Regarding the inductor, the datasheet specifies a 4.7uH inductor, so select a power inductor with a max current that is 1.5 to 2 times the peak current. To determine the Ipeak, use the following formula:
Ipeak = (((Vin - Vtran - Vout)Ton)/L) + Io
Where, Vin is the input voltage (in your case, probably the car's battery voltage), Vtran is the voltage drop accross the switching transistor when it's fully on (the on resistance is 650mOhm, so at 2A, the Vtran would be 2A * 650mOhm = 1.3V), Vout would be 3.9V. Ton is the time that the power transistor is on during one cycle--that is computed, below. Io is the max output current--in your case, 2A
To determine Ton we need to questimate the duty cycle that this chip will be using:
D = (Vout -Vd)/(Vin - Vtrans - Vd) where Vd is the voltage drop in the "free wheeling diode"
Once we know the duty cycle, we can apply that to the freq and get Ton:
D = Ton*f
Therefore:
Ton = D/f
So, if the minimum Vo is, say 11V (i.e. car ignition off, and battery a little bit run down), and the diode has a worse case forward voltage drop of 0.8V, then the Dutycycle is:
D = (Vout -Vd)/(Vin - Vtrans - Vd) = (3.9 - 0.8)/(11 - 1.3 - 0.8) = 0.34
So, since this IC runs at a worst case fixed frequency of 1.8MHz, Ton would be approximately:
Ton = D/f = 0.34/1.8 = 189ns
And, Ipeak will be approx:
Ipeak = (((Vin - Vtran - Vout)Ton)/L) + Io = ((11 - 1.3 - 3.9)189)/4.7) + 2 = 2.2A
So, select a Power Inductor with a max DC current around 3 to 4 Amp (saturation current) and a self resonate frequency well above 2MHz.
Of course, if the parameters that I used don't fit with your application, then be sure to recalulate using the formulas, before selecting an inductor.
And, anyone who wants to chime in and refine this (or correct it) please do. I'm not anywhere near as experienced at this as I'm sure some of my colleges, here, are
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