Overdriving the Cree R2 @ 1400mA??

[kurni]

The SSC P4 I plan on driving at 1,4A is going to be mounted to a H22A heatsink... will this do any difference? And will three Alkaline D-cells be able to sustain a regulated 1,4A to the LED through a 7135-circuit?

3 D Alks won't cut it.

1.4A LED current means at least 1.5A input.

Alks voltage will drop under such current, and the driver will try to suck more current resulting further voltage sag.

Another Alks drawback is capacity drop when discharged at high current.

See Energizer datasheet.

You better use D or even AA NiMH.

 

[saabluster]

You better use D or even AA NiMH.

energizer e2/ultimate lithium AAs will work for this as well.

 

[Scattergun]

Would it be better to run three Eneloop AA´s in D-A-converters instead of the three Alk D´s??? And what would runtime be in that case?

 

[kurni]

Would it be better to run three Eneloop AA´s in D-A-converters instead of the three Alk D´s??? And what would runtime be in that case?

The benefit of NiMH including Eneloop is its ability to raesonably sustain high current discharge; Alk D is not designed for high current discharge. Notice in the datasheet Energizer only tested Alk D at 500mA in comparison to 4000mA NiMH AA or 1000mA Lithium E2 AA.

• 3 Alk D capacity would be unpredictably low, note that at 25mA it holds 21,000mAh but at 500mA it holds only 11,000mAh; I have no idea how much it holds at higher current but I don't like the trend.

I have 6D M@g with BuckPuck pushing 1A through Q5; with 6 Alk D I see 500mA * 9V = 4.5W, 5 Alk D I see 600mA * 7.5V = 4.5W, 4 Alk D I see 800mA * 6V = 4.8W (BuckPuck efficiency went down), I can't test with fewer cells because BuckPuck minimum input is 5V.

2000mAh NiMH will give you at least 1 hour in practical use; the datasheet infers 1.5 hour runtime but notice the voltage sag; your circuit will draw more than 1500mA as the voltage drops (just like my BuckPuck setup).

I just did a naughty test, a fresh Alk D could actually push almost 9A, yes 9000mA Not so fresh Alk D could only push 2A. I wouldn't dare to do this with LiIon fearing explosion but Alk is safe I have no idea what the voltage was because I only have 1 DMM, theoretically it's near-zero volt because the DMM would have little resistance.

So in theory 3 Alk D should be able to sustain 1500mA draw but nobody know what sort of capacity it would have; note that at the capacity is halved when you increase the discharge from 25mA to 500mA; I can't imagine 1500mA because it's too far from the specification.

To get the most of your 3 D M@g body, I suggest you use 3 NiMH D, or 9 Eneloop paralel holder (3 * 3)

 

[Bullzeyebill]

The benefit of NiMH including Eneloop is its ability to raesonably sustain high current discharge; Alk D is not designed for high current discharge. Notice in the datasheet Energizer only tested Alk D at 500mA in comparison to 4000mA NiMH AA or 1000mA Lithium E2 AA.

• 3 Alk D capacity would be unpredictably low, note that at 25mA it holds 21,000mAh but at 500mA it holds only 11,000mAh; I have no idea how much it holds at higher current but I don't like the trend.

I have 6D M@g with BuckPuck pushing 1A through Q5; with 6 Alk D I see 500mA * 9V = 4.5W, 5 Alk D I see 600mA * 7.5V = 4.5W, 4 Alk D I see 800mA * 6V = 4.8W (BuckPuck efficiency went down), I can't test with fewer cells because BuckPuck minimum input is 5V.

2000mAh NiMH will give you at least 1 hour in practical use; the datasheet infers 1.5 hour runtime but notice the voltage sag; your circuit will draw more than 1500mA as the voltage drops (just like my BuckPuck setup).

I just did a naughty test, a fresh Alk D could actually push almost 9A, yes 9000mA Not so fresh Alk D could only push 2A. I wouldn't dare to do this with LiIon fearing explosion but Alk is safe I have no idea what the voltage was because I only have 1 DMM, theoretically it's near-zero volt because the DMM would have little resistance.

So in theory 3 Alk D should be able to sustain 1500mA draw but nobody know what sort of capacity it would have; note that at the capacity is halved when you increase the discharge from 25mA to 500mA; I can't imagine 1500mA because it's too far from the specification.

To get the most of your 3 D M@g body, I suggest you use 3 NiMH D, or 9 Eneloop paralel holder (3 * 3)

Check out SilverFox's Alkaline shoot out. Scroll down to D cells. https://www.candlepowerforums.com/threads/64660

Bill

 

[kurni]

Check out SilverFox's Alkaline shoot out. Scroll down to D cells. https://www.candlepowerforums.com/threads/64660

Bill

Thanks Bill, I missed the good-old thread Very informative.

