• State Resolved
  • Locked Locked
  • Replies 15 replies
  • Answers 2 answers
  • Subscribers 63 subscribers
  • Views 3113 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Problem with TPS61087

David Zhang16358
Expert 1860 points
Other Parts Discussed in Thread: TPS61087

 

My design is  based on figure 14 of TPS61087 datasheet and the input voltage is 3.3V

But TPS61087 can't generate 15V output, only 10V is generated.

Please help ASAP.

Thank you so much.

  • 0
    TI__Guru 65390 points

    Hello David,

    Every EVM of the TPS61087 generates 15V from an input voltage of 3.3V, therefore the part is able to generate 15V from a 3.3V source.

    One possibility I could think of is that you might overload the part. What is the load that you have connected at the output? With 3.3V input, the typically deliverable output current is roughly 500mA at high switching frequency.

    If you did not overload the output, please send me schematic and layout and if possible the BOM, that I can have a more detailled look on your design.

    Best regards,
    Brigitte

  • 0 in reply to Brigitte
    Expert 1860 points

    Thank you so much for your quick answer. I will send all to you ASAP.

  • 0 in reply to David Zhang16358
    TI__Guru 65390 points

    Hello David,

    What was the applied load during your tests?

    Best regards,
    Brigitte

  • 0 in reply to Brigitte
    Expert 1860 points

    Please check the attached files for my design.

    The 15V is going to feed two LT1763 LDO chips located on other board.

    On my test, i have disconnected those two boards, and TPS61087 can't generate 15V.

     

  • 0 in reply to David Zhang16358
    TI__Guru**** 150870 points

    David,

    How did you base your circuit on Figure 14 in the datasheet?  Figure 14 shows a circuit that converts 5V to 15V and runs at 1.2 MHz.  Your circuit is converting 3.3V to 15V at 650 kHz (FREQ is grounded)--so figure 15 is closer to what you want.  But even then, your inductor and output capacitor are different and this will affect things.  If you want to run at 650 kHz, please use the component values recommended in table 3 that show your exact conditions (3.3V to 15V at 650 kHz) or go through the detailed description to select your components.

    Another thing to try is powering the TPS61087 circuit from a known good power supply--a lab supply putting out 3.3V.

  • 0 in reply to Brigitte
    Expert 1860 points

    There is no load in my test, the circuit is open.

    the LT1763 and TPS61087 are disconnected during my test.

    I have used TPS61087 to replace MAX17067 for generating higher current.

    MAX17067 can generate proper +15V output.

    After changing R20 to 91K and C51 to 1.2nF, I have put TPS61087 on the footprint of MAX17067 by connecting pin4 and 5 together, connecting Pin6 and Pin7 together.

    I have tested today again, only 10V can be generated by TPS61087.

    So sad!!!

     

  • 0 in reply to Chris Glaser
    Expert 1860 points

    In my current circuit, I am using 1.2MHZ.

    I have selected those inductor(3.3uH) , resistor(91K), cap(1.2nF) values based on table 3 of TPS61087 datasheet  for 3.3V input and 1.2MHZ frequency.

    This 650KHZ mode is for getting more current.

     

     

     

  • 0 in reply to David Zhang16358
    TI__Guru**** 150870 points

    The schematic you posted has FREQ connected to ground.  This is 650 kHz operation.

    Could you post a scope picture of the SW pin, Vout, and Vin?

    Could you post your board layout?

  • 0 in reply to David Zhang16358
    TI__Guru 65390 points

    Hello David,

    After reviewing your schematic, I do not see anything that would lead to an output voltage of 10V.

    Could you please recheck the feedback divider and also the current rating of the inductor, please?

    I am asking you this, because normally I do not see any reason for the part to deliver the wrong output voltage. If you are measuring the voltage on the FB pin, it should be far below 1.238V and therefore the part should increase the delivered energy.

    Best regards,
    Brigitte

  • 0 in reply to Brigitte
    Expert 1860 points

    it is fixed,  I have resolderedTPS61087 then it is working and I got 15V.

    My final questions are:

    the input current of TPS61087 should be greater then 3mA when input voltage is 3.3V.

    when above requirements are meet, I can get 15V and around 600mA output.

    Thanks.

     

  • 0 in reply to David Zhang16358
    TI__Guru**** 150870 points

    Glad you got it working.

    I don't understand your question.  Can you please rephrase it into the form of a question?

  • 0 in reply to Chris Glaser
    Expert 1860 points

    My final questions is:

    The +15V output voltage and 600mA output current of TPS61087 is based on +3.3V input voltage and 3mA input current?

    Do you think my understanding is correct?

     

  • 0 in reply to David Zhang16358
    TI__Guru**** 150870 points

    I'm still not sure I understand what you are asking.  Input power will equal the output power plus the power lost in conversion.  15V and 600 mA is 9W output.  At 85% efficiency (a rough guess), this is 10.58W of input power.  At 3.3V in, this is 3.2A of average input current.  The peak current in the FET will be higher than this amount due to the ripple current in the inductor.  The minimum value of the current limit is 3.2A and this is the peak current in the FET.  So, 600 mA is more load than can be expected at 3.3V input.

  • 0 in reply to Chris Glaser
    Expert 1860 points

    Thank you so much.

     I am confused with the SUPPLY CURRENT vs SUPPLY VOLTAGE relation figure on page 7 of datasheet of TPS61087.

  • 0 in reply to David Zhang16358
    TI__Guru**** 150870 points

    Ah, figure 10.  This figure shows 3 different sets of information.  In all 3 cases, the supply current is the current into the IN pin of the IC--this is the current required to operate the IC.  It is not the input current to the system, as it does not include the current into the power stage which goes into the inductor.

    The bottom trace in the figure is the Iq of the IC--the non-switching supply current required to operate the IC.  This corresponds to the spec on page 3 of 75 uA typical.

    The top 2 traces show the total current into the IC when it is switching.  This is the IC's Iq plus the current required to operate the gate drive of the internal FET, which is very significant.  When the IC is operated at the higher frequency, the gate drive current increases.