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
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
Best Answer
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.
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).