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I am a (joung) programmeer and I recently began studying some electronics&AVR programming.

I am trying to use a step down voltage regulator (LM2678-5.0) to transform battery voltage (9/12V) into a suitable 5V. I am not using a simple LM7805 since this is part of a bigger robotic project, where I need to transform more than 1A, and it would be nice to spare energy (=less battery weight).

I built this schematic (photos below); with a 10kOhm load I get 5.02V; but as I attach a load to it, voltage drops. As example, as I attach a 20mA led, I get about 3V. I tried attaching a little (~400mA) motor, but it doesn't start (there's no voltage!).

Any suggestions? Thanks in advance

rear front schematic

P.S.: inductors and capacitors has been chosen accordnigly to Texas Instruments online simulation tool

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  • \$\begingroup\$ What sort of battery are you using? Is its output also sagging under the load? \$\endgroup\$ Jul 17, 2015 at 15:16
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    \$\begingroup\$ Also, what is its current rating for L1? \$\endgroup\$ Jul 17, 2015 at 15:18
  • \$\begingroup\$ No battery yet, tested with a regulated power supply. I set it 10V \$\endgroup\$ Jul 17, 2015 at 15:19
  • \$\begingroup\$ What is the current limit of the DC power supply set to? \$\endgroup\$ Jul 17, 2015 at 15:20
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    \$\begingroup\$ What is the current rating of the inductor? Not the inductance, the current rating? If it doesn't have a DC current rating, it's not an appropriate part for this sort of circuit. \$\endgroup\$ Jul 17, 2015 at 15:21

2 Answers 2

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It's a common trap for beginners and DCDC converters ! Your inductor is not up to the job.

These kind of DCDC converters there is always a LARGE INDUCTOR, it needs to be large so it can handle the current. Well actually it needs to be able to store a large amount of magnetic energy. I did not see such an inductor in your photos !

You small inductor can only store a small amount of magnetic energy and then it saturates, it cannot store more energy. As a result the output voltage of your DCDC converter drops.

If you replace the inductor with a more suitable one, you have a better chance it will work. Here are some pictures of suitable inductor types: toroid inductor or SMD power inductors.

Also study the datasheet, ideally you want an inductor that can handle at least 2x the current you need at the output.

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  • \$\begingroup\$ Ok, thanks a lot! I'll go buying a larger inductor. \$\endgroup\$ Jul 17, 2015 at 15:55
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That's definitely not an appropriate inductor to use in that circuit.
I'm assuming from the options presented in the datasheet you linked that the one you have is part number 00 6122 04 listed on page 2.
Its DC resistance is almost 1 ohm and its current rating is only 560mA (and we don't even know if that's average or saturation).
For an inductor used in a DCDC converter, I also usually like to see some data on how it would behave at my DCDC converter's switching frequency, but that datasheet is only giving info for 100kHz while your LM2678 is operating at 260kHz.
You're going to need something significantly bigger and better spec'ed to do the job properly.

You may also want to think about a better diode too as the 1N5819 is only good for an average current of 1A.

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  • \$\begingroup\$ Thank you. I didn't know those things about inductors. Learned something new. \$\endgroup\$ Jul 17, 2015 at 15:58
  • \$\begingroup\$ About the diode: I know the 1A problem, but I can't find a through-hole which can handle more current. Can I put it in parallel with other 1N5819? \$\endgroup\$ Jul 17, 2015 at 16:00
  • \$\begingroup\$ @GiacomoFabris: The 1N5822 is the 3A part, but will need a larger hole. \$\endgroup\$ Jul 17, 2015 at 16:10
  • \$\begingroup\$ Thanks. Yes, this was just a test board, it's the first time I use switching regulator. Obviously needs also better connectors, and maybe also better soldering ;) \$\endgroup\$ Jul 17, 2015 at 16:15
  • \$\begingroup\$ My first attempts looked very much like that too ;) \$\endgroup\$
    – brhans
    Jul 17, 2015 at 17:03

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