Mag hotwire question

[Justin Case]

IMO, a really nice advantage of using the JimmyM driver is that you can load up on a big battery stack to drive your hotwire, thus giving you lots of run time. For example, in direct drive, with something like the FM1909, you are limited to 3xLi-ions such as 3x18650, 3x26500, and 3x26650. But with the JimmyM driver, no problems loading up a 4D Mag, for example, with 5x26500 to run the 1909. You get regulated output and 5 cells vs 3 cells of battery power. Similarly, instead of running a WA1111 on 2xLi-ion, you can load a 2D Mag with 3x26500 for example and get 50% more battery capacity. Of course, this is not an perfect apples-apples comparison. You could run the WA1111 in a SureFire M6 using a 2x18650 holder, and that host is smaller than a 2D Mag. So to get the greater battery capacity, you also have to move up in volume.

I pack the top of my Mag switch with some ceramic fiber insulation. This is one source.

 

[charlestt]

How about using NIZN batteries go give a higher voltage ? in a 6aa configuration..

 

[Justin Case]

CPF reports from 2010 and 2011 indicate that the Powergenix AA NiZn cells seem to suffer from consistency and reliability issues. Battery Guy found consistent leakage under a 3C discharge rate. That doesn't bode well for powering a 5761. I'd stick with Eneloops or Li-ions.

 

[jcvjcvjcvjcv]

IMO, a really nice advantage of using the JimmyM driver is that you can load up on a big battery stack to drive your hotwire, thus giving you lots of run time. For example, in direct drive, with something like the FM1909, you are limited to 3xLi-ions such as 3x18650, 3x26500, and 3x26650. But with the JimmyM driver, no problems loading up a 4D Mag, for example, with 5x26500 to run the 1909. You get regulated output and 5 cells vs 3 cells of battery power. Similarly, instead of running a WA1111 on 2xLi-ion, you can load a 2D Mag with 3x26500 for example and get 50% more battery capacity. Of course, this is not an perfect apples-apples comparison. You could run the WA1111 in a SureFire M6 using a 2x18650 holder, and that host is smaller than a 2D Mag. So to get the greater battery capacity, you also have to move up in volume.

I pack the top of my Mag switch with some ceramic fiber insulation. This is one source.

You can run a WA1111 in a Maglite 6D with six Ni-MH D cells. 72 Wh
In fact, I have one laying in the closet, but I don't have enough D cells to power both my P7 and that WA1111 beast, so I might switch it to something else. But the reflectors with the smaller hole are hard to get last time I checked.

 

[charlestt]

You can run a WA1111 in a Maglite 6D with six Ni-MH D cells. 72 Wh
In fact, I have one laying in the closet, but I don't have enough D cells to power both my P7 and that WA1111 beast, so I might switch it to something else. But the reflectors with the smaller hole are hard to get last time I checked.

But i only have a 2d available, the only 3ds i have are sporting an SSt-90 and a triple XML-U2

 

[Justin Case]

You can run a WA1111 in a Maglite 6D with six Ni-MH D cells. 72 Wh
In fact, I have one laying in the closet, but I don't have enough D cells to power both my P7 and that WA1111 beast, so I might switch it to something else. But the reflectors with the smaller hole are hard to get last time I checked.

That's a nice way of getting a lot of Wh. But if I were to go to a 6D length, I'd favor more output than from a WA1111. Also, 7xIMR26500 is around 60 Wh, with a huge advantage in Vbatt. 6xD NIMH barely hits the Vbulb that I would prefer to use for the WA1111 when using a JimmyM driver. The voltage advantage of the IMRs gives me a lot more flexibility in selecting various incandescent bulbs. I could run a WA1111, WA1185, FM1909, or one of the many Osram superbulbs.

 

[Justin Case]

Post deleted.

 

[charlestt]

If you want cheap, bright, and fun, I'd say that your path sounds like a ROP High to me.

ok, now down to power i've just noticed that battery junction has some really good deals on AA. I notice they have Tenergy cells cheap, are they any good ? I have there C version and have been very impressed so far

 

[nighttrails]

That's a nice way of getting a lot of Wh. But if I were to go to a 6D length, I'd favor more output than from a WA1111. Also, 7xIMR26500 is around 60 Wh, with a huge advantage in Vbatt. 6xD NIMH barely hits the Vbulb that I would prefer to use for the WA1111 when using a JimmyM driver. The voltage advantage of the IMRs gives me a lot more flexibility in selecting various incandescent bulbs. I could run a WA1111, WA1185, FM1909, or one of the many Osram superbulbs.

