Sequence impedances and zero sequence comment

Sequence impedance of an element is the impedance it offers to the flow of currents of one particular sequence: $Z_1 = V_{a1}/I_{a1}$ (positive), $Z_2 = V_{a2}/I_{a2}$ (negative) and $Z_0 = V_{a0}/I_{a0}$ (zero sequence), when only that sequence is present.

Star connected load

Consider a balanced star load with impedance $Z_s$ per phase and neutral grounded through $Z_n$ (solid grounding: $Z_n = 0$).

Circuit and phase equations

   Ia -->  Zs
 a o-----/\/\---+
   Ib -->  Zs   |
 b o-----/\/\---+ n
   Ic -->  Zs   |
 c o-----/\/\---+
                |  In = Ia+Ib+Ic
               Zn
                |
              ground

Voltage of each phase to ground = drop in $Z_s$ + drop in $Z_n$:

$$\begin{aligned} V_a &= Z_s I_a + Z_n (I_a + I_b + I_c) \\ V_b &= Z_s I_b + Z_n (I_a + I_b + I_c) \\ V_c &= Z_s I_c + Z_n (I_a + I_b + I_c) \end{aligned}$$

In matrix form $V_p = Z_p I_p$ with

$$Z_p = \begin{bmatrix} Z_s+Z_n & Z_n & Z_n \\ Z_n & Z_s+Z_n & Z_n \\ Z_n & Z_n & Z_s+Z_n \end{bmatrix}$$

Transformation to sequence quantities

Put $V_p = A V_s$ and $I_p = A I_s$:

$$A V_s = Z_p A I_s \;\Rightarrow\; V_s = A^{-1} Z_p A\, I_s = Z_s^{seq} I_s$$

with $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$, $A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$.

Write $Z_p = Z_s U + Z_n J$, where $U$ is the unit matrix and $J$ is the matrix of all ones. Then $A^{-1}(Z_sU)A = Z_sU$. For $J$: every row of $J A$ is $[3\ \ 0\ \ 0]$ (since column sums of $A$ are $3, 1+a^2+a=0, 1+a+a^2=0$), and $A^{-1}$ times that gives

$$A^{-1} J A = \begin{bmatrix} 3&0&0\\0&0&0\\0&0&0 \end{bmatrix}$$

Hence

$$Z_{seq} = A^{-1} Z_p A = \begin{bmatrix} Z_s+3Z_n & 0 & 0 \\ 0 & Z_s & 0 \\ 0 & 0 & Z_s \end{bmatrix}$$

The sequence impedance matrix has non-zero elements only on its diagonal:

$$Z_0 = Z_s + 3Z_n, \qquad Z_1 = Z_s, \qquad Z_2 = Z_s$$

All off-diagonal elements are zero, which means a current of one sequence produces a voltage drop of the same sequence only. So the three sequence networks are decoupled and can be solved separately, which is the main advantage of symmetrical components.

The same holds if the load also has equal mutual impedances $Z_m$ between phases: then $Z_1 = Z_2 = Z_s - Z_m$ and $Z_0 = Z_s + 2Z_m + 3Z_n$, still diagonal.

Comment on zero sequence impedance

• $Z_0 = Z_s + 3Z_n$ is larger than $Z_1$ and $Z_2$, because zero sequence currents of all three phases are in phase and add up to $3I_{a0}$ in the neutral; the drop $3I_{a0}Z_n$ is felt by each phase.
• Solidly grounded star ($Z_n = 0$): $Z_0 = Z_s$.
• Ungrounded star ($Z_n = \infty$): $Z_0 = \infty$; zero sequence current cannot flow, and the zero sequence network is open.
• Delta connected load: no return path, so $Z_0$ seen from the lines is infinite.
• In a fault study, the method of neutral grounding therefore mainly changes the zero sequence network and hence the single line to ground fault current; this is why resistance or reactance grounding is used to limit earth fault current.