# Current Transformer danger when secondary is left open

Discussion in 'General Electronics Chat' started by MrBuggy, Sep 21, 2014.

1. ### MrBuggy Thread Starter New Member

Nov 7, 2013
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Hi guys,

I've learned from my seniors that when the secondary of CTs it poses a huge hazard (CT secondary voltage will shoot up and possibly destroy the device). I do not understand this concept as I know that its okay to leave the secondary of normal power transformers. Can anyone shed light into this issue?

Sincerely,
MrBuggy

Jul 18, 2013
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3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I've always assumed the danger is to people who may be (inadvertently) shocked or electrocuted by the open CT secondary. Measuring CT's are considered as isolated from the active voltage line which the CT is monitoring. The temptation may be to think they are therefore intrinsically safe on the secondary side.
One wouldn't intentionally touch the open or unloaded secondary of a live power transformer. One may however think the CT is safe because it is not directly connected to the active conductor. Beware of unforseen hazards.

Last edited: Sep 21, 2014
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4. ### MrBuggy Thread Starter New Member

Nov 7, 2013
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For Transformers in general I thought voltage comes first and then the current (the same for current transformers, the magnetic flux created by the AC current should induce a voltage in the secondary winding, only then when there is a load a current will flow and be provided). So in the case of the Secondary winding of the current transformer there is already a high voltage across the terminating leads whether there is a low impedance ammeter or not.

5. ### MrBuggy Thread Starter New Member

Nov 7, 2013
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Here attached is a snippet from the book Power Systems Analysis and Design. The steps outline how to determine the missing parameters. In a way I disagree to this as Step 1 starts with a known current rather than a known secondary voltage. In electromagnetics theory we have Lenz law that states there will be voltage when there is a change in magnetic flux instead of a current.

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6. ### #12 Expert

Nov 30, 2010
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The whole point of a current transformer is that it transforms current, not voltage. That's why it has a different name. If 500 amps flows through the core of a 500:1 current transformer, one amp is going to try to come out the secondary, and it will develop as many volts as necessary to do that.

It's like trying to turn off a relay coil. The magnetic field has to collapse, and that energy has to go somewhere, even if it has to arc through the transistor that tried to shut it off.

You just have to accept that this one works, "backwards". Do the math for the current and the voltage will fall into place.

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7. ### Brownout Well-Known Member

Jan 10, 2012
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There isn't any such thing as voltage first or current first. Voltage and current exists concurrently in any circuit. As a thought experienme, consider the following circuit: A current source of xA connects to a resistor of 1ohm. A voltage of 1V is measured across the resistor. Can you determine the value of the current source from the voltage measurement?

Now consider this: a wire loop is connected to a 1 ohm resistor, and the magnetic flux and number of turns are such that 1V is produced at the ends of the coil (per Lenz) Is there current flowing in the resistor currently with the voltage produced? Can you determine the voltage by measuring the current?

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8. ### subtech AAC Fanatic!

Nov 21, 2006
123
5
There is no magic here.

The CT transforms current, according to its ratio, primary to secondary, and the CT secondary voltage varies in direct proportion to the magnitude of the connected impedance. (up to it's design limit)

The PT transforms voltage, according to its ratio, primary to secondary, and the PT secondary
current varies in inverse proportion to the magnitude of the connected impedance.

A careful and detailed study of the physical relationship of windings to the core of each device

Last edited: Sep 22, 2014
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9. ### MrBuggy Thread Starter New Member

Nov 7, 2013
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There's a lot of things out there that is just the way it is, including CTs I guess. In my head, it doesn't matter where the flux comes from when it goes through the coils, whether the flux comes from the primary of a PT or the primary of the CT. A potential must always be induced in order to balance the flux inside the coils as much as possible, ie if the external flux is increasing inside the coils there will be induced flux such to decrease this phenomena. The concept of current transformation is still difficult for me to understand .

I get what you are saying. But what then would happen to a secondary of a CT? If there is infinity impedance (open circuit), there should be a finite amount of voltage across the secondary.

I have done this through Lenz Law. Voltage is produced across terminals when there is a changing flux inside a coil. So shouldn't it be safe to have the secondary open? It will be a high finite voltage, yes, but it shouldn;t destroy the CT...

10. ### #12 Expert

Nov 30, 2010
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Well...infinity is the theoretical limit if there was no impedance in the windings and there was no insulation failure inside the current transformer and there was no stray capacitance. Let's just say it tries to get to infinity.

With a common inductor, energy = 1/2 LI^2. If you try to stop the current instantly, the induced voltage would approach infinity except it finds stray capacitance to charge up to the tune of energy = 1/2 CV^2 and some of the energy is destroyed as heat in the resistance of the winding, and if that isn't enough it commits suicide by arcing in the coil. Much like a bullet, you can give it somewhere safe to go or get knocked on your @ss.

