Modeling single and 3 phase lc filters

We have already done modeling in a few blog posts before where we modeled a power supply and a bunch of cascaded LC filters as well as a pm motor in synchronous reference frame. The inductor current and capacitor voltage differential equations are given below

\begin{align} \dot i_L(t) &= u_{\rm L}(t)\cdot \frac{1}{L} \\ \dot u_c(t) &= i_{\rm C}(t)\cdot \frac{1}{C} \\ \end{align}

To model a simple LC filter with the equation, we simply collect the voltages in the left and right side of the inductor and subtract them to get the voltage over the inductor. Following the markings in the schematic below, we get the left side voltage as $u_{in}$ and the right side voltage is the voltage drop of the resistor and the capacitor.

Collecting these in a set of equations also known as state-space equations we get the dynamic model of an lc filter. The simplicity here is that we do not need to calculate anything, just collecting the information directly from the figure is enough.

\begin{align} \dot i_L(t) &= \overbrace{\bigg(u_{in}-i_L\cdot r -u_c\bigg)}^{u_{\rm L}} \cdot \frac{1}{L} \\ \dot u_c(t) &=\underbrace{ i_L}_{i_C}\cdot \frac{1}{C} \\ \end{align}

We can extend this to 3-phase quite easily since 3-phase lc filter is just three single phase lc filters coupled by the common capacitor voltage.

3 phase LC filter

When the capacitor connection point is not wired to ground, the 3 phase are linked together with the neutral voltage $u_n$ which is common to all 3 LC filters.

\begin{align} \dot i_L(t) &= \bigg(\overbrace {u_{in1}-i_{L1}\cdot R -u_{c1}}^{u_{L}} - {u_n} \bigg)\cdot \frac{1}{L} \\ \dot u_c(t) &= i_L\cdot \frac{1}{C} \\ \end{align}

We get a 3-phase lc filter by just writing down 3 of the single phase lc filter equations. The inductor voltage is the same as before, but we have the added neutral voltage in the inductor current differential equation.

\begin{align} \dot i_{\rm L1}(t) &= \left(u_{in1} -u_{c1}-i_{L1}\cdot r_1 - u_n\right)\cdot \frac{1}{\rm L1} \\ \dot i_{\rm L2}(t) &= \left(u_{in2}-u_{c2}-i_{L2}\cdot r_2 - u_n\right)\cdot \frac{1}{\rm L2} \\ \dot i_{\rm L3}(t) &= \left(u_{in3} -u_{c3}-i_{L3}\cdot r_3 - u_n\right)\cdot \frac{1}{\rm L3} \\ \dot u_c(t) &= i_{\rm L1}\cdot \frac{1}{\rm C1} \\ \dot u_c(t) &= i_{\rm L2}\cdot \frac{1}{\rm C2} \\ \dot u_c(t) &= i_{\rm L3}\cdot \frac{1}{\rm C3} \\ \end{align}

Note that we don’t assume balanced inductances, capacitances or voltages hence we use all 6 state variables in our calculations.

Neutral voltage

The neutral voltage $u_n$ can be solved by noting that since there is no alternate path for the current other than through the other two phases, hence the sum of the currents and by extension the sum of the derivatives has to be zero.

\begin{align} i_1+i_2+i_3 &= 0 \Leftrightarrow\dot i_1 + \dot i_2 + \dot i_3 &= 0 \end{align}

Using the equation (16) with (12)-(14) we get

\begin{align} \frac{u_{\rm L1}- u_n}{L1} + \frac{u_{\rm L2}- u_n}{L2} + \frac{u_{\rm L3}- u_n}{L3} &= 0 \end{align}

We can just put the equation (14) to Wolfram alpha and solve for $u_n$ to get the neutral voltage in a nice form that we can use.

\begin{align} u_n &=\frac{u_{\rm L1}\cdot L_2L_3 + u_{\rm L2}\cdot L_1L_3+ u_{\rm L3}\cdot L_1L_2}{L_1L_2+L_1L_3+L_2L_3} \\[1em] &=u_{\rm L1}\cdot a_1 + u_{\rm L2}\cdot a_2+ u_{\rm L3}\cdot a_3 \end{align}

With the neutral voltage, the 3-phase lc filter has the following form. Aside from the presence of the neutral voltage, the 3 phase model is the same as 3 single phase lc filters.

\begin{align} u_{\rm L1}(t) &= u_{in1} -u_{c1}-i_{L1}\cdot r\\ u_{\rm L2}(t) &= u_{in2}-u_{c2}-i_{L2}\cdot r \\ u_{\rm L3}(t) &= u_{in3} -u_{c3}-i_{L3}\cdot r\\ u_{n}(t) &= u_{\rm L1} \cdot a_1 + u_{\rm L2} \cdot a_2 + u_{\rm L3} \cdot a_3 \\ \dot i_{\rm L1}(t) &= \left({u_{\rm L1}} - u_n\right)\cdot \frac{1}{\rm L1} \\ \dot i_{\rm L2}(t) &= \left({u_{\rm L2}} - u_n\right)\cdot \frac{1}{\rm L2} \\ \dot i_{\rm L3}(t) &= \left({u_{\rm L3}} - u_n\right)\cdot \frac{1}{\rm L3} \\ \dot u_c(t) &= (i_{\rm L1}-i_{\rm load1})\cdot \frac{1}{\rm C1} \\ \dot u_c(t) &= (i_{\rm L2}-i_{\rm load2})\cdot \frac{1}{\rm C2} \\ \dot u_c(t) &= (i_{\rm L3}-i_{\rm load3})\cdot \frac{1}{\rm C3} \end{align}