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I'm trying to use the LMR51420 Simple Switcher IC to step down my 20 V DC to 5 V DC.

I have this circuit which is the typical application circuit from the datasheet: circuit

When I measure the voltage on the output with my multimeter, it says 6.55 V. The voltage on R8 is 0.8 V but it should be 0.6 V according to the datasheet.

This is the output which should be 5V

2 V, 10 us per division:

1

Also the output, 2 V, 200 us per division:

2

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    \$\begingroup\$ Do you have an image of your layout/construction? Additionally, do you have an oscilloscope trace or any other way of determining whether that 6.55 V is a stable 6.55 V or some kind of oscillating value that averages at 6.55 V? \$\endgroup\$
    – nanofarad
    Commented Jan 5, 2023 at 15:49
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    \$\begingroup\$ What's the load on your circuit? \$\endgroup\$
    – Hearth
    Commented Jan 5, 2023 at 15:53
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    \$\begingroup\$ @greybeard Those are the values the datasheet shows for 5V, using those values I get 0.602V across R8. Datasheet picks 13.7k for R8 and then calculates R7 as 100.4k, uses 100k as closest standard value. \$\endgroup\$
    – GodJihyo
    Commented Jan 5, 2023 at 16:02
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    \$\begingroup\$ @Andyaka I added a short description \$\endgroup\$
    – Linus
    Commented Jan 5, 2023 at 16:38
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    \$\begingroup\$ Sorry, but why does the box in the left corner say the offset is -8.00 volts? Have you tried adding a load to the output and seeing what happens? \$\endgroup\$
    – Andy aka
    Commented Jan 5, 2023 at 16:42

1 Answer 1

Reset to default
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Web bench says you need a 12uH inductor with 25mOhm DCR. The output capacitor also needs to have low DCR. If you aren't using the reference layout, make sure the PCB traces have lower or equal resistance and inductance. Also check the vias and make sure they are not contributing too much inductance or resistance in the layout.

enter image description here

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