Are integrated CTs any different from CTs in clamp meters?

  • #1
Hi!

I have question about current transformers (CT). I have BH-0.66 class 0.5 20/5 current transformer. According to my understanding 20/5 means 4 times current reduction. For some reason I get about 20 times current reduction. Now I am curious if the factory managed to label 100/5 CT as 20/5 CT or if those integrated CTs have some specific needs in order to work correctly that I am not aware of.
I know that it measures about 20 times less current because I compared it with multimeter. And also tried clamp meter. At least with clamp meter I get correct reading easily. And polarity or cable angle doesnt seem to affect reading too much - maybe +/- 10%. But with CT looks like no matter how I wire it the reading is always about 20 times less, not 5 like I would expect. It happened with both small and large loads. And that is why I was wondering if there is something specific that I need to know about CTs in order to get correct reading that normal clamp meter hasn't require me to learn yet.

20A_ 5A.png

Also I got one more question when I was reading about CTs from interenet:
https://testguy.net/content/190-Current-Transformer-Basics-Understanding-Ratio-Polarity-and-Class

Here it says that
It is very important to observe correct polarity when installing and connecting current transformers to power metering and protective relays.
I was wondering if there is any reason why it should be important in my case where I only have single phase AC and one meter? At least the clamp meter doesn't seem to care.
 

Answers and Replies

  • #2
DaveE
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This is nearly impossible to diagnose without more information about both the CT and how you are using it.

If you are measuring voltage at the output terminals, then check to see if you need/have the correct termination resistor. Also check that you aren't saturating it by putting too much current through it (really, too much voltage across it from the termination). Since these are pretty basic errors they probably aren't the problem. So, yes, it sounds like the transformer may be mislabeled.

I don't understand why they would refer to polarity in an AC circuit with an AC probe. Except that power flow could be measured as negative.

I would refer to the manufacturer for both questions. There should be a data sheet or documentation of proper use that you can check this stuff against.
 
  • #4
DaveE
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  • #5
anorlunda
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Yes, but it's not a power meter ???
The OP asked about this:
It is very important to observe correct polarity when installing and connecting current transformers to power metering and protective relays.
So yes it is power metering.

Some protective relays also care about the direction of power flow.
 
  • #6
DaveE
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The OP asked about this:

So yes it is power metering.

Some protective relays also care about the direction of power flow.
OK got it. This is the sensor for the power meter/protector then.
 
  • #7
Thanks for all the info!

If you are measuring voltage at the output terminals, then check to see if you need/have the correct termination resistor. Also check that you aren't saturating it by putting too much current through it (really, too much voltage across it from the termination).
This CT was around 240V cable. CT was 20/5 so I assume that it should boost the voltage 4 times if my understanding is correct. When I measured disconnected CT with multimeter it only showed few mV instead of 4*240V = 960V what I expected. The cable was running almost straight through - meanaing it wast not curved or anything.

There should be a data sheet or documentation of proper use that you can check this stuff against.
It only had chinese version of it on paper that was in the box of CT. didnt manage to find english from the internet.

I would refer to the manufacturer for both questions.
Just thought that I would get better quality response from this site than from them

Edit:
One more question. The meter itself also has 5/2.5 mA CT inside of it. Does anyone know if running CT through other CT reduces accuarcy? For example, if 0.5 class CT goes to other 0.5 class CT that will the end result become 1.0 class?
 
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  • #8
rbelli1
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This CT was around 240V cable. CT was 20/5 so I assume that it should boost the voltage 4 times if my understanding is correct.
No. It reduces the current by a factor of 4. Voltage will be increased by many more times if you draw any current through the mains conductor.

Please don't mess with this any more until you understand it better as you could hurt or kill yourself with this set up.

BoB
 
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  • #9
Averagesupernova
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Current transformers are certainly not intuitive. Always under every circumstance short the output leads until you have terminated them into the specified load whether it be a watt-hour meter or what I believe is called a burden resistor. Someone jump on if I have my terminology wrong. It is so easy to run the wire through the CT and simply say "I'll hook the rest up tomorrow". If the wire running through should carry a current, thousands of volts can build up on the CT output if left open. They are meant to transform current, not voltage.
 
