vetkaw63
Enlightened
Does anybody know how efficient an L1T circuit is?
Thanks,
Mike
Thanks,
Mike
Fenix L1T Circuit Efficiency?
- Thread starter vetkaw63
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Valpo Hawkeye
Flashlight Enthusiast
Do you mean does it drain while it's off like an HDS? In that case, no, because the tailcap cuts the current to the circuit.
vetkaw63
Enlightened
nerdgineer
Flashlight Enthusiast
There's no easy way to define efficiency for an entire light. Energy is lost in the boost circuit, in the LED, in the the optics, through resistance in the electrical path, and maybe elsewhere. Hard to define what 100% is.
I used to guesstimate "efficiency" of a light by eyeball integrating the area under the runtime curve for the light at FLR, scaling that by the peak output measured for the light, and dividing by the number of batteries that light used. Then you could sort of get an idea of how many photons the light could squeeze out of a battery.
Otherwise, I don't know how you'd deteermine overall efficiency. As luck would have it, FLR has stopped reviewing new lights, so you can only do this for lights up to the Fenix L2D and earlier.
I used to guesstimate "efficiency" of a light by eyeball integrating the area under the runtime curve for the light at FLR, scaling that by the peak output measured for the light, and dividing by the number of batteries that light used. Then you could sort of get an idea of how many photons the light could squeeze out of a battery.
Otherwise, I don't know how you'd deteermine overall efficiency. As luck would have it, FLR has stopped reviewing new lights, so you can only do this for lights up to the Fenix L2D and earlier.
2xTrinity
Flashlight Enthusiast
The only way to really tell would be to take the thing apart, measure current out of the batteries, and the terminal voltage across the batteries to find input power, then to measure output current, and voltage across the emitter to find the output power, then divide the output by the input. The problem is that this figure will be drastically different depending on the battery state-of-charge, and internal resistance. To accurately find the overall percentage efficiency, it would be necessary to find the average efficiency throughout the entire state-of-charge. In general though, switching regulators are more efficient when the voltage input is close to the voltage output. Running an LED off of 2xAA or 1xCR123 will usually be inherently more efficient than 1xAA. Also, buck conveters (voltage stepdown) are almost always more efficient than boost converters.
Also, another issue is that many boost circuits have a low-voltage cutoff where they may fail to start up, and other such problems. If a converter leaves battery capacity unused for some reason, even if it is more efficient while it is running, it may worse on the whole in terms of minimizing battery use.
Another issue is battery internal resistance. The more current is pulled from a battery, the more the voltage (and consequently the power that it delivers) will "sag". This is especially a concern in boost converter lights which operate at low voltage on the input side, they must draw a very high current to meet power demands, and as the battery's voltage starts to sag, the converter starts to demand more and more current to try to compensate for the lower voltage (and in many cases lower driver efficiency at that voltage as well), causing power to be lost due to internal resistance to increase as well -- which would not be taken into account in my efficiency calculations described above (power out/power in)
This is why in high power boost lights like the L0D-CE, or L1T, batteries with low internal resistance like NiMH will maintain maximum brightness for many many times longer than Alkaline cells which cannot keep up with demands. On lower brightness levels though where current pulled from the tailcap is not too great, NiMH and alkalines are about equal.
Also, another issue is that many boost circuits have a low-voltage cutoff where they may fail to start up, and other such problems. If a converter leaves battery capacity unused for some reason, even if it is more efficient while it is running, it may worse on the whole in terms of minimizing battery use.
Another issue is battery internal resistance. The more current is pulled from a battery, the more the voltage (and consequently the power that it delivers) will "sag". This is especially a concern in boost converter lights which operate at low voltage on the input side, they must draw a very high current to meet power demands, and as the battery's voltage starts to sag, the converter starts to demand more and more current to try to compensate for the lower voltage (and in many cases lower driver efficiency at that voltage as well), causing power to be lost due to internal resistance to increase as well -- which would not be taken into account in my efficiency calculations described above (power out/power in)
This is why in high power boost lights like the L0D-CE, or L1T, batteries with low internal resistance like NiMH will maintain maximum brightness for many many times longer than Alkaline cells which cannot keep up with demands. On lower brightness levels though where current pulled from the tailcap is not too great, NiMH and alkalines are about equal.
vetkaw63
Enlightened
Thanks guys,
I was really asking to see if anybody had checked the battery amperage draw compared to the amperage being delivered to the emitter. With a known "good" battery, that would give an estimate of how efficient the boost circuit is.
