Substation CT Saturation Calculation Made Easy

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CTs are critical for the operation of power systems. A CT saturation calculation will show how well a CT can operate in non-ideal conditions.

I’ll show you step by step how to do the calculation and the information you need to gather.

Before we go over the calculation, let’s discuss CTs, Current Transformers. You can then better appreciate the importance of this calculation.

What happens when a CT saturates?

current transformers used in metering equipment
Current transformers used in metering equipment (Photo Credit: Alistair1978)

When a CT saturates, the secondary current isn’t proportional to the primary current. The variance in the secondary current depends on the CT’s saturation level.

For example, let’s say you have a 100:5 rated CT. If 200A passes through the primary winding, you’ll expect to see 10A on the secondary winding. But with a saturated CT, you may only see 5A.

CT saturation conditions typically have one of the following causes:

  • High primary current sourced from a fault condition
  • Open circuit condition on the CT’s secondary side
  • The high burden on the CT’s secondary side
  • Improper tapping down of a CT
  • DC offset current
  • High x/r ratio in a circuit

The point where a CT saturates is also known as the knee point. You can see this knee point on a CT’s excitation curve.

To point out, a CT’s accuracy class is not the knee point. The accuracy class you’ll see shown in terms of a C rating. To illustrate, let’s go over an example of a case scenario for a 10P20 C400 CT rating.

C400  means the CT needs to develop 400V across a 4-ohm burden at 20 times over current. Thus, 20 x 5 Amps x 4 ohms = 400V. All the while, not exceed a 10% error, which is an excitation current below 10A.

Also, IEEE C57.13 tells us the knee point voltage is typically about 80% of the accuracy limiting voltage. So the knee point voltage of a CT rated C100 is about 80-volts.

Important Note: CT saturation causes poor operation of devices. For example, metering, measurement, and protective devices can read invalid values. This happens because the CT’s secondary current isn’t proportional to the primary current.

Incorrect CT settings can cause electrical devices to blow up due to miscoordination. Not only will you damage expensive equipment, but people can get hurt. 

Metering versus protection purpose CTs

You can use CTs for monitoring current, and protection. Each CT use case has a unique design though.

Metering accuracy CTs have ratings for standard burdens, loads. They’re highly accurate for low to maximum current readings. Due to their accuracy, utility companies use these CTs for customer billing purposes.

Protective relaying accuracy CTs aren’t as accurate as metering accuracy CTs. Because they’re meant to perform over a very wide current range. For example, to detect a high fault current.

To point out, metering CTs will have large errors during fault conditions. The currents can be several times greater than normal values for short durations. This isn’t a big deal though, as metering isn’t a concern when there’s a fault.

In short, the application of a CT is critical to know in your CT saturation calculation.

Real-world substation CT saturation calculation

SF6 110kV current transformer
SF6 110 kV current transformer (Photo Credit: Vivan755)

The CTs in our calculation will be for substation relay protection.

In our calculation, we’ll use the following information:

  1. The maximum available fault current at 115,000-volts is 24,900 amps
  2. CT secondary wiring is #10 AWG copper (1.3 ohms/1000′ resistance)
  3. The wiring run from breaker CT to relay meter is 500′ max. The total total circuit length is thus 2 x 500′ = 1,000′.

Important Note: a CT’s burden and the fault current determine the CT accuracy class. For example, whether to choose C100, C200,  or C400.

Selecting the proper CT accuracy class will ensure a CT can supply a linear output. This is critical when a CT needs to respond to a high fault current. 

To choose the correct accuracy class, you need to know the fault current and burden on the CT. For example, assume the secondary of a CT connects to a protective relay. The burden then includes the relay, CT wiring, and anything else between the CT and the relay. This can be test switches and terminal blocks.

