I don’t think I buy that one. If one cell is weak and two are strong, they will never be equal, but the strong ones will drain down much faster, trying to supply the weak one, while also trying to handle the load from the led at the same time. If one were to short, the others could overheat quickly, trying to replenish the shorted one. I think parallel is actually worse for Li-ion than series, since a short in series kills the circuit, where a short in parallel maintains the circuit and can lead to more damage. Sorry, I’m not an expert, but that’s my take on it.
Whatever energy goes into the parallel weak one will ultimately get used by the LED so nothing gets wasted, assuming it wasn’t completely dead to start with. Both cells are forced to balance while in use and both will always endure the same cycling and wear.
Series requires matched and balanced cells or some will get ruined eventually. They must also be monitored periodically as the slightly weaker cells will have accelerated wear because they go through more full cycles than their healthier neighbors.
Fortunately, the laws of physics don’t care if you buy them. A higher potential (votlage) battery (or batteries) connected in parallel to a lower potential (voltage) battery will yeild the higher potential battery charging the lower potential battery until equilibrium is reached.
PPtk
GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAHHHHHHHHHHHHHHHH! NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!
KVL anybody? (Kirchoff's Voltage Law)
The voltage sum around ANY loop of a circuit in either direction (clockwise or anti) must equal to zero.
Here I assume perfect DC, completely perfect. Other situations would get stickier.
Two possible scenarios:
1. The supplies are in series.
If the supplies are in the same direction, then you will have 17V across your load, or if they oppose each other you will have ±7V across your load.
Each supply needs to be able to handle the current from the respective voltage across your load.
ie. If you have 17V across a 17ohm load, both supplies need to be able to handle 1Amp (or 12W and 5W)
This is the case for the PSU listed above.
2. The supplies are in parallel
Diosmio. I hope you didn't do this.
Completely theoretically, any 12V voltage source directly in parallel with a 5V voltage source will drive the 5V with 7V. Also theoretically, there's no internal resistance of a power supply and the 7V will cause an infinite amount of current to flow. Boom.
I hope there is either a really high internal resistance across your 5V source or there is sufficient protection. Sufficient protection just means you end up with a redundant 5V supply, possibly sapping current from the 12V supply.
If the 5V supply is completely unprotected or not sufficiently protected for a higher voltage across its output, the 12V supply ends up driving the 5V supply (you've got 7V across the 5V supply if you connected them the same way ie. positive out 12V to positive out and vice versa.)
That 7V will do whatever it can to ruin your 5V supply because it's the bigger older brother and likes to bully the smaller one. Jokes. But seriously that remaining voltage can and will do soem serious damage.
If you connected them conversely, positive out A to negative out B, you basically have 17V across the wire that you got, and will soon have a wonderfully plastic smokey room. Some seriously hot stuff is bound to happen.
90% of the time you will end up with 2 smoldering hunks of metal and 0V across your supply.
If you don't know what you're doing. STOP. Just like using a power tool, you must not use it without knowing how to use it properly.
Few people will get my analogy, but this is basically twisting the signal core of two audio cables together and expecting to get a well combined signal out of it without one driving the other and phasing, etc.
Better analogy would be like tying a sled dog and an oxen together and expecting them to plough straight.
Here is a link to some(all) material which is the basic electrical unit I did in my first sem of uni, it basically explains everything I've said. -link
If you read it and understand it, you've saved yourself having to study a unit and over a thousand bucks in fees.
That was entertaining and informative, thank you. :D
Ok, thanks Ramblings. It should hypothetically never happen that they are both turned on at the same time while in parallel but I was just curious what would happen. Both have a switch to control each individual line so as long as I don't turn on both switches I will be fine. :)
Heheheh, no worries. No offense at all either, just don't want to see another great contributor on this forum disappear. :_(
If you only ever use one at a time, try a SPDT switch to toggle between the two. :)
To answer the original thread question:
[quote=PilotPTK] The raw amount of power they can deliver to the flashlight is the same. Watts is calculated as Volts * Current, so watts doubles when you double either Voltage or Current. [/quote]
Basically, this is true. But you have to keep in mind that 90% of drivers are buck which need ~0.5V over the LED voltage to operate.
This means that, unless the driver is boost-buck, it'll cut-off at around 3.3-3.6V depending on the LED. This is pretty high.
A battery can drain down safely to 3V, so you've got quite a bit of drain left, maybe up to 20-30%.
If you have two or more in series, this is gap is eliminated, providing a more complete battery drain.
What I don't understand is why batteries are listed in mAh, instead of Wh. A bit misleading if a battery can drain for ___mAh at 2.5V as opposed to 4V.
Something else to consider to get the most output in the smallest package possible. High voltage, low current (cells in series) is more efficient than low voltage, high current (cells in parallel).
