Carley 1057 pr based bulb

Does anyone know where I can get 2 or 3 of these bulbs??? I have a 2c m@glite with a (3 x cr123) adapter in it, and i've heard that the carley bulb is the best for this application... I think carley has a minimum $50.00 order... I guess I'll just have to use kpr112's till i find some...

You might want to post a WTB in the Buy/Sell/Trade Forum.

Wilkey

You'll need a different reflector, unless you only run it in *very* short bursts...

John

Illuminated suggested in his post that one would need a different reflector when using a 1057 bulb. I definitely respect what he's saying, but can anyone offer emperical evidence of that? I know people run some other fairly bright bulbs with the stock (plastic) Mag reflector, and up to some (unknown) point it does okay. The 1057 is rated 6V at 2A. Would someone care to offer a maximum amp guideline for using the stock plastic reflector? Is there a rule of thumb on this?

All I can say is that I tested the 1057 for maybe just a few minutes on 6 x 1/2D Powerstreams and the reflector started bubbling around the hole.

I also attempted to do a runtime test with the 805 on 3 x 1/2D's and the same thing happened - it just took a lot longer.

Both tests had UCL lenses.

John

Yikes! /ubbthreads/images/graemlins/mad.gif

Yeah, I've melted many mag reflectors doing tests with carley and welch allyn bulbs. It scared me the first time. The whole head clouded up with smoke!

The Mag reflector is supposed to be good to around 10W. If you're going much beyond that you might want to look into something like the PMR.

Can the PMR (of course with a different nomenclature) be gotten with the same order as a Carley 1057?

No. The PMR or Perfect Mag Reflector is a product of our very own Otokoyama. It is based on the Carley 1940 reflector. You can get the unmodified reflector from Carley as part of a group buy and then it is easy to modify to fit a Mag head. Here's a thread I posted on how to do it. Modifying the Carley Reflector to fit a Mag Head

Wilkey

Wilky -

Very kind of you to post that helpful link. I had lost it.

illumiGeek -

10 watts eh? That sounds right.

bryguy42 -

My apologies for this way off-topic. milkyspit's question requested information I'm very interested in.

Any -

Being in my sophomore year of flashology with no electronics background, I must admit I have yet to pass some of my freshman courses. Would any of you kind souls check my homework?

Voltage or Power = V
Amps or Current = A
Ohms or Resistance = R
Watts = W
True Watts = V x (A x R)

Driven at spec these bulbs might be rated at these watts?


WA01286: (4.2v .70A): 2.94W

KPR118: (7.2v, 700mA): 5.04W

WA01315: (6.27v 1.42A): 8.9W

WA01318 (9.6V, 1.93A): 18.53W

WA01274 (7.2V, 2.77A): 19.94W

WA01185 (9.6V, 3.15A): 30.24W

What is missing is the resistance.

The wildcards are the volts and amps. Folks using these bulbs use a plethora of power sources involving a myriad of watts, amps and resistance in tubes that include switches with their own resistance.

Some folks have driven a WA01185 @ 15v. Could this really be 47.25W? What were they thinking? Does my homework indicate even a miniscule bit of understanding?

At any rate, it sounds like illumaGeek's 10W rule of thumb is a correct measurement to avoid stock reflector meltdown.

- Jeff

What's missing is not resistance. Not really. The missing factor is voltage drop under load. I have run a WA01185 on a stack of 10xAA nimh, each with a voltage of 1.38V. You'd think the bulb would be pulling 13.8V but it's not because voltage sags under load. PaulW has done some definitive work measuring bulb draws under load and I would invite him to refresh us on his findings. The switches we use are not a likely factor in voltage reduction. Typical good industrial switches are 30 milliohms or less.

When I meant resistance is not really a factor, it sort of is. The internal resistance of the cell determines its ability to supply current and the associated voltage. For example, a 3 amp load will depress the hell out of a SF123 cell. High current nimh cells I use have a 6 mO internal resistance, many times lower than for the SF123 and the voltage sags very little even under a 10 amp load.

Wilkey

I see that with the Aurora project. OK. I'm getting it.

Are these assumed specked watts close?

Right on.

Wilkey

Icebreak,

The equations I think you're looking for are:

R = V / I (R in ohms, V in volts, I in amps)
V = I * R
W = V * I = I^2 * R = V^2 / R

And if you're going to calculate the power (watts) consummed by a bulb under various configurations, you need to use true values for voltage and current in the equations. True voltage is not the nominal, advertised voltage of either the battery or the bulb. Wilkey ably discusses the sag in a battery's voltage as current is increased. This is a property of the cell. How it "sags" is different for each battery chemistry. At Wilkey's invitation, let me amplify on that.

