CT Open Circuit 5

[charz]

I have read the old threads regarding the CT open circuit generating high voltages, but still I couldn't grasp it any easier. My questions are,
1. Why the acccuracy is high when the CT secondary is short circuited?
2. Why is the magnetic flux in a CT core is not constant (Varies with both open and short circuited CT secondary) as is the case with a VT?

I think mostly the answer to the question lies in the primary current of the CT not depending on the secondary of the CT? but still I'm not able to put it together.

 

[submonkey]

Hi

A CT behaves the same as pretty much any other transformer - the difference
is in how the CT is connected. Suppose you have a 2000/1 CT which is open
circuit and connected in series with an 11kV supply. The CT secondary is
open circuit, so the primary "tries" to behave like an open circuit, allowing
11kV / sqrt(3) = 6.35 kV to be developed across its primary winding (1 turn).
The secondary (2000 turns) "tries" to output 6.35 kV * 2000 = 12.7 Megavolts.

In reality, the core saturates long before 6.35 kV can be developed across
the primary, and the system voltage is instead dropped across the load, but
the CT will still output a high voltage (=N * d(phi)/dt) while the core moves
from forward saturation to reverse saturation each cycle.

Thanks,
Alan

 

[scottf]

submonkey-

The secondary open-circuit voltage of a CT is not at all a function of the system voltage on which it is applied.

In your example above, there is not 6.35kV across the primary winding of the CT. The CT's primary winding is in series with the line and there is very little voltage developed over the CT primary, basically just the primary current time the impedance of the primary winding, which is very low.

For a given core size and secondary winding design (number of turns), the secondary open-circuit voltage would be the same if the CT was applied on a 600V circuit or on a 500kV circuit.

While technically not complete, a simple way to think about secondary open circuit voltage is to think about the equivalent circuit of a transformer, which the magnetizing branch is in parallel with the "ideal transformer". For a CT, the primary is connected in series with the line and can be viewed as a constant current source. When there is a normal burden connected to the secondary of the CT, the current that flows through the ideal transformer is the primary current minus the excitation current, which flows through the magnetizing branch. When the burden connected to the secondary is removed, all of the primary current is "forced" through the magnetizing branch, which is a relatively high impedance. So you have high current through a high impedance, which develops a high voltage across the magnetizing branch, which is in parallel with the ideal transformer and, in turn, the secondary winding of the CT.

That is an overly simplistic explanation, but it illustrates that the secondary open circuit voltage of a CT is not a function of the system voltage on which it is applied.

 

[stevenal]

faq238-1189

 

[scottf]

That FAQ238 is patently wrong. Specifically (but not limited to) this statement:

"Zero current through the load means no voltage drop across the load. Since the sum of the voltages around the loop must equal zero, full system voltage must drop across the CT primary"

A core of cross section "x" with number of secondary turns "N" at a given primary current "Ip" will develop the same secondary open circuit voltage if inside a CT applied at 120V or a CT applied at 765kV. The system voltage has nothing to do with the secondary open circuit voltage.

It's wrong and potentially dangerous to say otherwise.

 

[charz]

In power transformer, if there is load in the secondary, the secondary mmf and the primary mmf cancels each other and in order to maintain the core flux constant,the primary compensates it by drawing from the primary.
But in the current transformer, if there is small load or (shorted) in the secondary, the secondary mmf and primary mmf cancels each other and there is a very small net mmf in the core and since the secondary does not draw any current but rather the current is forced in to the secondary circuit.
Why the acccuracy is high when the CT secondary is short circuited?
If the secondary is shorted, it means there is no load at the secondary of the CT. In other words,only the primary current has to be reflected according to the turns ratio, if there are small loads like relays and meters, it also would draw some small current in addition to the reflected primary current, thus a small error.

Is my understanding right?

 

[scottf]

Charz-

You have to remember that the primary winding of the CT is in series with the line, therefore, you model is like a constant current source.