I'm considering the JimmyM driver for a 5761 bulb, and 6 D nimh batteries does surpass the Vbulb. Or maybe I would use 7 D nimh or 6 26650 batteries. Would the Vlow setting be dependent on number and battery type, or would one setting work with any of those configurations. I'm not clear on how to arrive at the proper Vlow setting.

 

[Justin Case]

I'm considering the JimmyM driver for a 5761 bulb, and 6 D nimh batteries does surpass the Vbulb. Or maybe I would use 7 D nimh or 6 26650 batteries. Would the Vlow setting be dependent on number and battery type, or would one setting work with any of those configurations. I'm not clear on how to arrive at the proper Vlow setting.

Vlow is the voltage level that triggers a low voltage warning. Shortly after that, the light goes into shutdown based on the driver's low voltage shutdown setting.

Since Li-ions and NiMH have vastly different voltages, you can't use the same Vlow for 6xNiMH and 6xLi-ion. Think about it. NiMH cells can be discharged down to 0.9V. Do that with Li-ions and you'll kill them rapidly.

If you think your NiMH cells are solid, well-balanced, and unlikely to get unbalanced, then perhaps a Vlow of 6.0V (1.0V per cell) is the simple choice. It depends somewhat on your risk tolerance and on what you choose for your low voltage shutdown setting. if you stick with the default of 30 sec or 94%, then I'd say it is highly unlikely that good quality D NiMH cells will get overdischarged with an additional 30 sec of run time past Vlow. However, if you are using smaller cells like AA NiMH and you've chosen something like 5 min or even 10 of extra run time before shutdown, then I might estimate Vlow using the equation 5*(strong cell voltage) + (weak cell voltage). This equation assumes that one cell goes weak on you. The value for the weak cell voltage should probably be 0.9V, which is the typical terminating voltage for NiMH. The strong cell voltage could be the quasi-steady state voltage for the NiMH under the load imposed by the 5761 (i.e., about 5.5A). That voltage for Eneloops is about 1.15V. So Vlow becomes about 6.65V, or about 1.1V per cell.

For six 26650 cells, I think a similar approach can apply. For big IMRs and the default shutdown setting, perhaps the default Vlow could be 18V (3.0V per cell). If you are worried that you have a lot of cells in series such that the chances that one cell goes weak could be unacceptably high, and maybe your shutdown setting lets the light run for much longer than 30 sec, then maybe Vlow ~ 5*3.7V + 2.5V = 21.0V.

 

[darkknightlight]

I apologize if this is a stupid question! Does using a larger battery stick in conjunction with the JimmyM driver (as you've described below) create a lower battery current draw and so extend runtime? If so, is there a way to estimate current draw given a specific pack voltage and bulb voltage? Thanks for your help, and any information you can provide!

IMO, a really nice advantage of using the JimmyM driver is that you can load up on a big battery stack to drive your hotwire, thus giving you lots of run time. For example, in direct drive, with something like the FM1909, you are limited to 3xLi-ions such as 3x18650, 3x26500, and 3x26650. But with the JimmyM driver, no problems loading up a 4D Mag, for example, with 5x26500 to run the 1909. You get regulated output and 5 cells vs 3 cells of battery power. Similarly, instead of running a WA1111 on 2xLi-ion, you can load a 2D Mag with 3x26500 for example and get 50% more battery capacity. Of course, this is not an perfect apples-apples comparison. You could run the WA1111 in a SureFire M6 using a 2x18650 holder, and that host is smaller than a 2D Mag. So to get the greater battery capacity, you also have to move up in volume.

I pack the top of my Mag switch with some ceramic fiber insulation. This is one source.

 

[Justin Case]

The easiest way to think about how run time changes is to compare the bulb power draw to the energy of the batteries. For example, if you run a 5761 at 7.2V, the current draw should be around 5.5A, for a power draw of 40W. If you ran the light on 5xIMR26500 and use the nominal values for voltage and mAh, you would get 5*3.7V*2.3Ah = 42.6Wh. Thus, approximate run time is 42.6/40 = 1.1h. If you used only 2xIMR26500, you would have 2/5 of the energy, and thus 2/5 the calculated run time.

