Fenix P2D & L2D ?

[kts]

I just bought a P2D premium Q5, I read at the LED museum that the current draw is 346.2mA on turbo. (the P4 model)

On the L2D CE the current usage is 1,142mA on turbo.

They should put out the same amount of light, why is the L2D CE using so much more power? :shrug:

 

[eebowler]

You're sure Craig didn't mistakenly forget the '1' to make current draw of the P2D 1346.2mA? With both of them having a supply voltage around the same value, current draw should be about the same. (I believe both of them uses the same driver.)

 

[kts]

I dont think so, cuz the L2D also uses twice the power of the P2D on low

 

[divine]

I don't think those numbers are right. :shrug:

 

[Darkpower]

I just bought a P2D premium Q5, I read at the LED museum that the current draw is 346.2mA on turbo. (the P4 model)

On the L2D CE the current usage is 1,142mA on turbo.

They should put out the same amount of light, why is the L2D CE using so much more power? :shrug:

I own both a P2D and a L2D and right off the bat I will suggest that the difference in current draw is simply voltage. You mistook the term "power" for "current". I'm a retired electrical engineer and often non-electrical people will confuse voltage, current and power. All three are linearly related. The P2D uses 2 CR123s and the L2D uses 2 AA batteries. Assuming ideal conditions that an AA is 1.5 volt and the CR123 is 3.0 volt, to drawn the same power from a CR123 you need only half as much current. Power is measured in Watts and the equation is P=VI where Power=Voltage X Current. You can deliver more power by either increasing the current or the voltage. But higher voltage is much more efficient. That is why power companies transport power at very high voltage. You can deliver for example 1 million watts across transmission lines at 1amp X 1 million volts or you can TRY to transmit the same at 1000 amps at 1000 volts. A 1000 amp current will have much higher transmission losses then 1 amp. The voltage drop loss will be the resistance in the transmission lines time (x) the current. So that if the transmission line has a resistance of ½ ohm, the voltage drop across the transmission line is 500 volts for a 1000 amp load or ½ volt for a 1-amp load. I don't want to stray too off topic, but generally higher voltage is better and more efficient. However once you get to 80 volts and higher there are concerns of safety, and extremely large voltage, there is the danger of dielectric failures and arching.

Low voltage devices are notoriously high on current. The problem with high current is that it turns into heat when running through semiconductor P-N junctions and through resistive elements. Like the example with the high-tension lines. So if the P2D and L2D are running at 6 volts and 3 volts respectively, the L2D needs at least twice as much current delivered to the emitter, perhaps more because the Fenix have semiconductor switches. Each P-N junction there is an intrinsic voltage drop, plus these lights have DC-DC converters that push pull the voltage to keep the LED at its happy zone. The L2D not only has to delivery twice the current to achieve the same power, but the DC-DC converter is going to boast its voltage when the batteries sag below 1.5 volt, whereas, in the P2D, the batteries generally are on the high side like 3.2 volts and drop to 2.5 volts, so that the DC to DC will actually be chopping current to deliver less power. Plus on top of that you have different emitters which may have inherrently diffent efficiencies in terms of milliwatt in/ Lumens out.

That is my opinion and I make that opinion only from what I have read about the DC-to-DC converters used in the Fenix and with my experiences with these lights. I have not bench tests these lights nor have any equipment at my house to do a thorough round of testing.