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LM5023: Lm5023

john Pucci
Prodigy 50 points

Part Number: LM5023

Tool/software:

12V5A LM5023 B66229_87HE.pdf

The ripple in the circuit above is around 200 mvpp at 5 Ampere, without load 92 mvpp. Load regulation is very good, less than 0.5%.
How do I reduce the ripple to at least 1% or 120 mvpp?
If I increase capacitance, the regulation stops working correctly, and the same is true if I insert inductance.

Thanks

  • 0
    TI__Guru* 86286 points

    Hello,

    Your inquiry is under review and I will get back to you shortly.

    Regards,

  • 0 in reply to Mike O'
    TI__Guru* 86286 points

    Hello,

    Your high frequency output ripple voltage is generally caused by the I*R drop of the AC current passing through the capacitors equivalent series resistance.

    You have a few options for reducing this ripple.

    1. Select an output capacitor with a lower ESR

    2. Paralleling your output capacitors to reduce your ESR ripple.

    In your current design you might want to use two output capacitors instead of one.  I would just put a second capacitor of the same value and part number in parallel with the current one in your design.  This should cut the output ripple voltage in half from 200 mV to 100 mV.

    Thank you for interest in Texas Instruments (TI) products.  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.

    Regards,