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CT voltage 7
- Thread starter deltawhy
- Start date
Consider the secondary circuit - how much voltage does the CT have to produce to create the needed current to satisfy the turns ratio and balance the flux.
Now imagine the the resistance of the secondary circuit is slowly increased. The CT must produce increasing secondary voltage to create the same current. An open circuit on the secondary corresponds to the ultimate burden, so the CT secondary voltage will increase to its ultimate level to try to push the secondary current through the open circuit.
Now imagine the the resistance of the secondary circuit is slowly increased. The CT must produce increasing secondary voltage to create the same current. An open circuit on the secondary corresponds to the ultimate burden, so the CT secondary voltage will increase to its ultimate level to try to push the secondary current through the open circuit.
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5
- #4
That FAQ is fundamentally wrong.
The open-circuit voltage of a CT has absolutely nothing to do with the system voltage of the primary.
For a given core/secondary winding design, the secondary open-circuit voltage is the same, whether the CT is on a 120V circuit or a 765kV system.
The voltage developed on the secondary of an open-circuited CT is a function of the current flowing in the primary and the inductance properties of the secondary windings...which are a function of the core cross-sectional area, core material properties, and number of secondary turns.
For a simple explanation of open-circuit voltage of a CT, think of it this way:
Think of a CT trying to act as a constant current source. Since the primary winding of a CT is in series with the system, that primary current is not going to change appreciably regardless of what is connected (or not connected) to the CT's secondary.
Consider the transformer equivalent circuit model. Under normal circumstances, some current is lost to the ideal transformer based on excitation current flowing though the magnetizing branch of the circuit which is in parallel with the ideal transformer/secondary of the CT. The voltage developed across the secondary is a the excitation current times the magnetizing impedance. When the CT secondary is shorted or if there is a reasonable burden placed across it, the magnetizing impedance is much greater than the burden and most of the current flows through the ideal transform (transformed to the secondary by the turns ratio). If the secondary is open-circuited, that is seen as an infinite impedance, so all of the primary current is forced through the magnetizing branch. The voltage seen on the secondary is this primary current times the magnetizing impedance.
Note that this impedance changes as the core saturates and the resulting wave-form is very distorted. The peak voltages can be very high (50 kVp or so for large protection rated CTs) but the measured rms voltage would read much lower. People have been killed by CT open circuit voltage and people have been tricked in thinking the voltage is at safe levels by using rms volt-meters.
The open-circuit voltage of a CT has absolutely nothing to do with the system voltage of the primary.
For a given core/secondary winding design, the secondary open-circuit voltage is the same, whether the CT is on a 120V circuit or a 765kV system.
The voltage developed on the secondary of an open-circuited CT is a function of the current flowing in the primary and the inductance properties of the secondary windings...which are a function of the core cross-sectional area, core material properties, and number of secondary turns.
For a simple explanation of open-circuit voltage of a CT, think of it this way:
Think of a CT trying to act as a constant current source. Since the primary winding of a CT is in series with the system, that primary current is not going to change appreciably regardless of what is connected (or not connected) to the CT's secondary.
Consider the transformer equivalent circuit model. Under normal circumstances, some current is lost to the ideal transformer based on excitation current flowing though the magnetizing branch of the circuit which is in parallel with the ideal transformer/secondary of the CT. The voltage developed across the secondary is a the excitation current times the magnetizing impedance. When the CT secondary is shorted or if there is a reasonable burden placed across it, the magnetizing impedance is much greater than the burden and most of the current flows through the ideal transform (transformed to the secondary by the turns ratio). If the secondary is open-circuited, that is seen as an infinite impedance, so all of the primary current is forced through the magnetizing branch. The voltage seen on the secondary is this primary current times the magnetizing impedance.
Note that this impedance changes as the core saturates and the resulting wave-form is very distorted. The peak voltages can be very high (50 kVp or so for large protection rated CTs) but the measured rms voltage would read much lower. People have been killed by CT open circuit voltage and people have been tricked in thinking the voltage is at safe levels by using rms volt-meters.
Let me put it in a different way;
In transformers, primary ampereturn should be compensated by secondary ampereturns.In power transformers and voltage transformers,with open circuited secondary, the primary current is the negligible excitation current and that is the price that we pay for transferring voltage from primary to secondary.When secondary is shorted or loaded secondary AT will be equal to primary AT-excitation AT.