CT "error" is primarily a result of the excitation current needed to excite the core. When there is a short on the secondary, there is essentially 0 volts across the secondary, therefore the excitation current is very low. The CT does not have to "drive" any voltage on the secondary.

 

[davidbeach]

In very simple terms, consider the CT as an amp-turns device. The primary is connected in series with the load, but has no appreciable impact on load current. That load current, times the number of turns gives the amp-turns into the core. Now those amp-turns have to come out of the core, that's what the secondary is for. With the secondary shorted, there's a really easy place for those amp-turns to go, around that very low impedance loop. As the impedance (burden) of that loop increases the voltage across the burden, and thus the voltage across the CT secondary, increases. As that voltage increases more and more of the current gets pushed through the much higher impedance magnetizing branch (the non-ideal part of the transformer). The maximum open circuit voltage is related to the saturation voltage of the CT; a CT with a high saturation voltage, say a C800 CT, will produce a much higher open circuit voltage than will a CT with a lower saturation voltage. A C800 with a given current on the primary and a 120V primary circuit will produce a much higher open circuit voltage on the secondary than will a C100 of the same ratio on a much higher voltage circuit.

 

[stevenal]

The FAQ isn't wrong, it just uses a very simple model to illustrate the concept. This is explained in the last paragraph.

The FAQ was a response to similar discussions in the past that got too deep too quickly. I say begin with the simple ideal model prior to introducing non-ideal aspects such as the magnetizing branch.

 

[scottf]

Stevenal, not to seem argumentative, but the FAQ is indeed incorrect. The statement that the secondary open-circuit voltage is a function of the primary system voltage is wrong and the method used to reach that conclusion is wrong. I don't see how using the wrong explanation in the name of simplicity makes any sense. It surely doesn't make it easier to understand the concept in this case.

 

[stevenal]

The open-circuit secondary reliance on system voltage can be demonstrated easily enough. Take the system voltage to zero. The open circuit CT voltage will follow.

So David considered what I called the shunt admittance in the FAQ and what he called the magnetizing branch. Guess it's only wrong when I suggest it. Without having a C800 excitation curve handy, I'm guessing the voltage at the extreme right hand will be about 3000 or so. Assuming a ratio of say 1200:5, the voltage drop across the primary would be about 12.5 V. I would suggest you need a system somewhere in excess of 12.5 V line to ground to support that voltage drop. You will of course need a voltage sufficient to support the voltage drop across the load as well. This analysis doesn't consider frequency response of the saturated CT or hysteresis. All models have their limitations, but it doesn't make them wrong.

The point of the FAQ is to show that CTs are just transformers and share the same models. While CTs are designed to do what they do well, you would also have high voltages develop across the open leads of any transformer connected in series with a load when the higher number of turns is on the open side. It's not the device, it's the connection.

Yes, it makes it easier to understand. I learned the ideal transformer model first before that model was incorporated into the more complex models.

 

[scottf]

The open-circuit voltage is only a function of primary current and not system voltage. In your example above, the only reason the open-circuit voltage goes to zero when the system voltage goes to zero is that presumably no current would be flowing in the primary.

Since the open circuit voltage would be the same with 3000A in the primary with a system voltage of 1V or 800kV, we can say that the open circuit voltage is definitely NOT a function of the system voltage.

 

[HamburgerHelper]

Can't you get capacitive coupling to bring the voltage very high if the secondary is not grounded? This is separate from open ct.

 

[scottf]

Yes...there can be capacitive coupling that can cause a potential rise of the entire secondary winding...which is the reason a CT's secondary should always have a ground reference. The capacitive coupling certainly IS a function of the system voltage and CT design, but as you noted, it's not related to the secondary open-circuit voltage.