I'll check mine when I get it back from plating.
RV7 is going to make and sell a AA cell board that he says is %70 efficient. I was thinking of buying a couple if they were as efficient as a Fenix boost circuit.
Mike
I was really asking to see if anybody had checked the battery amperage draw compared to the amperage being delivered to the emitter. With a known "good" battery, that would give an estimate of how efficient the boost circuit is.
I'll check mine when I get it back from plating.
RV7 is going to make and sell a AA cell board that he says is %70 efficient. I was thinking of buying a couple if they were as efficient as a Fenix boost circuit.
Mike
vetkaw63
Enlightened
My salute 2xTrinity! You covered all aspects of the issue. I'd like to just add that when measuring battery current draw, multimeter internal resistance plays a big role due to low voltage and high current. If it's a low end MM, you may get a reading of 1.2A while the actual working current is 2A. If you don't have a high end MM, you can insert a 10 milliohm resistor to complete the loop and measure the voltage over the resistor to get a better idea of the actual current.
I'm glad to see people discussing my design. To get high output at 3.3V from a power source of 1.2V is not an easy task. In this situation, the requirements for high output and high efficiency actually overlap a lot. The circuitry and components were refined many times to reduce loss and output increased accordingly. The result is that despite the record high output, the efficiency is still maintained at a very good level.
Theoretically, boost circuits at lower output can more easily achieve better efficiency. But manufacturers seem to be more interested in reducing cost. It's actually rare to see a production AA boost with an efficiency better than 70%. So I too am very interested to see the efficiency figure of a Fenix 1xAA in high mode.
I'm glad to see people discussing my design. To get high output at 3.3V from a power source of 1.2V is not an easy task. In this situation, the requirements for high output and high efficiency actually overlap a lot. The circuitry and components were refined many times to reduce loss and output increased accordingly. The result is that despite the record high output, the efficiency is still maintained at a very good level.
Theoretically, boost circuits at lower output can more easily achieve better efficiency. But manufacturers seem to be more interested in reducing cost. It's actually rare to see a production AA boost with an efficiency better than 70%. So I too am very interested to see the efficiency figure of a Fenix 1xAA in high mode.
vetkaw63
Enlightened
RV7
Your boost circuit should be even more efficient at the lower levels? I would like the high, as high as possible but longer more efficient run times on the lower levels. I am just trying to figure out how many lights that I need to retrofit.
I am planning to buy several of your lithium ion boards. I can't wait.
Thanks,
Mike
Your boost circuit should be even more efficient at the lower levels? I would like the high, as high as possible but longer more efficient run times on the lower levels. I am just trying to figure out how many lights that I need to retrofit.
I am planning to buy several of your lithium ion boards. I can't wait.
Thanks,
Mike
Hi Mike,
At lower levels run times are longer due to lower current draws. But the efficiency remains roughly the same. These 1AA boards will be available very soon.
As for the Li Ion boards, I'm fighting with the inconsistency of the ICs. Outputs now vary from 850mA to a little more than 1A. May have to dump some chips.
Sorry for going off topic. Now let's hope someone will post the Fenix efficiency soon. At least wintermute will get one to test in a week.
At lower levels run times are longer due to lower current draws. But the efficiency remains roughly the same. These 1AA boards will be available very soon.
As for the Li Ion boards, I'm fighting with the inconsistency of the ICs. Outputs now vary from 850mA to a little more than 1A. May have to dump some chips.
Sorry for going off topic. Now let's hope someone will post the Fenix efficiency soon. At least wintermute will get one to test in a week.
Handlobraesing
Banned
- Joined
- Feb 14, 2006
- Messages
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You'll have to check out the input and output waveshapes. If there is a substantial amount of ripple, the only way is to use a dual channel power analyzer with a bandwidth at least a few times the highest order harmonic present in the current waveform.
