4sevens Quark AA2 R5 current draw

troelskc

Newly Enlightened
Joined
Dec 31, 2009
Messages
161
Location
Denmark
Hi all!
I got this light running @ 2.1 A on high mesured at the switch. They states the light produces 205 lumens OTF @ 0.7 A. why does my light draw so much current when they write a total different number on the website?
Anyone own this light who can measure the current at max?

Could it be the XP-G R5 is being overdriven? The R5 specs says max current is 1.5 A.
 
Last edited:

oKtosiTe

Enlightened
Joined
Jan 7, 2012
Messages
974
Location
Sweden
No one owns a Quark AA2 XP-G R5?
Perhaps no-one (who's seen this topic) could answer your question (yet). You should consider sending them an email; they usually respond quickly and satisfactorily.
Don't forget to post back if you get an answer from them.
 

Wiggle

Flashlight Enthusiast
Joined
Sep 19, 2008
Messages
1,280
Location
Halifax, NS
The current pulled from the batteries (a combined 2.4V) will have to higher than what is delivered to the emitter because the emitter has a forward voltage of closer to 3-3.5V. But regardless, 2.1A on Turbo is a little excessive, that would imply less than one hour runtime on an Eneloop. You may want to contact 47s it sounds like the driver is acting funny (assuming your ammeter is accurate).
 

reppans

Flashlight Enthusiast
Joined
Mar 25, 2007
Messages
4,873
deleted.. answered my own question
 
Last edited:

troelskc

Newly Enlightened
Joined
Dec 31, 2009
Messages
161
Location
Denmark
The current pulled from the batteries (a combined 2.4V) will have to higher than what is delivered to the emitter because the emitter has a forward voltage of closer to 3-3.5V. But regardless, 2.1A on Turbo is a little excessive, that would imply less than one hour runtime on an Eneloop. You may want to contact 47s it sounds like the driver is acting funny (assuming your ammeter is accurate).

Does this means the 2.4V (2xAA) option will draw a higher current than if the cells could provide 3.7V+?
 

reppans

Flashlight Enthusiast
Joined
Mar 25, 2007
Messages
4,873
Yes, think of it as the emitter needing a certain amount of watts (Volts x Milliamps), if it can't get the voltage, the driver will pull more ma. Except for boost/buck driver efficiency differences, it's not going to make a huge difference to runtimes since, for example, an Eneloop and a 14500 are pretty closely matched in total energy, or watt-hrs. 1.2Vx2000mah=2.4 w-h, and 3.7x750mah=2.8 w-h.

Course this is for the ML-High modes that both cells can power, one Eneloop doesn't have umph to power max.
 
Last edited:

Wiggle

Flashlight Enthusiast
Joined
Sep 19, 2008
Messages
1,280
Location
Halifax, NS
Yes thats exactly right. To get the same amount of power to the emitter, more voltage means less current. Remember, in this particular type of driver the current that is pulled from the batteries goes through DC to DC conversion and therefore cant directly predict the current at the emitter.

The AA to 14500 comparison is a good one. Comparable amounts of energy but due to the added circuitry required to boost the AA voltage up to LED voltage there are more losses and you will often find runtimes for 14500 is greater than AA for same brightness. For example my Quark AA Tactical runs nearly twice as long on high (not turbo) on 14500. But this gap is closing, look at the efficiency of the new Zebralight SC52. Also this gap matters less with 2 x AA cause now the amount of voltage modification needed is smaller.
 
Last edited:

troelskc

Newly Enlightened
Joined
Dec 31, 2009
Messages
161
Location
Denmark
Great info guys!

Does a sticky post about all this basic electronics knowledge exist?
 

reppans

Flashlight Enthusiast
Joined
Mar 25, 2007
Messages
4,873
The AA to 14500 comparison is a good one. Comparable amounts of energy but due to the added circuitry required to boost the AA voltage up to LED voltage there are more losses and you will often find runtimes for 14500 is greater than AA for same brightness. For example my Quark AA Tactical runs nearly twice as long on high (not turbo) on 14500.

I've always wondered if this is actually a double-edged sword. Yes the higher voltage Li-ion is more efficient powering high modes due to less efficiency loss of the boost driver.... BUT, conversely on the lower lumen modes, doesn't the Li-ion suffer greater efficiency losses from the buck driver? Looking at the 4/7s runtime testimonial thread, it looks like the Eneloop and 14500 crossover point is medium, and the Eneloops do better on low for the QAA (no ML test with the 14500, unfortunately). And the lowly alkaline batt. walks away with the record ML runtime record!

http://www.cpfmarketplace.com/mp/sh...ght-Run-Time-Testimonials-real-world-testing!

