Current Drawn at Buck Converter Load - Explanation Needed

[Harsh Chandola]

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

Input 220VAC

Output - 24 Volt, 350mA,

So after I finished designing I checked voltage across the output and it is close 24volts, but when I place a resistor as a load of a value that should give me 350mA everything changes. Output voltage becomes 80mv across the load and current resistor is drawing becomes 1.2mA.

I am not being able to understand that.

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?

What should I design if I wish to convert 220Ac voltage to say 24 volt but the current drawn depends on the load. what i mean is that I should get a fixed 24 volts at output and if I want 2 Amps I connect a 12ohm Resistor, if I want 1 Amps I connect a 24 ohm resistor, if I want 500mAmps I connect a 48ohm resistor etcetera.

Thank you 

[Rick Law]

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

Input 220VAC

Output - 24 Volt, 350mA,
...
...

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

...
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?
...

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. 

[Harsh Chandola]

I rectified AC voltage to DC first, so from 220V AC i got to 220V DC using full wave rectifier then stepped it down using the buck converter.
Your Analogy was on point. thank you so much.
Also if I use  a step down transformer say I step down from 220V AC to 24V AC and then rectify it to give 24V DC, would it be as good as the previous design I talked about? I am actually trying to design a driver for LED. So would like to know whether the transformer method is as good as the buck one.

[jeroen79]

How are you controlling the FET?
V2 is set up as pulse but I see no feedback from the output.
If you set the pulse rate and width just right for 24V unloaded then the voltage would drop when load is increased.

[Rick Law]

I rectified AC voltage to DC first, so from 220V AC i got to 220V DC using full wave rectifier then stepped it down using the buck converter.
Your Analogy was on point. thank you so much.
Also if I use  a step down transformer say I step down from 220V AC to 24V AC and then rectify it to give 24V DC, would it be as good as the previous design I talked about? I am actually trying to design a driver for LED. So would like to know whether the transformer method is as good as the buck one.

Good that you understood my initial reply, I was just re-editing it to correct some wording problem, and saw your post right after I saved the reedit.

I am not good enough to give you advice on your circuit...  I'm just a hobbyist kicking a few electrons around now and again.

[Harsh Chandola]

Currently I am only trying to get desired volts and desired amps at the load but I would use a 555 timer (maybe) or some IC to control the PWM, would use some current sensing method like monitoring voltage across a small resistor on load to make sure appropriate amp flows.
What exactly do you mean by increasing the load?

What i have in mind is that I'd use 4-7 number of 3.3v LEDs that draw 350mA in series. So what I was thinking was that voltage will not fall as long as I am not going above 7 LEDs at load. I am so confused here.

How should my load look like then? let me clearly explain what I was thinking : So I thought I have designed my circuit to provide an output of 24V and 350mA, if I connect 7 LEDs --> 7*3.3 = 23.1V in series with 2.57ohm resistor (remaining 0.9 volt drops on 2.57ohm resistor to facilitate 350mA current) I should be good and if I reduce the number of LEDs then increasing the resistance should do the job just fine.
So my basic idea is thatas long as all my components would not want to draw more than 350mA and would not have a collective voltage drop of more than 24 across them I could just do anything at load.
Am I wrong or am i correct. Clear my concept please if I am wrong.
Thanks a lot for your patience.

[jeroen79]

LEDs are components that have a fixed (ish) voltage drop and they will as much current as can be delivered once the voltage exceeds that value.
So if this is for LEDs then you would best add a constant current regulation.

Use a series resistor to compare the current with the set value and use that to drive the FET.
And give it no output capacitance.