1 Answer
Sizing a **Current Transformer (CT) ratio** correctly is essential in electrical systems for **accurate current measurement**, **metering**, and **protection**. If the CT ratio is too large or too small, it can cause inaccurate readings, ineffective protection, and even damage to the system. Letβs go through the concept and steps to determine the proper CT ratio.
---
### What is a CT Ratio?
A **CT ratio** is the ratio of the **primary current** (actual current flowing in the system) to the **secondary current** (measured current, usually 1 A or 5 A for standard equipment).
**CT Ratio = Primary Current / Secondary Current**
Example:
If a CT has a ratio of **300:5**, it means that when **300 A** flows in the primary conductor, the CT outputs **5 A** to the secondary circuit (for metering/protection).
---
## Steps to Size a CT Ratio
### β 1. **Determine the Load Current (Primary Side)**
Estimate the **maximum expected load current** that will flow through the conductor being monitored.
You can calculate it using:
$$
I_{\text{load}} = \frac{P}{\sqrt{3} \times V \times \text{Power Factor}} \quad \text{(for 3-phase loads)}
$$
Or directly from nameplate data or site conditions.
---
### β 2. **Choose a Standard Secondary Current**
Standard secondary currents are:
* **5 A** (common in older or heavy-duty systems)
* **1 A** (preferred for modern digital meters and protection relays due to lower copper loss)
Pick one depending on what your meters or relays are rated for.
---
### β 3. **Calculate the Required CT Ratio**
Now use:
$$
\text{CT Ratio} = \frac{I_{\text{primary}}}{I_{\text{secondary}}}
$$
Example:
If your full load current is **240 A** and your relay/meter accepts 5 A:
$$
\text{CT Ratio} = \frac{240}{5} = 48 \rightarrow \text{Use nearest standard size, e.g., } 50:5
$$
---
### β 4. **Select Nearest Standard CT Ratio**
CTs are manufactured in standard ratios like:
* 50:5
* 100:5
* 150:5
* 200:5
* 400:5
β¦ and so on.
**Always round up** to the nearest standard CT to avoid saturation at full load.
---
### β 5. **Check CT Burden and Accuracy Class**
#### βοΈ Burden:
* This is the total resistance (in ohms) on the CT secondary circuit (including meters, relay input impedance, and wiring).
* Ensure the burden does **not exceed** the rated burden of the CT (e.g., 10 VA, 15 VA).
#### Accuracy Class:
* For metering: Use **class 0.2, 0.5, or 1.0**.
* For protection: Use **class 5P, 10P** (the βPβ stands for Protection).
Example: A CT labeled **100:5, 15 VA, Class 5P10** means:
* 100 A primary, 5 A secondary
* 15 VA burden
* 5P accuracy class, 10 times accuracy limit factor
---
### β 6. **Consider Overcurrent Protection Requirements**
Protection CTs must handle fault currents without saturation. You may need to:
* Evaluate the **CTβs Knee Point Voltage** (especially for differential protection).
* Check the **Accuracy Limit Factor (ALF)** for protection CTs β it defines how many times the rated current the CT can handle while staying within accuracy limits.
---
## Summary Table
| Parameter | Consideration |
| ----------------- | -------------------------------------------- |
| Primary Current | Based on load or cable rating |
| Secondary Current | 1 A or 5 A (standard) |
| CT Ratio | Primary / Secondary, round up to standard |
| CT Burden | β€ rated VA of CT (include wiring and device) |
| Accuracy Class | Metering: 0.5/1.0; Protection: 5P/10P etc. |
| Saturation & ALF | For protection β avoid under-sizing |
---
### Example Problem:
**You have a 3-phase 400 V system, 200 kW load, 0.9 PF. Choose a CT ratio for protection.**
1. Load Current:
$$
I = \frac{200,000}{\sqrt{3} \times 400 \times 0.9} = 320.7 \text{ A}
$$
2. Choose secondary: 5 A
3. CT Ratio = 320.7 / 5 β 64.1 β Use **75:5** or **100:5** for headroom
4. Choose class 5P15 (if it's for overcurrent protection)
---
Would you like a downloadable CT sizing calculator or a diagram to visualize this better?
