Does high amp battery push the LED to hard?

Let's say I have a light that draws 15amp, and I get a 4200mah battery with 45amp continuous....and a 5200mah with 20amp continuous let's say..... Will the 4200mah battery last longer since it's not being required to release amps near its 45amps compared to the 5200mah that's close to its 20amp ability? Hope that makes sense thanks

Your question kinda makes sense. But no, the 4.2Ah battery will not last longer than the 5.2Ah battery it is simply capable of delivering more of it at once.

lightfooted said:

Your question kinda makes sense. But no, the 4.2Ah battery will not last longer than the 5.2Ah battery it is simply capable of delivering more of it at once.

Click to expand...

Thanks. And more at once because less internal resistance? So basically the more capacity the more dense thus less internal resistance? Am I understanding it correctly? Thanks again

There are three basic ways of driving a LED: direct drive, inline resistor, and a variety on constant-current circuits (often referred to as "drivers").

Direct drive demands careful matching of power supply to the Vf of the LEDs such that the otherwise unregulated cell voltage(s) and cell current characteristics do not overdrive the LED. This is difficult to achieve and rarely done in the market: the last direct drive flashlight I've seen was the very old CMG Reactor 2xAA, which had horrible performance since the nominal peak 3V of two alkaline cells was well below the then-common >3.6Vf of LEDs at the time.

The resistor uses a supply voltage greater than the led Vf and regulates current against the remaining voltage drop.from the supply. The wider the gap, the greater range of varying supply voltages the system can handle ... at the expense of efficiency. Cell performance can come into play with these circuits at high currents, but the resistor is generally the limit.

Far and away the most common is the driver circuit (in its many incarnations), which within its rated voltage input range Should™ never overdrive the LED regardless of the current capacity of the cells supplying it. An issue with high performance flashlights can be using cells with enough current capacity to meet demand, but otherwise the driver will handle the rest and doesn't care if a cell can deliver 15A or 40A when it always draws <15A.

Speaking in direct drive terms:
Does the light source (LED) pull the electricity from the fuel source (battery) or does the fuel source push electricity to the light source?

I had always thought the light bulb was like a vacuum in a sense where it sucks electricity in order to light up when a circuit +/- is completed. Yet amps available can push electricity beyond said bulbs ability to handle? Kinda like a radio speaker where the magnet sucks electricity towards it but the volume knob can cause more electricity in the form of sound amplified waves to overdrive things? Where does resistance play a role?

The power source is the source of electrical potential, thus "pushes" the current through the circuit. This diagram (humorously) illustrates the concept:
https://image.slidesharecdn.com/pre...160329171331/95/unit-4-electricity-15-638.jpg

Voltage and resistance are, broadly speaking, independent constants while current is a dependent variable. Real life is of course more complicated than this (i.e. incandescent filaments' resistance varies greatly with temperature and LED Vf also varying with temperature) but the principle is sufficient to estimate many situations.

Hypothetically let's say a special fet type driver was made to try to get the most out of an 18650 cell size (max of around 20-30 amps), and you forced a 21700 down the throat of the flashlight with modified hardware.

If anything broke, it would probably be the driver itself. The LED would have to be overdriven to begin with, because and LED has a maximum drive which can be put into a supermax state.

The typicall tolerance of an LED is 180% to 220% of the maximum. For example the CREE LED xhp50.2 has a stated max of 2000 lumens. But it can be driven to nearly 4000.

As you can imagine to is not going to have anywhere near the suggested 50,000 hour runtime at that supermax. The runtime will be reduced. On top of this a failure curve needs to be writen in.

Once in a blue moon some overdriven LED lights fail, especially when around 200%. This overwork can cause a spontaneous high temperature, which if you know your physicas happens all the time with everything. I fact in the room you're sitting in there are enough nitrogen atoms at a temperature hundreds of degrees below zero that if they all stayed around you it would freeze you to death !

On that scale temperature anomalies are rare. LEDs are small, I'm sure anomalies add up and can cause a failure but the point is statistically these over driven LEDs do fail at a rate of 2-4 percent vs much less than 4/1000 for max driven or less LEDs.

But the chance you will hurt the LED in your light is zero. The strongest batteries definitely have a contingency in place and manufactures are aware of all batteries.

However generally speaking most lights don't need the highest output batteries they just need enough amps. Typically you wont see much of a brightness difference with high amp cells but the energy loss and runtime sacrificed is huge.

And it is not easy to calculate. A 3300mah 18650 doesnt last only 10% longer than a 3000 mah cell, it may last 13 or even 15 percent longer in time because of energy thresholds and wasted amperage that just gets resisted by the driver anyway.

+

nope,
it's controlled by the driver.

Any driver capable of overdriving a LED or LEDs to the point that it can easily ruin them Should™ have thermal protection and/or current sensing such that better-than-planned on current capabilities or mere extended-duration operation don't easily wreck the LED ... or the driver.

Don't forget that the production of a voltage potential and the creation of amperage capacity are a function of the chemical composition of the cell. Cells that are capable of providing a higher amperage flow will have a different composition than cells that have a lower constant draw capability. In practical term, an LED that needs a low current flow to produce the desired output can use any cell that can supply that amount of current on a continuous basis. An LED/driver combo that wants to produce more light will need to use cells that can sustain a higher current flow over the capacity of the cell.

I'm going to reach back to the days of Morse telegraphy. A skilled telegrapher can easily do more than 20 word per minute. The minimum speed required to qualify for a Novice ham radio license is I believe 5 wpm. Put those two in a 'conversation' and the ability to actually impart information is controlled by the speed of the slower telegrapher. The more skilled telegrapher has to limit his output to match the ability of the less skilled one's ability to receive the data. He is only using a portion of his ability to carry on the conversation. On the other hand, the less skilled telegrapher is being pushed to the max of his ability to keep up with the conversation.To put that in terms of our lights, an LED/driver combo that need a high current flow to produce the desired quantity of lumens can only do that if the cell(s) providing the current can continuously provide that amount of current until their total capacity is reached and the cell is exhausted. In a setup requiring a lower amount of current cells that can only supply continuously a lower current will do fine, while the cells capable of sustaining a higher current will just be idling along. In a light requiring five continuous amps to produce its maximum output, any cell capable of producing that continuous current will work fine. But put that cell in a light requiring twenty continuous amps and the output of the light will never reach the designed value. But a cell capable of producing twenty or more continuous amps can drive either light to its design limits, and will just be idling along in the lower requirement circuit.
I hope this make sense, and I am sure if it doesn't I will be told so.