So I had designed a Buck Converter on LT Spice IV.

Input 220VAC

Output - 24 Volt, 350mA,

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If your input is VAC, you are not bucking. Buck is DC to DC conversion.

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Also I have one major confusion or doubt or lack of knowledge about one thing which is: If I design a converter for specific voltage and current, say for example 24volt and 2 amps, that means that load should have a resistance of 6ohms but if I replace thatr resistor with say 2 ohms, would the current become 24/2 = 12 Amps? Could I get an explanation for that please?

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If you designed a converter for specific voltage and current, it means it can supply

**at or up-to** that voltage

**at or up-to** that current. In the case you described, it is

**at or up to 24 volts** and

**at or up to 2 amps**. Depending on the buck design, it will try to drive the load at 24V and let it draw however much current it needs - up to 2 amps.

So, if you have a load that draws less than 2 amps, it just supply what the load will take. If the load tries to draw more than 2 amps, you exceeded the designed capability. The result will depends on how your circuit handles the overload - unless overload is properly managed, something will probably burn out and die. A

**buck converter with current-limited (CC)** will typically handle the overload by

**lowering the output voltage** to the load so the load draws equal-to or less-than the current limit you selected.

Like having a 2 gallon bucket, it could carry up-to 2 gallons. If all you have a single gallon of water in the 2 gallon bucket. It is still a 2 gallon bucket. It is still just one gallon of water. If your buck converter has 2 amps limit but the load needs 12 amps, that would be like trying to use your 2 gallon bucket to carry 12 gallons , your feet will get wet and your floor will get flooded. How your floor will handle it will depend on how the floor is designed.