So, if you have long CT wiring, you’ll require a higher-class CT. This is because of the increased resistance. Now, assume the burden remains fixed at 1-ohm for a C200 CT. Let’s see what happens when the fault current increases from 100A to 400A:

R = \dfrac{V}{I} = \dfrac{200V}{100A} = 2 \Omega

R = \dfrac{V}{I} =\dfrac{200V}{400A} = 0.5 \Omega

Thus, we need a C400 rated CT to accommodate the 1-ohm burden. A C400 CT calculates to, \dfrac{400V}{400A} = 1 \Omega

CT (1x, 3x, 5x) to existing 115,000 differential relay GE L90

CT = 2000:5A MRCT connected 2000:5A (400:1)

Important Note: the fault current is 24,900 Amps. So we want an accurate operating CT when the fault current is 24,900A.

Assume we use a 10P20 CT. This means the CT will have an accuracy of 10% maintained up to 20 times its primary rated current. Using this metric, we can calculate the CT ratio, which is  24,900/20 = 1,245A. Thus, we need to use a CT ratio of 2000:5A. 

Another way to view this calculation is, a 2000/5 CT is 20 x 2,000 = 40,000 Amps. This shows the CT will remain accurate in the field if there’s a 24,900A fault. 

To point out, the burden heavily impacts the CT accuracy rating. The lower the burden on the CT secondary, the higher the saturation point for a given CT. For instance, imagine a 10P20 50 VA CT has a burden of 25 VA. This means the CT shouldn’t saturate 40 times, 50/25 x 20, over the rated primary current, not 20.

GE L90 relay burden = 0.2VA at 5 Amp input

0.2 VA/5A = 0.04 ohms

Important Note: typically, old electromechanical relays have high burdens. Thus, they require high CT accuracy classes. While new-age microprocessor relays have a very little burden. 

CT Burden = CT lead resistance + relay burden

= (2) x (500′) x (1.3 ohms / 1000′) + 0.04 ohms

= 1.3 ohms + 0.04 ohms = 1.34 ohms

Secondary amps at maximum fault duty

I round the 24,900A fault current to 25,000A to account for any errors.

\dfrac{\text{Primary Amps}}{\text{CT Ratio}} = \dfrac{\text{25,000A}}{400} = 62.5 Amps

= 62.5 x 1.34\Omega = 83.8 volts

The minimum accuracy class of the CT should be C100.

Important Note: U.S. utilities commonly use C800 for high voltage equipment. For example, using C800 accuracy class CTs for breakers and transformers in substations. This is regardless of whether the available fault current is high or not. 

BUT, for retrofit projects, it’s always important to check the physical available space for installation. Because C800 rated CTs may not fit in an existing enclosure because they’re large in size. This is why there’s an art to CT sizing.  

CT (1y, 3y, 5y) to existing 115,000 differential relay ABB 350

CT = 2000:5A MRCT connected 2000:5A (400:1)

ABB 350 relay burden = 0.45VA at 5 Amp input

0.45 VA/5A = 0.09 ohms

CT Burden = CT lead resistance + relay burden

= (2) x (500′) x (1.3 ohms / 1000′) + 0.9 ohms

= 1.3 ohms + 0.09 ohms = 1.39 ohms

Secondary amps at maximum fault duty

\dfrac{\text{Primary Amps}}{\text{CT Ratio}} = \dfrac{\text{25,000A}}{400} = 62.5 Amps

= 62.5 x 1.39\Omega = 86.9 volts

The minimum accuracy class of the CT should be C100.

CT saturation calculation wrap up

CT saturation calculations are very straightforward and important. These calculations will ensure your power system will properly operate.

This means all your metering and protection schemes will be reliable.

On the flip side, using incorrectly rated CTs can easily cause power system failures. This can lead to millions of dollars in damages, and people can get hurt.

In summary, total your CT loads and use the max fault current to size your CTs.

Do you have any CT saturation calculation tips?  Do you find sizing CTs is like an “art”?

Featured Image Photo Credit: Vivan755 (image cropped)

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