Critical Maximum current for a DC-DC Buck Converter

I was wondering how one would go about the derivation for the output to input voltage ratio for a DC-DC Buck converter given the critical max current that is found to be $$I_{crit} = \frac{V_{in}T}{8L}$$ The final result of the output to input voltage ratio is $$\frac{V_{out}}{V_{in}} = \frac{D^{2}}{D^2 + \frac{I_{load}}{4I_{crit}}}$$

Edit: For clarification: T is the period and L is the inductance and D is the duty cycle and we assume the system to be in discontinuous current mode

• DCM or CCM? What is T and what is L and why is there a critical value of output current. Many things you should explain. – Andy aka May 7 '18 at 12:38
• T is the period and L is the inductance and D is the duty cycle and we assume the system to be in discountinous current mode. – Sam May 7 '18 at 13:23
• And $I_{crit}$? Is it at the boundary condition? – Andy aka May 7 '18 at 13:55
• @Andyaka $$I_{crit}$$ is at the boundary condition – Sam May 7 '18 at 14:51