Help find-best "cheap" led flashlight !

[ossumguywill]

Make sure that the current draw in your flashlight is reasonable. 5A from a single 18650 will be extremely dangerous.

AW 18650s are 5A rated, but then again, those are "expensive". Sorry guys, if you want quality, brightness, etc, you have to fork over a bit. No way around it.

 

[randomlugia]

AW 18650s are 5A rated, but then again, those are "expensive". Sorry guys, if you want quality, brightness, etc, you have to fork over a bit. No way around it.

Sorry, I didn't know that. I remember some instances when people tries to power 5A SST-50 lights with protected 18650's, but it wouldn't work, because the batteries were cutting off. They were then informed that the batteries were cutting off because 5A was too high a current for the batteries and it could get dangerous.

Also, the SST-90 is capable of 9A, so it seems a waste to put one in a single 18650 build and only drive them at half of their potential.

 

[Prototype]

DX has a single 18650 SST-50 light and a two 18650 SST-50 light, the second is claimed driven at 5A, DJNY bought one of these a while back, it includes two 18650s and a charger for ~$73 I believe. The single 18650 light is ~$57.

Single 18650 light: DealExtreme: $57.06 Aurora SH-40 Luminus SST-50 5-Mode 1000-Lumen Memory LED Flashlight (1*18650)

Two 18650 light: DealExtreme: $75.20 TrustFire ST-50 SST-50 5-Mode 1300-Lumen Memory LED Flashlight (2*18650)

And I consider them cheap, which is why I'm posting them.

Of course if you don't mind waiting and saving, a 50W HID can be had for ~$180 shipped in the US with a case, battery and charger.

 

[Toke]

I see, my 18650's are still in the mail and I am not familiar with their discharge characteristics. Looking at all the curves only gets me so far, the resistance of the host will be very important in the balance between battery voltage and diode Vf.

A sst-90 at 4-5A* on a single 18650 does appear wasteful, but I will be telling myself that it gives more lumen pr. Watt than e.g. a sst-50.

*If that is possible? I do not know yet.

It looks like I can get lipo's capable of 1100mA 3,3V 30A discharge current. They appear to work for RC people, so maybe for flashlights too.

 

[randomlugia]

I see, my 18650's are still in the mail and I am not familiar with their discharge characteristics. Looking at all the curves only gets me so far, the resistance of the host will be very important in the balance between battery voltage and diode Vf.

A sst-90 at 4-5A* on a single 18650 does appear wasteful, but I will be telling myself that it gives more lumen pr. Watt than e.g. a sst-50.

You should really consider a 2x18650 host. 5A shouldn't be a problem with two cells, and a host that small even at 4A will heat up to the point of burning you within minutes.

Prototype - I read a review on the SH-40, that's actually where a read about the discharge problems. The guy said the light was pulling 5A. Thread here.

 

[ossumguywill]

You should really consider a 2x18650 host. 5A shouldn't be a problem with two cells, and a host that small even at 4A will heat up to the point of burning you within minutes.

Prototype - I read a review on the SH-40, that's actually where a read about the discharge problems. The guy said the light was pulling 5A. Thread here.

Umm.... 5A is a problem no matter how many cells you use (in series), LOL... the amps stay the same, the volts double, right?

 

[Toke]

Umm.... 5A is a problem no matter how many cells you use (in series), LOL... the amps stay the same, the volts double, right?

Depends, I read it as wiring them in parallel.
It is perfectly possible even if they are end to end.

 

[randomlugia]

Umm.... 5A is a problem no matter how many cells you use (in series), LOL... the amps stay the same, the volts double, right?

Really? I figured if you only need half of the voltage from each battery, then you're dividing the power draw from each battery by two.

 

[ossumguywill]

Really? I figured if you only need half of the voltage from each battery, then you're dividing the power draw from each battery by two.

Yes, you are. But power is watts, not amps. 5A draw with one battery and 5A draw with two batteries is just as bad for each battery.

2 batteries in series at 5 amps is bad.
2 batteries in parallel at 5 amps is safer.

 

[randomlugia]

Yes, you are. But power is watts, not amps. 5A draw with one battery and 5A draw with two batteries is just as bad for each battery.

2 batteries in series at 5 amps is bad.
2 batteries in parallel at 5 amps is safer.

I was thinking that 5A & 2.1V from each battery would be safer than 5A & 4.2V from a battery. Well thanks for telling me before I tried and blew something up. lol.

 

[ossumguywill]

I was thinking that 5A & 2.1V from each battery would be safer than 5A & 4.2V from a battery. Well thanks for telling me before I tried and blew something up. lol.

Well, that would work beautifully if batteries could change voltage-here's the problem-voltage is pushed, current is pulled. A 4.2V battery will never deliver less than 4.2V (or whatever its current charge voltage is) (well, there are certain scenarios where it's possible but it results in explosion.) So basically you have current and voltage mixed up, you can draw whatever current you want (below a certain safe level) but voltage is not up to you. Glad nothing blew up!

 

[randomlugia]

Well, that would work beautifully if batteries could change voltage-here's the problem-voltage is pushed, current is pulled. A 4.2V battery will never deliver less than 4.2V (or whatever its current charge voltage is) (well, there are certain scenarios where it's possible but it results in explosion.) So basically you have current and voltage mixed up, you can draw whatever current you want (below a certain safe level) but voltage is not up to you. Glad nothing blew up!

But don't flashlight drivers regulate the current coming out of the batteries? Pretty much any 2 cell LED light is 8.4V, and I know a P7 can't handle that much.