With CT primary, being in series with supply voltage( in VT it is parallel to supply)the primary current is the load current and when secondary is open circuited, the primary AT will generate high voltage on secondary proportional to primary current.When secondary is loaded, primarY AT will be same as secondary AT.
In transformers, primary ampereturn should be compensated by secondary ampereturns.In power transformers and voltage transformers,with open circuited secondary, the primary current is the negligible excitation current and that is the price that we pay for transferring voltage from primary to secondary.When secondary is shorted or loaded secondary AT will be equal to primary AT-excitation AT.
With CT primary, being in series with supply voltage( in VT it is parallel to supply)the primary current is the load current and when secondary is open circuited, the primary AT will generate high voltage on secondary proportional to primary current.When secondary is loaded, primarY AT will be same as secondary AT.
- Moderator
- #6
A CT follows all the transformer laws UNTIL THE TRANSFORMER SATURATES. When the CT saturates, the secondary voltage of many CTs will be at a lethal level. An open circuit secondary will allow the CT to saturate at a fairly low primary current. The knee point voltage given in the transformer specs is an indication of the secondary voltage at or near the saturation point.
As for a multi-meter indicating RMS voltage, many multi-meters will not survive connection to an open circuited CT with normal primary current flowing.
Bill
--------------------
"Why not the best?"
Jimmy Carter
As for a multi-meter indicating RMS voltage, many multi-meters will not survive connection to an open circuited CT with normal primary current flowing.
Bill
--------------------
"Why not the best?"
Jimmy Carter
waross-
While what you said is basically correct, the saturation voltage is not directly comparable to the open circuit voltage. A CT with a knee-point voltage of 800V (as an example) could likely have an open-circuit voltage of 50 kVp.
I'm looking at a CT design now for a bid I'm working on...the Vkp is approx. 400V, but the calculated open-circuit voltage is 32 kVp.
While what you said is basically correct, the saturation voltage is not directly comparable to the open circuit voltage. A CT with a knee-point voltage of 800V (as an example) could likely have an open-circuit voltage of 50 kVp.
I'm looking at a CT design now for a bid I'm working on...the Vkp is approx. 400V, but the calculated open-circuit voltage is 32 kVp.
I believe that the closer approach to answer the question comes from “dpc” in his 16 Mar 12 10:51 post.
The equation that explains the principles of a CT is the Ampere’s law (generalized by Maxwell). It states the mutual association between the current that cross a surface and the magnetic field across the perimeter of that surface. Because the relation is direct between the current and the magnetic field, and because this is a law of physics, the voltage on the secondary (the CT winding itself) will adapt to maintain the current associated to the given magnetic field.
So, if the impedance is minimum (the CT winding is short-circuited), then the voltage necessary to maintain the current associated to the magnetic field is very low. This would just be by Ohm’s law.
Try to be perfect. You'll never make it, but you will be very, very good.
The equation that explains the principles of a CT is the Ampere’s law (generalized by Maxwell). It states the mutual association between the current that cross a surface and the magnetic field across the perimeter of that surface. Because the relation is direct between the current and the magnetic field, and because this is a law of physics, the voltage on the secondary (the CT winding itself) will adapt to maintain the current associated to the given magnetic field.
So, if the impedance is minimum (the CT winding is short-circuited), then the voltage necessary to maintain the current associated to the magnetic field is very low. This would just be by Ohm’s law.
Try to be perfect. You'll never make it, but you will be very, very good.
Please look at the FAQ again. It focuses on the simplest ideal transformer model, where the voltage across the secondary has everything to do with the voltage across the primary. Saturation and other non-ideal effects only lesson the voltage developed. The FAQ is not fundamentally wrong, just the opposite. It focuses on the fundamentals without delving into things like core properties.
The FAQ was written following a similar discussion where members immediately began talking about all the complexities following a question that could be answered simply with Ampere's and Ohm's laws.
The ideal model has its limitations, discussed in the final paragraph. This, however, does not make it wrong.
Perhaps I will enhance the FAQ with a diagram.
I have yet to receive a single comment on the FAQ.
