# 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:

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