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TPS61201: Understanding this Part

Part Number: TPS61201

Hi, I am designing a solar powered device that can power a micro-controller. I have a 3V 70mA solar panel that I want to connect to a TPS61201 to get a steady 3.3V output. I wanted to get the most output current from the TPS61201 device to power my MCU. I wanted to know a typical design to have a 0.5 - 3V input and get a 3.3V output without any fail. 

  • Hi Sherman,

    you can apply and TPS61022EVM for reference. that is the quick and easy way to evaluate if the device is OK in your system.

  • Jasper Li, thank you.

    Lastly, I would like to understand if I use this module or device, will I be able to output at least 500mA of current? 

  • Also, I found the TPS61021AEVM which I need for a 3.3V output.

  • Hi Sherman,

    do you really need 500mA 3.3V output power? that would mean at least 3.5A current from input side if the input voltage is 0.5V.

  • I have a solar panel that only can output max 70mA of current with a load. This evaluation module requires a high input current?

    I need a minimum output load from the module of 500mA and 3.3V.

    I have a 3V solar panel, so I can have an input between 1-3V with the 70mA current max to the module to produce my requirement.

    Could this work?

  • Hello Sherman,

    This will not work. Your panel needs to deliver the energy that the load needs. The maximum power of your panel is 3V × 70mA. This is a lot less than 3.3V × 500mA.

    Only if you do need the 500mA for a very short time, you could charge a big output capacitor from the panel as energy storage and draw then 500mA from that cap for a short time.

  • Hey Brigitte, I was thinking the same thing about the capacitor. I'm pretty sure I only need 500mA to start up my MCU.

    For any boost converter, I need to provide the same amount of current at the input to get the same amount of current at the output?

    Boost converters cannot increase current?

    So this is impossible? For any boost converters?

    Solar panel (1-3V × 70mA) ---> Boost Converter  ---> MCU (3.3V × 500mA)

    Sorry, trying to understand the logic behind power electronics.

    I'm doing R&D work.

  • Hello Sherman,

    Please do not forget the law for conservation of energy. The energy created by a source can only be converted, but not increased by a converter.

    Your source creates in the best case 3V * 70mA = 210 mW

    Your load needs 3.3V * 500mA = 1650mW.

    So if your load needs the 500mA only for a short period of time you need to charge a capacitor for a longer period of time. In your case to store the needed energy in a capacitor you need at least 8 times as long to get the energy from the source (1650mW/210mW = 7.86).

    Therefore you need a big energy storage element to get this energy from the source to the load. This could be a capacitor on the input or output then you can draw a higher current from the capacitor than from the source for 1/8th of the time.

  • Wow Brigitte, this makes sense. I completely disregarded the law of energy, sorry about that. I seen the converter and thought I had a solution other than a capacitor.

    To store this energy, can I design a capacitor bank or just one big capacitor that can source an rush of current at the output of the converter to my MCU?

  • Hello Sherman,

    Normally I would recommend a capacitor bank to reduce the ESR of the storage element. Nevertheless, often supercapacitors are used for storing energy and then sometimes only one is used.