Open Circuit a Current Transformer

Dhamdev... Further to my earlier comment, the application of 10% primary current the applied ampere-turns (AT) is about 30-45 times the nominal exciting-current AT.

Phil Corso
 
S
To get the answer for this let us understand the working principal of power transformer first. If you check the power transformer, when you apply voltage initially currents are very high (Magnetising current), but as soon as core get magnetised counter EMF induced in secondary winding which will oppose the primary EMF.

In power transformer counter EMF in secondary winding is playing role in controlling primary current and complete supply voltage to appear across primary winding and power transformers are designed for Higher VA rating.

Let us imagine the same power transformer without Core and secondary winding, now if you energise the transformer it will become short circuit and short circuit current will flow through primary winding, why because there is no Counter EMF induced in secondary winding which can oppose primary.

If you talk about CT it works same like transformer but it has designed with very small VA rating hence it cannot control primary current because of weaker counter EMF induced in secondary winding. and if you keep open secondary winding there is no current to flow and voltage will rise (Counter EMF) which will try to counter primary EMF but its is weaker as per design hence it will end with insulation failure or some time CT may blast.

As per my knowledge the basic working principle of Power transformer and C.T are the same, but they are designed as per the applications.
This is just my opinion, it may not be correct always welcome for the discussion which can bring better result.

Thank you
Sunil Kumar
 
C
Conventional xformer theory explains and fully predicts why large secondary voltages are generated if the secondary winding of a current xformer is open circuited.

A good place to start is to consider the properties of an 'ideal' xformer, noting that there is no fundamental difference in the operating theory of voltage and current xformers.

As explained in any textbook, an ideal xformer has infinitely high winding inductance, zero winding resistance, perfect magnetic coupling between windings, and 100% efficiency. The impedance transformation ratio is equal to the square of the turns ratio. At all times the primary amp-turns is equal and opposite to the secondary amp-turns, so that the net amp-turns is always zero. This is all true for a 'perfect' xformer, and a high quality real xformer will closely approximate this behavior.

In an ideal voltage xformer, we apply a fixed voltage to the primary winding, and if the secondary is unloaded, then the primary current is zero because the primary inductance is infinite. No nasty surprises, unless we short-circuit the secondary, resulting in very large winding currents.

Now let's look at so-called current xformers, noting that an xformer is an xformer, and all of the ideal properties are identical. Normally with a current xformer, a resistive load is connected across the secondary, and this resistance is 'seen' on the primary side, multiplied by the square of the turns ratio. For example, if Nsec/Npri=100, and the secondary is loaded with 10 ohm, then the resistive load seen on the primary side is 10/10000 = 0.001 ohms. In effect, the primary current is in series with a 1 milliohm shunt resistor. If the primary current is 100 amps, then the voltage developed across this primary 'shunt' resistance is 100x0.001=0.1 Volts. This primary voltage is stepped up by the turns ratio, so the secondary voltage is 100x0.1 = 10 volts. As a check, note that the primary and secondary powers are equal, ie, 100A*0.1V = 1A*10V. You can also check that the primary and secondary amp-turns are equal. (They are) So far, no problems, and all is easily understood in terms of well-known xformer theory.

NOW CONSIDER WHAT HAPPENS IF THE SECONDARY LOAD RESISTANCE IS INCREASED. Instead of Rsec=10 ohms, what if Rsec=1000 ohms?? The primary current is still 100A, because that is set by the external circuit, not by the current xformer. In our example, with Rsec=1000 ohms, Rpri=0.1ohms, Vpri=10V and Vsec=1000V. Hmmm. Clearly, for an ideal current xformer, as the secondary load resistance is increased towards infinity (=disconnecting the secondary load), the primary and secondary voltages will rise towards infinity. Real xformers approximate ideal xformers quite well so that, my friends, is why you should never open-circuit the secondary of a current xformer.

There is a clear analogy here with voltage transformers. With a voltage xformer, where a constant voltage is connected across the primary, never short circuit the secondary, or dangerously large currents will be generated. With a current xformer, where a constant current flows through the primary, never open-circuit the secondary, or dangerously large voltages will be generated.

