I'm trying to design a circuit that would boost a Li-Po battery voltage to 6V with 1A of current. I'm planning to use LMR62421 but I'm stuck what should be the inductance value for an inductor...

Datasheet has a detailed section for choosing an inductor, but it is too complicated for me to be honest: datasheet

Could I go with a rule of thumb to whack the biggest inductor that I have when in doubt? Or are there any downsides of using bigger inductors then you need in boost applications (other then the cost of components)?

  • \$\begingroup\$ For a buck converter, you can simply increase the inductance. Not for a boost converter though. \$\endgroup\$
    – user16324
    Mar 6, 2017 at 12:25
  • \$\begingroup\$ @Brian: You can't just keep making the inductor bigger. That has implications to stability, and also increases the transient response time. \$\endgroup\$ Mar 6, 2017 at 12:30
  • \$\begingroup\$ @Olin : in extremis, you are of course correct. But a buck converter doesn't have the energy storage considerations you have with a boost converter, so has a wider range of choice of inductor value than is practical here. \$\endgroup\$
    – user16324
    Mar 6, 2017 at 12:33

2 Answers 2


If the datasheet doesn't explicitly give you the value of inductor to use, then go back to first principles and decide for yourself.

You say you want 6 V at 1 A out. You didn't specify the input voltage other than with the vague "Li-Po battery voltage", so I'll arbitrarily pick 3.5 V to use as example. From the datasheet, you can see that the IC switches at 1.6 MHz.

For efficiency, you want to run in continuous mode at the 1 A output current, and you want the ripple to be reasonably small relative to the average. However, if you make the ripple too small, then the controller will have a hard time regulating the output in response to transients. For sake of example, let's see what 100 mA peak to peak gets us.

When the switch is on, there will be just the 3.5 V input voltage applied to increase the current in the inductor. When the switch is off, the inductor is conducting current from the input to the output thru a Schottky diode. (6 V) - (3.5 V) = 2.5 V reverse voltage. However, the forward drop across the diode is also in there. Let's say that adds another 500 mV to the reverse voltage on the inductor, for a total of 3 V. So the inductor is being driven forwards with 3.5 V and reverse with 3.0 V.

Since the we are assuming 1 A average steady state, the current rise during the switch on time must equal the current drop during the switch off time. The switch on time will be (3.0 V)/(3.0 V + 3.5 V) = 46% of the total switching cycle, and the switch off time the remaining 54%.

The switching frequency is 1.6 MHz, so 625 ns period. From above, we find that the switch on time is 288 ns, and the off time 337 ns. Now we can simply plug in the numbers to get the inductor value that satisfies the constraints. Using the particulars from the switch on time:

  H = V s / A = (3.5 V)(288 ns)/(100 mA) = 10.1 µH

As a sanity check, the particulars from the switch off time should yield the same answer:

  (3.0 V)(337 ns)/(100 mA) = 10.1 µH

So, if you're OK with 100 mA peak to peak ripple at 1 A out, then get a 10 µH inductor.

You also have to decide the inductor current rating. In this example, the current will go from 950 mA to 1.05 A each cycle. That means the saturation current should be somewhat higher than 1.05 A, like maybe around 1.2 A at least.

You then get to trade off DC resistance with size and price. That's a judgement call you have to make. Lower DC resistance will yield higher efficiency. For example, the 10 µH part of the Coilcraft MSS6132 series has 70 mΩ resistance and a 10% drop saturation limit at 1.36 A. At 1 A, that's a extra 70 mV out of 3.0 to 3.5 V, or about 2% extra loss. Only you can decide whether that's good enough.

  • \$\begingroup\$ Thanks for this elaborate answer! You did provide the answer, but more importantly you taught me how to approach solving this problem! Additionally, I plan to power inductive loads (DC motors) from this converter. Would the type of load play any role in choosing components with boost converters? \$\endgroup\$
    – StjepanV
    Mar 6, 2017 at 15:41
  • 1
    \$\begingroup\$ @stj: No, 6 V at 1 A is 6 V at 1 A no matter what produces it or comsumes it. \$\endgroup\$ Mar 6, 2017 at 15:43

Or are there any downsides of using bigger inductors then you need in boost applications (other then the cost of components)?

Certainly (aside from size and weight), an inductor with higher inductance will typically have more turns – and thus, more ohmic resistance, and thus, more loss (unless you also make the wire diameter larger, which makes the inductor even bigger and so on).

TI has an excellent design calculator on its homepage. Use it. The "Mobile/iPad version" works without adobe flash. I don't know why someone would use adobe flash for this, but TI does. When you use it, it will give you the component values as well as component recommendations.


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