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I'm making a buck regulator using L4970A using circuit from this application note figure 60 , except for the inductor. I change i to 180 uH

I wanna make it to be a power adaptor for laptop with solar cell source (about 100 W, Voc 27 V, Isc 5A), and the regulator itself can hold current to 10 A max, said the datasheet

As you know, usually laptop needs current from about 1.5 A-4 A, depends on it's using battery or not and how the laptop works. I've succeeded to make my regulator give supply to the laptop, but the current is too small to make it becoming primary supply.

Does anyone know what's wrong with the design? What actually limits the current of buck regulator??

I test this circuit using Asus A43SJ

For Russel, this is my inductor

Inductor I use

I don't know either the wire type or the core type exactly, but because it's not magnetic, I think it's iron powder. Outer diameter 2.3 cm and inner diameter 1.2 cm

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    \$\begingroup\$ Slowest link ever? 6.1 KB/sec. I'll be back after my cup of tea. \$\endgroup\$
    – stevenvh
    Jun 27 '12 at 13:31
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    \$\begingroup\$ This is related to your previous question, where in the comments I stated that most laptops use a specific protocol to enable charging. Remove the laptop from the equation and test your power supply with resistors before concluding that 'the buck doesn't work'. \$\endgroup\$ Jun 27 '12 at 13:43
  • \$\begingroup\$ @Madmanguruman Yap, you're right... Okay, I'll try and tell the results... \$\endgroup\$
    – Wardhana
    Jun 27 '12 at 13:47
  • \$\begingroup\$ @Madmanguruman it is the SMBus protocol, right? \$\endgroup\$ Jun 27 '12 at 13:58
  • \$\begingroup\$ Side note, please try not to sign your posts with a thank you line. \$\endgroup\$
    – Kortuk
    Jun 27 '12 at 17:16
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L4970A data sheet here

Your cited application note here - veryy extensive and useful.

Fig 60 circuit that you say you used is shown below.
The circuit looks potentially OK as is.

Frequency about 227 kHz.
I made my inductor myself, but I'm not sure about it's characteristic.

The ability of the inductor to carry the current required without saturating is crucial. Can you provide a picture of the inductor and describe the core material used. At a minimum if a picture is not available please provide construction details with core material and dimensions.

What voltage does the laptop require.
If 19V then at 1A it looks like R = V/I = 19/1 ~= 20 Ohms load and at 4A it looks like ~= 5 ohms.
At say 10 Ohms you'd get ~= 2A.
Power = V x I ~= 40W

When you load it with a 10 Ohm, 40W rated resistor, what happens? Use an app[ropriate length of Nichrome wire (eg from an old heater or toaster element)or say 4 x 10W, 10Ohm resistors in series parallel (2S2P).

enter image description here

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  • \$\begingroup\$ Would the user who UP voted this post just now (1st upvote)(about 1110 UTC June 29th) please let me know who they are. I of course have no problem with the upvote. It happens to be a "special vote" for reasons that are essentially irrelevant but it would be fun/nice to know who the voter was. As upvoters often enough identify themselves this will hopefully not be a problem. RM \$\endgroup\$
    – Russell McMahon
    Jun 29 '12 at 11:14
  • \$\begingroup\$ It's me who upvote it, haha... *sorry if it's wrong, I'm a newbie... \$\endgroup\$
    – Wardhana
    Jun 29 '12 at 11:24
  • \$\begingroup\$ I've uploaded the picture above. \$\endgroup\$
    – Wardhana
    Jun 29 '12 at 11:32
  • \$\begingroup\$ @Wardhana - no problem with the upvote :-). It increased my user reputation to 40,003 FWIW. (Possibly very little indeed :-) ) \$\endgroup\$
    – Russell McMahon
    Jun 29 '12 at 12:11
  • \$\begingroup\$ I've tested it with power supply 24 V and variable resistor 40 ohm, set to 10 ohm. When I set the output to 19 V, The current output is good, it's about 1.80 A. So, is the unmatched current in my test before because I use laptop as my load? \$\endgroup\$
    – Wardhana
    Jun 30 '12 at 4:18

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