Hope this helps.
Colin
 
S
To better understanding of CT let us connect two single phase power transformers TrA & TrB, in series which are equal rating (for example 11kv/33kv, 1000/100Aamps and %impedance is equal) in series to 11kv supply.

Now let us consider both transformer are equally loaded, then voltage accross each transformer should be half of supply voltage;5.5kv aprox (equal impedance)and current is equal at both transformer because connected in series.

Now let us consider both transformers at different load, now the voltage drop across each transformer is depends upon their respective circuit impedance but current through both transformers are same and depends on total impedance of the circuit I=V/ZTrA+ZTrB(series circuit).

Now let us open circuit one transformer then the impedance of that transformer become infinity and the voltage acrross that transformer will rise upto the rated voltage(Supply voltage 11kv/33kv) but here transformer has designed to operate at rated voltage, hence there will not be any insulation failure.

Now let us consider CT inplace of one of two trasformers, Now when the CT is connected to certain load. the impedance of CT is negligible compare to the power transformer connected in series. Hence voltage drop accross CT is negligible. Current in the circuit is I=V/ZTrA+Zct( connected in series)Here Impedance of Current transformer is negligible hence the load current is totally depends on ZTrA, ie.I=V/ZTrA.

Now let us consider CT open circuited. Now CT Impedance become high and it will try to increase its voltage, but far before insulation fails, because it has desinged to operate at certain load or with short circuit condition.

Hope this explanaiton may help someone.

Thank you
Sunil Kumar (Mangalore, India)
 
All of the above comments explains what happens to the voltage when a ct circuit is opened, but what happens to the current? Doesn't the current need to somehow find a path back to the ct it came from? Does it just circulate inside the winding of the ct? I don't have a handle on what happens to the current.
 
When current flows in the primary of a CT with an open-circuited secondary, the first thing that happens is it hums--very loudly. And the second thing that happens is that it basically explodes and any flammable components burn. This doesn't take long, and the time is inversely proportional to the magnitude of the primary voltage--the higher the voltage the faster the explosion/fire occurs.

I've heard CTs with open-circuited secondaries hum very loudly when the primary was energized but no current was flowing, and the breaker that was closed to energize the primary is quickly opened, manually. I've never seen on energized for more than a couple of seconds without current flowing in the primary, but they don't last that long--even at 440 or 480 VAC primary voltage when current is flowing in the primary.

So, the current doesn't do anything for very long. First of all, it never had anywhere to go if the secondary was open-circuited. Second, when current does flow (usually at the secondary terminals of the CT which are very close together), there is a large spark and near simultaneous explosion.

That's why most CT secondary circuits have shorting means that are supposed to be inserted in the circuit before the secondary circuit is ever opened for any reason. And the shorting means are usually located very close to the CTs themselves.

The upshot of all of this is: Don't ever energize the primary of a CT if the secondary is open-circuited. Someone will be, at least, be buying and installing a new CT, and possibly cleaning up a large mess and replacing some wiring, maybe even the primary wiring/bus bar.
 
Colin,

This is the most concise, clear, explanation that I have read. It makes perfect sense to me. Only having what I consider basic electronics knowledge, this explanation has gone a long way for me. I kept saying .... but they are both just transformers...how can a potential transformer be so predicable, and a current transformer be a magical box? Thanks for clearing that up. From a theory standpoint, they really are the same ... imagine that! Just matters whether the supply is constant voltage (like the ones we use all the times in our daily lives) or constant current - I guess that could even be my little clamp on amprobe meter?

Thanks,
Bob
 
> I know you should never open circuit a CT because it will lead to massive voltages being developed in the
> secondary and possibly fire. But why does it do this? Why doesn't a power transformer destroy itself if you open
> circuit it? What makes the 2 different? Any information I would be hugely grateful for.

Part 1 of your question:
A Current transformer designed to withstand open circuit would be more expensive, and may have worse characteristics in normal operation.

Explanation - (Warning: Technical)

What is a CT?
Usually a current transformer consists of a single 'turn' of wire through an iron core for the primary, and a few hundred turns for the secondary. If you understand what an idealised component model is, you can think of it as an inductor in parallel with an ideal transformer.

