# Initial Short-Circuit Current

The general ohmic network equation is given as:

The SC is calculated in two steps:
• calculate the SC contribution $$I''_{kI}$$ of all voltage source elements

• calculate the SC contribution $$I''_{kII}$$ of all current source elements

These two currents are then combined into the total initial SC current $$I''_{k} = I''_{kI} + I''_{kII}$$.

## Equivalent Voltage Source

For the short-circuit calculation with the equivalent voltage source, all voltage sources are replaced by one equivalent voltage source $$V_Q$$ at the fault location. The voltage magnitude at the fault bus is assumed to be:

$\begin{split}V_Q = \left\{ \begin{array}{@{}ll@{}} \frac{c \cdot \underline{V}_{N}}{\sqrt{3}} & \text{for three phase short circuit currents} \\ \frac{c \cdot \underline{V}_{N}}{2} & \text{for two phase short circuit currents} \end{array}\right.\end{split}$

where $$V_N$$ is the nominal voltage at the fault bus and c is the voltage correction factor, which accounts for operational deviations from the nominal voltage in the network.

The voltage correction factors $$c_{min}$$ for minimum and $$c_{max}$$ for maximum short-circuit currents are defined for each bus depending on the voltage level. In the low voltage level, there is an additional distinction between networks with a tolerance of 6% vs. a tolerance of 10% for $$c_{max}$$:

Voltage Level

$$c_{min}$$

$$c_{max}$$

< 1 kV

Tolerance 6%

0.95

1.05

Tolerance 10%

1.10

> 1 kV

1.00

## Voltage Source Contribution

To calculate the contribution of all voltage source elements, the following assumptions are made:

1. Operational currents at all buses are neglected

2. All current source elements are neglected

3. The voltage at the fault bus is equal to $$V_Q$$

For the calculation of a short-circuit at bus $$j$$, this yields the following network equations:

$\begin{split}\begin{bmatrix} \underline{Y}_{11} & \dots & \dots & \underline{Y}_{n1} \\[0.3em] \vdots & \ddots & & \vdots \\[0.3em] \vdots & & \ddots & \vdots \\[0.3em] \underline{Y}_{1n} & \dots & \dots & \underline{Y}_{nn} \end{bmatrix} \begin{bmatrix} \underline{V}_{1} \\ \vdots \\ V_{Qj} \\ \vdots \\ \underline{V}_{n} \end{bmatrix} = \begin{bmatrix} 0 \\ \vdots \\ \underline{I}''_{kIj} \\ \vdots \\ 0 \end{bmatrix}\end{split}$

where $$\underline{I}''_{kIj}$$ is the voltage source contribution of the short-circuit current at bus $$j$$. The voltages at all non-fault buses and the current at the fault bus are unknown. To solve for $$\underline{I}''_{kIj}$$ , we multipliy with the inverted nodal point admittance matrix (impedance matrix):

$\begin{split}\begin{bmatrix} \underline{V}_{1} \\ \vdots \\[0.4em] V_{Qj} \\[0.4em] \vdots \\ \underline{V}_{n} \end{bmatrix} = \begin{bmatrix} \underline{Z}_{11} & \dots & \dots & \dots & \underline{Z}_{n1} \\ \vdots & \ddots & & & \vdots \\ \vdots & & \underline{Z}_{jj} & & \vdots \\ \vdots & & & \ddots & \vdots \\ \underline{Z}_{1n} & \dots & \dots & \dots & \underline{Z}_{nn} \end{bmatrix} \begin{bmatrix} 0 \\ \vdots \\[0.25em] \underline{I}''_{kIj} \\[0.25em] \vdots \\ 0 \end{bmatrix}\end{split}$

The short-circuit current for bus m is now given as:

$I''_{kIj} = \frac{V_{Qj}}{Z_{jj}}$

To calculate the vector of the short-circuit currents at all buses, the equation can be expanded as follows:

$\begin{split}\begin{bmatrix} \underline{V}_{Q1} & \dots & \underline{V}_{n1} \\[0.4em] \vdots & \ddots & \vdots \\[0.4em] \underline{V}_{1n} & \dots & \underline{V}_{Qn} \end{bmatrix} = \begin{bmatrix} \underline{Z}_{11} & \dots & \underline{Z}_{n1} \\[0.8em] \vdots & \ddots & \vdots \\[0.8em] \underline{Z}_{1n} & \dots & \underline{Z}_{nn} \end{bmatrix} \begin{bmatrix} \underline{I}''_{kI1} & \dots & 0 \\[0.8em] \vdots & \ddots & \vdots \\[0.8em] 0 & \dots & \underline{I}''_{kIn} \end{bmatrix}\end{split}$

which yields:

$\begin{split}\begin{bmatrix} I''_{kI1} \\[0.25em] \vdots \\[0.25em] I''_{kIn} \\ \end{bmatrix} = \begin{bmatrix} \frac{V_{Q1}}{Z_{11}} \\ \vdots \\ \frac{V_{Qn}}{Z_{nn}} \end{bmatrix}\end{split}$

In that way, all short-circuit currents can be calculated at once with one inversion of the nodal point admittance matrix.

