So regarding efficiency... I just did some math, assuming we are using this to power a 12x BR diode at 500mA with a 3.7V li-ion.
The equation for efficiency is as follows, where n = efficiency.
n = (1-D2*Vd/Vin)/(1+(Rdcr+D*Rdson)/(D2^2*Rout))
Vd = .35V, which is the typical forward drop of the Schottky.
Vin = 3.7V because that's the nominal voltage of a li-ion
Rdcr = .047 Ohms, which is the resistance of the inductor (I used another inductor for my version of this design - smaller)
Rdson = .17 Ohms, which is the on-resistance of the switch (stated in the datasheet)
D = Duty-cycle, which is calculated as (Vout-n*Vin)/Vout, so I substituted that into the equation.
D2 = Defined in the datasheet as (Vin*n)/Vout, so I substituted that into the equation.
Rout = equivalent resistance of the load diode - Vf/I.
That left me with this:
n = (1-0.05833333334*n)/(1+(.2191380569*(.217-.1048333333*n))/n^2)
Then, using Maple to solve it (I did all of this in Maple, by the way, because it's such a great program
), I got three values of n = 0, .917, and .049.
Obviously, the value we will be using is .917, because the value of 0 is trivial and the value of .049 is the conjugate of the efficiency (other side of the curve).
So, assuming the values obtained for all parts are equal to the typical value stated in data-sheets, we are looking at a minimum efficiency of 89.7% (the formula stated above says that it does not account for losses totaling less than 2%, so we can take that into account).
Not bad for such a simple driver. Often better than the Flex!
I redid the math for one driver set at 900mA outputting 4.6V for a 445nm diode (assuming we would parallel these two). The minimum efficiency comes out to be 88% even. Turns out, paralleling drivers shouldn't reduce the efficiency of the entire system. So, we are looking at an input of 2.4A at 3.7V to power a 445nm diode at 1.8A.
Not bad at all!
(Sorry for the long post!)
