Current transformer

[alphacat]

Hello.
I got the following specifications of a current transformer i'd like to order.

I dont understand two things there:

1. What does it mean 'Rated Load'?
Can I only connect a 100Ω load to the secondary coil?

2. There are 4 graphs in the picture, each belongs to a certain resistance.
What does each resistance mean?


Thank you very much.**broken link removed**

 

[Hero999]

1) Yes, 100R is the rated load resistance, meaning it's designed to be connected to a 100R resistor.

2) The graphs are the different secondary voltages vs primary current with different resistors connected to the secondary.

I agree, it does seem odd to give a rated resistance, then provide a graph showing the voltages with different resistors connected. I assume that 100R is the recommended resistance, rather than the rated resistance.

Note, that make sure that you never use the transformer without a resistor and that the resistor has an adequate power rating.

 

[antknee]

Where are you ordering?

 

[alphacat]

Thank you very much Hero.

Why shouldnt I use the transformer with no load?
What could happen?

Antknee, i'm ordering it from China.

 

[antknee]

I thought you might provide a link to a website where it might be possible to see a bit more information.

I've never had the confidence to order from China myself, I've always wondered if the goods I ordered would actually turn up. Reading Chinese websites may be difficult too...

 

[crutschow]

If you use a current transformer without a load, the output voltage can go high enough to break down the winding insulation and short the transformer. Remember its a current transformer so the output secondary will try to maintain a current proportional to the input current independent of the secondary load. With no load, that can lead to very high secondary voltages.

 

[Reloadron]

I think what your chart is getting at is what is called 'burden". A CT is a ratio device and in your case passing a current of 50 A through the primary of the donut will yield 5 A on the secondary. That 5 A is based on a 100Ω or less load.

Note on the graph how a 10Ω resistor provides 200 mV, the 100Ω provides 2 V. However what happens when larger load resistances come into play? You would expect 200Ω to provide 4 V but it falls well short at about maybe 3.3 V and the 1KΩ load is a disaster.

This means a CT with a burden of 100Ω can tolerate a load resistance up to 100Ω before the current ratio begins to deteriorate significantly. The 50/5 CT you are looking at needs to work into a resistance of 100Ω or less. Also if working into 100Ω lead length from the CT to the measuring device becomes critical.

Ron

 

[Boncuk]

DCR 510±10%Ω!

Why not use that value?

The 'imaginary' 500Ω line is settled at approximately 45degrees giving the optimum input current to output voltage ratio.

I doubt that 9000mV will break through the transformer insulation.

Those current transformers are not connected to any primary load, but measure the magnetic field strength around the conductor (cable).

Boncuk

 

[MikeMl]

My understanding is that CTs need to be terminated into a resistance called a burden. I interpret the posted data sheet as recommending that a burden of 100Ω is the maximum that should be used consistent with keeping this CT within its linear range. There is nothing to prevent you from using a smaller resistance, except that you will get a smaller output voltage.

Think of a CT as a transducer that produces an AC voltage which is proportional to the line current multiplied by the burden resistance.

 

[Reloadron]

DCR 510±10%Ω!

Why not use that value?

The 'imaginary' 500Ω line is settled at approximately 45degrees giving the optimum input current to output voltage ratio.

I doubt that 9000mV will break through the transformer insulation.

Those current transformers are not connected to any primary load, but measure the magnetic field strength around the conductor (cable).

Boncuk

No there is no Primary in a sense of a winding. The primary is a steel core with a secondary winding wrapped around it. Thus my refrence to donut as CTs of the type I suspect the poster is ordering are frequently called donuts. I attach a few images. The 50:5 I believe the OP is ordering is one I had lying around. The 100:5 image is of one at work monitoring a 3 phase 480 volt panel for some heaters.

Using a 500 Ohm load will result in accuracy problems. The problem with this Chinese stuff is they actually don't mention burden in the data provided. The burden can be and is usually specified in VA. It can be specified in resistance but I have never seen a rating of 100 Ohms. That is exceptionally high for a CT, even a 50:5 CT. They don't exactly say that 100 Ohm number is burden.

Voltage or voltage breakdown have nothing to do with it. When the main current carrying conductor passing through the CT (donut) is carrying 50 amps the CT will output 5 amps into a load. That load needs to be a very low resistance (burden) for the 50:5 ratio to be maintained. When using a CT even the lead length (resistance) is critical. The 50:5 ratio depends very much on burden. Note the 10 Ohm load in the chart.

On a safety note:

Care must be taken that the secondary of a current transformer is not disconnected from its load while current is flowing in the primary, as the transformer secondary will attempt to continue driving current across the effectively infinite impedance. This will produce a high voltage across the open secondary (into the range of several kilovolts in some cases), which may cause an arc. The high voltage produced will compromise operator and equipment safety and permanently affect the accuracy of the transformer.

Also since this is likely a donut type CT a mains line will pass through it. Caution should be exercised when passing the mains power line through the CT. Make sure, extra sure, power is not present.