The voltage drop across the wire 'primary' in an unloaded CT will be V=L*(di/dt).
Where the inductance L is determined by properties of the wire and the core of the CT, and (di/dt) is change in current with respect to time.

This voltage will, by transformer action occur at the secondary. This unloaded voltage will be proportional to the current, frequency and turns ratio. The constant of proportionality depends on the core material and the wire used to form the ‘primary’. As such, the open circuit method a not very useful way to measure current.

In normal operation a low impedance is placed on the secondary, this appears as a very low impedance across the primary - much lower than that of the ‘inductor'. The output current will now be (mostly) directly proportional to the input current and the turns ratio.

Part 2:
Simply, a power transformer is designed to withstand the full supply voltage, while a current transformer is designed for a tiny fraction of that supply voltage.

If the current transformer were of ideal construction, and the secondary went open, then the infinite secondary impedance would be reflected into the primary. The 'primary' would then appear open circuit and be subject to the full supply voltage.

Source: Electrical Power Engineering lecture notes - Current Transformers.
 
Rama...

I suggest you search Control.Com Archives using the term +"Current Trans", in the SEARCH THE SITE box. There are some 35 CT threads!

Regards,
Phil Corso
 
R
> I know you should never open circuit a CT because it will lead to massive voltages
> being developed in the secondary and possibly fire. But why does it do this?

I think about a CT in simple transformer turns e.g. you have a 500:5
so that's 100 turns on the secondary for 1 turn on the primary.
Now suppose the CT is measuring the incoming buss on a 400 Volt switchgear and the secondary becomes open, the impedance of the transformer shoots up and with a substantial load on the switchgear it tries to drop the whole 400 Volts across the CT, that's 400x100 across the secondary or 40 kV.

I know this is not strictly correct but it explains it well enough for me.

Many years ago I was given the task of checking the overload relays on some medium Voltage switchgear, the engineer insisted we inject the equipment right on the busbars, we did this with a 1 kVa transformer with about 4 parallel turns of welding cable for a secondary, it would easily put out 2,000 Amps. On injecting one set off gear we couldn't get more than a few hundred Amps, we found there was a wiring mistake on the CT secondary side and the CT impedance was limiting the current. This would have gone un-noticed using the usual relay testing equipment.
 
In all comments there's no mention of core saturation.

The field strength has limited value. So, while the waveform will be messed up the CT open circuit voltage has limits governed by the size of core.
 
the core is rated for so many ampere-turns without saturation. you need secondary current to get saturation effects. basically saturation under an open-circuit secondary is not likely, until secondary arcing take place.

ferro-resonant failures are unlikely with current transformer installations. that only occurs in large sub-station transformers under low loads, and with power correction caps in the primary that have not been switched out of service.
 
I've been reading this thread with some interest because the danger of an open circuit CT figures largely where I work. I've puzzled it over and come up with this explanation. Hope it makes sense:- When a CT is loaded the primary and secondary ampere turns are balanced. A small amount of primary current (~1%) provides CT excitation. If the secondary is opened, all the primary current now drives the excitation. The flux rises to approx. 2 tesla and then the flux waveform flattens as the core becomes saturated. The flux waveform has changed from a smooth sinusoidal to more of a square wave where the rate of rise and fall is roughly the same as the primary current about it's zero crossover. So, if we take the gradient of the primary current at it's zero crossover we get dI/dt=(Imax)X(2xpixf). This steep rise and fall corresponds to a large dflux/dt at the start and end of the 'square' flux waveform. The secondary voltage is Vs=-Nxdflux/dt. So,for a 1000:1 CT, using the Primary current's crossover gradient and 1000 turns, the secondary voltage is roughly -1000x(Imax)x(2xpix50) which is 314200xImax. If Imax is 100 amps, we get a very large secondary voltage.
 
Basically, the voltage developed in a coil, from a fundamental point of view, is directly related to the rate of change of flux. So, a high rate of change gives a high voltage.

With a CT secondary open circuit, and the CT excitation driven by the Primary current, the rate of change of flux is very high at the zero crossover. The core then becomes saturated for most of the sinusoidal half cycle. The voltage output will be seen as spikes at the zero crossovers.
 
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