Got my MTE P7 - initial impressions and a couple of questions

shuffles

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Oct 6, 2008
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Received my 5 mode MTE P7 yesterday (12060), a pair of blue 2500mah Trustfires, and the no name charger.

sku_12060_1.jpg


I really like the looks of the light, seems well made, tough, smaller than expected, and lightweight. Fits on my helmet very well. Contrary to some reports, the reflector does seem to be aluminum, not plastic, and the lens seems to be glass.

I charged up the batteries and fired it up last night. I will use it for cycling, so comparing it to my upgraded MiNewt USB (U bin P4), it's definitely brighter with a much larger light area (spill?). They seem to be about the same brightness when the P7 is on medium, but my observations were no where near scientific just firing the lights at the inside of my garage door, down the driveway, and at the fence in the backyard. It did not get noticeably warm while I was using it, but I only played with it for about 15 minutes and it was 25F outside.

The mode changes seemed a bit flaky, though. when tapping the clicky, the modes did not always change in the same order. It's supposed to be Mid/Lo/High/Strobe/SOS, but it definitely jumped all around. One time it would go from Med straight to High, and other times it would go in other orders.

I did not get to take it on a ride this morning as I ride on the road and there were snow squalls and ice on the road.

With all the modes, I'm really wishing I just had two modes: Medium (maybe about 60% of max), and high. I will never need the low or either of the strobes. And so, now for my questions to all you gurus.

I've seen another thread here at CPF that talked about removing the driver and just using the two modes built into the clicky switch, and replacing the resistors on the clicky to give me the high and medium mode. I may do that, but the EE math is a little over my head. Can someone help with the resistors I'd need to use on the clicky? Also, do I just remove the whole driver disk and wire the hot lead presently on the driver stright to the emitter? (I haven't taken it that far apart yet). Finally, are all the clicky switches the same? I.e., will the clicky that comes with this light have the ability to do the two modes and places to connect those resistors?

Thanks for all your help.
 

waddup

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Oct 29, 2008
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1,269
fully charge the batteries and run the light on medium non stop till it switches off, if when you charge the batteries again it works normally,

i will buy your 5 mode and you can go buy a 2 mode :)

i want a cheap p7 , but have seen multiple threads about problems with them, if you happen to have got a light that is working well
:takeit:
 

shuffles

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Oct 6, 2008
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Thanks Waddup, but I think I'll keep it. If I hose it trying to take out the driver, then so be it.

But I'd love some input from someone on what resistor size to use to get full power with 2500mah 18650 battery, and what size to use to get 60-70% power.

Thanks.
 

Jarl

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zero (full power) and 2 ohms (1 watt or more, 60 to 70% power). I'd suggest you won't notice the difference with this resistance, personally I'd go for about 5 ohms for 200 lumens or so for the lower level; this is still brighter than most single LED lights.
 

shuffles

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Thanks Jarl! Actually, that's what I'm hoping for, a small difference in output, but a longer run time. Right now the medium is (from what I can gather) about 35%.

Are all these clickie's the same? That is, do they all have positions for two power levels?
 

shuffles

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zero (full power) and 2 ohms (1 watt or more, 60 to 70% power). I'd suggest you won't notice the difference with this resistance, personally I'd go for about 5 ohms for 200 lumens or so for the lower level; this is still brighter than most single LED lights.

How does one do this calculation? If 2.8 amps is the max to drive the P7 (is that right?), and the voltage is 4.2 (18650), and Ohms is Volts / Amps, then the P7 circuit itself must be 1.5Ohms. If so, do I calculate the Ohms for 60% as of 2.8 - 1.68 amps (the difference needed in Ohms). If Ohms is Volts/Amps, then 4.2/(2.8-1.68) = 3.75 Ohms.

Am I thinking this correctly? When selecting a resistor, do they differ by voltage?
 
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Jarl

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Bear in mind the resistance of the LED increases a bit as drive current drops. I got 1.5 ohms at 2.8A for the P7. To get 60%, then you want about 1.7A, so total resistance is 2.5 ohms (4.2/1.7; 2.8-1.68=1.12A, which is 40% or so), so no, I'm not sure how I got to 2 ohms; try just under 1 ohm, as then the resistance of the P7 at 1.5 ohms should give a total resistance of 2.5 ohms for 60%. I suspect I got my numbers muddled- it should be 1 ohm, 2 watts, rather than 2 ohms, 1 watt.

Armchair EE at work! Keep well back! ;)

You need to select what wattage of resistor as well; this is voltage dropped*current; in this case, 4.2V to about 3.2V (Vs to Vf), and current of about 1.7A, so 1*1.7= 1.7W, then factor in a bit of leeway so it doesn't melt if I've been an idiot again and miscalculated something, and get 2W.

Voltage just kind of sorts itself out; due to variation in Vf and Vs, choose the worst case (Vs max=4.2, Vf min~3.2, probably more like 3.5V though) and go from there to get power.
 

shuffles

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Oct 6, 2008
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Thanks Jarl! I think I get it now, you get the resistor to match the difference between direct driven circuit voltage and the total resistance of the circuit that you want. So I think I want to try it with a .3 Ohm 2W resistor for 90-95% or rated power (so as not to burn out the LED), and 1 Ohm for my 60% or so.

I'm assuming I can get these resistors at Radio Shack or something.

Thanks again for your help.
 

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