11. ### crutschow Expert

Mar 14, 2008
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There really is no difference between an ideal voltage and and ideal current transformer in theory. In each one the secondary voltage is equal to the primary voltage times the turns ratio and the secondary current is equal to the primary current divided by the turns ratio.

So lets take an ideal 1:100 transformer (it's immaterial whether it's called a current or voltage type). Then if you have 1A going through the primary, there will be 10mA going through the secondary. If the secondary has a 1k load then the secondary voltage drop will be 10V and the primary voltage drop will be 100mV. Now suppose you open circuit the secondary. The secondary voltage will rise as determined by the available primary source voltage and the turns ratio. If the primary source generating the current is 110V then the secondary peak voltage will rise to 110V x 100 x 1.4 = 15,400V which is obviously a very dangerous voltage and will likely arc over and break down the secondary insulation. That's why you don't ever want to open circuit the secondary of a transformer used to measure current.

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Apr 28, 2014
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13. ### HighVoltage! Member

Apr 28, 2014
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Nice explanation! However, where did the 100 & 1.4 from (110 X 100 X 1.4) come from? Is 100 the turns ratio? How about the 1.4?

14. ### crutschow Expert

Mar 14, 2008
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Yes, 100 is the turns ratio and 1.4 is simply the peak voltage of a sinewave whose EMF is given in RMS volts.

15. ### xiaotanmao New Member

Jan 12, 2017
2
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Nice explanation. For a typical high voltage current transformer, current ratio may be 3000:5. If 3000 A flows on the conductor, 5 A flows on the CT secondary side. If the resistor at the CT secondary side is 0.2 ohm, a typical setting, you see 5 A x 0.2 ohm = 1 V. However, then secondary side is open, the only way this 5 A current can flow is the stray capacitance. The stray capacitance

16. ### xiaotanmao New Member

Jan 12, 2017
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For a typical high voltage current transformer for a 500 kV transmission line, current ratio may be 3000:5. If 3000 A flows on the transmission line, 5 A flows on the CT secondary side. If a typical resistor at the CT secondary side is 0.2 ohm, you see 5 A x 0.2 ohm = 1 V on the secondary side. However, when secondary side is open, the only path this 5 A current can flow is the stray capacitance. A typical stray capacitance for CTs is 200 pF (IEEE Standard C37.011), or 1.3e7 ohms at 60 Hz. So you see 5 A x 1.3e7 ohm = 65 million volts on the secondary side.

This assumes that 5 A has to flow on the secondary side. I guess if the CT is not saturated, the magnetizing current has to be close to 0. So if 3000 A flows on the primary side, we need 5 A to flow on the secondary side to make the magnetizing current 0. However, if voltage on the secondary side is as high as 65 million volts, the CT must have been saturated. I found a CT saturation curve in this paper: https://cdn.selinc.com/assets/Liter...ers/TP6038_19920817_Web.pdf?v=20150812-085648

According to the saturation curve table in this paper, when magnetizing current is about 5 A, the magnetizing impedance is only about 100 ohm, much smaller than the impedance of the stray capacitor, which means that the 5 A current will flow into the magnetizing branch, instead of the stray capacitor. So the voltage at the secondary side is about 470 V. Note that even if the current on the transmission line is only 30 A, it means 0.05 A on the secondary side, or 300 Volts if secondary side is open.

17. ### leckyman New Member

Mar 6, 2017
1
0
Hello,
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 primary current is determined by load on the power system and not influenced by CT secondry loading. On each half cycle of the AC primary current the flux density in the CT iron core rises to approx. 2 tesla and then the flux waveform flattens as the core is saturated. The flux waveform has changed from a smooth sinusoidal with a relatively low peak value to a rapidly rising square wave where the rate of rise and fall for the flux is steep. As an estimate of how steep the flux rises to saturation we can compare it to the rate of rise of the primary current about it's zero crossover. After all, in the linear region the flux is proportional to the excitation current. 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 voltage in a coil depends on the relationship Vs= -N x dflux/dt. So,for a 1000:1 CT, using the Primary current's crossover gradient and 1000 secondary turns, the secondary voltage is roughly -1000 x (Imax) x (2xpix50) which is 314200xImax. If Imax is 100 amps, we get a very large secondary voltage. This model is based on the assumption that the primary current is unaffected by the secondary open circuit and voltage in a coil depends on the rate of change of magnetic flux. Also that the core rapidly saturates due to the extremely high excitation provided by the full primary current. The resulting voltage waveform will consist of large spikes corresponding to the high rate of change of flux at zero crossovers and no voltage during the periods of core saturation where dflux/dt=0. In a practical situation stray capacitance and arc resistance will limit the secondary voltage, but, the voltage and damage will still be significant and the magnetic properties of the core can be severely changed.

18. ### schmitt trigger Active Member

Jul 12, 2010
57
8
A succint and accurate analogy.