dc dc converter - LT3652 Inductor Selection

First question here

In my system, a LT3652 IC should drive a solar panel with these specifications:

Voc = 9,94V
Vp = 8,54V
Vmp = 7,95V
Isc = 0,90A
Wp = 7,15W

to charge a 3,7v Li-Ion battery at 2A max.

To select the inductor, it has been suggested me from a user:

"Select an inductor of 10uH capable of 2.5 to 3A. E.g., Bourns p/n SDE0805A-100M. It is tricky to calculate ripple current for a solar panel, because Vin max does not occur during Iout max. But 10 uH should give you low ripple current"

I would like to know if this formula in the datasheet can help me how to calculate L.

It is:

$$L= {\frac{10\cdot R_{SENSE}}{\Delta_{I(MAX)}}}\cdot V_{BAT(FLT)} \cdot \left[1-\left(\frac{V_{BAT(FLT)}}{V_{IN(MAX)}}\right)\right] (μH)$$

In the above relation, $\Delta_{I(MAX)}$is the normalized ripple current, $V_{IN(MAX)}$is the maximum operational voltage, and $V_F$ is the forward voltage of the rectifying Schottky diode. Ripple current is typically set within a range of 25% to 35% of $I_{CHG(MAX)}$, so an inductor value can be determined by setting $ 0.25 < \Delta_{I(MAX)}< 0.35 $

$R_{SENSE} = 0,05 \text{ }\Omega$

$I_{CHG(MAX)} = 2 \text{ }A$
$\Delta_{I(MAX)} = 0,5 \text{ }A$ (25% of $I_{CHG(MAX)}$)
$V_{BAT(FLT)} = 4,2 \text{ }V$
$V_{IN(MAX)} = 9,94 \text{ }V$

Hence:

$$L=\frac{0,5}{0,5} \cdot 4,2 \cdot \left[1-\left(\frac{4,2}{9,94}\right)\right] (μH)$$

$$L= 2,44 \text{ }μH$$

Is this formula correct to estimate the inductance?

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- EDIT - BOUNTY REWARD

In the datasheet it is also mentioned that inductors also need to meet the volt-second product requirement. Is this parameter important? I calculated:

$$V_{SEC}= -1,34 V\cdot \text{ }μS$$

from this formula:

$$V_{BAT(FLT)} \cdot \left(\frac{1-V_{BAT(FLT)}}{V_{IN(MAX)}} \right) (V\cdot \text{ }μS)$$

Why this specification is not listed in the vendors' datasheets?

Answer

I believe there is an error in your calculations and the actual value is 2x your calculated value.

The datasheet gives

$$L = \frac{10\cdot R_{sense}}{\frac{\Delta I_L}{I_{CHG(MAX)}}} \cdot V_{BAT(FLT)} \cdot \left(1-\frac{V_{BAT(FLT)}}{V_{IN(MAX)}}\right) \text{(uH)}$$

where $\Delta I_L = 0.25 \cdot I_{CHG(MAX)} = 0.5 \text{A}$.

With $R_{SENSE} = 0.05\text{ } \Omega$, $V_{BAT(FLT)} = 4.2 \text{V}$, $I_{CHG(MAX)} = 2.0\text{A}$, $V_{IN(MAX)} = 9.94\text{V}$, we have:

$$L = \frac{10\cdot 0.05}{\frac{0.5}{2}} \cdot 4.2 \cdot \left(1-\frac{4.2}{9.94}\right) = 4.85 \text{ uH}$$

if my calculator is correct. So a 5.1 uH inductor would probably be a good choice. On page 22 of the datasheet there is a 1 cell LiFePO4 reference design, which has a 5.6 uH inductor (similar to your design). However, as your maximum input voltage is lower, your inductor can be smaller. Of course choosing a larger inductor (within reason) will probably not hurt the circuit, and improve the ripple. So consider 5.1 uH as more of a lower bound, anything between 5 uH and 10 uH is probably appropriate.

Of course, you should build a prototype with your chosen inductor value, and then measure voltage ripple, battery charging current, and so on, and make sure that these look tolerable for your application.