Thanks, I think I see what I need. And, “Noctigon” turns out to be the name of the copper “star” that I mentioned in my previous post.
Sounds interesting. Can you provide a height dimension? I presume it is 17mm in diameter. Some easily understandable (even by me) performance specifications such as how many watts (amps at 7.2 volts?) I can expect from 6 Eneloops?
I have looked at the thread linked in your “signature” line for “Released Drivers:” But, it was not very clear to me. It seems you are offering the PC board only, but you can also provide one with all the necessary components soldered in place?
I do “understand” that a DD driver will “put out whatever it can get” which is fine with me, but I cannot say I truly understand all the implications. I certainly do not wish to “blow up” the MT-G2 emitter by pushing too many watts into it…
Unfortunately, I think 6 Eneloops is the limit. 2 D-Cells are already a pretty tight fit, so I do not think that 8 Eneloops (7 + dummy) can be made to fit easily. Certainly not in any convenient battery carrier that I know of. Why might I prefer 8.4v over 7.2v? (Isn’t 6v enough for the MT-GT emitter that you are suggesting?) Is there some kind of risk of rapid and serious damage to the Eneloops if I try to only use 6 of them?
Thank you for these details. Your suggestion remains attractive. I will certainly consider trying your suggested method once I have built the LED and heat sink assembly. No need to purchase a driver if I truly don’t need one.
I do see advantages in a buck driver now, but I am still not sure it is needed.
Starting from the beginning, voltage is like water pressure, electric current is analogous to water current and resistance is like a narrow place in a hose that resists flow. If something has constant resistance, then I = V/R. The current is equal to the voltage divided by the resistance. That is also V = IR, the voltage drop across a resistor is equal to the current times the resistance. Power is given by P = VI. There aren’t any constants in these formulas, because they are used so often that the units are chosen to make the formulas simple. So the power VI = V**2 / R goes into the resistor and turns to heat.
An incandescent light is basically just a resistor that gets white hot because the heat doesn’t have anywhere else to go but escape as visible or infra red light. It is not efficient because most of the power become infra red light instead of visible.
An LED or light emitting diode converts electric energy to light directly. The heat generated is an imperfection that the engineers try to reduce. Therefore an LED does not make a good resistor. To make light it needs a voltage that gives one electron the energy it takes to make one photon. So it has a minimum forward voltage below which it makes no light at all, and an increase in forward voltage with current that may act like a resistor in series with a perfect LED.
Incandescent lights usually don’t have drivers, but LED lights usually do have drivers for several reasons:
The current in an LED is more sensitive to the voltage than that in an incandescent light bulb.
Incandescent lights come designed for a large variety of voltages, but an LED’s minimum voltage depends on a constant of nature, plank’s constant.
LED’s work well over a wide current range while incandescent light do not.
So a driver may raise or lower the voltage of the source, hold current at a constant level and change the level at which it holds the current to dim the light.
Cheap 3.7 V drivers usually use resistors to hold the current something like constant and soak up the small voltage difference between the source and the LED’s forward voltage at the desired current. They sometimes also change modes by changing the fraction of the time the light is on “pulse width modulation”. More expensive ones actually regulate the current actively to keep it more constant. Some cheap one mode drivers are just simply resistors, sometimes discrete resistors and sometime distributed resistance in the wiring.
It is unusual for a large* 7.4 V (really 6 V to 8.4 V) light to have only a resistor for a driver because more than half the energy gets lost in the resistor, making a lot of heat and a sort run time. But this light may be better simple than efficient. It seems to suit its character for the driver to be a piece of wire from the back of an old picture frame or something like that, instead of a printed circuit board with surface mounted resistors or other components. Wire cut from the heating element of an old electric heater might last longer if it is available, but I think steel will do if it doesn’t get past orange.
Three nickel metal hydride cells in series seems too little voltage. I can’t get over around two amps with a single lithium cell myself. Six in series puts a lot of heat into the driver unless it is a buck driver that lowers the voltage and increases the current (like a transformer does with alternating current). So maybe 4 or 5 would be optimal if you can arrange it. Otherwise it may take a lot of wire to distribute the heat generated with 7.4 V. I really like using two 3 x AA D cell adapters, so maybe 7.4 best, and maybe it will be hard dumping that much heat in a wire. I am not sure.
*Button cell lights often use two 3 volt lithium primary batteries in series. Button cell lights never use buck drivers.