The FAQ was written following a similar discussion where members immediately began talking about all the complexities following a question that could be answered simply with Ampere's and Ohm's laws.
The ideal model has its limitations, discussed in the final paragraph. This, however, does not make it wrong.
Perhaps I will enhance the FAQ with a diagram.
I have yet to receive a single comment on the FAQ.
stevenal-
Not to keep splitting hairs, but the voltage on the secondary of a transformer is not a direct result of the voltage on the primary, but rather the current through the primary winding (which is why we speak to ampere-turns). Whereas the current in the primary induces a flux in the core and the flux in the core induces a voltage (emf) on the secondary. The ideal transformer models typically used show both primary voltage AND primary current and some just use primary current for this very reason.
We can agree to disagree, but I stick by my opinion that the FAQ is fundamentally wrong, in that it gives the reader the impression that the open-circuit voltage of a CT is at all related to the system voltage of the primary. This can be a dangerous mistake to make.
Not to keep splitting hairs, but the voltage on the secondary of a transformer is not a direct result of the voltage on the primary, but rather the current through the primary winding (which is why we speak to ampere-turns). Whereas the current in the primary induces a flux in the core and the flux in the core induces a voltage (emf) on the secondary. The ideal transformer models typically used show both primary voltage AND primary current and some just use primary current for this very reason.
We can agree to disagree, but I stick by my opinion that the FAQ is fundamentally wrong, in that it gives the reader the impression that the open-circuit voltage of a CT is at all related to the system voltage of the primary. This can be a dangerous mistake to make.
electricpete
Electrical
The bolded part is a little off-base imo (and the stuff that results from following the implications of that statement).FAQ said:
Opening secondary of CT is not going to bring the primary current to zero. I think you’ll quickly realize this if you think about it.
Scott gave a good explanation imo.
=====================================
(2B)+(2B)' ?
electricpete
Electrical
I obviously (?) meant to bold the statement that there will also be zero current in the primary.
Now I see Scott already responded. I didn't see that when I posted.
=====================================
(2B)+(2B)' ?
Now I see Scott already responded. I didn't see that when I posted.
=====================================
(2B)+(2B)' ?
Did anyone even read the the last paragraph of the FAQ?
The ideal transformer model will have zero and zero, the real transformer will not. This does not make the model wrong, it is simply a first approximation. I'm not yet ready to throw out the ideal gas law either, even though it models gases that don't exist in nature.
I've no problem with adding the complexities when needed, but the question can be answered with simple circuit analysis.
Funny that folks apparently have no problem with zero and zero current on the windings of a VT or power transformer model. Same model can be used for CTs.
The ideal transformer model will have zero and zero, the real transformer will not. This does not make the model wrong, it is simply a first approximation. I'm not yet ready to throw out the ideal gas law either, even though it models gases that don't exist in nature.
I've no problem with adding the complexities when needed, but the question can be answered with simple circuit analysis.
Funny that folks apparently have no problem with zero and zero current on the windings of a VT or power transformer model. Same model can be used for CTs.
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1
- #15
So here's the diagram I promised.
It is true a real transformer will not function without current. The idealized version works very well without current and illustrates that the main reason an open circuit CT develops high voltage while a VT doesn't is that the primary is connected differently. This very fundamental difference is the first thing missing in the knowledge of those asking the question.
Throw your stones, but read through to the very end of the FAQ first please.
It is true a real transformer will not function without current. The idealized version works very well without current and illustrates that the main reason an open circuit CT develops high voltage while a VT doesn't is that the primary is connected differently. This very fundamental difference is the first thing missing in the knowledge of those asking the question.
Throw your stones, but read through to the very end of the FAQ first please.
Stevenal-
A few comments:
- In my opinion, the "ideal transformer" is best used as PART of an equivalent circuit and not on its own. You state it's based on an "ideal transformer model", but the ideal transformer isn't a "model" at all, but rather part of a model.
- I think it's very dangerous to give the impression that the open-circuit voltage of a CT is in any way related to the system voltage across the primary. (You state that the secondary voltage is the system voltage stepped up by the turns ratio.
)
It's fundamentally false and could lead to a false sense of security. For instance, someone opens a CT on a 5V system, thinking the open-circuit voltage is 5V or thereabouts. In reality, it turns out to be 5 kV and someone gets killed or injured.
