# Silvio

#### Current transformer calculation

by

, 10-05-2017 at 08:57 PM (7323 Views)
To determine how to calculate the number of turns according to the current handling for a SMPS.

1) Determine the voltage needed for sensing and add 1 diode forward voltage in a centre tapped configuration, or 2 diode

forward voltage drop for a single secondary winding with a bridge rectifier.

2) Calculate the peak current for the SMPS in the primary winding.

3) Determine the turns ratio for the current transformer. Please note that the higher the number of turns the lower the current

will be in the secondary winding and the higher will be the load resistor. However one must keep the sense voltage at the

output of the transformer not more than 2 volts as it will become inaccurate. The lower the output voltage the more

accurate the current sense will be

4) Selecting a ratio it must be calculated with the primary coupling loop as 1.5 turn. This is a more convenient way as usually

with just looping to the core will only bring the effect of 1/2 a turn rather than one

Example:- Half bridge uses 1/2 the DC bus voltage so for a rectified and smoothed mains grid voltage of 220VAC it will be equal to around 310v and half will be 155vdc.

The wattage in the input will be according to this voltage multiplied by the maximum current in the primary. For this example we are taking 500watts as the peak power for tripping the smps. We also have to add a bit more due to efficiency loss and this calculation is for 80% efficiency. The total power absorbed by the smps will be 500w X 100 / 80 =625 watts

Voltage needed to sense the tripping current is 1Volt and we are using a single winding.

For the use of a bridge rectifier with 4 diodes we add 1.2v (0.6v X 2)

Watts = V X I I = W/I so 625w / 155 =4.03Amps@ 220vac input

The wattage will always be 500watts and if the input voltage varies the tripping current will adjust on its own until 500 watts are reached, thus the current will rise a little with a lower input and lowers with a higher input voltage.

We usually take ratios at 1:15 , 1:30, 1:50 and 1:100.

For this example we are taking a 1:50 ratio as the current is not so large.

Example.

1) So with a coupling loop of 1.5 turns and a ratio of 1:50 we calculate 50 X 1.5 =75 turns.

2) To generate 1volt we add 2 diode voltage 1v + 1.6v = 2.6v

3) we calculate the current first and this will come 4A / 50 turns =0.08A

4) The bridge resistor for this voltage will be 2.6v / 0.08A =.32.5 Ω

However we will find the nearest value resistor which will be 33Ω. This will give the sense voltage needed.

In general 1volt is more than enough to trigger a transistor base or an SCR of which these vary between 0.6 and 0.9

volts respectively. This voltage can be put across a 1KΩ preset and the exact tipping current can be set.

So our CT is calculated. We havearound the core for secondary and 1.5 coupling loop75turns

The Shunt resistor is to beand we have a full bridge with 4 fast diodes.33Ω

END