Direct Drive vs. Buck Driver

[WeatherB]

Looking to build myself a P7 driven Mag light and was wondering what are the advantages of using a buck driver as opposed to a direct drive?

I am only looking for one mode, full power.

Tnx

 

[Greg G]

To be able to DD a P7 the bin of the LED should be an "I". You can use that with a single Li Ion or 3 C or D NiMH. A resistor *may* still be needed, depending on the front voltage of the actual LED.


Is your Mag a 2 or 3 cell light?

 

[HarryN]

Hi - welcome to CPF. Lots of fun, lots of light

I am assuming the configuration you have in mind is something like 1 x 18650 protected driving each led die ?

So, there are of course 4 main modes to drive an LED
- Direct drive - hook up the wires and let it rip.
- Resistored - 1 - 2 ohm resistor in series
- Buck - theoretical constat current if the Vf of the LED is less than the V bat.
- Boost - use when the LED Vf is always higher than the V Bat.

So, here is my simple opinion - which is somewhat subjective of course.

Battery - always use protected cells or Ni based cells. I cannot stress this enough. If you use unprotected Li based cells - well - please don't.

So - I don't recommend a complete direct drive setup ever. There is just too much risk of burning out the LED, the battery, and the light output fall off is just silly.

Resistored - I really like these - cheap, fast, simple - IF - the battery is say .5 - 2 volts above the LED Vf. 1 ohm, 2 max. Safe, simple, can do with kids, etc.

Buck driver - advantage - more control, and more range of battery input voltage. You can put several cells in series and have a much higher battery voltage. (more run time) Downside - there is usually a minimum voltage "overhead" between the LED Vf and the V bat. So - very difficult to run a power LED from a single Li Ion cell this way.

Boost - some concept as buck - just in reverse.

 

[WeatherB]

Thanks for the replies.

I haven't put together the Mag yet, but am thinking of either a 2 x 18650 or 2 x 32600, protected of course. This way I can use the Der Wichtel buck driver if I decide to go with a driver.

How can I find out which bin my P7 emitter is? The flashlight that came with the emitter was advertised as a "C" bin.

 

[Bullzeyebill]

Using direct drive, I would think that the drive current of the P4 would sag the voltage of an 18650 low enough that there would be no damage to the LED. Built in resistance of the flashlight, including switch should handle any over voltage issues too.

Bill

 

[bikenjam]

Any recommendations for a resistor size if i wanted to run 3 D alkaline batteries direct drive?

 

[rmteo]

Resistor value = (Vbat - Vf)/I where:

Vbat = Battery voltage
Vf = Forward voltage of LEd
I = Drive current

Example: (4.5-3.6)/1.2 = 0.75 ohms.

You also need to take into consideration power rating of the resistor:
P = IxIxR

Example: 1.2x1.2x0.75 = 1.08W minimum - use a 2W resistor.

 

[JamisonM]

Resistor value = (Vbat - Vf)/I where:

Vbat = Battery voltage
Vf = Forward voltage of LEd
I = Drive current

Example: (4.5-3.6)/1.2 = 0.75 ohms.

You also need to take into consideration power rating of the resistor:
P = IxIxR

Example: 1.2x1.2x0.75 = 1.08W minimum - use a 2W resistor.

rmteo, you really helped clear up some lingering questions about resistors in your post. I don't don't understand why you multiply the drive current twice though? Do you mind checking my math for a derect drive P7 with 3 alkalines?

(4.5*3.5)/2.8 = 0.36 ohms
2.8*2.8*0.36 = 2.82 watts This leads me to guess that I should get a 3 watt resistor?

 

[rmteo]

Your calculation is correct and yes, you should use a 3W (or larger) resistor. The reason why the current is multiplied twice is because:

V = I * R (Ohm's Law)
P = I * V

Substituting V with I*R in the second equation gives P=I*I*R.

One thing you will notice is that with direct drive, the current (brightness) will gradually diminish as the battery is depleted (voltage decreases over time). This is one the main reason to use a regulated driver as it will maintain constant output over time until the battery is depleted.

BTW, thanks for answering my question about removing the Rebel switch assembly in the other forum.

 

[baterija]

<Beat to the first part so it's snipped>

Jamison,
Your math works. I think the assumption that you'll have 4.5 volts under load is overstating voltage, which will lead you to much lower current than you are thinking. If you look at this graph from silverfox's battery shootout at 3.0 amps you are probably lucky to get more than 1.3 Volts even briefly under that load.

If you solved the first equation for current using 3.9 volts in and .36 ohm resistance you get 1.1 amps. 3 of the 4 cells tested in the graph don't even show a value above 1.3 volts per cell. The one that does drops below it almost immediately. I think you would be very disappointed.

 

[bikenjam]

Does this mean that a direct drive to a P7 with 3 alkaline D batteries would not fry the led?

 

[HarryN]

Does this mean that a direct drive to a P7 with 3 alkaline D batteries would not fry the led?

That is correct - A standard alkaline D cell has almost no ability to deliver those currents.

As a practical matter, you will be tempted very fast to replace the D cells with AA cells in parallel and / or go to NiMH D cells for that setup. At that point, there will be plenty of current for frying LEDs. :)

 

[wquiles]

The only other aspect that I think it is very important to mention is efficiency - to talk about DD and drivers and not talk about efficiency is like telling just the good side of a story and leaving out all of the "bad" things.

Only a DD solution is 100% efficient, everything else is a compromise. Outside of a DD solution, you are using battery power to warm up the driver - this extra power never reaches the light source. So, I basically use a DD solution whenever I can, sometimes combined with the TaskLED D2Flex, which gives you DD with dimming, at 98% or better efficiency.

