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I'm trying to pick the best (cost-effective, small footprint, reliable) dc-dc converter, to convert 24V -> 5V for my device. In the future I'm planning to switch to 3.3V.

The device consists of microcontroller, some logic ICs, RS485 transceiver and a few leds.

I need at most 300mA at 5V.

So far I tested three setups:

LM2576, running at 57kHz. 680uH inductor, 330uF input capacitor, 1000uF output capacitor. However, this setup is pretty big. Big inductor, two very large capacitors, and D2PAK converter use too much space on the board. Original IC is pretty expensive and designed for much higher currents (3A) than my needs (300mA).

Mornsun B2405S-1W, integrated DC-DC, 100kHz, 220uF input and output capacitor. It's almost ideal, but the input voltage may not vary more than 10% from 24V. There is also no 3.3V version availible. It is isolated (what I don't need) and quite hard to buy. The datasheet recommends 22uF electrolytic capacitor at the output, which also consumes some space.

RT8259, this was my last try. It's designed to be quite cheap and efficient. It is elastic enough to provide 5V and 3.3V. The datasheet recommends 22uF output, 10uF input ceramic capacitors and 6.8uH inductor. I used 10uH inductor. No electrolytic capacitor was an advantage for me. It runs at 1.4MHz. I used small CD32 inductor and SS31 diode.


The problem is, that with RT8259, I'm getting really unstable output voltage:

1

To compare - from LM2576:

2

and from B2405S-1W:

3


What I tried to do:

I tried adding 100nF ceramic, 4.7uF tantalum caps at the output of RT8259, but there was NO change.

The layout is exactly as in the datasheet.

The questions:

  • How bad is such voltage for powering digital circuits, like my? How can I improve the output voltage?
  • If I increase the inductance, should it help?
  • What is going to happen when the inductance is to high / to low?
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  • \$\begingroup\$ It looks like you are getting charge current leaking into your feedback voltage. Did you notice the recommended PCB layout in the datasheet? How and where did you tie the feedback ground to the high current ground? And I'd suggest you do need an output capacitor. richtek.com/assets/product_file/RT8259/DS8259-03.pdf \$\endgroup\$ Mar 22, 2017 at 23:58
  • \$\begingroup\$ You need to list your load and current \$\endgroup\$
    – Voltage Spike
    Mar 23, 2017 at 6:46
  • \$\begingroup\$ @Jack Creasy. As I said before, the layout is almost exactly the same as in the datasheet. There is an output capacitor of course \$\endgroup\$
    – peku33
    Mar 23, 2017 at 8:36
  • \$\begingroup\$ @laptop2d I said I need up to 300mA at 5v. I tested the converter with 0 to 100mA load, there is almost no change to output voltage \$\endgroup\$
    – peku33
    Mar 23, 2017 at 8:38
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    \$\begingroup\$ Please show the Richtek layout. Even if you say it's identical to datasheet, show it. Also, how are you probing it? If you are using long ground alligator clip on your scope probe, it may pick up noise. \$\endgroup\$
    – bobflux
    Mar 23, 2017 at 22:05

3 Answers 3

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Update

The problem is most likely not caused by components selection, but by noise from unshielded inductor.

I tried to keep the layout as small as possible and chose 3mm x 3mm unshielded coil.

The layout:

Layout

I connected the inductor with some wires and turned it in different direction. The output ripple changed to:

Which gives less than 10mV ripple.

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  • 2
    \$\begingroup\$ Yes, unshielded inductor could be a problem. Your layout is good, but there is one mistake: you should take the 5V output directly on the output capacitor (top left corner) and not on the inductor. In your layout, trace inductance in series with the cap will greatly diminish its ability to filter out the HF ripple. \$\endgroup\$
    – bobflux
    Mar 24, 2017 at 12:35
  • \$\begingroup\$ What is the shape of inductors magnetic field? Shoud I rotate it, move it away or choose fully shielded inductor? \$\endgroup\$
    – peku33
    Mar 24, 2017 at 14:08
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    \$\begingroup\$ Shielded is better. Can you show your datasheet? \$\endgroup\$
    – bobflux
    Mar 24, 2017 at 14:23
  • \$\begingroup\$ tme.eu/pl/Document/e69e78dcc0747fec4cfda0221e0a906a/… \$\endgroup\$
    – peku33
    Mar 24, 2017 at 14:27
  • \$\begingroup\$ I'm still looking for an option to fix current device \$\endgroup\$
    – peku33
    Mar 24, 2017 at 14:28
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OK... I've done some shopping for you: the following shielded inductors are available at Mouser and should radiate less than the one you got.

http://www.mouser.com/ds/2/54/RU2016-778241.pdf http://www.mouser.com/ds/2/281/product-1023196.pdf

However, there are problems with the layout:

1) As I said in my comment, placement of input cap, diode and output cap is very good, as their GND pins are very close to each other and will keep HF currents into GND plane in a very tight loop.

However, the 5V output is taken right after the inductor, which is wrong. It should be taken after the capacitor. The trace inductance from inductor to output cap adds up and reduces its efficiency as a filter, which will add noise.

The first trace you posted with noisy output voltage doesn't look much like inductive coupling to me. Since inductor current is a sawtooth, it should couple sawtooth-y stuff into your measurement, but it doesn't.

It is possible that the reduction in noise that you see by adding wires to turn the inductor around is actually due to adding a few nH on top of the inductor's parasitic capacitance, which would make the output cap's filtering better.

2) Let's have a look at the feedback. I don't know if the 5V trace at the bottom right keeps going to the right... but if it does, then the voltage will be 5V at this point, however it will be higher at the bottom left 5V trace.

3) Feedback resistors are right above noisy data lines... I'm going to bet sending digital signals in the vertical blue traces will upset your regulator output.

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  • \$\begingroup\$ 0) I already bought some new inductors from mouser, but I have to wait until monday to test them. 0) When I move or bend the inductor towards the RT8259, the ripple begins to raise. When I move it away, it reduces to 10mV. 1) All parts that require +5V are below the inductor. If I take the power directly from the capacitor I would have to route it below the inductor. My next idea is to swap the places of inductor and output capacitor, should be better. \$\endgroup\$
    – peku33
    Mar 24, 2017 at 22:26
  • \$\begingroup\$ 2) Yes, it goes to the end of the board. I did it that way to allow ground plane flow towards the DCDC 3) Those lines drives shift registers that sets relays. 99.999% of time there would be no clock signal / no data changes \$\endgroup\$
    – peku33
    Mar 24, 2017 at 22:27
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As long as the circuits are purely digital a couple 10s of mV should be just fine. Its when you have analog circuits that a ripple like that can have a serious effect.

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