Sizing a metering current transformer (CT)
Tags: electrical, transformer
Published: 2024-07-12 04:46:26

Sizing a metering current transformer (CT) while taking the Instrument Security Factor ([ISF of CT](https://www.electricalvolt.com/instrument-safety-factor-of-current-transformer/)) into account involves several steps. The goal is to ensure accurate measurement and protect the connected instruments under fault conditions. Here’s a step-by-step guide:

**## Step 1: Determine the Application Requirements**

**Rated Primary Current (Iᵣₚ)**: Determine the maximum current that the CT will need to measure under normal operating conditions.
**Rated Secondary Current (Iᵣₛ):** Commonly, this is 1A or 5A, depending on the metering equipment specifications.
**[Accuracy Class](https://www.electricalvolt.com/accuracy-class-of-current-transformer/):** Choose the accuracy class required for the application, e.g., Class 0.2, 0.5, etc., which indicates the maximum permissible error at rated conditions.

**## Step 2: Select the Appropriate CT Ratio**

The **[CT ratio](https://www.electricalvolt.com/ct-ratio-calculator-ct-ratio-calculation-formula/)** is chosen based on the maximum expected load current. For example, if the maximum load current is 1000A, a CT with a ratio of 1000:5 might be appropriate.

## **Step 3: Consider the Instrument Security Factor (ISF)**

ISF Specification: Check the ISF rating for potential CT candidates. A typical ISF might be 5 or 10, indicating the CT can handle up to 5 or 10 times the rated current without causing significant errors or damage.
Fault Current: Determine the maximum fault current that could occur in the system. This is typically provided by system designers or utility standards.

## **Step 4: Calculate the Maximum Allowable Fault Current**

Using the ISF, calculate the maximum current the CT can handle safely without exceeding its specified error margin:

Maximum Fault Current=ISF×𝐼𝑟𝑝

For example, if the CT's ISF is 5 and its rated primary current is 1000A, the maximum fault current the CT can handle safely is:
5×1000𝐴=5000𝐴


## **Step 5: Evaluate the CT's Performance Under Fault Conditions**

Ensure that the CT can handle the calculated fault current without saturation and while maintaining the accuracy required for protection and measurement.

## **Step 6: Verify the Burden**

The burden is the total impedance (in ohms) connected to the secondary of the CT, including the metering instruments and the connecting leads. Ensure that the CT can drive the required burden without exceeding its accuracy limits.

## **Step 7: Compliance with Standards**

Ensure that the selected CT complies with relevant standards (e.g., IEC 61869-2 or ANSI/IEEE C57.13) for metering CTs.

**Example**
Application Requirements:

Rated Primary Current (Iᵣₚ): 1000A
Rated Secondary Current (Iᵣₛ): 5A
Accuracy Class: 0.5
CT Ratio: 1000:5

ISF: 5

Maximum Fault Current Calculation:

ISF = 5
Iᵣₚ = 1000A
Maximum Fault Current = 5 × 1000A = 5000A
Verify Performance:

Ensure the CT can handle 5000A without significant error or damage.
Check the CT’s datasheet to confirm it can handle the calculated fault current.
Verify Burden:

Calculate the total burden and ensure it is within the CT’s specified limits.
Conclusion
By following these steps, you can size a metering CT that meets the application requirements while ensuring the safety and protection of connected instruments through appropriate consideration of the ISF. This ensures reliable operation and accurate measurement in both normal and fault conditions.