 

[TigerhawkT3]

AFAIK, the AMC7135 and other LDOs and linear regulators have Iin=Iout. The energy loss comes from reducing the Vin down to Vout without reducing Iin below Iout (which is what a buck converter would do). So, a Vin of Vf+Vdropout+Vdrop, with Vf being the LED's Vf at whatever current you chose, Vdropout being the regulator's minimum voltage drop (which is very small for LDOs like the 7135), and Vdrop being any additional dropped voltage, would waste (Vdropout+Vdrop)*I watts for an efficiency of Vf/Vin. As Vin falls due to sagging battery voltage, Vdrop will decrease, resulting in a more efficient regulator but, of course, reduced cell capacity. If Vin falls below Vf+Vdropout, the light will basically DD with as much voltage as it can muster, with Vout (no longer a fixed value, but determining the emitter's Vf, and therefore its If, as well as Iin from the battery) being equal to Vin-Vdropout. At that point, efficiency would be Vout/Vin or, equivalently, (Vin-Vdropout)/Vin.

This is AFAIK and I could be wrong, but I hope not.

 

[kurni]

AFAIK, the AMC7135 and other LDOs and linear regulators have Iin=Iout. The energy loss comes from reducing the Vin down to Vout without reducing Iin below Iout (which is what a buck converter would do). So, a Vin of Vf+Vdropout+Vdrop, with Vf being the LED's Vf at whatever current you chose, Vdropout being the regulator's minimum voltage drop (which is very small for LDOs like the 7135), and Vdrop being any additional dropped voltage, would waste (Vdropout+Vdrop)*I watts for an efficiency of Vf/Vin. As Vin falls due to sagging battery voltage, Vdrop will decrease, resulting in a more efficient regulator but, of course, reduced cell capacity. If Vin falls below Vf+Vdropout, the light will basically DD with as much voltage as it can muster, with Vout (no longer a fixed value, but determining the emitter's Vf, and therefore its If, as well as Iin from the battery) being equal to Vin-Vdropout. At that point, efficiency would be Vout/Vin or, equivalently, (Vin-Vdropout)/Vin.

This is AFAIK and I could be wrong, but I hope not.

Hmm, you might be right; I've never understood what "linear" regulator means.

Any regulator will cause power loss; hence Pin > Pout; I won't be wrong about this

P = V * I; this must be correct also

Do you mean that "linear" regulator simply "chops off" excess voltage to maintain constant I?

nFlex and bFlex are not "linear", are they? You can feed 20V to power 4V LED without having to heatsink nFlex and bFlex; if it's "linear" then 16V * 1A = 16W has to be converted to heat, which has to be released through a massive heatsink.

 

[TigerhawkT3]

Yep, that's right.

If a circuit is "buck" or "boost" (or both) then it's not linear. For an explanation of buck and boost, see the Electronics/Electrical section of the Welcome Mat.

Was my explanation of linear regulators clear and accurate? If it was, I might add it to the Welcome Mat.

 

[kosPap]

I am thinking of building a D26-module with a Cree R2. I plan on using the AMC1735 1400mA driver since I want all light possible from this setup.

hmmm shouldn't we also discussing proper module-body contact? one cannot 100% expect proper fit so heat dissipation-transfer should be a better problem that heat-sinking of the board-LED.

I recall a member using tight wound aluminum foil between module and body and he got much better heat tranfer.

till latter, kostas

 

[Scattergun]

From post #18 in this thread

You can greatly enhance the heat transfer of the cheap modules by using foil.

 

[Drewfus2101]

I have done a test or two on a P4 where I ran it at 700mA and 1050mA (through AMC driver) and I got a higher value in my light box with the 700mA one. This was after they warmed up a little. I imagine the problem was the heat of the 1050mA one. So more current isn't always better.

 

[FLT MEDIC]

hmmm shouldn't we also discussing proper module-body contact? one cannot 100% expect proper fit so heat dissipation-transfer should be a better problem that heat-sinking of the board-LED.

I recall a member using tight wound aluminum foil between module and body and he got much better heat tranfer.

till latter, kostas

We use aluminum foil tape for roofs sometimes called roof flashing tape (Sello brand from New Zealand). It's as wide as duct tape but made of aluminum. Just cut to about 7 inches and about the same width as the flat portion of the R2 drop in reflector. Peel off tape backing and stick aluminum foil around the reflector until it's snug inside the flashlight.

It really transfers a lot of heat from the drop in reflector to my C1's bezel, right up to the tailcap switch! :twothumbs