Ultrafire MCU-C7 cree R5 flashlight

Hi there,

Anyone out there have any experience with the Ultrafire MCU-C7 (R5) flashlight.

Saw it on Manafont website and feel tempted to get one. I'm looking for a EDC light and the brightest ones all seem to be CR123/RCR123 powered. I like the look and colour of the C7 and it's got the R5 emitter .

Generally, what are your opinions on these sort of EDC lights? I noticed some discussions on the Ultrafire C3 recently.


Sorry. Forget to include the link


i have a BUNCH of the q5 version of this light

powerful for the size

great price for the q5, but to me, the r5 is a little pricey.

recently, the newer MCU-C7s Q5s have been MUCH cheaper quality. with greenish bin leds (probably p4s being overdrived), cheaper reflectors, crappy drivers.

good for the price of course (about 12 bux on DX)

If you want a solid R5, i would go with the 14500 platform. the Trustfire R5-A3. while i plan on testing an MCU-C7s R5, i am waiting for the price to come down some first.

the R5-A3 is about 18, runs pretty good off a AA, and puts out a TON of light. also is an even BETTER thrower than the MCU-C7 q5 (including the older higher quality version i have!)

for 18 bux, its the BEST pocket light you can ask for.

versitile, more comfortable to hold (the longer profile is more appealing to me), and built like a TANK!

go to DX and type R5-A3. comes right up!

I have a old MCU C7s with a R5 (made myself) and is my EDC.

For EDC format IMHO,has more utility than a thrower a flooder flashlight.

If you like the CR123A format (the CR123A has little to envy 14500) may be a good option.

Perhaps the size is slightly low for the heat to be dissipated ...

I do not know if endure continuous use, but for occasional use is a light cannon.

For me, the CR123 sized lights are a better size for stowing in a pocket than the 14500/AA ones. I like the look of that Ultrafire light, but have bought enough lights to last me for a while. There are several needing modding though.

Anyone know what sort of resistance it would be a good idea to fit in order to tame the current for a direct-driven R5 - I'll probably use a 17670 to power it in an old light with a fried driver.

Use these: http://en.wikipedia.org/wiki/Schottky_diode, with appropriate vf drop, but you need to keep them reasonably cool.

otherwise use ohm's law V=I/R, or R=I/V, or 1A/0.4V= ~2-3ohm.

Surely 1.5A and about 3.7V or around 15/37 or around 0.4 ohms. Which I ought to have remembered by myself. Power would be voltage times current. I want 1.5A so power is around 1.5x3.7 or around 5.5W. That means hefty resistors - may have to stuff in a driver as it'd be difficult to fit an appropriate power handling resistor into the head. 1 ohm sounds like a decent starting point. 5 4.7 ohm quarter watt resistors in parallel should work - and will fit in the head.

I'm pretty sure you don't want to drop 3.7v across the resistor, unless you want a flashlight that doesn't put out light.

Those are about right for that light emitter resistor thing, though ;)

Nah - I need 1.5A to the LED and it is direct drive so assuming I have 3.7V available, I need to get 1.5A in the whole circuit which means around 0.4 ohms. But the LED will have some resistance of its own. By paralleling more resistors, you can start with a fairly high resistance and check as you go along - I'd probably start at about 2 ohms and add resistors in parallel till I got the desired current. This way it is possible to balance heat and output.

"3.7v" is total voltage across the system. The resistor is in series with the emitter. Vf of the emitter is known (3.3-3.7v). This is why I estimated about 0.4v across the resistor.

What you've calculated is the power of the led. Much less is dissipated from the resistor.

you guys love to thread jack lol


I was being an idiot and remembered ohm's law wrong. Guess that's empirical evidence that drinking makes you stupid. Taking the simplist view above, I=V/R, or R=V/I, or R=0.4/1.5, or ~0.25ohm, or ~0.6w. Likely 0.4v is an overestimate, according to Cree's spec sheet.

Except that's really not the proper way to solve it. The system is nonlinear, and you need to take--the graphs for Vf vs I for the emitter, I=V/R for the resistor and additional resistive loss in the system, and V vs I for the battery--and find the point at which they all meet.

As an example for the above, assume 1.5A through everything, which will take the battery from about 3.8-3.4v full to discharge (through it's not going to be 1.5A fairly quick in a resistive circuit). Assume you'd only want 1.5A max, we'd take the 3.8v. Also at 1.5A, the emitter according to cree has ~3.5Vf. At 0.3v through everything else, that's only 0.2ohm. There's going to be probably that much resistance through everything in the system (including the battery), so chances are you may be alright without anything.

However to illustrate the problem with these kind of estimates is for instance, this guy only measured 3.36 vf at 1.5a for his sample http://media.photobucket.com/image/xp-g%20vf/jtr1962/CPF/Cree_XP-G_bin_R5_Lumens_Chart.gif

if that's true for the specific emitter you use, the "extraneous" voltage drop will be 3.8-3.36 = .44v. If indeed you only had 0.2ohm in the system, ~1.8A through the system would be likely given the same assumptions (say 3.78v in battery - 3.40 from chart = 0.38v, 0.38/1.8 ~ 0.2ohm).

By the same token, when the battery starts getting drained, it'll get to be around 3.5v at ~1a. Using the graph above, that'll be around 1.1A @ 3.28v for the emitter. However, again, if we use the cree sample data, it's around 900ma.