# 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?

- 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?

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.

• Thank you @uint128_t. I can't understand why ∆IMAX became ∆IL / ICHG(MAX). – Nic1337 Jan 31 '17 at 17:57
• I found the problem: there is an error in the datasheet version LT 1212 REV D. In the LT 1215 REV E the formula is like yours ;) – Nic1337 Jan 31 '17 at 19:37
• In the datasheet it is also mentioned that inductors also need to meet the volt-second product requirement. Is this parameter important? I calculated -1,34 V*uS – Nic1337 Jan 31 '17 at 21:41

The recommended L for the LC filter is adequate in most application from the document to avoid input ripple

The LT3652 is biased directly from the charger input supply through the VIN pin. This supply provides large switched currents, so a high-quality, low ESR decoupling capacitor is recommended to minimize voltage glitches on VIN. The VIN decoupling capacitor (CVIN) absorbs all input switching ripple current in the charger, so it must have an adequate ripple current rating.

10μF is typically adequate for most charger applications

Increasing L value reduce ripple and also modify the response time of the output, so the 10u henry recommended is adequate for most cases

• The paragraph you have quoted is discussing the input filter capacitor, not the inductor! – uint128_t Jan 31 '17 at 16:26
• yes, that is what I was referring to, since the input supply is direct from solar panel, which is subjected to have more variation, more over the IC regulation circuity is biased from this voltage it is more important to suppress the ripple at the VIN rather than VOUT since he is already using the stated L value recommended in the document – Raj Jan 31 '17 at 16:31
• Do not flag 'wrong' answers as low quality. Low quality flags are intended only for spam, comments posted as an answer, and questions posted as an answer. Use the downvote flags for wrong answers or edit the answer to make it right. – Voltage Spike Jan 31 '17 at 16:48