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I'm very new to electronics, this is the most complicated circuit I've tried to create thus far. I'm attempting to step down a 24V DC input to 5V output. I'm skipping the output filter for right now just until I get the buck to work. I know that in a real circuit I'll want the traces to be shorter and the components as close together as possible.

I'm having 2 problems.

  1. The output (red trace from C33) is about 3.5V
  2. I'm unclear if I should have the common ground across the breadboard.

enter image description here Sorry if my question is unclear or the image is too cluttered. Component List Below:

Here's another picture with labels.

enter image description here

Here's the schematic from the datasheet. It's almost exactly what I'm trying to do.

enter image description here

Edit 1 Taking the suggestions of both letsfetz and Yoshimitsu I've made some changes.

  1. I'm now using this inductor
  2. I've moved the diode and rerouted the line coming from pin 2 (see image) enter image description here

I'm still unclear on if I should connect both ground rails. If I do connect both grounds then I read about 23V on the output. If I do not I read about 4.5 V on the output (for both reading I leave the ground of my multimeter on the J side ground rail).

Edit 2 I'm still trying to figure out why I'm getting about 23V on the output. Below is a screenshot from my oscilloscope. 5v/div & 1ms/div. Reading output of pin 2 with respect to ground rail. I have a load of a LED with a 1K0 pull up resistor. The voltage measured after the LED is about 2.8V

enter image description here

Edit 3 I never figured out why it wasn't working on the breadboard. I created a schematic and board design file in Eagle, sent it to a PCB mill and after soldering all the same components, it works. The only differences are that I added the output filter and a couple LEDs. Wish I knew why the breadboard didn't work.

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  • \$\begingroup\$ for anybody curious here is the link to the project in Asiler. aisler.net/p/QWSVBPHO \$\endgroup\$ – Sean Mar 7 at 6:30
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The link to the inductor data, that you provided, shows that the inductor has a series resistance of 6.8 Ohm and is rated for 180 mA only. If you wanted to draw 0.5 A from the output, the voltage drop across that resistor would be 3.4 V. Furthermore the inductor would saturate. I suggest you choose an inductor that can handle the output current (plus the ripple).

Grounding: unless you really know what you are doing, I highly recommend you to use a solid ground plane for the hole circuit and connect all ground connections as short as possible to it. Generally the connections to the capacitors should be minimized. You can have longer connections to the inductor (but only the inductor - not the 0.33 Ohm sense resistor).

Good luck!

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  • \$\begingroup\$ Try an inductor similar to this: Murata 12RS224C , 220 uH, 1.1 A, 0.3 ohms \$\endgroup\$ – Mattman944 Feb 9 at 13:32
  • \$\begingroup\$ +1 for ground connections. Also, anything more than a few hundred milliamps will give you trouble on cheap solderless breadboards. \$\endgroup\$ – Mattman944 Feb 9 at 13:34
  • \$\begingroup\$ thanks for the tip. I've switched inductors to another one I had one hand. I'm still unclear if I should be connecting the two ground rails of the breadboard. \$\endgroup\$ – Sean Feb 9 at 22:18
  • \$\begingroup\$ Yes, yes, yes. Connect all grounds together. And I fully agree with @Mattman944, that the boards you are using are pretty much worthless for Switch-Mode Power Supplies and other electronics, which are containing rather high speed signals (especially their edges). Solid copper boards are ways better. For fast prototyping you can mill out some traces / bars and for the ICs you can buy some glue on footprints. \$\endgroup\$ – letsfetz Feb 9 at 22:35
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    \$\begingroup\$ thanks for all the help @letsfetz after shipping off the PCB to a mill the end board works just fine. I'm getting a 5-5.3VDC output and the ripple is so small I can't zoom in enough on the DSO Nano to view it. \$\endgroup\$ – Sean Mar 7 at 6:21
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Looks like you've connected your inductor to gnd.

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  • \$\begingroup\$ ... instead of to pin 2 of the IC \$\endgroup\$ – Huisman Feb 9 at 13:12

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