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I'm trying to use TI's LMR16006 Simple Switcher Buck Converter to regulate 48V down to 5V on my circuit.

I used TI's WebBench tool to calculate component values for their reference design, and had a board fabricated according to the design:

Buck Converter Circuit Schematic

I hand-soldered a prototype of the board, and was able to get it working after some finicky hand-soldering. I had some trouble, but assumed that it was just due to poorly soldered joints (since I was hand-soldering a bunch of small parts for the prototype).

Then I had 10 boards professionally made, and I'm finding that they don't function at all. V_out (across C6) is 1.26 V, and I'm expecting to see 5V!

I checked a couple of obvious things on the nonfunctioning boards:

  • I checked connectivity across the board to see if there were any bad joints and/or if anything was wired incorrectly, but that doesn't seem to be the issue.
  • The bootstrap capacitor (C4) needs to have a voltage of at least 3V; in both of my boards (working and not working), the voltage is around 6-7 volts.
  • I was concerned that perhaps the SHDN pin was being pulled low, but on both boards the value is hovering around 5V, which should be high enough to ensure that the regulator itself isn't shut down.

After looking at them on scope, it looks like the on the working board, the SW pin is switching at the correct frequency (~700 kHz), while on the non-working board, the frequency is MUCH slower (~150 kHz).

I'm trying to figure out why the converter's not working as expected, but I'm honestly not sure what to investigate next. Any suggestions for how to proceed troubleshooting would be super helpful.

Edit: Board design uploaded below. enter image description here

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    \$\begingroup\$ Stick a current sense resistor on the input and scope it and check the input current waveform. \$\endgroup\$ – DKNguyen Apr 8 at 20:29
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    \$\begingroup\$ Are there more layers to the PWB? Your ground plane is cut in half on the right side. \$\endgroup\$ – Mattman944 Apr 8 at 22:16
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    \$\begingroup\$ And to what extent your layout have resemblance to TI suggestions? This is ridiculous. I see no resemblance whatsoever, only brutal violation of all PCB design rules for power switchers at every possible point. In no way this construction can be salvaged. You need to start over, and don't be overly inventive. Their layout done for a reason, and the reasons are explained in many places, e.g. here, rohmfs.rohm.com/en/products/databook/applinote/ic/power/…. \$\endgroup\$ – Ale..chenski Apr 8 at 22:31
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    \$\begingroup\$ Nathan, majority of the switch-mode powre supply ICs (including our LMR16006) have evaluation boards. Next time you want to add a switcher to you board, be sure to look at the layout of the evaluation board. Switchers can be layout-sensitive. Try to copy the layout of the eval board to the extent that it's practical. That can save you some tough troubleshooting. \$\endgroup\$ – Nick Alexeev Apr 8 at 22:32
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    \$\begingroup\$ @NickAlexeev, "Switchers can be layout-sensitive." They ARE sensitive, all MHz-range switchers are fundamentally sensitive to implementation of "high-current" loops. You can't get decent (advertized) efficiency (nor even basic functionality, as this example shows) without reasonable implementation of these loops. Especially in 10:1 designs. \$\endgroup\$ – Ale..chenski Apr 8 at 22:38
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I used TI's WebBench tool to calculate component values for their reference design, and had a board fabricated according to the design:

Allow me to express a serious doubt about the board fabrication statement. Here is your layout, and the recommended layout, side-by-side:

enter image description here

Their layout is done that way for a reason, and the reasons are explained in many places, e.g. here. All MHz-range switchers are fundamentally sensitive to implementation of "high-current" loops. You can't get decent (advertised) efficiency (nor even basic functionality, as this example shows) without reasonable implementation of these loops. Especially in a 10:1 design. Manufacturers spent a lot of application resources to make right layouts, and design of internal pads on IC are usually optimized for particular fan-out. There is no reason whatsoever to deviate from the reference designs unless you can propose a better layout (sometimes it may happen). In this design the most glaring mistake is the routing of high-current switch node (around SW pin). Huge parasitic inductances of skinny traces likely change the switching curve of internal transistor, which likely causes some odd internal behavior and refusal to start the switcher properly.

Also note the tight low-inductance placement of Cin and Cout caps. And the designs usually use no thermal relief, only solid connections, all for the same reason - to exclude any parasitic inductance in high-current switching loops.

I am afraid this construction can't be salvaged and needs total re-design, preferably in full accord with manufacturer's suggested layout.

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  • \$\begingroup\$ Thank you for your help -- I saw the recommended schematic and component values, but somehow I missed the recommended design, and this makes a lot of sense given the switching speed. The points you made really help me understand some not-obvious parts of the reference board layout, and I'll be sure to capture those in my next design. thank you! \$\endgroup\$ – nathan lachenmyer Apr 9 at 18:22
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I agree that layout is not good. I also agree that there is no reason not to copy the demo board. At the very least look carefully at demo boards and suggested layouts in datasheets as a guide. The key is to keep the high current loops small. Remember that vias have inductance; do not use them in the switching current loops.

  • Check that the polarity of D4 is correct. Although if it was wrong a ~0.5 V output is what you should get, so this is probably not the solution.
  • Check that the voltage divider resistors are the correct values or measure the voltage at the feedback pin. It should be 0.765 V (typ) according to the datasheet. if R1 and R2 are swapped Vout should regulate at 0.9 V.
  • Try applying a load. For low power loads a switchable resistor box works well. Maybe a load will overcome some instability that is causing the switcher to shutdown.
  • When I bring up a switcher I like to apply a square wave pulse to the input. That way I can capture just the startup behavior on an o'scope. The pulse duration limits the energy available to damage any components if there is a problem. I suggest you do the same. Put probes on the INPUT, SW, FB, OUTPUT nodes. Trigger on rising edge of INPUT. Only apply 12 V or so to the input for a few ms (since there is no soft start time in the datasheet). Go from there. Also start with a board you haven't powered up yet. It may be that poor layout is generating ringing that trips up the IC or some protection circuits within it.
  • Order one of the TI demo boards. From the tools and software section of the product page. Either the LMR16006XEVM which is designed to operate at 700 kHz or the LMR16006YEVM which is designed for 1.2 MHz switching. Swap out the inductor for yours. Your should be a a higher inductance than the demo board as that is designed for 12 V nominal.
    • Consider buying an off the shelf module or a system in package switcher instead of brewing your own. I assume you are new to switching converters. Learning them by yourself is a hard way to do it. This search yielded the LMZ36002 and LMZ35003 as possibilities for you to look at.
  • Consider using a part from Analog Devices power by Linear line. They have excellent field support. This is probably the best chance to get an outsider to help you out.

Your schematic is missing a value for L1, what is it?

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