Now, I will be the first to admit that a DD solution only works "well" in a few cases whre matching the vf with the loaded cell voltage works - most of the time you really need and should use a driver of some sort. So, the goal is find a driver that gives you the most efficiency as possible - I personally like 85% or better as a goal, which is important when dealing over 1Amp of current. Again, the goal is to waste as little energy as possible in the driver, so that the light source can have a cooler operation (LED's give less output as they warm up) and longer runtime (better use of the battery's energy storage).

In general, for applications less than 1000-1500mA, a boost driver is still fairly efficient enough. However, the higher the current, the more you want to have a buck driver - battery voltage being higher than the vf of the LED(s). This is specially the case when dealing with the 2.5-3 Amps for the newer P7 and MC-E lumen monsters.

To end my rant, remember this one single equation:
Power In (battery/pack voltage x Amps) - Power Out (operating voltage/vf * current) = wasted energy :eek:

Just my 2 cents ...

Will

 

[JamisonM]

Your calculation is correct and yes, you should use a 3W (or larger) resistor. The reason why the current is multiplied twice is because:

V = I * R (Ohm's Law)
P = I * V

Substituting V with I*R in the second equation gives P=I*I*R.

One thing you will notice is that with direct drive, the current (brightness) will gradually diminish as the battery is depleted (voltage decreases over time). This is one the main reason to use a regulated driver as it will maintain constant output over time until the battery is depleted.

BTW, thanks for answering my question about removing the Rebel switch assembly in the other forum.

Thanks for answering my questions. You don’t know how much this helped me. Right now, I got a DD 3D P7 maglite that I put together late last year. I’m running it with 3AA NiMH in D adapters, but hope to get my hands on some D NiMH some time in the future. Using the same formula, I figure I’ll need a 0.036 ohm, 0.3 watt resistor. I’ve gone through and redone the calculations over and over, but why is the resistor needed here so much lower then the last one? Is this because of how close the vf and battery voltage are? One other thing I’d like to ask if you don’t mind, even if I resistor my DD 3D P7 maglite, this doesn’t guarantee that It will be bright does it? To my understanding, a resistor is just a limiter correct? If I wanted it to be brighter, I should get a driver or use batteries that can provide more current without one? Thanks.

 

[rmteo]

You correct that the resistance is lower because the difference between Vbat and Vf is only 0.1V (I assume that you are using 3x AA for a total voltage of 3.6V). However, fresh off the charger, 3x AA can be as high 4.5V - it will quickly settle to about 4V once you turn on the light.

Depending on how much voltage your battery pack can sustain, there is just so much current you can push through the LED. You should be able to get 2.8A from a 3D config depending on the internal impedance of the D cells. Looking at the datasheet, the internal impedance of a typical D NiMH cell is 0.007 ohms. At 2.8A, the voltage drop across each cell is 0.02V, for a 3D configuration, the total voltage drop across the battery is 0.02*3 = 0.06V. The LED Vf is 3.6v so the total voltage required is 3.6+.06 = 3.66V. As long as your battery is able to sustain at least 3.66V under load (2.8A), the resistor method will work. I believe that you should be able to get close without needing a driver.

 

[Superdave]

whenever i've direct driven anything from an 18650 i get a weird flickering.. i use drivers now with the same batteries and don't have problems.

Might just be the Ultrafire cells though. :thinking:

 

[JamisonM]

You correct that the resistance is lower because the difference between Vbat and Vf is only 0.1V (I assume that you are using 3x AA for a total voltage of 3.6V). However, fresh off the charger, 3x AA can be as high 4.5V - it will quickly settle to about 4V once you turn on the light.

Depending on how much voltage your battery pack can sustain, there is just so much current you can push through the LED. You should be able to get 2.8A from a 3D config depending on the internal impedance of the D cells. Looking at the datasheet, the internal impedance of a typical D NiMH cell is 0.007 ohms. At 2.8A, the voltage drop across each cell is 0.02V, for a 3D configuration, the total voltage drop across the battery is 0.02*3 = 0.06V. The LED Vf is 3.6v so the total voltage required is 3.6+.06 = 3.66V. As long as your battery is able to sustain at least 3.66V under load (2.8A), the resistor method will work. I believe that you should be able to get close without needing a driver.

Thanks rmteo. I hope I answered your question as well as you've answered mine.

 

[ace0001a]

Does this mean that a direct drive to a P7 with 3 alkaline D batteries would not fry the led?

From my experience, direct driving with Alkaline Ds won't fry a P7. Alkalines have alot of internal resistance themselves. Under the load of a P7, I've usually seen 3 Alkalines drop to around a combined 3.6V which is were you want the voltage to be...but since Alkalines are unable to sustain high current output, the P7 will usually run with around 1.8A from them as opposed to the 2.8A+ you would get running on 3 NiMH batteries. I have built a few Direct Drive P7 3D Maglites and they have never had any problems running off of Alkalines...they're simply not going to be as bright as running off of NiMHs. Personally I prefer having a driver in my flashlights, but I do understand and appreciate the simplicity of direct drive too. For one thing, your LED will run for a very long time as the batteries deplete (allowing for "moon mode") and there's no circuit to worry about failing. As a flashlight enthusiast, I think it's good to be able to appreciate both methods of driving the emitter.

 

[Bullzeyebill]

+1

Bill

 

[bikenjam]

This might go a little off topic, but i am seeing that alkalines would be a bad choice for a P7 power source. Considering i just acquired 60 industrial rayovac D batteries for my maglites, what would be a good led selection for alkaline D batteries?