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  • #10
berkeman
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Edit:
One more question. The meter itself also has 5/2.5 mA CT inside of it. Does anyone know if running CT through other CT reduces accuarcy? For example, if 0.5 class CT goes to other 0.5 class CT that will the end result become 1.0 class?
You need to figure out if the CT has an internal burden resistor or expects an external burden resistor. See if you can find a friend who reads Chinese so they can translate that information sheet for you. The CTs that I use have internal burden resistors, and are labeled "5A/333.3mV" to indicate that a 5Arms current generates 333.3mVrms across its internal burden resistor (and the datasheet specifies at least a 10kOhm input impedance needed at the voltmeter used to read that CT output voltage).

Other CTs that I have seen do not include an internal burden resistor, and are often just labeled with the current step-down ratio. That ratio plus the max line current would be used to calculate a reasonable value of burden resistor that will not overvoltage the voltmeter to use for the reading. I haven't seen a case where an ammeter (whether CT based or shunt resistance based) is used to read the output current of a CT, but I suppose it could be done.

How much experience do you have in working with AC Mains circuits? What is your background?
 
  • #11
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Calling @anorlunda for @berkeman will you accept the charges or should we charge @berkeman instead?

With respect to reading Chinese, google has a translate app that can translate Chinese to English for Chinese phrases picked up by your phone camera. I use it when I’m trying to help my mother-in-law with her computer problems and a message pops up that I can’t read.
 
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  • #12
Averagesupernova
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I haven't seen a case where an ammeter (whether CT based or shunt resistance based) is used to read the output current of a CT, but I suppose it could be done.
I guess the definition of ammeter is up for discussion? Lol.
 
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  • #13
anorlunda
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This CT was around 240V cable. CT was 20/5 so I assume that it should boost the voltage 4 times if my understanding is correct. When I measured disconnected CT with multimeter it only showed few mV instead of 4*240V = 960V what I expected. The cable was running almost straight through - meanaing it wast not curved or anything.
Your numbers are for an open circuit ideal transformer, where V2=n*V1, I2=I1/n, V1*I1=V2*I2.

But the CT wired in series, so that V1 is not the 240V line voltage to ground, but rather the series voltage drop across the CT. That voltage drop should be as close to zero as possible. Remember that the 240 V drop in the circuit should be measured across the load, not measured across the current measuring device upstream. Also, many CTs use a burden resistor R, which is internal and invisible as @berkeman said. So the way to think about it is:

Assume I1=100 A
Then I2=I1/20 = 5 A
We can scale R so that we get 100 mv when I1=100 A, so R=V2/I2, 0.100/5= 0.02 ohms.
Then V1=100/20 = 5 mv when I1=100 A.

Typically, the 100 mv secondary voltage is then measured with a high impedance voltmeter not an ammeter.

V1 is typically not measured. 5mv is close to zero when compared to 240V, and that is what we want. Assuming unity power factor and an ideal transformer, 100 A * 5 mv = 0.5 watts power lost in the current measurement, and 100 A * 239.995 V = 23999.5 watts of power delivered to the load. In reality, CTs are far from ideal transformers, so the real numbers will be different.

Always remember that to measure voltage, the voltmeter always has two lead wires, and the answer depends on where you connect them. The dotted lines in the sketch below show where to put the voltmeter leads in this case. I think your confusion in this case was mis-imagining where to put the two leads for the CT.


1593900272331.png


Note that you could accomplish the same 100 mv measurement by inserting a shunt resistance with R= 0.1/100 ohms, in series with the cable. But that would requires you to cut the cable to install it, it requires the shunt to carry the entire load current, and it exposes you to more safety hazards and practical obstacles than using a CT. So in AC circuits, CTs are better than shunt resistances. (But in my boat, I use a shunt resistor in the 12 V power circuits, to measure current.)
 