 

[ossumguywill]

But don't flashlight drivers regulate the current coming out of the batteries? Pretty much any 2 cell LED light is 8.4V, and I know a P7 can't handle that much.

Yes-but that's a bit different. let's say a LED driver works at 90% efficiency, and gives the LED 18.9 watts. With a single battery at 4.2v, the driver would need to draw 5 amps from the battery. With two batteries at 4.2v, the driver would only need to draw 2.5 amps from each battery. (That is if the driver can handle the voltage range). So if there are fewer batteries, the driver needs to run each battery harder.

 

[BShanahan14rulz]

random,

The flashlight driver is what bucks the voltage. The batteries are still totalling 8.4V across both of them.

The way that using two cells is safer than one is that with two cells and a DC-DC converter, Pin=Pout (minus efficiencies, but basics for now).

For example:
We want to power a device that needs 6 amps, and will drop 5V at 6A. The two Li-Ion cells in series feed a DC-DC converter, which powers the device.

V(in)=8.4V
I(in)=?
V(out)=5V
I(out)=6A
P(in)=P(out)

Well, P=I*V, so:
I(in)*V(in)=I(out)*V(out)

Plugging in the values we know, we have:
I(in)*8.4V=6A*5V
I(in)=(30A*V)/(8.4V)
I(in)=~3.6A :wave:

You'll notice that 3.6A is not as many amps as what's going to the load, and it is within the safe current discharge range of the Li-Ions.


Ex. 2a
Single Li-Ion powering a 6V@3.5A load, via a theoretical 100% efficient boost-driver

V(in)=4.2V
I(in)=?

V(out)=6V
I(out)=3.5A


P(in)=P(out)
4.2*I(in)=6*3.5
I(in)=21/4.2
I(in)=5A :wave:

You'll notice that in this example, while a 5A draw from a single Li-Ion is kind of high, it's still within the 2C discharge specification for the higher capacity name-brand cells.

Ex. 2b
Same as example 2a, except the battery has been discharged some, to say, 3.4V.

V(in)=3.4V
I(in)=?

V(out)=6V
I(out)=3.5A


P(in)=P(out)
3.4*I(in)=6*3.5
I(in)=21/3.4
I(in)=6.2A :wave:

You'll see that as the battery depletes, it's voltage gets lower but the load doesn't change. The boost converter is then forced to pull more current from the battery. 6.2A is too much for a cobalt Li-Ion

 

[randomlugia]

random,

The flashlight driver is what bucks the voltage. The batteries are still totalling 8.4V across both of them.

The way that using two cells is safer than one is that with two cells and a DC-DC converter, Pin=Pout (minus efficiencies, but basics for now).

For example:
We want to power a device that needs 6 amps, and will drop 5V at 6A. The two Li-Ion cells in series feed a DC-DC converter, which powers the device.

V(in)=8.4V
I(in)=?
V(out)=5V
I(out)=6A
P(in)=P(out)

Well, P=I*V, so:
I(in)*V(in)=I(out)*V(out)

Plugging in the values we know, we have:
I(in)*8.4V=6A*5V
I(in)=(30A*V)/(8.4V)
I(in)=~3.6A :wave:

You'll notice that 3.6A is not as many amps as what's going to the load, and it is within the safe current discharge range of the Li-Ions.


Ex. 2a
Single Li-Ion powering a 6V@3.5A load, via a theoretical 100% efficient boost-driver

V(in)=4.2V
I(in)=?

V(out)=6V
I(out)=3.5A


P(in)=P(out)
4.2*I(in)=6*3.5
I(in)=21/4.2
I(in)=5A :wave:

You'll notice that in this example, while a 5A draw from a single Li-Ion is kind of high, it's still within the 2C discharge specification for the higher capacity name-brand cells.

Ex. 2b
Same as example 2a, except the battery has been discharged some, to say, 3.4V.

V(in)=3.4V
I(in)=?

V(out)=6V
I(out)=3.5A


P(in)=P(out)
3.4*I(in)=6*3.5
I(in)=21/3.4
I(in)=6.2A :wave:

You'll see that as the battery depletes, it's voltage gets lower but the load doesn't change. The boost converter is then forced to pull more current from the battery. 6.2A is too much for a cobalt Li-Ion

Thanks for the in-depth answer, though I didn't really need something that exact...

Yes-but that's a bit different. let's say a LED driver works at 90% efficiency, and gives the LED 18.9 watts. With a single battery at 4.2v, the driver would need to draw 5 amps from the battery. With two batteries at 4.2v, the driver would only need to draw 2.5 amps from each battery. (That is if the driver can handle the voltage range). So if there are fewer batteries, the driver needs to run each battery harder.

Ok I get it now, when you change the electrical power (assuming resistance isn't a factor) , all you're changing is current; voltage stays the same no matter what. But why isn't 2.5 amps from each battery any safer than 5 amps from one? (as you said a few posts back)

 

[ossumguywill]

Ok I get it now, when you change the electrical power (assuming resistance isn't a factor) , all you're changing is current; voltage stays the same no matter what. But why isn't 2.5 amps from each battery any safer than 5 amps from one? (as you said a few posts back)

2.5 amps from each battery IS safer... :thinking:
Either I made a mistake, or you misunderstood it.

Also, a while ago we were talking about AR coatings on both sides of a lens-turns out it does help total transmittance. Oops! There is a little bit of internal reflection between the outside of the lens and the air.