Suggestion to Breaker (Electrical) Mar 3, 2004 marked ///\\Why does leaving a CT secondary open while passing primary load current, cause high voltages in the CT secondary windings and wiring?
///Visit
for:
Chapter 7 Current Transformers
OVERVOLTAGE IN SATURATED CT SECONDARIES
Although the rms magnitude of voltage induced in a CT secondary is limited by core saturation, very high voltage peaks can occur. Such high voltages are possible if the CT
burden impedance is high (e.g. open circuit on the secondary), and if the primary current is many times the CTCs continuous rating.\\
"I don't know what you mean by saying: "Put an open circuit secondary power or voltage transformer in series with a load, and you will nearly disconnect the load.""
Try drawing it out: source connected to VT H1. VT H2 connected to load (two bushing VT required). Load return connected back to source. VT X1 and X2 disconnected. Replace VT primary with an open switch and the load will not see much difference.
When I speak of load I am refering to the load on the primary system like motors and lights.
"Neither is the primary winding in a CT connected any differently than any other transformer, there is simply just one turn."
I don't know how you do it, but I usually connect power transformers and VT primaries across the line while CT primaries are connected in series with the line. Seems to work better that way.
stevenal, when you say 'Put an open circuit secondary power or voltage transformer in series with a load'. It sounds like you are connecting the open circuit secondary in series with the load. if you mean 'connect the primary winding in series with a load', why mention the open circuited secondary? Surely you realize that the transformer primary, CT or PT, is a load? 'nearly disconnect the load'? Do you mean that the voltage across the load will be reduced because of the series circuit?
Obviously, the external primary circuit load current flows through the primary winding of the CT. If there is no primary current flow in the CT, then there is no voltage drop across the CT due to its burden. If you can imagine the CT burden on the primary winding as a voltage drop, then you should understand how the simple transformer model can apply to both a CT or a PT.
Maybe you can understand it this way: Of course, if you open a switch in a circuit with a load(and a voltage applied), the voltage will drop across the open switch, and none will drop across the load, because no current is flowing through it. If you then close the switch and allow current to flow through the load, the voltage drop will then be across the load, because current is now flowing through it. I'm not sure how you get: "By that line of reasoning there can be no voltage drop across an open switch, nor in an open CT secondary" from that.
Let me simplify further: The CT imposes a burden(by induction) on the primary circuit. This burden is a load, however, it has a very low impedance, which changes based on the load(burden) imposed on the secondary of the CT(meter, relay, etc). The current flow in the primary circuit is based on the total impedance of the circuit, including the normal primary curcuit load(motors, lights, etc). The CT primary load is very small compared to the normal circuit load. The voltage drop in the primary circuit is mostly across this load, but a small amount of voltage is dropped across the CT primary. This provides the power(IxV) to drive current(or spike voltage) in the CT secondary circuit.
If there is no current flow, there can be no voltage drop across the impedance, the same as any other load.
Comment on ScottyUK (Electrical) Mar 3, 2004 marked ///\\A basic principle of any transformer is ampere-turn balance. Thus, primary amps x primary turns = secondary amps x secondary turns.
///Agree\\By opening the secondary with the primary energised, you are upsetting the ampere-turn balance relationship.
///Not true. See Reference:
Gordon R. Slemon "Magnetoelectric Devices Transducers, Transformers, and Machines," John Wiley & Sons, Inc., 1966,
page 201, Figure 3.17 a) Current Transformer and b) Equivalent Circuit,
where the ampere-turn balance still exists; however, it is changed due to the burden circuit branch being open.
Clearly, Fig. 3.17 b) indicates a current I1'=N1xI1/N2 flowing through the CT transformer secondary winding to the point where it branches to the CT burden branch and to the CT magnetization branch with magnetization admittance Y1m. If the burden circuit is open, then the current I1'=N1xI1/N2 flows through Y1m only causing the high voltage on the CT secondary dependent on complex number value of Y1m.\\ In order to maintain ampere-turn balance, the secondary voltage rises until insulation breaks down allowing current to flow and restore balance,
///Why would the insulation break down if it happens to be rated higher than the secondary CT open circuited voltage?\\ or until the core saturates.
///Yes, the core usually saturates with high voltage across the Y1m branch due current I1' being high through Y1m.\\ Saturation of the iron core keeps the voltage well below its theoretical limit,
///Please, clarify the "theoretical limit."\\ although a big CT will easily develop several kV open circuit, more than enough to kill you.
///Yes, certainly it could.\\\
1) Insulation breakdown:
A CT secondary is usually constructed with relatively light insulation, unlike the primary which is obviously constructed for service on the line. If it is open circuit and developing several kV across the terminals, the voltage stress between turns and to the core can become excessive, leading to insulation failure. I'm not an expert on breaking CT's - perhaps ScottF as a transformer manufacturer can elaborate on the failure modes?
2) 'Theoretical limit'
An open-circuit CT without the real-world restrictions of core saturation would develop a voltage large enough to break down the insulating material between the secondary terminals. Depending on the insulating material, this could in theory approach infinite voltage.
Re. Ampere-turn balance not being upset: it depends on the model for the CT that you use. If you consider the actual CT behaviour, the physical primary is carrying current and the physical secondary is not. Thus the ampere-turn balance is indeed upset. The model provides a simplified means of understanding what is going on in the CT - it is not perfect in all circumstances.
Comment: Often, CTs with a defective shorting switch on the CT secondary do not result in the CT being decommissioned because of damaged insulation. Had it been so, many CT manufacturers would be very busy and very rich.
1) The failure mode depands a bit on the type and design of CT. Generally, damage is done to the secondary terminal strips, as they are typically rated for 600 V continuous. Turn-to-turn failures are rare in CTs and are mainly a function of the winding design. IEEE multi-ratio designs which have each segment equally distributed may be more prone to turn-to-turn faiures than other designs. Shorts to the core are even rarer, at least those caused by overvoltages.
Not to beat a dead horse here, but as for the modelling and simple explanations above (not neccessarily your's), the best way to model a CT is to treat the primary as a constant current source. The voltage drop across the primary winding discussed above, really has nothing to do with the open circuit voltage discussion. The open circuit voltage is only a product of the fact that when the secondary is open, all the primary current is forced through the magnetizing branch, which is a high impedance. High current, through a high impedance gives a high voltage that is capped by the core saturation characteristics. The primary winding impedance, in all practicality, has nothing to do with it.
Comment: The CT ratio error in % or Ratio Correction Factor (RCF)
%error=100x(NxIs-Ip)/Ip
actually suggests how much of current flows in the magnetization branch. The smaller the error, the higher open CT voltage will be.