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  • #14
Tom.G
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I haven't seen a case where an ammeter (whether CT based or shunt resistance based) is used to read the output current of a CT, but I suppose it could be done.
Yup, done quite regularly. If the CT the OP has is labelled 20/5 then it most likely expects to be connected to a 5Amp meter whose scale is calibrated 0 to 20A. Note that those Ammeters are extremely low resistance, usually just a Copper busbar next to a magnet on a pivot for DC.
For some examples see:
https://www.google.com/search?q=current+transformer+usage

A burden resistor and voltage measurement came into widespread use when we started using electronic devices to measure stuff.

Cheers,
Tom
 
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  • #15
No. It reduces the current by a factor of 4. Voltage will be increased by many more times if you draw any current through the mains conductor.
How many more times?
I read this page: https://www.electronics-tutorials.ws/transformer/current-transformer.html
Here it explains that when secondary winding becomes open-circuited CT secondary V = primary V times (turns on secondary divided by turns on primary). In case of 20/5 turns on secondary would be 4. And if I use V_secondary = V_primary * (4/1) = V_primary * 4. Or what's the exact formula?

You need to figure out if the CT has an internal burden resistor or expects an external burden resistor. See if you can find a friend who reads Chinese so they can translate that information sheet for you. The CTs that I use have internal burden resistors, and are labeled "5A/333.3mV" to indicate that a 5Arms current generates 333.3mVrms across its internal burden resistor (and the datasheet specifies at least a 10kOhm input impedance needed at the voltmeter used to read that CT output voltage).
Is manual the only source to get that info? Or can it also be assumed from some specs? For example that if it mentions output voltage then it must be with resistor? And that the one that just states 5A/25mA does not have resistor?

With respect to reading Chinese, google has a translate app that can translate Chinese to English for Chinese phrases picked up by your phone camera.
I think it helped me to figure it out. The CT wanted 5 turns of primary. I am wondering if there is any difference between 100/5 CT with 5 turns of primary and 20/5 with 1 turn of primary? Is the first one more precise or is it just so they could manufacture the same unit for both cases?

Always remember that to measure voltage, the voltmeter always has two lead wires, and the answer depends on where you connect them. The dotted lines in the sketch below show where to put the voltmeter leads in this case.
Maybe what other options are there than two lead wires? I measured the output voltages the way you showed. That is why I was wondering why I didnt get high voltage out of CT. I expected to get high voltage according to formula that V_s = V_p *4 which didnt happen. I now measured the resistance of CT secondary. 0.0 - 0.1 Ohm which makes me think there is no resistor inside.

Btw does anyone know how bad is it for the accuracy when amp meter uses external 0.5 class CT? If amp meter has internal 5/2.5 mA CT and the secondary of 0.5 class CT is connected to primary of 0.5 class 5/2.5 mA CT that is inside amp meter, is the accuracy of entire thing much worse than it would be if it was single 20/2.5 mA CT instead of 2 CTs?
 
  • #16
anorlunda
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How many more times?
When I measured disconnected CT with multimeter it only showed few mV instead of 4*240V = 960V what I expected.
Maybe what other options are there than two lead wires? I measured the output voltages the way you showed. That is why I was wondering why I didnt get high voltage out of CT.
How many times indeed. A transformer has a primary coil and a secondary coil. What is the primary coil of a CT? One could argue that it is the entire closed circuit. But a 3D spatial view considering Maxwells equations would show that most of the flux lines intersect only the cm segment of wire penetrating the CT. That segment is where the tiny primary side EMF would be induced. A short segment of wire that would be considered a short circuit if the CT was not there.

So the primary side voltage is not 240V, it would be closer to 5 mv for a 20:1 ratio and 100 mv on the secondary.

This thread is an example of a circuit that cannot be correctly described with ordinary circuit analysis. You need 3D modeling with Maxwells equations to model the CT's effect on the primary circuit.
 
  • #17
Thanks, that explanation on how mV come was good. I just dont understand how the CT tutorial that I linked above got 76.8 kV and why some people on youtube warn against high voltage when CT is left open circuit? Are these cases just mistakes of tutorial makers or is there something that I may not have noticed when reading/watching?
 
