Drivers vs. Resistors

I'm confused on exactly how drivers like the Downboy work, I see that they use resistors to regulate current so how do they improve efficiency over using a simple resistor?

T-Bone, Welcome to CPF!

Downboy, BadBoy, NextGen, Shark, Fatman nFlex, MaxFlex.
Are all pretty much constant current regulators. Whether they are buck (lower) or boost (raise). As for EXACTLY how they function you need to look into the data sheet.

Your resistor does not control the current and as the battery capacity drains so will the output of your light. Resistor "typically" is less efficient. The only time when they are more efficient is when you put a really really REALLY small resistor in the chain which kinda defeat their purpose.

Thanks for the info, I''ll be sure to check out the data sheets. Am I right to assume that both buck circuits and resistors will essentially burn off the excess current in the form of heat?

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oops

Using a resistor to limit current does waste power in the form of heat, but it is not all wasted. Since it also limits the current throughout the whole circuit, the reduced current also lowers the power in the other components. This reduced power is what is protecting the LED. But it is also reducing the power lost in the switch resistance and other such accidental losses.

A driver uses a more complicated circuit to accomplish the same current limiting function. The efficiencies of these cicuits can vary a lot. Most of the numbers I have seen have been in the 70 to 90+ percent range. The "lost" efficiency is going to show up as heat in the driver.

Generally, in these circuits the resistors don't just "dump" the excess power but are used more to set the control point of the circuit. For example: a 350ma, 700ma, and 100ma converter may all use the same actual "driver chip" but by changing what resistors are connected to it you change what current it operates at.

The converters are really switching regulators. That's as far as I can explain. Beyond that detail you'll need to do your own research.

Most of the inefficiency will result in heat.
But like everything else "the Devil is in the details"

Lets take a look at the following examples to give you a better idea.

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#1 12V battery driving 3 LEDs @ 700mA / 3.5Vf

Solution #1 (DD with resistor).
The effective forward voltage of the 3 LEDs in series is 10.5V, so you decides to get a small resistor to drop the voltage from 12V to 10.5V
Your efficiency is 10.5/12 = 87.5% (pretty good huh!)

Solution #2 (Downboy @ 700mA).
DB efficiency is 75-85% (not too shabby either).


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#2 12V battery driving 1 LEDs @ 700mA / 3.5Vf

Solution #1 (DD with resistor).
You'll need a resistor to drop the 12V to 3.5V.
Efficiency is then 3.5V / 12V = 29% (ouch! better get a pretty hefty heatsink to come along with that resistor)

Solution #2 (Downboy @ 700mA).
DB efficiency is 75-85% (not too shabby either).
this is where you pay $13 for the converter, instead of the $1 power resistor

Ah I see thanks for the examples that clarifies things very much. Aside from being more efficient during some applications a driver will also give you constant output, that definately seems the way to go. I have a DB k2 drop in replacement it is rated at 6-3/4 Watts does this mean at 4.5 V it requires 1.5A? So if i'm running at 6V will it only require 1.125A for the same output?

As for power requirement here's the equation

Pin = Pout / Efficiency

So for your case
Pout = 6.75 W
Efficiency = 0.85
Therefore Pin = 7.94W

P = V * I

If your power source is 4.5V then your Iin is 1.76A.
If your power source is 6.0V then your Iin is 1.32A.

Hope this helps.

Oups! How did I end up here?