# Capacitor currents equations for Buck Converter with Input Filter

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

• Why can't you use a simulator? – Andy aka Nov 5 '20 at 18:03
• 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. – Teo Protoulis Nov 5 '20 at 18:09
• 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. – Andy aka Nov 5 '20 at 18:11
• 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? – TimWescott Nov 5 '20 at 18:12
• 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 $P_{out} = I_RV$ ? – 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:
$$V = DV_g$$ $$I_2 = \frac{V}{R} \Rightarrow I_2 = \frac{DV_g}{R}$$ $$I_1 = DI_2 \Rightarrow I_1 = \frac{D^2V_g}{R}$$