running a P7 on 2 D alkalines?

[big vin]

Just finished building a cutdown 1D mag, with a D size li ion, works like a charm.

I have some spare parts left (heatsink, led, switch etc), and wondered if it is possible to direct drive a P7 in a standard 2D maglite, using normal alkaline batteries. I know that i won't get full power out of it. Seems to me it would however be a improvement of the standard bulb.

Is this possible, and how many lumens/runtime should i be able to get?

Thanx, Vincent

 

[Justin Case]

At least based on the P7 datasheet, 3V nominal input could result in a forward current of about 525mA (130mA per core). That could give you a relative luminous flux of about 0.4, or about 180 emitter lumens for a D flux bin P7.

The challenge is that even a very small voltage drop due to contact resistance, switch resistance, etc can drop the 3V nominal input voltage by a few tenths. That is enough to really reduce your output. The datasheet shows zero output at Vf~2.5V. So, you have a fairly small window (even smaller if the alkalines can't hold their voltage under load).

 

[big vin]

mmmm, any suggestions for another more suitable led, in the 300 lumen range perhaps that i can use? and that will fit on my h22a P7 heatsink?

 

[big vin]

and what would it do in a 3D?

 

[gswitter]

Does it have to be alkaline D cells?

If you already have the 2D host, just run the P7 direct on 3x NiMH C cells.

 

[Justin Case]

and what would it do in a 3D?

Look at the P7 datasheet, available on the web. Examine the Forward Current vs Forward Voltage graph. For 3D, you will have a nominal Vf~4.5V, assuming no voltage sag for the 3xalkalines. Realistically, however, this is going to be the problem right off -- at that voltage, you are going to demand over 3A from the alkalines, and alkalines have a hard time with high discharge rates. Thus, the 3xalkalines will no doubt quickly sag down to 1.2V per cell (3.6V total) and continue to head downward. But you at least will get some high initial brightness, followed by acceptable output as the alkalines drain and sag in voltage.

Pick off the corresponding forward current for that forward voltage of 3.6V. That's probably going to represent your initial, best case output. Basically, you will be close to full power output since 3.6V is essentially the Vf at 2800mA drive current.

It's hard for me to say what the quasi steady state Vbatt will be as the alkalines sag in voltage, but let's guess 1.1V per cell under load, or 3.3V total.

Look at the graph of Forward Current vs Foward Voltage for the P7's datasheet. Find the value for forward current, given a forward voltage of 3.3V. That looks like about 1050mA drive current to me (about 260mA per core).

Then look at the Relative Luminous Flux vs Forward Current graph. Pick off the relative luminous flux value for the given forward current.

Next, reference the P7 binning and labeling datasheet. For a D flux bin P7, the LED is rated at 800-900 lm at 2800mA total (700mA per core). De-rate that lumens value to get the lumens output at 1400mA drive current (350mA per core) by dividing 800-900 lm by the value for relative luminous flux at 2800mA (about 1.75 by my eyeball). That should give you about 450-510 lm at 1400mA. Finally, multiply 450-510 lm by the relative luminous flux value that you picked off for your specific drive current (looks like about 0.75 to me). I get about 340-380 emitter lumens.

Here is the easy way -- search CPF for Daekar's Excel spreadsheet tool. Then all you have to do is figure out how much drive current per core your battery config might deliver to the LED. Plug that into the Excel tool and read off the estimated lumens value.

Doing it that way, I get 350-395 emitter lumens. Seems like reasonable agreement.

If you look at the alkaline shootout data for D cells, at 1A draw, D alkalines show a slow, steady voltage decline vs time. But you'll probably be close to 1.1V per cell for a while.