I’m putting together a Mag 2C with an xhp70.2 but don’t know which driver/battery package will offer better overall performance - more volts/less amps or less volts/more amps?
Looking for some feedback. Thanks.
8.4 volts
- 6v, 5amp, 6-12.6v buck driver.
OR
4.2 volts
- 6v, 4.8amp, 3-4.2v driver.
I would go with the 4.2V/1X 21700.
The first option with the buck driver will have a slight edge in output. For comparison in terms of output here, all that matters (assuming batteries, LED, driver all match, specs. are accurate, etc.) is the amperage rating of the two drivers. The first does 5A and the second 4.8. This is negligible though and would not be noticed without a meter. For runtime though, compare watt hours. So you look at voltage of each cell multiplied by amp hours, and then multiply that by the number of cells. So (3.7x2)x2 = 14.4 for the two 18500’s or (3.7x5)x1 = 17.1 for the 21700. If the two drivers are equally efficient then you will have better runtime with the 21700 option. I would do the 21700 but go with a higher drain battery. I think most boost drivers are below 90% efficiency. That means that 4.8A driver is going to pull over 10A off a fresh cell and as the battery drains, it is only going to get worse. I’d be concerned about those 18500’s too. Even with fresh cells and a 90% efficient driver you are going to probably be pulling close to 4A.
I’d do the 21700 boost build and get a higher drain battery like a Molicel p42.
A pair of 18650 cells will fit in a incan style 2C mag,
A pair of 18650 cells and a Pellican brand hologen bulb was a very popular drop in mod back in the day.
It was called a ROP which was short for Roar of the Pelican.
For a 2C the anodizing inside the tailcap was removed and the spring was reversed which allowed the batteries to use some of the space in the tailcap where the spare stock krypton bulb would have been stored.
Superstocker
Wow, this is great stuff! Ok, I read through it, I got most of it, but I have just a few questions.
- If I understand you correctly, you’re saying that as far as output goes, amperage rating of the driver will determine final output. My question is, does voltage (4.2 vs 8.4) have any bearing on output?
- How do I find out how many amps a particular emitter (3v or 6v) can safely be driven?
- “That means that 4.8A driver is going to pull over 10A off a fresh cell”. I don’t understand what this means.
- You recommend higher amp, high drain cells. Is it possible to go too high based on the other components in the build? For example, could I use 2 high drain cells in series rated at 50amps each with a 6v, 5amp driver and a 6v emitter?
I have a driver question.
If I change from a 6v to a 3v xhp70.2 with a single cell, do I even need to use a driver? Can the emitter handle a fresh 4.2v cell? Assuming I can direct drive the light, are there performance differences going with or without?
Glad to be of help. I owe a lot to a whole host of people on here. I numbered the questions above and I’ll try and answer them here.
(1) The best thing for you would be to try and learn a little bit about voltage and amperage. That will help a lot. The short of it is that neither of those are a power rating. Watts is a unit of power. Voltage times amperage equals watts. So that means 2V times 3A, and 3V times 2A are both 6 watts. So for power, output, whatever you want call it, what matters is the voltage and the amperage at the emitter (now of course emitter efficiency, light transmission through a lens, etc. effect things but for now we will disregard all those). The drivers you are looking at work like this: send a set power rating to the LED and manipulate the output from the battery to get what’s needed. So let’s say, just to make the math easy, driver A sends 6V and 3A to the LED. It is designed to run off one 3V battery. So it draws 6A from the battery. It converts the 6A at 3V to 3A at 6A. Viola! Just what your LED needed. This is a boost driver because it boosts the voltage it is able to do this because it draws “extra” amps. Driver B does the same thing but it decreases the voltage from the battery to the LED. So it wants to send 3A at 6V to the LED, same as before. But it runs off of a 12V battery. No problem. It draws 1.5A from the battery and converts that 1.5A at 12V to 3A at 6V. Now the input voltage is what the driver receives or can receive from the battery(s) and the output is what it send to the LED. So for the drivers you are looking at it doesn’t matter what input voltage is (as long as it is in spec.). What matters is the power sent to the LED. Since both send 6V, it just comes down to current, amps. Hopefully you can see how asking about the input voltage here really does not effect output now.
(2) One of the easiest and best ways to get started here is to just google whatever the name of the emitter is and “test” and look for a BLF link! Lots of really great people have done quality work here to help us all out. So for an XHP70.2 just google “XHP70.2 Test” and Texas ace’s test here on BLF should be one of the first few results! You can search emitter name here in the forum too. One thing to consider is that LEDs and heat do not get along well. So we always want to get heat away from the emitter. So if you are putting the LED in a small host with little mass or in something that does not have the best thermal path, then know you probably don’t want to push things to the ragged edge. You can also look up the emitter and host in the forum and see of someone has done a similar build.
(3) Go back to what we said about question 1. See how the driver in the boost version pulled “extra” amps and converted those to the needed volts? Well, that means the driver being discussed here is going to do the same thing. So it needs to send 6V and 4.8 Amps to the emitter. That’s 6Vx4.8A so 28.8 watts and that last number is what we need. On the battery side it is seeing 4.2V, with a fresh cell, and needs to make that 28.8watts. So 4.2V x ?amps = 28.8 watts. Answer: 6.8Amps. But there are two problems. first this is only true if the driver is perfectly efficient. It is not. It is probably best to guess it is about 80% efficient. That means it really needs closer to 36 watts since 20% of the power is just converted to heat. Using the same formula as earlier we get 4.2V x 8.5A = 36 watts. So you are already looking at pulling 8.5A from the cell. The second problem is that batteries sag under load and voltage decreases as the battery is used. So say you charge your battery to 4.2V and then draw 5A from it. While it is putting out those amps the voltage dips down, so it becomes, say 4V. If you stop drawing current from it, it will hop back up. Look at the second graph in the first post of this thread. You will see that at 15A the cell instantly drops from 4.2V to 3.9V because of it being under load. Not all batteries handle this load the same. Some will dip more and some less. If everything else is kept the same, i.e. same cell size, quality etc., then as capacities increases, the dip under load increases too and as cell resistance decreases the dip under load decreases, more amps can be safely pulled, but the capacity decreases too. So if your cells are only rated at 10A and are 5,000mah then they will probably dip a lot and you wind up with something more like 3.8V x 9.5A = 36 watts. So your fresh battery is pulling around 10A, putting out 3.8V and the driver converts this 20% of the is power to wasted heat and the other 80% goes to the Led in the form of 6V at 4.8A. This problem is continuously exacerbated as the cell voltage constantly decreases as it is used. So the driver pulls more and more current from it. That’s why you want a cell rated for a much higher current.
(4) No. The driver will regulate the current from the battery. As long as the driver is rated for the input voltage from the cell(s) you will be fine. The batteries will just dip less underload and probably make things more efficient. The driver must be rated for the Voltage from the cell(s). The batteries must be rated for the current that the driver will draw. But the driver does not need to be rated for what the batteries can put out. In that way the cells tell you what can be safely pulled from them not what they will “push out.”
There is no 3V XHP70. There is a 3V XHP50 and that would be fine to run direct drive from a high drain 21700 cell.
as mentioned no 3v xhp70,but a 3v xhp50 would be an easy build
here’s my MAG 3D xhp70 build for some ideas.
mag 3D xhp70