 

[darkknightlight]

Thank you for the explanation! It all makes sense now

Sent from my DROID RAZR using Tapatalk 2

 

[jcvjcvjcvjcv]

The easiest way to think about how run time changes is to compare the bulb power draw to the energy of the batteries. For example, if you run a 5761 at 7.2V, the current draw should be around 5.5A, for a power draw of 40W. If you ran the light on 5xIMR26500 and use the nominal values for voltage and mAh, you would get 5*3.7V*2.3Ah = 42.6Wh. Thus, approximate run time is 42.6/40 = 1.1h. If you used only 2xIMR26500, you would have 2/5 of the energy, and thus 2/5 the calculated run time.

When talking high current, nominal figures are a bit too rosy. More of the battery's energy will be turned into heat in the cell itself.

 

[BVH]

Look up Peukerts Factor/Peukerts effect.

 

[Justin Case]

When talking high current, nominal figures are a bit too rosy. More of the battery's energy will be turned into heat in the cell itself.

Yes, of course battery discharge performance varies with load. Sure, I can point to a 26500 discharge graph, wax eloquently about integrating the area under the curve, etc etc. For what? A small gain in accuracy? Certainly, not for any gain in clarity. I can linearize the non-flat part of the 26500 discharge curve (as seen here, for example), take the mid-point voltage value, which looks like it is about 3.6V, and multiply it by about 2.3Ah (since the drop-off is quite steep) to get a slightly more accurate estimate of the energy. So instead of 3.7V*2.3Ah, I now have 3.6V*2.3Ah. One part in 37 difference. BFD. The point was to provide a conceptual example of how to think about how run time improves with a bigger battery stack. It varies as the battery energy capacity varies, which the sample calculation clearly shows. Five nines accuracy is not necessary. However, if you think that the added accuracy adds to the conversation, I look forward to seeing your more refined analysis.

 

[jcvjcvjcvjcv]

Yes, of course battery discharge performance varies with load. Sure, I can point to a 26500 discharge graph, wax eloquently about integrating the area under the curve, etc etc. For what? A small gain in accuracy? Certainly, not for any gain in clarity. I can linearize the non-flat part of the 26500 discharge curve (as seen here, for example), take the mid-point voltage value, which looks like it is about 3.6V, and multiply it by about 2.3Ah (since the drop-off is quite steep) to get a slightly more accurate estimate of the energy. So instead of 3.7V*2.3Ah, I now have 3.6V*2.3Ah. One part in 37 difference. BFD. The point was to provide a conceptual example of how to think about how run time improves with a bigger battery stack. It varies as the battery energy capacity varies, which the sample calculation clearly shows. Five nines accuracy is not necessary. However, if you think that the added accuracy adds to the conversation, I look forward to seeing your more refined analysis.

With some cheaper cells the difference in the percentage of rated capacity you get between a load of C and 2C is much bigger then it is with AW IMR cells. If you look at the two last graphs in that post you see what I mean. A load of 4C on the 16340 cell gives ~92% of rated capacity, a load of 5C only gives 77%; that's 16% less. Said otherwise: an hour versus 50 minutes.

 

[Justin Case]

With some cheaper cells the difference in the percentage of rated capacity you get between a load of C and 2C is much bigger then it is with AW IMR cells. If you look at the two last graphs in that post you see what I mean. A load of 4C on the 16340 cell gives ~92% of rated capacity, a load of 5C only gives 77%; that's 16% less. Said otherwise: an hour versus 50 minutes.

Yes, that's nice, but I specifically cited the IMR26500 in post #32 on this topic. Why would I want to mess around with cheaper cells? As for the smaller 16340 cells, of course they will suffer even more from higher discharge rates. But again, all along I have been discussing the bigger IMRs like the 26500 and the 26650. Also, you may note that the IMR16340 cell is rated only to 4A continuous, so it isn't a good cell to use to drive a 5761 anyway.

Also, even if the capacity decreases with discharge rate by more than a few percent vs the nominal ratings, does the capacity still scale reasonably closely with the number of cells? That, after all, was the point. Two IMR26500s will give roughly 2/5 the run time of 5xIMR26500. If you are off by 10%, do you really think that that changes the overall concept of estimating run time vs battery stack size? I don't think so. Frankly, IMO 10% is probably within the noise of the manufacturing variation for the cells themselves, convolved with variations in parasitic resistance from flashlight to flashlight that also wastes battery power.