I am working on a IoT project that involves an arduino and a 4G LTE module. The device will be operated with two Alkaline batteries. Both the arduino and LTE module require 3.3V input so I want to design a boost converter that reliably delivers 3.3V from a voltage range of 1.6 to 3.1. The device will be in a deep sleep for most of the day and only wake up 2-3 times a day for a short period of time, so an important converter requirement is low quiescent current. While the arduino doesn't draw much current (50mA-100mA), the LTE module requires up to 1.2A of current when awake.

Doing my research for suitable boost converters, I came across the TPS613221ADBVR. It has low quiescent current (6.5µA) and can provide up to 1.6A of current. But given my limited knowledge in electrical circuits, I'm a bit overwhelmed when it comes to choosing the right inductors and capacitors. The datasheet is excellent and gives two examples on how to choose inductor/capacitor for the given requirements of 2.2V output @ 50mA (example 1) and 5V output @ 500mA (example 2).

The requirements in my case are as follows:

  • Input Voltage: 1.6V ~ 3.1V
  • Output Voltage: 3.3V
  • Output Current: 1.2A
  • Output Voltage Ripple: ?? (how to find out?)

What inductors and capacitors do I choose? Do I need a Schottky Diode and if so, which one? What Output Voltage Ripple can be tolerated?

  • \$\begingroup\$ The datasheet has all the needed formulas and values. At what step do you need help? Or let a computer do it for you. \$\endgroup\$
    – CL.
    Commented Jan 1, 2019 at 21:38
  • \$\begingroup\$ Also read the spec VERY carefully. If you need 1.2A at 3.3V from 1.6V in, its switch needs to handle 3.3/1.6 * 1.2A, about 2.5A. If that switch is rated at 1.6A, ... too bad. \$\endgroup\$
    – user16324
    Commented Jan 1, 2019 at 22:04
  • \$\begingroup\$ Thanks @BrianDrummond! I didn't think of this limitation. \$\endgroup\$ Commented Jan 1, 2019 at 23:03
  • \$\begingroup\$ @CL. the power designer looks like a great tool. Unfortunately it errors out when I use an output current > 0.2A. Maybe to help me get started, how do I know what output voltage ripple a load can tolerate? The LTE module for example can tolerate input voltage from 3.0V to 3.6V. Does that mean I can have output voltage ripple of ±0.3V for 3.3V output? \$\endgroup\$ Commented Jan 1, 2019 at 23:40
  • \$\begingroup\$ Generally speaking anything digital will perform much better the less and the slower the ripple is. Larger ripples at closer to it's operating speed would be worse I would imagine. Because you're at the start of your design process and using a calculator, I would start with "reasonable ripple" and see what components you would need, and then incrementally decrease the ripple and recheck components. If you can find a premade power supply (with any input voltage) that you know for sure is OK to use, you could check the output ripple on the known good supply and use it as a maximum. \$\endgroup\$
    – K H
    Commented Jan 2, 2019 at 1:06

2 Answers 2


The TPS613221A has a typical switch current limit of 1.2 A, and the guaranteed minimum is 0.75 A. Anyway, the maximum output current is different (as shown by formula (1)).

At least for an inductor of 2.2 µA, trying to get 1.3 A out of it would be a bad idea:

TPS613221A output current

You need a more powerful boost converter, like the TPS61021A:

TPS61021A output current

And putting your requirements (1.6­–3.1 V, 3.3 V, 1.3 A) into TI's Webench tool indeed results in a design based on the TPS61021A, with a 1 µH inductor (SRU1028-1R0Y), and 10/30 µF capacitors:

1.6-3.1 V to 3.3 V, 1.3 A boost convert

  • \$\begingroup\$ This is exactly what I needed @CL. I know my initial question sounded like I'm just to lazy to do the math myself. But I was just so overwhelmed that I didn't even know where to start. I'll give this a try and see how well it performs. Do I need a Schottky diode with this circuit? Also, I was curious, will the quiescent current vary based on the inductors/capacitors/resistors I use? \$\endgroup\$ Commented Jan 3, 2019 at 6:26
  • \$\begingroup\$ If a Schottky were needed in this circuit, the tool would have added it. Iq is determined by the chip, the current through the feedback divider, and any leakage currents. \$\endgroup\$
    – CL.
    Commented Jan 3, 2019 at 7:22
  • \$\begingroup\$ Got it. Now comes the fun part to figure out how to hand-solder a WSON-8 package (2 mm x 2 mm) like the TPS61021A... sigh \$\endgroup\$ Commented Jan 3, 2019 at 7:54

I'm afraid you fell into a common trap related to Boost converters by forgetting that:

Input Current = Output Current * Boost Ratio / Efficiency

1.2 Amps at 3.3V is 4 Watts. The converter cannot create energy, so it has to draw at least 4 watts from the input. At 1.6V, this is 2.5 Amps. If we factor in a realistic 80% efficiency, this will be close to 3.1 Amps.

Problem: AA or AAA Alkaline batteries won't be able to deliver this kind of current. Let's have a look at their internal resistance, here the data is from Energizer.

enter image description here

So, two fresh batteries will total 0.3-0.4 ohms between 10°C and 25°C, which already results in a substantial voltage drop... but internal resistance also increases as the batteries discharge:

enter image description here


Around the center of the graph, a 500mA current drops battery voltage by 0.2V, so its internal resistance is 0.4 ohms. Two batteries in series would be 0.8 ohms. It isn't realistic to expect 4 watts from these, voltage will drop way too much.


Try to get a good idea of the current your module needs using an oscilloscope, it will most likely be a low average current made up of a number of high current spikes. Measure the height and width of the current spikes.

Then you can decide if you can use a capacitor to provide energy during high current spikes and smooth that to a lower average current that your battery can handle, or if you actually need another battery which is capable of high transient current, like lithium. Then you can decide which converter to use, but not before you are sure the battery can deliver the required power.

  • \$\begingroup\$ Yes, thanks @peufeu for your thoughts on the batteries. I am aware of the limits of two AA Alkaline batteries, which is why I want to use six AA Alkaline batteries (3 x 2). Two batteries in series have a capacity of about 2500mA. Using 3 of those in parallel will increase the capacity to about 7500mA. This should reduce the total internal resistance. I only mentioned two batteries in my initial question to help understand the voltage range I am dealing with (1.6 - 3.1). The reason I want to use Alkaline is because they have much lower self-discarge rate than NiMH, LiIon, or Lipos. \$\endgroup\$ Commented Jan 3, 2019 at 6:02
  • \$\begingroup\$ @DanielSchnabel If you have 6 batteries then wire them in series and use a Buck converter instead of a Boost, it will be more efficient. \$\endgroup\$
    – bobflux
    Commented Jan 3, 2019 at 10:16
  • \$\begingroup\$ You're giving me a new idea @peufeu. Indeed buck converters are a bit more efficient than boost converters. So the voltage range we are talking is 4.8 V - 9.3 V (6 Alkaline in series). It seems though that buck converters generally have a higher quiescent current than boost converters (~1-5 mA). Maybe I just haven't found one that has an acceptable low quiescent current (~5-20uA). Anyway, thanks for the tip! \$\endgroup\$ Commented Jan 4, 2019 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.