problems related to buck converter/voltage regulator
Here's a starter for a synchronous buck converter: -



It would typically use MOSFETs to rapidly switch between Vinput and 0 volts. The output of the switch is a square wave that has a duty cycle. That duty cycle is controlled by a circuit that measures the output voltage (Voutput) to see if it is at the desired DC level.

So, if the switch is operating at 100 kHz, to avoid a lot of ripple on the output, you want the cut-off frequency of the LC filter to be low: -



Picture source.

The filter above has a cut-off frequency of about 1 kHz and this means that the 100 kHz square wave will have its switching edges attenuated by about 80 dB. So, if the input voltage is 12 volts, roughly and simply speaking, the ripple voltage will be 80 dB lower than 12 volts at about 1 mV p-p.

1 mV p-p is very low but, to obtain this low level of ripple, you will need a big value of L and C so, it's probably quite acceptable to have a cut-off frequency of about 10 kHz (40 dB attenuation). It now partially becomes a question of economics to get lower ripple. 40 dB attenuation produces about 100 mV p-p ripple in this simple explanation.

So you need to develop an idea for where you want the LC output filter to have its cut-off frequency. You should also consider the maximum output load current and the increased ripple it might cause in discharging the output capacitor. 

For instance, if the output load current can be (say) 1 amp, using the formula I = C.dV/dt you can estimate how much the capacitor discharges by (dV) in time (dt). If the switching frequency is 100 kHz then dt could be in the realm of 5 us. 

If dv is assumed to be comparable to the natural change in voltage when unloaded (calculated above at 100 mV for a 10 kHz cut-off frequency) then: -

\$C = \dfrac{I\cdot dt}{dv}\$ = 50 uF

Again this isn't a hard and fast formula but it is useful as a rule of thumb.

So now, rearranging the formula for the cut-off frequency for an LC low pass filter: -

\$Fc = \dfrac{1}{2\pi\sqrt{LC}}\$

You can work out that L should be 5.1 uH.

Bear in mind this is just an approximate taster to help you understand the general ideas. I would urge you to do some reasearch to acquire better formulas that save you having to laboriously go through things as I have done above. Here's what the output voltage will look like for 12 volts in, 9 volts out with the values above (duty cycle of the square wave is 75%): -



Ripple is about 130 mVp-p so not a million miles off. I would also recommend that you acquire a simulation tool for experimenting with.