BTW, anyone ever see a ZL runtime testimonial on low/ML modes? I'm quite skeptical.
 

troelskc

Newly Enlightened
Joined
Dec 31, 2009
Messages
161
Location
Denmark
I've always wondered if this is actually a double-edged sword. Yes the higher voltage Li-ion is more efficient powering high modes due to less efficiency loss of the boost driver.... BUT, conversely on the lower lumen modes, doesn't the Li-ion suffer greater efficiency losses from the buck driver? Looking at the 4/7s runtime testimonial thread, it looks like the Eneloop and 14500 crossover point is medium, and the Eneloops do better on low for the QAA (no ML test with the 14500, unfortunately). And the lowly alkaline batt. walks away with the record ML runtime record!

http://www.cpfmarketplace.com/mp/showthread.php?236290-4Sevens-Flashlight-Run-Time-Testimonials-real-world-testing!

BTW, anyone ever see a ZL runtime testimonial on low/ML modes? I'm quite skeptical.
'

When running on low the emitter still needs 3-3.5 V to operate? When dimming the light is it the current or voltage that's being lowered or both?
 

AnAppleSnail

Flashlight Enthusiast
Joined
Aug 21, 2009
Messages
4,200
Location
South Hill, VA
'

When running on low the emitter still needs 3-3.5 V to operate? When dimming the light is it the current or voltage that's being lowered or both?

Both. LEDs have a nearly exponential current response when voltage is changed. So dropping by 0.1V might half current. Most white LEDs will glow, extremely dimly, on about 2.6v at less than 1 mA. This is a few mW of power. Also, most white LEDs have very unpredictable outputs in the hundred microamp range, and are not 'rated' for a certain output. Hook 2 "dead" alkaleaks up to a white LED in a very dark room to see this.

At very low drain, most batteries have higher available power than at higher drains. Alkaleaks have more apparent gain here because they suffer so much under high drain. Duracell alkaleak AAs, for example, are certified for power on an 8 ohm resistor. This leads to a 188mA 'drain test,' and gives about 1500-2000mAh available. At higher currents they may only have 200 mAh available, but at lower currents this can approach 3000 mAh. Sing it with me, Eneloops get 2000-2500 mAh, so the alkaleak just barely edges it out in very low-drain apps. Of course, it might leak during the month-long use, in which case you've gained nothing.

The driver in the Quark is a buck/boost driver. In boost mode it's probably 85% efficient, and in buck mode around 95% efficient. Since the LED takes "power" (Not amps OR volts, but BOTH), then we can calculate battery power, efficiency, and get LED power to estimate runtime. Say each eneloop is 1.2v@2000 mAh, each 14500 is 3.6v@750mAh, and we have a light that wants 3.2v@100mA. How long do we run on 1 or 2 eneloops, or 1 14500?

1 eneloop: 1.2v*2Ah = 2.4Wh. 85% efficient = 2.04 available watt-hours. The LED wants 320 mW, so expect to run for 6.3 hours.
2 eneloop: 2.4v*2Ah = 4.8Wh. 85% efficient = 4.08 available watt-hours. The LED wants 320 mW, so expect to run for 12.6 hours.
1 14500: 3.6v*.75Ah = 2.7 Wh. 95% efficient = 2.57 available watt-hours. The LED wants 320 mW, so expect to run for 8 hours.

These are only estimates, but show the idea.
 

troelskc

Newly Enlightened
Joined
Dec 31, 2009
Messages
161
Location
Denmark
Really enlightened me there sport! I'm still encouraging to put together a sticky containing all this great info.
 

reppans

Flashlight Enthusiast
Joined
Mar 25, 2007
Messages
4,873
....At higher currents they may only have 200 mAh available, but at lower currents this can approach 3000 mAh. Sing it with me, Eneloops get 2000-2500 mAh.

Thanks for explaining AnAppleSnail, makes sense to me and is consistent with why 4/7s specs 1xAA vs 2xAA runtimes as something closer to 1+1=3. I had a suspicion that it would actually be the batt chemistry differences explaining why Alk and NiMh did better than Li-ions on the lower modes, as seen in the testimonial thread.
 

troelskc

Newly Enlightened
Joined
Dec 31, 2009
Messages
161
Location
Denmark
Another question. I have this Dereelight DBS V3 XM-L T6, 2xRCR123. When measuring the amps at the switch using 2 batteries (8,4v) the current shows only about 1,66 A, it should be about 2.5A. The actually current being drawn is 2 x 1,66 A = 3,32A because the current was measured using 2 cells?
 