The bottom line point is that I think you're using an ideal transformer alone to explain something incorrectly, which leads to the reader drawing a false and dangerous conclusion.
A few comments:
- In my opinion, the "ideal transformer" is best used as PART of an equivalent circuit and not on its own. You state it's based on an "ideal transformer model", but the ideal transformer isn't a "model" at all, but rather part of a model.
- I think it's very dangerous to give the impression that the open-circuit voltage of a CT is in any way related to the system voltage across the primary. (You state that the secondary voltage is the system voltage stepped up by the turns ratio.
)
It's fundamentally false and could lead to a false sense of security. For instance, someone opens a CT on a 5V system, thinking the open-circuit voltage is 5V or thereabouts. In reality, it turns out to be 5 kV and someone gets killed or injured.
The bottom line point is that I think you're using an ideal transformer alone to explain something incorrectly, which leads to the reader drawing a false and dangerous conclusion.
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1
- #17
As a suggestion, it isn't simply the model of the transformer which you need to consider, but also how you model the source.
A VT is usually connected to a much larger system whose properties equate reasonably well to an ideal voltage source because the VT has very little influence on the source regardless of how you connect it.
Conversely a CT connected to much larger system will be driven by something resembling a current source. One of the key things about a current source is that the current is independent of the burden.
I do see what stevenal is trying to say, which (I think) is that a CT on a conductor does cause a voltage drop in the primary, and when the CT is connected to a burden then the volt drop in the primary has the usual turns-ratio relationship between the voltage across the burden (relay input or whatever) and the primary. It's just too small to be of any practical consequence. The simple 'series R-L network plus parallel R-L network plus ideal transformer' model works pretty well in the linear region as long as the core doesn't saturate. The simple model breaks down when the flux density in the core causes the transformer to operate its non-linear region, because the turns-ratio relationship no longer holds true.
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If we learn from our mistakes I'm getting a great education!
A VT is usually connected to a much larger system whose properties equate reasonably well to an ideal voltage source because the VT has very little influence on the source regardless of how you connect it.
Conversely a CT connected to much larger system will be driven by something resembling a current source. One of the key things about a current source is that the current is independent of the burden.
I do see what stevenal is trying to say, which (I think) is that a CT on a conductor does cause a voltage drop in the primary, and when the CT is connected to a burden then the volt drop in the primary has the usual turns-ratio relationship between the voltage across the burden (relay input or whatever) and the primary. It's just too small to be of any practical consequence. The simple 'series R-L network plus parallel R-L network plus ideal transformer' model works pretty well in the linear region as long as the core doesn't saturate. The simple model breaks down when the flux density in the core causes the transformer to operate its non-linear region, because the turns-ratio relationship no longer holds true.
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If we learn from our mistakes I'm getting a great education!
- Moderator
- #18
How about considering the voltage drop across the primary times the turns ratio.
Up to saturation the voltage drop across the primary times the turns ratio should give the open circuit voltage. Much past saturation/knee point, the transformer acts as an air core reactor and the voltage drop is the current times the impedance. The voltage drop above the knee point is still a straight line but at a different slope. The increase in primary voltage drop as a function of the increase in primary current is much less than it is below the knee point.
The voltage drop across the CT primary at normal loads may easily be higher than the knee point but not much as the dynamic impedance is much less above saturation than below saturation.
I will leave it to someone else to comment on how much above saturation and how much primary voltage drop a normal load current will cause.
Let's not discuss fault conditions just now. Most issues with open secondary CTs are associated with normal load currents, not fault currents.
Bill
--------------------
"Why not the best?"
Jimmy Carter
Up to saturation the voltage drop across the primary times the turns ratio should give the open circuit voltage. Much past saturation/knee point, the transformer acts as an air core reactor and the voltage drop is the current times the impedance. The voltage drop above the knee point is still a straight line but at a different slope. The increase in primary voltage drop as a function of the increase in primary current is much less than it is below the knee point.
The voltage drop across the CT primary at normal loads may easily be higher than the knee point but not much as the dynamic impedance is much less above saturation than below saturation.
I will leave it to someone else to comment on how much above saturation and how much primary voltage drop a normal load current will cause.
