Stupid question from a newbie!

[Dan2503]

If I am to drive an LED with a rating of. 3-18v , will I only get maximum lumens and throw by putting 18v into it.
Thanks
Dan

 

[moderator007]

Hi Dan2503,
Do you mind telling us which led or dropin you are referring to.

 

[Dan2503]

I'm looking at the XML t6 but the range is 3-8.4v. Does this mean I have to try and achieve 8.4v for max output. I'll be honest I don't really understand the physics. Do two 3.7v make 7.4v or just a bigger capacity 3.7v. Also the mAh rating for the drop in is 500 to 1000 ma. Surely a 18650 @ 2400mAh is not suitable then.
Hope I've made sence
Thanks

 

[ryguy24000]

Do two 3.7v make 7.4v or just a bigger capacity 3.7v.
Thanks

Yes and Yes
3.7 x 2 = 7.4 in a series circuit and 3.7 in a parallel circuit.
Series= one path of current flow
Parallel= multiple paths of current flow

 

[Dan2503]

Now I'm getting somewhere. If I'm going to drive the drop in mentioned, will I need to use two cr123 @ 3.7v then to achieve near full potential

 

[moderator007]

Two connected end to end (series) to give you 7.4v nominal voltage, yes.
Most buck drivers work best with voltage well above the led's vf.
They usually have a sweet spot (voltage) for the best efficiency.
If its a constant current driver the brightness should be about the same as long as your voltage is in range of the driver. The efficiency will greatly vary with the voltage though.

 

[45/70]

Most buck drivers work best with voltage well above the led's vf.
They usually have a sweet spot (voltage) for the best efficiency......The efficiency will greatly vary with the voltage though.

It depends on the actual driver, but generally buck drivers are most efficient with a voltage about 0.5 Volts above the Vf of the LED. The more a driver has to "buck" down the voltage, the less efficient they are. That's not to say that adding more cells (more voltage, as well) to power the light won't increase runtime, it certainly will, but it'll actually be slightly less efficient.

It works the same way at the other end. With boost circuit drivers, the closer the source voltage is to the Vf of the LED, the more efficient the circuit. For example a 3.6/4.2 Volt Li-Ion cell will be much more efficient than a 1.2 Volt NiMh cell. One difference here, is that with many drivers, the LED will be in direct drive at cell voltages above ~3.8-4.2 Volts, due to the fact that the voltage is higher than the actual LED's rated Vf. While still more efficient, the cells will be drained faster due to over driving the LED during the first part of the discharge, thus affecting runtime. Again though, the more the circuit has to "boost" the voltage, the less efficient it is. Of course, buck circuits are much more efficient than boost circuits to begin with. It's much more efficient to reduce voltage, than it is to multiply it.

Dave

 

[Dan2503]

I'm a little lost. I'm not worried about run time just maximum performance from the led. I think what your saying is some reduce voltage whilst some boost it. The more favourable is to reduce. So if the drop in I'm looking at says 3- 8v will this reduce the voltage down if I use two 3.7v cr123's
Please keep it simple.......I'm very new to this :-)

 

[Mr Happy]

OK, as with many questions the answer is "it depends". As far as I can tell from scanning the thread you have not told us what drop in you are considering, so it is hard to give you an better answer. The XML T6 is not a drop in, it is an LED emitter and there are many lights and drop ins using that emitter.

Most LEDs are driven with a regulator. The regulator sits between the battery and the emitter and controls the flow of power to the emitter to regulate the brightness. In general, a regulator that can take a wide range of voltages will give maximum brightness at the higher voltages, and the maximum brightness might fall off at lower voltages. This may also depend on the specific battery used. It may also depend on the heat sinking arrangements of the drop in and the thermal management of the regulator. What's more, it can be hard to tell differences in brightness by eye, so you may not notice anyway.

If you have a drop-in that says 3-8 V you may expect it to work best in the 6-8 V range and to be less bright at 3 V. But that is just a general assumption.

 

[moderator007]

It depends on the actual driver, but generally buck drivers are most efficient with a voltage about 0.5 Volts above the Vf of the LED. The more a driver has to "buck" down the voltage, the less efficient they are. That's not to say that adding more cells (more voltage, as well) to power the light won't increase runtime, it certainly will, but it'll actually be slightly less efficient.


Dave

Here's a buck driver that doesn't seem to follow that. Like 45/70 said it depends on the actual driver.