In case a fault impedance is specified, it is added to the diagonal of the impedance matrix. The short-circuit currents at all buses are then calculated as:

$\begin{split}\begin{bmatrix} I''_{kI1} \\[0.25em] \vdots \\[0.25em] I''_{kIn} \\ \end{bmatrix} = \begin{bmatrix} \frac{V_{Q1}}{Z_{11} + Z_{fault}} \\ \vdots \\ \frac{V_{Qn}}{Z_{nn} + Z_{fault}} \end{bmatrix}\end{split}$

## Current Source Contribution

To calculate the current source component of the SC current, all voltage sources are short circuited and only current sources are considered. The bus currents are then given as:

$\begin{split}\begin{bmatrix} I_1 \\[0.2em] \vdots \\[0.2em] I_m \\[0.2em] \vdots \\ I_n \end{bmatrix} = \begin{bmatrix} 0 \\[0.2em] \vdots \\[0.2em] \underline{I}''_{kIIj} \\[0.2em] \vdots \\ 0 \end{bmatrix} - \begin{bmatrix} I''_{kC1} \\[0.2em] \vdots \\[0.2em] \underline{I}''_{kCj} \\[0.2em] \vdots \\ I''_{kCn} \end{bmatrix} = \begin{bmatrix} -I''_{kC1} \\[0.2em] \vdots \\[0.2em] \underline{I}''_{kIIj} - \underline{I}''_{kCj} \\[0.2em] \vdots \\ -I''_{kCn} \end{bmatrix}\end{split}$

where $$I''_{kC}$$ are the SC currents that are fed in by converter element at each bus and $$\underline{I}''_{kIIj}$$ is the contribution of converter elements at the fault bus $$j$$. With the voltage at the fault bus known to be zero, the network equations are given as:

$\begin{split}\begin{bmatrix} \underline{V}_{1} \\ \vdots \\[0.4em] 0 \\[0.4em] \vdots \\ \underline{V}_{n} \end{bmatrix} = \begin{bmatrix} \underline{Z}_{11} & \dots & \dots & \dots & \underline{Z}_{n1} \\ \vdots & \ddots & & & \vdots \\ \vdots & & {Z}_{jj} & & \vdots \\ \vdots & & & \ddots & \vdots \\ \underline{Z}_{1n} & \dots & \dots & \dots & \underline{Z}_{nn} \end{bmatrix} \begin{bmatrix} -I''_{kC1} \\[0.2em] \vdots \\[0.2em] \underline{I}''_{kIIj} - \underline{I}''_{kCj} \\[0.2em] \vdots \\ -I''_{kCn} \end{bmatrix}\end{split}$

From which row $$j$$ of the equation yields:

$0 = \underline{Z}_{jj} \cdot \underline{I}''_{kIIj} - \sum_{m=1}^{n}{\underline{Z}_{jm} \cdot \underline{I}_{kCj}}$

which can be converted into:

$\underline{I}''_{kIIj} = \frac{1}{\underline{Z}_{jj}} \cdot \sum_{m=1}^{n}{\underline{Z}_{jm} \cdot \underline{I}_{kC, m}}$

To calculate all SC currents for faults at each bus simultaneously, this can be generalized into the following matrix equation:

$\begin{split}\begin{bmatrix} \underline{I}''_{kII1} \\[0.5em] \vdots \\[0.5em] \vdots \\[0.5em] \underline{I}''_{kIIn} \end{bmatrix} = \begin{bmatrix} \underline{Z}_{11} & \dots & \dots & \underline{Z}_{n1} \\[0.3em] \vdots & \ddots & & \vdots \\[0.3em] \vdots & & \ddots & \vdots \\[0.3em] \underline{Z}_{1n} & \dots & \dots & \underline{Z}_{nn} \end{bmatrix} \begin{bmatrix} \frac{I''_{kC1}}{\underline{Z}_{11}} \\[0.25em] \vdots \\ \vdots \\[0.25em] \frac{I''_{kCn}}{\underline{Z}_{nn}} \end{bmatrix}\end{split}$