Yes, I missed that. So if he uses six AA in series he may have to copper braid things instead of adding resistance.
Did I also miss a reason to use a driver?
Right. In terms of the driver - I’ve been recommending a DD driver which can handle either stepdown or rampdown. We’re talking about a host with fairly bad thermal properties:
steel tube with ridges - very difficult to interface with well
shallow pill - low mass
So we probably want to allow a high initial mode for the “shock and awe” effect. After some short period of time we can then bring the output level back down, hopefully imperceptibly. That way if the hotrod is shown-off for 10, 15, or 20min it can survive and not be scalding hot to hold.
@Rosoku Chikara - I forgot to mention to you: At this drive level your stock switch assembly is NOT OK. That’s another hurdle, it may be a major or minor problem in your design. BTW many of the custom drivers around here can take a momentary input. (So can several of the commercial Chinese drivers.) That may be the best option. Typically this type of driver is not appropriate for long-term storage of cells though.
Yes, I knew the switch would eventually become an issue. Despite my lack of experience with such matters, the thought had occurred to me that the existing switch might not be “adequate” for my needs. However, I was hoping that I might be able to “get away” with using it. It is of very simple design and uses very robust looking materials. Like many electrical devices from this era, it seems to be over-engineered with respect to the amount of copper(?) conductor being used. It looks to me like it could potentially handle a lot of current despite its primitiveness.
- It there absolutely no chance that it would work? (How to easily check or verify?)
What alternatives can you suggest? (As you point out, this could be a major problem…)
Thanks again for all your thoughts and comments.
In the meantime, I figure I can get started on the mechanical construction of the “pill.” Once it fits as tightly as possible within the allocated space, I will mount an emitter star on the front, and attach a driver to the back. (I presume it can be easily “wired” by drilling a couple small hoes through the heat sink plate.) Then, I can do my early electrical testing with 2 x 18650 in series (without fitting the Li-Ion cells inside the body of the flashlight at all). If everything seems to work, then I will get some serial D-cell adapters and test it on Eneloops… I guess at that time I would have an accurate picture as to my real current requirements? Anyway, I expect to enjoy myself with this project for a considerable period of time. I see the switch issue as simply another a part of that “enjoyment.”
You may measure the resistance of the switch & voltage drop of the switch under load. If too much energy is dissipated in the switch it will be damaged. Regardless of the robust nature of the switch, I speculate that it will be high resistance - at low currents a high resistance is fine, but at higher currents this high resistance turns into a high power dissipation, weakens the metals, and causes damage. djozz has a thread where he tests clicky switches and posts his testing methodology. I do not suggest that you try this switch with high current, but you can easily measure voltage drop at 500mA and maybe 1A in order to determine what’s going on.
It’s unclear to me how the stock switch is positioned inside the light electrically. Does it interrupt the GND connection for the light bulb? Please sketch how this works.
I guess theoretically, that should give you something like 3 - 4 amps (~.5 volts/0.5 ohms to ~2 volts/0.5 ohms) into the emitter, depending on the battery voltages and the wiring, etc. resistances.
That was one of the first “mods” that I did, with a DST, awhile ago, but I eventually ended up removing the 0.5 ohm resistor completely because the wires and the switch gave me more than enough resistance, even with 2 x 16340 batteries (this was in a DST with a “shorty” battery tube).
I still have that light now, and it’s working fine.
First about the switch, my solid copper RAY-O-VAC has a steel switch. Try it with a magnet. But maybe you can “braid” all but the contact surfaces with copper. Small bits of steel should work. Of course it would be easier to put in a modern switch.
It’s probably much less easy to put in a modern switch if you try to keep the stock appearance. I’d just worry about this being a one-way trip for the old switch…
My understanding is that in these low-voltage applications arcing isn’t the issue, it’s fatigue of the metal/springs. They get crushed due to losing their springyness from the high temperatures due to all the voltage drop (which is due to them being made of steel, which is where the springyness comes from of course…. cyclic, eh?).
I don’t know how the old switch is made, so I don’t know if there will be room. But, what about a momentary toggle switch/slide switch like these ? If you can find one that will fit under the original “switch” and just change it out for the guts, then you may use it as an e-switch. Your driver would have to be capable of using an e-switch. I don’t know if it will work or not, because my understanding of drivers is still not very high.