---
### What is a CT Ratio?
A **CT ratio** is the ratio of the **primary current** (actual current flowing in the system) to the **secondary current** (measured current, usually 1 A or 5 A for standard equipment).
**CT Ratio = Primary Current / Secondary Current**
Example:
If a CT has a ratio of **300:5**, it means that when **300 A** flows in the primary conductor, the CT outputs **5 A** to the secondary circuit (for metering/protection).
---
## Steps to Size a CT Ratio
### β 1. **Determine the Load Current (Primary Side)**
Estimate the **maximum expected load current** that will flow through the conductor being monitored.
You can calculate it using:
$$
I_{\text{load}} = \frac{P}{\sqrt{3} \times V \times \text{Power Factor}} \quad \text{(for 3-phase loads)}
$$
Or directly from nameplate data or site conditions.
---
### β 2. **Choose a Standard Secondary Current**
Standard secondary currents are:
* **5 A** (common in older or heavy-duty systems)
* **1 A** (preferred for modern digital meters and protection relays due to lower copper loss)
Pick one depending on what your meters or relays are rated for.
---
### β 3. **Calculate the Required CT Ratio**
Now use:
$$
\text{CT Ratio} = \frac{I_{\text{primary}}}{I_{\text{secondary}}}
$$
Example:
If your full load current is **240 A** and your relay/meter accepts 5 A:
$$
\text{CT Ratio} = \frac{240}{5} = 48 \rightarrow \text{Use nearest standard size, e.g., } 50:5
$$
---
### β 4. **Select Nearest Standard CT Ratio**
CTs are manufactured in standard ratios like:
* 50:5
* 100:5
* 150:5
* 200:5
* 400:5
β¦ and so on.
**Always round up** to the nearest standard CT to avoid saturation at full load.
---
### β 5. **Check CT Burden and Accuracy Class**
#### βοΈ Burden:
* This is the total resistance (in ohms) on the CT secondary circuit (including meters, relay input impedance, and wiring).
* Ensure the burden does **not exceed** the rated burden of the CT (e.g., 10 VA, 15 VA).
#### Accuracy Class:
* For metering: Use **class 0.2, 0.5, or 1.0**.
* For protection: Use **class 5P, 10P** (the βPβ stands for Protection).
Example: A CT labeled **100:5, 15 VA, Class 5P10** means:
* 100 A primary, 5 A secondary
* 15 VA burden
* 5P accuracy class, 10 times accuracy limit factor
---
### β 6. **Consider Overcurrent Protection Requirements**
Protection CTs must handle fault currents without saturation. You may need to:
* Evaluate the **CTβs Knee Point Voltage** (especially for differential protection).
* Check the **Accuracy Limit Factor (ALF)** for protection CTs β it defines how many times the rated current the CT can handle while staying within accuracy limits.
---
## Summary Table
| Parameter | Consideration |
| ----------------- | -------------------------------------------- |
| Primary Current | Based on load or cable rating |
| Secondary Current | 1 A or 5 A (standard) |
| CT Ratio | Primary / Secondary, round up to standard |
| CT Burden | β€ rated VA of CT (include wiring and device) |
| Accuracy Class | Metering: 0.5/1.0; Protection: 5P/10P etc. |
| Saturation & ALF | For protection β avoid under-sizing |
---
### Example Problem:
**You have a 3-phase 400 V system, 200 kW load, 0.9 PF. Choose a CT ratio for protection.**
1. Load Current:
$$
I = \frac{200,000}{\sqrt{3} \times 400 \times 0.9} = 320.7 \text{ A}
$$
2. Choose secondary: 5 A
3. CT Ratio = 320.7 / 5 β 64.1 β Use **75:5** or **100:5** for headroom
4. Choose class 5P15 (if it's for overcurrent protection)
---
Would you like a downloadable CT sizing calculator or a diagram to visualize this better?