  • #18
Averagesupernova
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Thanks, that explanation on how mV come was good. I just dont understand how the CT tutorial that I linked above got 76.8 kV and why some people on youtube warn against high voltage when CT is left open circuit? Are these cases just mistakes of tutorial makers or is there something that I may not have noticed when reading/watching?
A CT is designed to transform current. So, if you pass 1 amp through a current transformer that steps down the current to 20% the current transformer tries to drive 200mA into an infinite resistance if the secondary is left open. You should know ohms law well enough to realize how high the voltage will get across the secondary. Not sure how many ways there are to say this. A current transformer with a secondary winding left open is asking for trouble.
 
  • #19
anorlunda
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Especially if that burden resistor is your own body.

https://en.wikipedia.org/wiki/Electrical_injury said:
If the current has a direct pathway to the heart , a much lower current of less than 1 mA (AC or DC) can cause fibrillation.
 
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  • #20
arydberg
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Hi!

I have question about current transformers (CT). I have BH-0.66 class 0.5 20/5 current transformer. According to my understanding 20/5 means 4 times current reduction. For some reason I get about 20 times current reduction. Now I am curious if the factory managed to label 100/5 CT as 20/5 CT or if those integrated CTs have some specific needs in order to work correctly that I am not aware of.
I know that it measures about 20 times less current because I compared it with multimeter. And also tried clamp meter. At least with clamp meter I get correct reading easily. And polarity or cable angle doesnt seem to affect reading too much - maybe +/- 10%. But with CT looks like no matter how I wire it the reading is always about 20 times less, not 5 like I would expect. It happened with both small and large loads. And that is why I was wondering if there is something specific that I need to know about CTs in order to get correct reading that normal clamp meter hasn't require me to learn yet.

View attachment 265568

Also I got one more question when I was reading about CTs from interenet:
https://testguy.net/content/190-Current-Transformer-Basics-Understanding-Ratio-Polarity-and-Class

Here it says that

I was wondering if there is any reason why it should be important in my case where I only have single phase AC and one meter? At least the clamp meter doesn't seem to care.
Could you please tell us what exactely you are trying to do. My understanding of a current transformer does not fit most of what has been said here.
 
  • #21
Could you please tell us what exactely you are trying to do.
Just trying to understand how they work.

You should know ohms law well enough to realize how high the voltage will get across the secondary.
while someone else says this:

How many times indeed. A transformer has a primary coil and a secondary coil. What is the primary coil of a CT? One could argue that it is the entire closed circuit. But a 3D spatial view considering Maxwells equations would show that most of the flux lines intersect only the cm segment of wire penetrating the CT. That segment is where the tiny primary side EMF would be induced. A short segment of wire that would be considered a short circuit if the CT was not there.

So the primary side voltage is not 240V, it would be closer to 5 mv for a 20:1 ratio and 100 mv on the secondary.

To me it looks like contradiction between 2 views here. One says that leaving CT open ended generates dangerously high voltage and while other explains why it would stay in mV scale. And it's bit confusing.


Also an other question that came to my mind: If there are 2 CTs: 1) 400A/50mA (1:8000) and 2) that is 20A/2.5ma (1:8000). Can 1st CT be used in place of 2nd CT? I mean if both are 1:8000 then option nr 1 should be able to replace option nr 2 right?
 
  • #22
Averagesupernova
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A CT is designed to transform current. So, if you pass 1 amp through a current transformer that steps down the current to 20% the current transformer tries to drive 200mA into an infinite resistance if the secondary is left open. You should know ohms law well enough to realize how high the voltage will get across the secondary.
Didn't this cover it? Pass a current through a CT and compute the voltage on the output of the CT based on the ratio of the CT and the size of the burden resistor. If you are unable to do this math for whatever reason, I don't see how you can expect to understand it. So, let's go from there. Pick your hypothetical values and let's work through a hypothetical example.
 
  • #23
Averagesupernova
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Also an other question that came to my mind: If there are 2 CTs: 1) 400A/50mA (1:8000) and 2) that is 20A/2.5ma (1:8000). Can 1st CT be used in place of 2nd CT? I mean if both are 1:8000 then option nr 1 should be able to replace option nr 2 right?
Yes, but not necessarily the best choice. Do you routinely use a 100 ft long tape measure to measure lengths less than one inch?
 
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