I have not checked that (I suppose you have?), but does that really happen? I understand the theory: PWM-ing 15A output to get the medium modes has the potential to get under LVP voltage early because of the voltage sag under that load, but has anyone checked that this really happens at the point that the battery still has a significant charge?
I donāt think this will happen. I donāt think it will draw 15A. Theoretically if you want half the output using PWM, it should draw about half the amps, so about 7.5A (slightly non linearity). At least, thatās my guess.
Edit:
- 200 lumens with FET drive using PWM means youāre getting the low efficacy of the led at the max output of ~3500 lumens. Efficacy of the same led at 200 lumens using current control is much better.
- 200 lumens with FET drive using PWM means youāre getting the tint that you get at max output of ~3500 (which could mean bad news). Tint of same led at 200 lumens using current control is āregularā tint.
Ampdraw, and voltage sag when getting 200 lumens with FET drive using PWM is not the same when getting max output of ~3500 lumens, but is directly proportional to the amount of time FET is open. So 50% open means 50% of max drawof ~15A.
Well thatās how I see it. Ok maybe Iām not understanding something?
Exactly. But this occurs so fast, that what the battery actually feels is the effective value. So when the FET is 50% open, then the draw will fluctuate between 350 mA and 15A, but effectively will be around 7.5A. And this value determines voltage sag, and not the value at 15A.
D4 set at output 200 lumen (leds are 219C 3500K 90CRI so the current is a bit higher than with stocks leds at 200lm), on a 30Q battery starting at 3.54V (so well over half drained). No protections kicking in yet, waiting for things to happen
I just found out that all my INR18650-30Q cells have only around 3,40-3,50V. Now they are all in charger.
Will report back if this will fix the problem. Most likely it will.
Another way of looking at this phenomenon is by analysing what exactly current is. 1A = 1 Coulomb of electrons/sec, which is an amount of particles per unit time.
This should mean that if the FET is opened 50% of the time, then only half the amount of electrons pass through. This should result in half of the current, hence 7.5A, hence voltage sag when 7.5A.
Edit:
Analogue. Itās actually the same thing with why PWM is used in the first place. Max output of letās say 1000 lumens, and if we use PWM to reduce output to 500 lumens, then the max output is still 1000 lumens, except we see only 500 lumens. 500 lumens is the effective value, although max output is 1000 lumens.
Therefore max draw with PWM is 15A, but it alternates between this and 350mA so fast that the effective valuer is 7.5A.
You are right, sadly buck drivers all have big inductors, I donāt think it is possible to replace the driver with a buck in D4, unless putting shorter batteries in tubes meant for longer ones.
Half hour into the test, output is back to 175 lumen, resting battery voltage is 3.27 V now.
I have reset the output at 210 lumen (sorry, setting an exact output using ramping is not easy).
I have to leave in 10 munutes (family related stuff: dinner in restaurant ) so next report is then. Further testing later this evening.
Have to go, at this monent no LVP has kicked in yet, the D4 is down to 193 lumen, with a rested battery voltage of 3.18 V. A li-ion at 3.18 V is as good as empty (less than 5%? Who knows exact values?).
Hold on, maybe Iām not getting it, but we want to know what voltage sag is when output is a fairly low 200 lumens, and whether LVP is going to kick in or not right? The question is whether voltage sag is the same high value when current draw is 15A or not (and I claim itās not: much lower)?
Ok, so why not just measure voltage directly? Letās say resting voltage is 3.3V (example), and we connect battery to D4ās head without battery tube using low resistance wire. Then we measure voltage directly when output is around 200 lumens, so we know voltage sag immediately or not?
That requires a DSO.
Edit: Even if we can see the voltage, we donāt know how the protection circuit works, maybe it is triggered with one low voltage measurement, maybe it is triggered when seeing consecutive measurements for who knows how long. Wait, my bad, its open source.
Of course. This is the kind of mistake I would have made if I were testingā¦ ughā¦ :person_facepalming:
Edit:
Itās frequency/algorithm related that the DMM is likely not able to give the correct value (I think). But would an analogue volt meter give the correct value?