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I have made a PCB with the LM5013 buck regulator, but can't seem to get the output voltage high enough. The goal is to be able to adjust the output voltage using the LOAD_REGULATE pin, for a variable input voltage, but for now I have LOAD_REGULATE=0 V and V_IN=40V. I have a 20 ohm, 300 W load resistor at the output.

According to the datasheet, LM5013 regulates the output voltage so that the voltage at the FB pin is equal to an internal reference of 1.2 V. I am measuring 15 V at the output, and a switching duty cycle of 38 % and frequency 148 kHz. However, from the resistor values of the FB pin voltage divider, I would expect a duty cycle of 100 % as V_FB should be lower than the 1.2 V reference:

V_FB = V_OUT * (R5||R7) / ((R5||R7)+R6) = 15 * 6667 / (6667 + 432000) = 0.23 V

If V_OUT=V_IN: V_FB = 40 * 6667 / (6667 + 432000) = 0.61 V

The actual V_FB that I am measuring is 1.96 V.

My questions are: How can V_FB be 1.96 V when V_OUT if 15 V? Where does the "extra" voltage come from? And why isn't the output voltage regulated such that V_FB=1.2 V?

Edit:

Added pictures of the layout and oscilloscope. Ch.3 (blue) is SW pin, ch.4 (red) is V_OUT, and ch.2 (green) is FB pin.

I tried to remove C6, making the circuit similar to type 1 in fig 7.1 in the datasheet. This resulted in a higher V_OUT (28 V), but the inductor made a lot of weird high-pitch sounds. The sound changes as I hover over or touch the FB pin with a probe or finger.

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  • \$\begingroup\$ Few buck converters can do 100% duty. This one has a minimum off time of 50 ns, as given in the datasheet. \$\endgroup\$
    – Hearth
    Commented Jul 5 at 20:50
  • \$\begingroup\$ Datasheet also states "...a minimum forced off time of 50 ns provides extremely high duty cycles, allowing VIN to drop close to VOUT", which is what I want \$\endgroup\$
    – leheim
    Commented Jul 5 at 20:53
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    \$\begingroup\$ Yes, that's the part I'm talking about. But that minimum forced off time is going to prevent you from reaching 100% duty. It'll let you get close to 100%, especially running at such a low frequency, but it won't let you get to 100%. \$\endgroup\$
    – Hearth
    Commented Jul 5 at 20:54
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    \$\begingroup\$ C6 in your design does what? \$\endgroup\$
    – fraxinus
    Commented Jul 5 at 21:13
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    \$\begingroup\$ I would expect less than 200 pF for C6 here, calculated from the datasheet. \$\endgroup\$
    – Jens
    Commented Jul 5 at 23:55

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