If you match Vin to Vf then you don't really need a regulator or resistor. But because Vf is not exact and Vi changes over time, that is impossible to do accurately.
Again, incorrect. Practically speaking for a battery-powered flashlight, you will want to start with Vin > Vf (@whatever target drive current is desired, e.g., 1A for a Cree XR-E). But if you are too close to Vf, then the batteries will most likely fall below Vf sooner than is preferred, meaning shorter run time for the required light output. So, now maybe you increase Vin to get some headroom/additional run time. The problem with that is for direct drive, you now overdrive your LED.
With an LDO regulator, driven by 1xLi-ion, I can drive an LED with a constant current for virtually the entire life of the Li-ion. The driver might briefly start at 80% efficiency (e.g., 4.0V in vs 3.5V Vf), but it should spend most of its time at 90% or above (e.g., 3.8V in).
Thus, using a driver even when matching Vin to Vf does provide important, tangible benefits.
Here is my plain simple point.
If you have a 6 volts supply.
A 3 Vf LED that you want to run at 2 amps.
If you use a resistor you will need 6v-3v=3v/2A=1.5 ohms.
So we take our 1.5 ohm resistor and put it in our circuit.
The power used just by the resistor is equal to the power output by the LED! We have a 6 watt LED and we are wasting 6 watts through our resistor. So your wasting just as much power in the resistor as in the LED. That means you need a 6 watt resistor and your only 50% efficient.
Your driving 12 watts to use 6.
Now if we want to change the resistor to dim the LED to 1 amp.
6v-3v=3v/1A=3 ohms
Power in the LED and resistor again equals 3 watts.
Again, our efficiency is 50% and we are driving 6 watts to get three.
Now you can see that our efficiency stays the same even though we dim and you can see in this case we are wasting just as power in the LED as we do in the resistor as heat.
You've confused the proverbial forest from the trees.
You claimed that the light would draw the same amount of power as running without a resistor, as well as claiming that you would not get any run time increase. All the while, you'd get lower lumens output. This is a direct quote: "With a resistor the light will still use the same amount of power, but output less light. So your burn time wont increase and your light output will decrease." Sure sounds bad, right? But those claims are simply wrong. The light does not use the same amount of power (it can use far less), and the burn time does increase (by an order of magnitude in my case).
As I explained previously, actual experience shows your claims to be wrong. I have a triple XP-E Q2, wired in parallel, and driven by a 6xAMC7135 LDO regulator driver (a nominal 2100mA total drive current, or 700mA per XP-E). And in fact, I measure 2.1A current draw at the flashlight tail. At that current draw, my 1x18650 cell is going to last about 1 hr and the flashlight gets hot quite quickly. When I cut in a 60 ohm resistor, output drops to something dim and I've measured a run time that exceeds 24 hours. The light also runs completely cool.
Second, LED Vf is not a constant. It increases as drive current increases.
So, your first example (which appears to assume direct drive, not regulated by a driver) is probably more realistically closer to Vf = 4V (or more) at If = 2A. That gives 8W to the LED (most of which is shed as heat). Thus,
6V-4V = 2V/2A = 1 ohm -> 4W dissipated in the resistor
Efficiency is thus 67%, not 50%. Yes, this is quite poor. But you are missing the point (forest vs the trees).
If you direct drive the LED at 6V (and assuming that the LED would actually survive and not fry), you'd pull much more than 2A from your batteries. That's a lot more heat, a lot more current draw, and vastly shorter run time.
As you increase the resistor value (and drop the driver out of regulation, if you are using a driver), you correspondingly cut the Vf and the matching If.
Using my integrating sphere Mark I eyeball and comparing the low mode output of my triple XP-E Q2 to my Blackhawk Gladius, I estimate that my triple XP-E Q2 puts out about 10 lumens. That's about 3 lumens per XP-E. Referencing the XP-E datasheet, you need about 20mA drive current to get about 3 lumens, or about 60mA total drive current to the triple XP-E. Vf at that level looks to be around 2.8V.
Thus,
LED power draw = 3 * 2.8V * 0.02A ~0.2W
Resistor power dissipated = I^2 * R = 0.06A^2 * 60V ~ 0.2W
So, efficiency is about 50%. Yeah, that is poor. But you are drawing only 60mA from the 18650 and sending only 20mA drive current per XP-E.
That's the point. When you are dealing with very low current draws and very low drive currents, 50% vs 90% efficiencies don't really matter all that much. You will still get long run time and cool running. And actual, real world experience bears this out.
The reason your only getting 80% efficiency out of a switching regulator is because you are running at to low of input voltage. Switching regulators tend to get more efficient as the input and output voltage delta increases.
I typically get about 85%, not 80%, from my various SOB buck drivers. It's not because the input voltage is too low. How would you even know that, since I never stated what the input voltage is? FYI, I've tested various SOBs out to the max voltage spec'ed by The Shoppe (16V).
