# What will happen if I use a air core for 50Hz current tranformer

• chudiandeyu
In summary, the conversation discusses the possibility of using an air core transformer for a 50Hz current probe and the differences between high and low frequency transformers. It also explores the equation for transformer primary voltage and current, as well as how to compensate for changes in frequency by adjusting inductance or number of turns. The conversation concludes with the suggestion of using a commercially available current transformer for the desired specifications.
chudiandeyu
I am thinking to use the air core to make transformer to get better linearity behaviour, but I found air core often use in high frequency,so can I use air core for 50Hz? what is the difference between high frequency and low frequency?
Thank you!

The equation relating voltage and current for a transformer primary is

V = L dI/dt, or

V∫dt = L∫dI

If you increase the period ∫dt by lowering the frequency, you will need to increase the inductance L in order to maintain the same maximum reactive input current Imax = ∫dI.

If you do not add iron, you will need to add turns to the primary. For example, if you remove iron from a 50Hz transformer, the inductance will drop by a factor of ~4000. You will therefore need to increase the number of turns by a factor of ~√4000 = 63 to compensate.

Bob S

@Bob S
my transformer is a part of a current probe, so do you think it is reality to use air core for it?
and the 63 is 63 times the turns of the previous winding? can I decrease the secondary turns?
Thank you :)

@Bob S
What is the formula for the inductance and factor?

This is some more discrabetion of my question

I have a current transformer, it is soft iron core.
I want to change the core to air core to improve the linearity, but according to the information I checked, the output current will become very small.
I don't understand how will the output current change, it become small only because the inductance become small? And how can i calculate how small the output current will be?

The primary winding is a straight wire, and this current transformer is a part of a current probe. working in 50Hz.

Are you talking about a current probe that is measuring a steady ac current in a single-turn transformer primary (current probe) wire? For example a N-turn secondary on a transformer toroid that has a single-turn primary carrying I(ω) = I0sin(ωt) ? Please describe the situation in more detail. I have ordered special gapped toroids (to limit permeability) to measure 100 amp, 500 Hz signals on single-turn primary conductors. Is this what you want?

Bob S

My situation is a current probe for measuring ac current in single turn primary winding, and the current probe is a current transformer with secondary winding 500 turns, the current in primary wire is 10A and 50Hz.
So how can i calculate the current in secondary winding if i use a air core, and is it practical to use the air core?

Thanks !

This is an interesting problem. I simulated a 1-turn-primary, 500-turn-secondary Rogowski coil wound on a toroid with permeability = 1. The toroid is the only shape that will give a high mutual coupling inductance. The dimensions of the toroid were outer radius 4 cm, inner radius 2 cm, length 4 cm. This gives a primary inductance of ~5 nanohenrys. See model in thumbnail. The simulation shows that for 10 amps rms 50 Hz input, the peak open circuit output voltage is ~ 7 volts. This is limited by the shunt input inductance. To get 0.1% accuracy, you need to keep the output voltage to less than 7 mV and the output load to less than 0.2 ohms. So the maxumum output voltage is ~ 4 mV rms at 10 amps rms input. Normally, Rogowski coil output signals go into a transimpedance amplifier. To increase the maximum recommended output load (and voltage), the shunt input inductance should be increased either by using a longer toroid, or a coil form with a limied amount of permeability. How do you plan to process the toroid output signal? I hope this helps.

Bob S

#### Attachments

• Current_trans_Rogowski.jpg
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imranmwt
There are current transformer toroids available that nearly exactly match your requirements. See the model TR-5025 in

http://www.toroid.com/standard_transformers/current_sensing_transformers/current_sensing.htm

It has a single turn primary, 500 turn secondary, 90 amp peak primary. The cost is probably ~\$20.00 in unit qty (see bottom of page). It is not worth the effort to make your own at this price. I am not advertising this brand, but just letting you know that inexpensive ac current transformers with 1:500 turns ratio are available.

Bob S

Last edited by a moderator:
@Bob S
Thank you for your answer i decided to use a Hall effect sensor to make the current probe:)

## 1. What is an air core for a 50Hz current transformer?

An air core for a 50Hz current transformer is a type of transformer that uses air as its core material instead of a solid material such as iron. This allows for a higher degree of precision in measuring current.

## 2. How does using an air core affect the performance of a 50Hz current transformer?

Using an air core in a 50Hz current transformer can result in a more accurate and efficient measurement of current. The absence of a solid core material reduces energy losses and allows for a higher frequency response.

## 3. Can I use an air core for a 50Hz current transformer in all applications?

No, an air core for a 50Hz current transformer is typically only used in high frequency or precision applications. In lower frequency applications, a solid core material may be more suitable.

## 4. Are there any disadvantages to using an air core for a 50Hz current transformer?

One potential disadvantage of using an air core for a 50Hz current transformer is that it may be more susceptible to external magnetic interference. This can affect the accuracy of the current measurement.

## 5. How do I choose between an air core and a solid core for a 50Hz current transformer?

The choice between an air core and a solid core for a 50Hz current transformer will depend on the specific application and the desired level of accuracy. It is best to consult with an expert in the field to determine the most suitable option.

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