209

Diversified Electronics

800.727.5646

atcdiversified.com

Diversified Electronics

800.727.5646

atcdiversified.com

Example:

1.

Measure the current on each leg.

Assume

A = 10 amps

B = 12 amps

C = 9 amps

2.

Find Average 10 + 12 + 9 = 31

31

= 10.33

3

I max = 12

I avg = 10.33

3.

Apply formula 12 - 10.33 = 1.67

1.67

= .139 x 100 = 13.9% unbalance

12

THREE PHASE CURRENT UNBALANCE

EXTERNAL CURRENT TRANSFORMERS

Balanced or matched currents on a three phase system are difficult

to maintain because of the many varying factors involved such as,

unequal single phase loading, poor connections and cabling and/or

dirty or burnt starter contacts.

Although these varying factors can be controlled to maintain as close

as possible a balanced line, the unseen conditions such as overheated

motor windings, burnt bearings, low voltage, high voltage and single

phasing need to be constantly monitored to protect your valuable

equipment.

The CLB Series Three Phase Current Unbalance and Over Current

Monitor (page 176) offers this protection.

To determine the condition of your three phase line and to properly

select the CLB Series percent unbalance setting a simple calculation

formula is needed as follows:

(I max - I avg)

x 100

I max

The load or burden that can be connected to the secondary of the

Current Transformer is usually specified in VA. The rated accuracy of

the Current Transformer is guaranteed only when the sum of the VA

ratings of all devices (ampmeters, wattmeters, current monitors, etc.)

connected to the secondary windings does not exceed the specified

VA rating.

The interconnecting conductor resistance must also be considered,

especially when the Transformer is installed at some distance from the

Current Monitor or other load.

For the wiring, the VA can be calculated using Ohm’s Law:

VA = E x I = (I x R) x I = I2R

Where I = 5 Amps and R is the DC resistance of the wire.

All of the standard DE Current Transformers have a rating of 2 VA

except the 2500/5 version which has a 5 VA rating.

From the above formula we can also calculate the maximum resistance

that can be connected to the secondary of a Current Transformer:

VA=I2R, Hence R = VA

I2

Example:

VA = 2

Rmax = VA = 2 = 0.08 Ohms = 80 milliohms

I2

(5)2

VA = 5

Rmax = VA = 5 = 0.2 Ohms = 200 milliohms

I2

(5)2

If the only load on the Current Transformer is a DE Current Monitor,

its VA rating (approx. 0.15 VA) is small compared to the Current

Transformer rating and can be neglected. This means that the resis-

tance of the wiring can be 80 milliohms max. for the 2 VA units and

200 milliohms max for the 5 V A unit.

Gauge

Ohms per 1000'

Milliohms per foot

AWG 14

2.5

2.5

AWG 16

4.0

4.0

AWG 18

6.4

6.4

AWG 20

10.0

10.0

Example:

For a 2 VA Current Transformer, the length of AWG 16 wire would be:

80 milliohms

4.0 milliohms/ft

= 20'

Because we are dealing with a pair of wires, the maximum distance

from the Current Transformer to the Current Monitor can be only 10

ft.

As we can see, it is important to keep the wire length to minimum,

use heavy wire, and keep all connections clean and tight.