Let's not discuss fault conditions just now. Most issues with open secondary CTs are associated with normal load currents, not fault currents.
Bill
--------------------
"Why not the best?"
Jimmy Carter
bacon4life
Electrical
What is the highest voltage that might be produced with a system voltage of 765 KV and a 1000:5 turns ratio transformer? 153,000 kV
How about 120V? 24 kV
Although the first answer is clearly not physically possible, I don't why either is a particularly dangerous conclusion. The ideal transformer model answers why transformer secondaries normally have a low voltage when they have a small burden and why the voltage is very large when they are opened. Nothing else is needed to explain why the transformer could cause some havoc.
In the second example at 120V, the CT could not reach the typical open circuit values Scottf mentioned.
All of the complications such as flux, saturation, complete transformer models and Ampere law are irrelevant to the op's question. Those complications go into answering exactly how high the voltage will get before something saturates, fails, or flashes over.
I find thinking about the secondary voltage as the primary voltage times the turn ratio just like stevenal has drawn is very useful. Labeling something a CT or a PT initially trapped me into only considering them in their normal application and just memorizing the rules of thumb like "don't open a CT". Once I started picturing each one as a transformer model, it became intuitive why PTs are dangerous when back-fed, why CTs should usually be shorted, and why CT's could be open circuited during ratio testing.
How about 120V? 24 kV
Although the first answer is clearly not physically possible, I don't why either is a particularly dangerous conclusion. The ideal transformer model answers why transformer secondaries normally have a low voltage when they have a small burden and why the voltage is very large when they are opened. Nothing else is needed to explain why the transformer could cause some havoc.
In the second example at 120V, the CT could not reach the typical open circuit values Scottf mentioned.
All of the complications such as flux, saturation, complete transformer models and Ampere law are irrelevant to the op's question. Those complications go into answering exactly how high the voltage will get before something saturates, fails, or flashes over.
I find thinking about the secondary voltage as the primary voltage times the turn ratio just like stevenal has drawn is very useful. Labeling something a CT or a PT initially trapped me into only considering them in their normal application and just memorizing the rules of thumb like "don't open a CT". Once I started picturing each one as a transformer model, it became intuitive why PTs are dangerous when back-fed, why CTs should usually be shorted, and why CT's could be open circuited during ratio testing.
bacon4life...you're kind of proving my point for me.
The open-circuit voltage of a CT has absolutely nothing to do with the system voltage of the system the CT is being applied to.
A CT on a 120V system most certainly can reach 24 kVp open-circuit voltage (and higher).
A CT's primary winding is in series with the system and not across the system. The open circuit voltage produced by a CT is a function of the primary current level and the magnetic properties of the CT (number of secondary turns, core size, core material) and nothing else.
The same physical CT applied on a 120V system and a 765 kV system will have exactly the same open-circuit voltage.
Waross-
Thinking about the voltage drop across the actual CT's primary is a very difficult concept. Consider a head-type high-voltage CT, with a 2000:5A ratio, RF2.0, C800. That CT will have a primary conductor of approx. 1m in length and about 100mm diameter solid aluminum. The voltage drop across that is in the mV/micro-voltage range. During a transient, when the voltage rises to 800V on the secondary, there will not be 2V (800V / 400) across the actual primary winding.
The open-circuit voltage of a CT has absolutely nothing to do with the system voltage of the system the CT is being applied to.
A CT on a 120V system most certainly can reach 24 kVp open-circuit voltage (and higher).
A CT's primary winding is in series with the system and not across the system. The open circuit voltage produced by a CT is a function of the primary current level and the magnetic properties of the CT (number of secondary turns, core size, core material) and nothing else.
The same physical CT applied on a 120V system and a 765 kV system will have exactly the same open-circuit voltage.
Waross-
Thinking about the voltage drop across the actual CT's primary is a very difficult concept. Consider a head-type high-voltage CT, with a 2000:5A ratio, RF2.0, C800. That CT will have a primary conductor of approx. 1m in length and about 100mm diameter solid aluminum. The voltage drop across that is in the mV/micro-voltage range. During a transient, when the voltage rises to 800V on the secondary, there will not be 2V (800V / 400) across the actual primary winding.
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