# What factors do I need to consider when selecting an inductor to use in a circuit with the MT3608?

I am trying to use the MT3608 chip to boost the power output of a 3.7V lipo battery from 3.7V to 5V, which will ultimately power an ATMEGA32U4 chip. Below is a picture of my circuit where I have already selected resistors and capacitors to output a voltage of 5.01V.

The MT3608 datasheet states:

The recommended values of inductor are 4.7 to 22μH. Small size and better efficiency are the major concerns for portable device, such as MT3608 used for mobile phone. The inductor should have low core loss at 1.2MHz and low DCR for better efficiency. To avoid inductor saturation current rating should be considered

My questions are/what I need help with is:

• How (and where) do I calculate the current of my circuit so that I can determine the appropriate current rating of inductor L1?
• What inductance should be used for inductor L1 between 4.7 and 22μH and why?
• How do I calculate the best ratio of inductance to DCR to minimize power core loss?
• I think this is a good question (well-phrased, includes datasheet link and schematic, etc.), so I'm not sure why it was down-voted. My only guess would be that it asks several questions at once, instead of a single focused one. (The latter is a better fit for the site.) Commented Oct 18, 2019 at 17:54
• "How (and where) do I calculate the current of my circuit" - First you need to know how much current your ATMEGA32U4 etc. will draw. Without that you cannot calculate the 'best' inductor properties. Commented Oct 18, 2019 at 19:05
• Iin mean = Pout/Vin x 1/efficiency. Start with about 1.25 to 1.5 for 1/efficiency. PEAK Iin is higher (typically) by a factor of about (Vin + Vo) / Vin. || Inductor size is often a rough indicator of goodness - larger = better. At 5 to 20 uH you can wind an aircore inductor with no core losses. Commented Oct 19, 2019 at 6:13

Consider the operation of this circuit. When the switch (SW) is "on," (tied to ground) the inductor has Vin across it, so the current in the inductor is increasing. Since $$\V = L\times di/dt\$$, you can calculate the rate of change of current for whichever value of inductor you choose. Since the frequency is fixed, and the maximum duty cycle is 90%, you can also calculate the maximum amount that the current can increase on each cycle. So, a larger inductor value means that the circuit may take more cycles to adjust to a sudden change in load (poorer load regulation) and may have a longer start-up.
When the switch is open, (assuming that the power supply is working) the energy in the coil is $$\0.5 \times L \times {i_{peak}}^2\$$. This is how much energy you could theoretically get from each cycle (your worst case), so a larger inductance value will result in lower peak current from your battery for a given load. Multiply this energy (joules) by the number of cycles per second, and you get the power in watts. Your inductor current rating should be this calculated peak current, which will be different for different inductor values. In practice, your circuit will not charge/discharge the inductor completely on every cycle, but this will be a safe value.