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TPS92515-Q1: Please review the schematic

Chase Jeong
Genius 5685 points
Part Number: TPS92515-Q1

Hello TI experts,

My customer drew their first schematic with TPS92515QDGQTQ1. Could you review the schematic?

- Vin: 8V
- Vout: 2.8V
- Output Current: 2A
- 2 LEDs will used. (Forward Voltage 2.5V, Total Power 1340mW )

Best regards,

Chase

  • 0
    TI__Mastermind 20845 points

    Hello Chase,

    Can you give me the Vin range?  What is the desired ripple?10%, 20%, 40%????  Is there a desired switching frequency?  Also, I can not read the component values.  Please give a .pdf or something with better resolution.  The connections look correct, but I can't do calculations to verify correct values.  Have you gone through the design calculations outlined in the datasheet?  I don't see your calculations here.  

    -fhoude

  • 0 in reply to fhoude
    Genius 5685 points

    Dear fhoude,

    Thank you for your support.

    Here is the design calculator and latest schematic. Please give me some advice about it.

    - Input voltage is 8V constant. it is from POC filter.

    - Desired ripple is 10%.

    TPS92515_sch2.pdf

    TPS92515 Calculator.xls

    Best regards,

    Chase

  • 0 in reply to Chase Jeong
    TI__Mastermind 20845 points

    Hello Chase,

    I think you entered all the information correctly in the spreadsheet.  I was able to look at your schematic and look up parts.  Most everything looks good. 

    • The diode (D1) might get a bit hot based on the operating conditions and expected power dissipation, but it should work.  Depends on your heat sinking and temp requirements.  I would go with a larger diode like a 3 or 4A diode just so that we aren't right up to the max rating of the part and the max dissipation of the part. 
    • I also saw you were using an inductor that seems a bit poor based on current rating and series resistance.  The inductor will see the full 2A of the output and you are using a 1.1A temp rated inductor which also has only a 1.2A rms current.  The series resistance alone is 0.183 ohms, but at 2A that is 732mW of power dissipation which is just conducted loss let alone core losses.  You need to check the saturation current rating too.
    • You haven't decided on the resistor divider (0 ohms for both resistors right now) for the PWM signal.  what will be driving that?      
    • Your Roff resistor doesn't quite match your calculations (R36).  Not sure if that was an accident or on purpose.

    -fhoude

  • 0 in reply to fhoude
    Genius 5685 points

    Dear fhoude,

    Thank you for your support.

    I got feedbacks from my customer, here are some questions.

    - We will check D1 in 1st PCB sample. <- done.

    - We will re-check the inductor with over 2A saturation current. <- done.

    - We will check it in 1st PCB. <- done.

    - Actually R36 is already decided in the reference (which made from other customer, and they drew their schematic almost same as this reference.)

    the value was 0.33 ohm in that reference, but they could not find that value with design calculator. Could you help me about this for finding 0.33 ohm or other right value?

    and here is another question.

    - they set output voltage as 2.8V, and they measured real value of it on PCB (L1.1) it is about 5V, and each LEDs (in series) had 2.8V each. We thought that each LEDs would have 1.4V because output voltage is 2.8V and use 2 LEDs. Is it right operation?

    I attach a PDF review for your convenience.

    TPS92515-Q1_Review1.pdf

    Best regards,

    Chase

  • 0 in reply to Chase Jeong
    TI__Mastermind 20845 points

    Hello Chase,

    I don't understand why you can't find a 0.33 ohm resistor.  This is readily available.  for example:

    https://www.digikey.com/en/products/detail/te-connectivity-passive-product/TL3AR033F/3673492 

    I don't know what they are using for LEDs, but there should be a Vf vs Current curve.  Look on the curve to see the Vf at the constant current setpoint you set for the 515.  That should show you what Vf (output voltage) you should expect for that condition.  Where did you get your assessment from?  Why 1.4V????  Output of this converter is a constant current source.  The voltage is a function of the Vf of the diode at that current.  

    -fhoude