AnAppleSnail

Flashlight Enthusiast
Joined
Aug 21, 2009
Messages
4,200
Location
South Hill, VA
Another question. I have this Dereelight DBS V3 XM-L T6, 2xRCR123. When measuring the amps at the switch using 2 batteries (8,4v) the current shows only about 1,66 A, it should be about 2.5A. The actually current being drawn is 2 x 1,66 A = 3,32A because the current was measured using 2 cells?

With cells in series, they are both delivering power. This usually manifests as "same current, higher voltage." Think of each battery as a 'pump' that can move electrons up (x) volts. For Li-Ions, this is 4.2v per cell when fresh. A series of 10 Li-Ions would give 42v. At 1 amp, it would still be 1 amp, but 42V means it's 42W.

Some lights can accept higher voltages. They then suck lower amperage out of the cells, getting the same power out. You could slurp in 2A @ 4.2v, or 1A @ 8.4v and have the same power delivered.
 

troelskc

Newly Enlightened
Joined
Dec 31, 2009
Messages
161
Location
Denmark
With cells in series, they are both delivering power. This usually manifests as "same current, higher voltage." Think of each battery as a 'pump' that can move electrons up (x) volts. For Li-Ions, this is 4.2v per cell when fresh. A series of 10 Li-Ions would give 42v. At 1 amp, it would still be 1 amp, but 42V means it's 42W.

Some lights can accept higher voltages. They then suck lower amperage out of the cells, getting the same power out. You could slurp in 2A @ 4.2v, or 1A @ 8.4v and have the same power delivered.

Argh! This is not getting any easier. How will I know how much power the emitter/driver wants on max?
Is it impossible to measure the exact current the emitter draws?
 

AnAppleSnail

Flashlight Enthusiast
Joined
Aug 21, 2009
Messages
4,200
Location
South Hill, VA
Argh! This is not getting any easier. How will I know how much power the emitter/driver wants on max?
Is it impossible to measure the exact current the emitter draws?

Well, you can easily* measure the voltage. But measuring current requires either:

1. Break the circuit and insert probes.
2. Characterize the LED for Vf at varying currents and replace in circuit.
3. Estimate driver efficiency and calculate from power in

So you can estimate it, or measure it, but you get mushy results, or messy lights.

The driver will draw what it does. The emitter is probably good to double its rated current, but you get very diminishing returns over rated current except in specially-designed applications (Soldered to copper slab, etc).
 

Wiggle

Flashlight Enthusiast
Joined
Sep 19, 2008
Messages
1,280
Location
Halifax, NS
As a special case, some types of buck drivers using a specific current control method deliver the exact same current to the LED as they pull from the batteries. The catch is that any voltage beyond the Vf for that current must be burned off in the electronics to limit this current. This method was/is very popular for single Li-ion use with single LED though because a Li-ion has most of it's discharge curve (say 3.5 to 4.2V) within 1V or less of most LED voltages for good drive levels. Many lights you see rated as 18650 or 2xCR123 lights only are of this type and you can determine the LED current by measuring what gets pulled from the batteries. They are much less efficient with 2xCR123 because a voltage closer to 6V means significantly more must be "thrown away" to get down to 3.0-3.7V of the LED.

You can picture this as the batteries delivering current like water through a hose. To limit how much water (current) leaves the hose, your thumb can restrict the flow. There is energy wasted as your thumb restricts the flow but since the hose (hopefully) has no other holes, the amount that leaves the nozzle (LED) is the exact same as is being supplied by the pump (battery).

This type of driver has another advantage in that it historically it offered a long period of regulation with a period of dimming so you get the advantages of flat output but with some warning on a dead battery. This occurs because as the cells voltage lowers down to that of the LED, the driver essentially "steps out of the way" and, aside from some losses on a polarity protection diode, the cell will directly drive the LED. That being said, with newer LEDs have lower Vfs, this dimming period may not exist for as long before a Li-ion protection circuit cuts off the flow.
 
Last edited:

troelskc

Newly Enlightened
Joined
Dec 31, 2009
Messages
161
Location
Denmark
"I still don't get it" (entirely.)

I made this table.
I stated = the stated emitter current @ 2xAA's by 4sevens.
U=actual voltage used to the test.
I=current measured at tailcap-switch.
Calculated I @ emitter = I - 85% or 95% (buck vs. boost)
BrandModelEmitterBinU rangeI statedUICalculated I @ emitter
4sevensAA2XP-GR50,9-4,20,71,331,331,1
4sevensAA2XP-GR50,9-4,20,72,691,751,5
4sevensAA2XP-GR50,9-4,20,74,110,830,8

Now I just need the missing link here guys! When the voltage goes higher than Vf the current @ tailcap drops but the light gets brighter. Explain that in detail please.
 
Last edited:
Top