25 each with tax. Shipping would be from $2.25 to $3.50 depending on Regular vs. Priority (to Tampa).

No big deal either way.
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Anyone else want one, feel free to drop me a line. I'll need $5.25 plus postage. There's nothing in this for me. Just want to help out our little "community".

These make excellent "utility" meters. I only got mine so I could measure volts and amps simultaneously.... on 2 different circuits at once.
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Hmm, that does look nice, Doug, Mike what do you guy`s think?...ok...I`ll take one!

Now...should I go for the backlight? Hmm...any thoughts...well any way e-mail me at [email protected]

I`ll send cash...it`s a small amount and I`ll trust the postal folks not to loose it.

I`ll send $12 or $15 (backlight,if you or others think it`s worth it) to help pay for the gas a little.

Thanks

Ken

PS: you may have to make a lot of trips as that is a great price, if other show intrest, it would be ok with me for you to wait and only have to make one trip...no hurry, but I`ll get the money to you right away
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