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Looking for 5 Volt Regulator Recommendations

Other Parts Discussed in Thread: TPS61032, TPS65023, TPS61030, TPS2513EVM-527, TPS61030EVM-029, TPS2513

Hello PMIC Forum,

I am seeking a 5 Volt regulator for an energy conversion project.

My project converts mechanical energy in to electrical energy used to power consumer electronics. After the conversion, the output voltage from the converter is between 2 to 4 Volts. I need to regulate this converted voltage to 5 Volts that can source 200 mA to 500 mA. I am wondering if there is a TI product that can achieve this. I am looking for a regulator demo board that would help in my application.

Best regards,

Frank Hayden

  • Hi Frank,

    It sounds like you need a Boost product like TPS61032. I'm routing your post over to the non-isolated DC/DC forum where they should hopefully be able to provide more help.

    (PMICs in our context refer to ICs that integrate multiple power rails and help to manage the sequencing as well, products like the TPS65023)
  • i would suggest TPS61030 for this application.
  • Hi Jasper,

    I am working on a project that takes mechanical energy and converts it to electrical energy. I want to use this converted energy to charge batteries and drive other consumer electronics devices. The conversion yields a voltage anywhere from 2 to 4 volts. To test the system I am currently using a DC power supply to drive the TPS61030EVM-029 which drives the TPS2513EVM-527. I have independently verified that the TPS2513EVM-527 can charge a battery when driven by a DC power supply capable of supplying 5 Volts at anywhere from 250-500 mA. 

    Today I received the TPS61030EVM-029 board and used this to drive the  TPS2513EVM-527. I am using a DC power supply to drive the TPS61030EVM-029 set to 3.3 volts. But I am unable to charge a battery with the TPS2513EVM-527 when it is driven by the TPS61030EVM-029 module. Below are the conditions of the TPS61030EVM-029.

    Vbat=3.3 VDC, JP1 has SYNCH connected to Vbat = 3.3 VDC, JP2 has ON connected to Vbat = 3.3 VDC. Vout=4.83 VDC, and LBO is at 2.24 VDC.

    The TPS2513EVM-527 is powered up but and has the battery "thinking" it is charging, but the charge percentage is decreasing rather than increasing. Also the current I measure from the TPS61030EVM-029 to the input of TPS2513EVM-527 is only 6 mA. Can you offer any suggestions? I am under the impression that the TPS61030EVM-029 can source up to an amp. I am attaching a picture of my setup, if this helps.

    Thanks and best regards,

    Frank Hayden

  • 1. what is output voltage of TPS61030 if the TPS2513 is disable or disconnected.
    2. please change the SYNCH pin to low logic
  • Hi Jasper,

    Thanks for your reply.

    I had already switched the SYNCH pin to be low.

    The output of the regulator is 4.97 Volts at no load and with the load.

    Today I used a more powerful source to drive the TPS61030. I was able to successfully charge the battery, albeit slowly. I suspect that the source of my problem is that I am making the mistake that the source driving the TPS61030 does not have to have the same source current capability that the TPS61030 is supplying to the load. In other words if the load is 250 mA the input source to the TPS61030 should have a 250 mA capability. There is no current amplification in TPS61030, correct?

    Thank you and best regards,

    Frank Hayden

  • Hi Jasper,

    I am wondering what input current is required by the TPS61030 so that the regulator can source 250 to 500 mA?

    And also, I am wondering if the TPS61030 is capable of current amplification.

    I would appreciate a response regarding these questions and any other advice you might have for me.

    Thank you and best regards,

    Frank Hayden

  • the input current of a boost converter is higher than the output current, because the input power should be equal to output power even the efficiency is 100%. the input current calculation method IIN=VOUT*IOUT/(VIN*0.8)
  • to further explain the last post. if require 500mA@5V output power from 3.3V input, the input current will be 0.5*5/(3.3*0.8)=0.95A.
    let me know if there is further question