<EDIT> **broken link removed** is a good example of what a data sheet should look like for a CT. Note how the uncertainties are spelled out for a 2 VA burden. The link also gives a few thoughts as to CT use. </EDIT>


Ron

 

[kinarfi]

In my days of working with CTs, #1 rule was never power up a CT without a load, dead short is just fine. A CT will try to pass the secondary current and generate as much voltage as it needs to pass that current, theoretically, it will generate enough voltage to arc across the gap, but I never saw it happen. Another thing I experimented with was a low voltage, high current circuit breaker tester. I put 100 amps through CT and left it open to see what would happen. With the low voltage capability of the tester, with the CT shorted, it did pass the correct current, set at 100 amps, with the CT secondary open, it couldn't generate enough voltage to pass the current, so, if the secondary is open, it creates a significant impedance to current flow. On you 50/5 CT with a 100 ohm load, at a primary current of 50 amps,you should have 500 volts across the 100 ohm resistor and 5 amps flowing through it, generating 2500 watts of heat. If you have a 1 ohm load, you will still have the 5 amps flowing through the resistor, but just 5 volts and 25 watts of heat. and most of the instruments I ever worked with that used CTs had 0 ohms resistance. The power has to come from somewhere and it comes from the primary lead being measured.
Kinarfi
Another thing about CTs is they all have a ratio of xxx/5.

 

[alphacat]

Hello.

Thank you so much for your helpful posts.

How did you conclude that its a 50:5 CT?
According to the 100Ω curve:
Vout = 2V @ Iin = 60A
=>
Is = 20mA @ Ip = 60A
=>
Ns / Np = 3000

Meaning, each 3000A at input produce 1A at output. isnt it?

 

[MrAl]

Hi there,


There's also the core saturation issue. As the core gets closer and closer to saturation, the output voltage starts to flat top and that causes non-linearity which appears to the measuring equipment as a decrease in the perceived current level but that decrease is not really there. That happens because the flux density goes up with output voltage and as the voltage goes up the core gets closer and closer to saturation.
There are ways to approximate this behavior, but it's easier just to use a resistor that isnt too large. That prevents the output voltage from going too high. What is too high? That depends on the core cross sectional area and the number of turns, as well as the operating frequency and the core material. For most material used for line operated equipment you should probably stay under 10kGauss or maybe even less to assure good linearity.

The equation looks something like this:
Bmax=E*10^8/(4.44*F*A*N)

If you need more details for that equation (it's probably all over the web by now though) i'll post some.
I remember the last part of the denominator because it spells out "fan", but in some literature the letters are ordered differently like AFN or NAF, etc.

To find out the core cross sectional area (A) you could simply measure the core stack height and the width of the center leg.

Another trick:
Since many times current transformers are used to measure current in some process and the resulting voltage analog has to be converted to a DC voltage, standard rectifiers are used. The trick is to connect the resistor after the rectifiers so that
the voltage drop and non-linearity of the rectifiers becomes minimal.

 

[alphacat]

Thanks MrAl.
I understand it now.

Perhaps you could tell me how it was conducted that the CT has a ratio of 50:5?

 

[Reloadron]

How did you conclude that its a 50:5 CT?

I guessed.

The specification sheet reflects a Rated Current of 50 Amps with a Max Current of 75 Amps. Therefore I assumed 50:5 because the secondary current is generally 1 or 5 Amps. Actually with newer and better measuring systems for CTs the 1 Amp out is gaining in popularity, especially in Europe.

Also, read MrAl's post about saturation to understand what is going on.

Try not to think in terms of Voltage Out as the voltage seen will be a function of the load. Think along the lines of a current ratio with current out. Read the link I posted as to a CT Data Sheet also.

I also guessed at the donut design as they do come in a few flavors, including split core but I think the donut is the most common type.

Ron

 

[alphacat]

Hey.

Actually, I read MrAl's post and the great datasheet you attached.

Am I correct by concluding from the graphs that this is a 3000:1 CT?

 

[alphacat]

In addition to whether its a 3000:1 CT, i'd like to ask one more thing please.

If I measure the current of the entire house (by having the Current Transformer to "wrap" the Live wire of the house), an also measure the voltage at one of the sockets of the house, will I be able to compare the two waveforms and calculate the power factor of the entire house?
I already have such IC that compares voltage and curren waveforms and tells that power factor, however will it work?
Could I really tell the power factor of the entire house by doing this?

Thanks a lot.

 

[Hero999]

I agree with you about 3000:1 but don't take my word for it as I don't have much experience with current transformers.

Yes you should be able to work out the power factor by looking at the waveforms.

The trouble is, you'd need to measure the voltage at a part of the circuit as physically close to where you're measuring the current, otherwise voltage drop could be a problem.

 

[alphacat]

Thanks Hero!

Once the CTs are here, i'll check up its ratio by measuring two different input currents and their corresponding output currents.

The CT has two wires in its output which I will connect to the power sensor I got.
I might demand these wires to be 1m-2m long.
Will this cause a problem to measure the power factor then?
Meaning, will the wires create a delay which will affect the power factor measurement?

Thank you again.

 

[Hero999]

The delay don't be a problem because the frequency of the mains is too low to matter.

The resistance might be a problem but at 2m of cable it isn't going to be noticeable.