In this boost converter , when i increase the load resistance to very high value ,it acts as a boost converter. Otherwise it wont.
Can anyone tell what is the effect of load resistance on this circuit ..??
When you reduce load resistance, it causes output voltage to drop.
Your converter cannot supply much current. Apparently it has high resistance in the first half of the cycle.
Perhaps your switching device (mosfet) does not turn on sufficiently? Current does not build in the coil. As a result it delivers a weak pulse to the output stage (second half of the cycle).
What will happen to the output , when i increase/decrease the value of L or/and C ??
Greater Henry value, longer time constant, less output.
Greater Farad value reduces ripple voltage more.
If u give the design equations or a model circuit along with the answer , it will be more helpful to me..
thanks...
I have a Youtube video consisting of an animated simulation of simple switched-coil converters (buck, boost, buck-boost). It portrays current moving through wires, capacitors charging and discharging, flux fields building and collapsing.
www.youtube.com/watch?v=FT_sLF5Etm4
Even more helpful is an interactive simulation. The link below is to my boost converter, drawn in Falstad's simulator.
The component values are the same (give or take) as in your schematic.
The link opens the website
www.falstad.com/circuit...
loads my schematic into the simulator...
and runs it on your computer.
https://tinyurl.com/lnt9eza
You can change values by right-clicking on a component, then select Edit.
Select the clock input, or select user-switched input. Press the switch to start the cycle. Let up on the switch to see the second half of the cycle.
Watch the waveforms to see the effects of switching speed, duty cycle, etc.