I have discussed Li-Ion properties in the Pila current & voltage question thread. Below, I have pasted and modified that text to show the properties of NiMHs. It describes my findings for my PowerEx 2200 mAh AA NiMHs. You can probably extend the results to NiMH cells with similar capacities

I ran a battery of a few AA NiMH cells into a load at about 2.5 Amps. I continued the test for 42 minutes, at which time the voltage per cell had dropped to 0.97 volts under a load of 2.34 Amps.

Open Circuit Voltage (V.oc). I stopped the current drain at the one-minute mark and every 10-minute mark to test the no-load voltage.

t = 00 . . . . 1.43
t = 01 . . . . 1.37
t = 10 . . . . 1.28
t = 20 . . . . 1.27
t = 30 . . . . 1.25
t = 40 . . . . 1.20
t = 42 . . . . test ended

Voltage Under Load (V.cell & V.bat). During those pauses in the test above, at each mark (including the t = 0 mark), I tested the battery under varying load-- measuring its voltage and current simultaneously. I measured voltage under loads of approximately 0.0, 0,5, 1.0, and 2.5 Amps. From these data I determined that the voltage sag from the open circuit voltage is:
<ul type="square">[*]directly proportional to the current
[*]independent (within the accuracy of my measurements) of cell freshness or depletion[/list]
For these reasons, the voltage under load could reasonably be described by a model of the battery as a voltage source equal to the open circuit voltage in series with a constant resistance of 0.070 ohms. The under-load voltage of the AA cell can be given by the following:

V.cell = V.oc - 0.070 * I, where I is the current.

To get the total battery voltage, just multiply by the number of cells:

V.bat = #(cells) * { V.oc - 0.070 * I }

Bulb Voltage (V.bulb). It is possible to also derive an equation for the WA 01185 bulb, as I did in the 500 Lumen Mag 3D thread. This reference also describes the characteristics of the 123 cell.

The equation for one of the 1185 bulbs I have is

V.bulb = 5.3 * I.bulb – 6.95 (for current between 2.5 amps & 3.5 amps)

If you're into solving simultaneous equations, or if you're into charting (which is more fun), you can determine exactly the current and voltage under which the WA 01185 will operate. Plot the equation for V.bat. The specific equation will depend on V.oc, as given by cell depletion in the chart above. Then plot V.bulb. Where they intersect is where the bulb will operate. If you plot a series of equations for V.bat (for a series of depletion levels), you'll get a series of intersections, which show how the voltage and current drop with time.

While these equations work for me, and probably will work for you, I have to admit that there is some error in using them. This is because during my the tests there were other resistances in the circuit. The cells were tested within the Elektrolumens 3-to-D holders. I had modified them so that the resistance of each holder itself was 0.048 ohms for 3 cells. That would account for 0.016 ohms per cell. In addition, there was contact resistance between the cell and holder, and between the holders and the clip leads attached to them. These were the only high-current connections. I am guessing (and the guess is a little on the wild side) that the stray resistance in a flashlight would be about the same. This is my way of saying that I will continue to use an internal resistance of 0.07 ohms in my calculating gyrations.

I mention all of this because Wilkey has stated that his NiMHs have an internal resistance of 0.006 ohms. I'm hoping that's a typo. I have read that AA NiMH resistances are on the order of 0.020 to 0.060 ohms. I trust Wilkey will comment.

Icebreak, I hope this has been of help and not just a confusing waste of your reading time. /ubbthreads/images/graemlins/smile.gif

Paul

Paul,

The high current Sanyo AUP and KAN nimhs are in the 6 milliohm range. The very high current GoldPeak sub-Cs are in the 3-4 mO range. This is why 30 amps can be pulled out of a roughly AA-sized cell without it exploding. It's the equivalent of a carbuerated vehicle running a Weber 4-barrel or six-pack versus a standard nimh cell which is running a plain old 2-barrel.

Wilkey

Wilkey,

Good. I see we are talking about different classes of batteries. I think you're refering to a higher-current, lower-capacity, different-sized version of an AA. I haven't seen anything really definitive on these cells, although I have read tempting bits here and there in your Aurora thread.

I did a google search but couldn't find specs too easily. But I'm going to keep my eye open for information about size, capacity, cost, etc. I don't want to take us too far OT here, so if you happen to have any specs or links to them, perhaps we could talk off line. Meanwhile, I think the best I can put in my Mag 3D to run a WA 01185 (or in any sized Mag for other WA or Carley bulbs) is standard AA NiMH cells.

Paul

you may also want to replace the lens to a glass one like ucls.

you may also want to replace the lens to a glass one like ucls.

PaulW -

That is a great piece on understanding how all this works. I've read it twice and I think I'm getting the majority of what you are saying. I need to do a little more research to understand some things. Some of the math is past me right now.

I can say I now know much more than I did before I read it. I'll save this, chew on it and use it as a reference. It was somewhat complicated but the flow of the explanation was easy to follow.

Thank you very much for posting this work. It is a little jewel of knowledge.

- Jeff