TK's Emisar D4 review

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tocirahl
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Man this data is really making me reconsider my change to a FET + n + 1 driver…

It doesn’t seem like it would actually be much more efficient, but I guess the regulation would at least be nice.

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tocirahl wrote:
Man this data is really making me reconsider my change to a FET + n + 1 driver…

It doesn’t seem like it would actually be much more efficient, but I guess the regulation would at least be nice.


The problem is not at all the amount of regulation but the sheer output: you want 4000 lumen?; then your pour single battery is sucked dry in 3 minutes! You want 4 lumen to read your book? (The beam is great for reading a book) You have 50 hours, enough to finish half the Potter series.
tocirahl
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I’m actually not too interested in the 4000 lumen output, but I was hoping to be able to get say 1000 lumens of stable output with significantly longer battery life at ~300-400 lumens.

Looks like the 7135s may put out too much waste heat unfortunately.

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@djozz
Want 200 lm? On a weak battery you can’t have it.

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tocirahl wrote:
I’m actually not too interested in the 4000 lumen output, but I was hoping to be able to get say 1000 lumens of stable output with significantly longer battery life at ~300-400 lumens.

Looks like the 7135s may put out too much waste heat unfortunately.


What D4 does now puts out even more heat at the same lumen output.
tocirahl
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Right I’m aware that a FET + 1 is horrendously inefficient at medium modes, but it looks like a FET + n + 1 isn’t that much better.

I was hoping to jump from 67 lm / W to around 110 lm / W at ~400 lumens, but it looks like it’s going to be more like 80 lm / W.

I guess on the plus side, efficiency will improve as the battery drains. Maybe I should look into IFR batteries?

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Agro wrote:
@djozz
Want 200 lm? On a weak battery you can’t have it.

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?
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djozz wrote:
Agro wrote:
@djozz Want 200 lm? On a weak battery you can’t have it.
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 don’t think this will happen. I don’t 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, that’s my guess.

Edit:

- 200 lumens with FET drive using PWM means you’re 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 you’re 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 “regular” 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 that’s how I see it. Ok maybe I’m not understanding something?

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djozz wrote:
Agro wrote:
@djozz
Want 200 lm? On a weak battery you can’t have it.

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 have not tried, will later today.
hIKARInoob wrote:
djozz wrote:
Agro wrote:
@djozz Want 200 lm? On a weak battery you can’t have it.
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 don’t think this will happen. I don’t 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, that’s my guess.


No. It will keep shifting between 350 mA and 15A.
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Agro wrote:
No. It will keep shifting between 350 mA and 15A.

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.

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Ok, started a check.

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 Smile

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Agro wrote:
What batteries are you using and are they fully charged?

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.

Thank you for your help.

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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. It’s actually the same thing with why PWM is used in the first place. Max output of let’s 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.

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zak.wilson wrote:
7135+219C is a pretty inefficient combination. The larger the difference between forward voltage and battery voltage, the more the 7135 just burns off the excess as heat, and the 219C has very low forward voltage. Combined with a buck driver, the 219C’s efficiency is much better.

You are right, sadly buck drivers all have big inductors, I don’t think it is possible to replace the driver with a buck in D4, unless putting shorter batteries in tubes meant for longer ones.
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hIKARInoob wrote:
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. It’s actually the same thing with why PWM is used in the first place. Max output of let’s 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.

Current is instantaneous, PWM is to trick our eyes, not making the circuit “feel an averaged current”.

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leestrong wrote:
Current is instantaneous, PWM is to trick our eyes, not making the circuit “feel an averaged current”.

This makes sense. But even if current is instantaneous, the amount of electrons passing through the circuit is the average value.

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BTW, who’s gonna be the first to run the emitters 2s2p with a pair of 18350? Evil

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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 Smile ) so next report is then. Further testing later this evening.

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hIKARInoob wrote:
This makes sense. But even if current is instantaneous, the amount of electrons passing through the circuit is the average value.

I don’t really know how battery voltage reacts to 15kHz PWM , I am just guessing here. I think the experiment will be more conclusive.
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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?).

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djozz wrote:
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 Smile ) so next report is then. Further testing later this evening.

Hold on, maybe I’m 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 it’s not: much lower)?

Ok, so why not just measure voltage directly? Let’s say resting voltage is 3.3V (example), and we connect battery to D4’s 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?

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leestrong wrote:
hIKARInoob wrote:
This makes sense. But even if current is instantaneous, the amount of electrons passing through the circuit is the average value.
I don’t really know how battery voltage reacts to 15kHz PWM , I am just guessing here. I think the experiment will be more conclusive.

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hIKARInoob wrote:
Ok, so why not just measure voltage directly? Let’s say resting voltage is 3.3V (example), and we connect battery to D4’s 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 don’t 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.

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leestrong wrote:
hIKARInoob wrote:
Ok, so why not just measure voltage directly? Let’s say resting voltage is 3.3V (example), and we connect battery to D4’s 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.

Of course. This is the kind of mistake I would have made if I were testing… ugh… Facepalm

Edit:

It’s 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?

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hIKARInoob wrote:
It’s 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?

To see the voltage fluctuation under 15kHz PWM, we need something with sampling frequency higher than 30kHz, store the samples, and show the curve on a screen.

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leestrong wrote:
Edit: Even if we can see the voltage, we don’t 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.

Yes we do, and I see you figured that out. The most relevant section of the code is here

To sum up for those who don’t read C, it checks for low voltage every 16ms and steps down if it detects low voltage four times in a row. That’s many times slower than the PWM and should be sufficient to prevent spurious stepdowns.

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leestrong wrote:
zak.wilson wrote:
To sum up for those who don’t read C, it checks for low voltage every 16ms and steps down if it detects low voltage four times in a row. That’s many times slower than the PWM and should be sufficient to prevent spurious stepdowns.
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Ok, but the next question is what exactly is observed: what exactly does the hardware see? In reality you can have high frequency fluctuation of voltage due to PWM, but what does the voltage sensor read? Is it able to extract the peak low voltages in a tiny interval, so LVP kicks in, or maybe it only sees an average, so LVP does not kick in? So, something checks voltage, but the voltage is not a fixed value (PWM)…

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gregor wrote:
Agro wrote:
What batteries are you using and are they fully charged?

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.

Thank you for your help.

That will do it! I find a couple of spare ready to go 30Q’s work with this light………….does not take long to get the V down into the mid 3’s. Swap out and carry on the fun Smile
What would work better is having 3 x D4’s , that way the spare cells are housed for instant WOW Big Smile

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My little test was based on this:

Agro wrote:
@djozz
Want 200 lm? On a weak battery you can’t have it.

Result: at least when using a high drain battery (that is the least what the D4 deserves) you can run rhe D4 at 200 lumen until almost empty.

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