Ohm's law:
V / R = I (Volts / Ohms = Amps)
Examples:
4V / 4Ω = 1A
8V / 4Ω = 2A
4V / 8Ω = .5A
8V / 8Ω = 1A
4V / 16Ω = .25A
8V / 16Ω = .5A
Joule's Law:
V * I = P (Volts * Amps = Watts)
4Ω circuit:
4V * 1A = 4W
8V * 2A = 16W
8Ω circuit:
4V * .5A = 2W
8V * 1A = 8W
16Ω circuit:
4V * .25A = 1W
8V * .5A = 4W
Please, correct me if I've missed something or wanna elaborate.
Not a clue what you’re saying here… 1 /1 = 1, 2/2 = 1, 4/2 = 2… They’re just random numbers…
You don’t ‘gain’ or ‘lose’ efficiency by running in parallel or series. Some driver/emitter combos might me more efficient at higher or lower voltages or the higher voltage of series may allow you to drain the batteries further - but the parallel or series arrangement doesn’t change efficiency in and of itself…
Series wiring of the cells may show a VERY slight gain in efficiency because the losses in wires and springs will be a small percentage of the total watts consumed. For our purposes, however, these numbers are small enough to be irrelevant.
PPtk
Not at all. Their “examples” showing that an increase in voltage decreases the need for big, low-resistance components or allow them to handle higher output without a loss in efficiency. I would think there would be a big difference between a 4Ω circuit and 16Ω with equal performance.
Surely, it would have to be designed for the higher voltage. If anything, series cells (unless perfectly matched) can’t drain as far since one will reach it’s lower limit before the other.
Maybe not as important for flashlights since the extra voltage never makes it to the LED. Idunno.
Lightme - I think you are pointing to the fact that power lost across the resistor is
P = I ^2 * R (Just substituting for V in the power definition using Ohm’s Law)
That works out in all your examples; the double current quadruples the number you have for power. You missed one effect though. Resistance of the battery (combining the individual cells) will be lower in the parallel light. The cell internal resistance and protection circuit resistance are part of that parallel connection
Assuming the resistances of each cell are the same R
Battery Resistance(cells in series ) = Number of cells * R
Battery resistance (cells in parallel) = R / Number of cells
Too hard to do to work around lots of notation issues so I’ll leave the next piece for you. Bottom line is for the power through the battery doubling the current but hooking the cells in parallel is a wash; power loss is the same regardless. Obviously there are other resistances that are connected in series to the battery like the switch, springs, etc. That piece still has the current squared relationship and it tends to be a bigger part of the inefficiency. It’s not the entire story though.
Admittedly, for our purposes here, there’s no significant distance between battery and load so it was probably not worth mentioning. However, circuits that have a relatively long path, the loses can be great. Where bulky electronics with large wires and connectors/contacts become necessary to handle the current, higher voltage/lower current would be the best option.
I hate the idea of running rechargeables in series with the likely risk of killing the weakest cells prematurely but primaries only go thru one cycle so there’s no long term problems. I think a good multi-cell D-size flashlight and extra batteries is the best thing to have for emergencies like long power outages or any situation where recharging isn’t an option. If all you have is rechargeables, a car charger is a something you should own. Sorry for straying off topic a bit. Jus’ my 2¢.
Multi D-cell lights are old technology and much too heavy as compared to today’s new lights!
And they do not make lumens like the newer lights on the market.
All correct - however, in an emergency, D Cell batteries are plentiful and easy to get. Not having at least one or two flashlights that use readily available primary cells would be a missing link in your disaster planning.
PPtk
Besides, newer, lighter, smaller lights are nowhere near as effective as an improvised night stick as a good 3+ d cell Maglite. And with a plastic lense, you don’t have to worry about cut feet if you apply the wrong end to an intruders skull…. :bigsmile:
Why couldn’t they have the same output given the same LED? The voltage and capacity of 3 Ds are more than enough for a single LED.
They could, but at a much heavier carry. I am generalizing. I have many lights that I have options with. I prefer to use 2-AA options with low voltage drop ins for my primary batteries, along with CR123’s. And I do have a 4 D cell Maglight. 
I just don’t use it anymore due to the size and weight. The newer, smaller lights are much more convenient and provide far more light for my needs. I also have a Magcharger light, Ultra Stinger, and a Stinger LED as well. I was in law enforcement for 24 years. I just prefer the newer technology smaller, lighter hosts with Li-Ion batteries. Just my preference. I really have no use for the 4-D cell light anymore.
But with more mass, you get far greater capacity and (as pointed out) a better swing! I see more advantages with D cells in bad situations and convenience for the average person. My 2-D woot-lite (mag clone?) is the one I usually reach for when going outside. I only wish it was up to today’s standards. I don’t think MagLite’s much better and there’s not a lot of other choices.
The voltage would still be 12v, however, the current will be the total of the 2 sources.