 

[waross]

Look at a power transformer. It has a saturation voltage limit. As you raise the applied voltage the secondary voltage rises, until the transformer saturates.
The applied voltage to the primary of a CT is not related to the system voltage but is the primary voltage drop across the primary of the CT. This is very low. As the resistance of the secondary increases the secondary voltage increases and the voltage drop across the primary increases until the transformer saturates. As the resistance increases further the transformer starts to act as an air core transformer. Not a very effective mode of operation, but with an open circuit the voltage may go somewhat beyond the saturation level. The limit may be related to the insulation resistance of the transformer secondary winding.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[scottf]

waross-

That is correct, except that with an open-circuit on the secondary, the secondary peak voltage goes well above the saturation voltage. For example, for a C800 rated CT, the knee-point voltage is in the range of ~680V, but the secondary open-circuit peak voltage could be in the range of 50kVp with the wave-form being highly distorted and the rms voltage possible reading much lower.

 

[submonkey]

Hi Scottf,

If the CT, the load, and the system were ideal (which they are not), the
output voltage of the CT would be proportional to the system voltage.

The primary of the CT would behave as an open circuit, because the CT
would have a infinite magnetising impedance and all of the system volts
would be dropped across the CT primary (because the CT magnetising impedance
would be infinite and the load impedance would be finite).

In practice, this does not happen because an open circuit CT saturates, and
the non-ideal properties of the CT dominate the behaviour (actual output
voltage determined by the core parameters). The principles described in my
explanation are still completely correct, and they need to be understood
before the non-ideal behaviours are covered.

Thanks,
Alan

 

[scottf]

Alan-

As described above, the voltage across the CT's primary terminals is NOT the system voltage nor is it related to the system voltage. The voltage across the CT primary terminals is a function of the primary current and the primary winding impedance (which is normally very very low).

The system voltage could be 15kV, 69kV, or 500kV and the voltage across the primary terminals of a given CT would not change.

 

[waross]

Hi scottf;
I realise that the voltage is above the saturation voltage.
I understand that there is a parallel with an iron core reactor. Above the saturation current the reactor acts as an air core reactor to limit further increases in current.
The CT core acts as a reactor. Above saturation current it acts as an air core reactor. Now add the secondary winding. We have an air core transformer. Not very effective and not able to drive an appreciable burden above saturation voltage.
The open circuit voltage is not related to system voltage but is limited by the burden represented by the load caused by losses through the resistance of the winding insulation.
If my understanding is faulty, please correct me.
Yours
Bill

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

In the past, I've used a typical C10 75:5 3" window CT installed on a wire loop. The wire loop was powered using another 200:5 15kV rated bar CT. The experiment was the initial testing to eventually build a 15kV isolated power supply with multiple CT's on the wire loop. It was fairly easy to make these CT's work just like voltage transformers. I simply put the correct input voltage into the 200:5 CT which in turn produced the correct voltage across the 1-turn primary loop passing through the 75:5 CT and then the 75:5 CT in turn produced the desired output voltage.

Try this as an experiment. Get yourself 2 matching clamp-on ammeters. Put 1 meter around 1 wire of a loaded parallel set of wires and observe the current. Then, clamp the second meter around the other wire and watch the current rise on the first meter.

Point of the above being that a CT certainly does require a certain voltage drop across the 1-turn primary to develop the secondary voltage. The argument that the CT secondary voltage is not a function of the system voltage being flogged over and over again is getting tiresome. No-one is suggesting that the CT secondary voltage is a direct mathematical relation to system voltage. But, multiple people have correctly posted that the CT 1-turn primary circuit requires a certain voltage - NOT the system voltage, just a certain voltage - across it to produce saturated secondary voltage levels.

Of course, the above agreement that the CT secondary voltage is not a function of the system voltage is based on the CT being used in a typical manner. If you did control the 1-turn primary voltage down to a low enough level, the CT secondary voltage certainly does become a function of the primary or supply voltage.

This means I also have to call BS on the argument that a system voltage of 1V will produce the same open circuit voltage as a higher system voltage. 1V is simply not capable of supplying the required CT primary circuit voltage required to drive the CT to saturation. It you could actually create a 1V, 3000A circuit, the current would significantly drop when you installed an open secondary 3000:5 CT and the CT would start operating as a voltage transformer.