I am trying to analyze the capacitor currents for the below buck converter with an extra L-C input filter:

enter image description here

  • Switch in Position 1

\$ i_1(t) = i_{c1}(t) + i_{τ}(t) \$

\$ i_2(t) = i_{c2}(t) + i_R(t) \$

\$ i_τ(t) = i_2(t) \$

  • Switch in Position 2

\$ i_1(t) = i_{c1}(t) \$

\$ i_2(t) = i_{c2}(t) + i_R(t) \$

\$ i_τ(t) = 0 \$

These are the equations I am obtaining but I know I am missing something because after this analysis stage, I am applying the capacitor charge balance in order to come up with the equations regarding the DC components \$I_2, I_1\$ of the inductor currents and the results, which are the following, are wrong.

\$ I_2 = \frac{V}{R} \$

\$ I_1 = DI_2 = D\frac{V}{R} \$

where \$ V = V_R \$ is the output voltage of the converter. As for the inductor voltages, I get them right and the application of the volt-second balance yields the correct results regarding the DC components of the output voltage (\$ V = V_R = V_{c2} = DV_g \$) and the \$ C1 \$ capacitor voltage (\$ V_{c1} = V_g \$).

  • \$\begingroup\$ Why can't you use a simulator? \$\endgroup\$ – Andy aka Nov 5 '20 at 18:03
  • \$\begingroup\$ Didn't know I could use a simulator to get the equations of currents or voltages. Been a while since I started diving into electronics. But I would also like to learn these basics quite well and be able to perform these analysis by hand before going to the simulators. \$\endgroup\$ – Teo Protoulis Nov 5 '20 at 18:09
  • \$\begingroup\$ You said you were missing something and maybe the sim could help you understand what it is. A sim won't deliver equations but it does provide a sanity check. \$\endgroup\$ – Andy aka Nov 5 '20 at 18:11
  • \$\begingroup\$ Looks right to me. You've defined \$D = V / V_g\$, and assuming no losses in the circuit \$I_2 V = I_1 V_g\$. Your expression for \$I_2\$ is correct by Ohm's law, and your expression for \$I_1\$ is correct by energy conservation. So where's the problem? \$\endgroup\$ – TimWescott Nov 5 '20 at 18:12
  • \$\begingroup\$ That's how I thought it but putting these equations for testing, the answer is that these results are incorrect. So, I thought that I am missing something anf that's why I asked here. However, should maybe the output power be \$ P_{out} = I_RV \$ ? \$\endgroup\$ – Teo Protoulis Nov 5 '20 at 18:23

Eventually, I figured it out. All the equations at the original post are correct. However, I had to express the inductors' currents \$ I_1, I_2 \$ as a function of duty cycle \$ D \$, input voltage \$ V_g \$ and load resistor \$ R \$. To sum everything up:

$$ V = DV_g $$ $$ I_2 = \frac{V}{R} \Rightarrow I_2 = \frac{DV_g}{R} $$ $$ I_1 = DI_2 \Rightarrow I_1 = \frac{D^2V_g}{R} $$


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