Open Circuit a Current Transformer


Thread Starter


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 greatful for.

Many Thanks.
Josh Banks

Phil Corso, PE

Responding to Josh's Mar 10, 5:45pm query... without getting too technical:

Q1) Why does opening a CT's secondary cause dangerously high secondary voltage?

A1) With the flow of both primary and secondary currents the transformer's exciting current is very low. The secondary current mmf serves to keep the magnetizing flux in check! If the secondary now opens the primary current mmf produces an exciting current 'orders of magnitude' greater than normal. And, the resultant large increase in flux density produces an extremely high voltage in the secondary.

Q2) (Paraphrasing ) Why don't power transformers exhibit the same phenomena?

A2) While current transformer (CT) and power transformer (PwrT) theory is about the same the salient difference is their Volt-Ampere characteristic. The PwrT is a constant-voltage (shunt-loaded) device whose primary and secondary VAs are equal. Conversely, the CT is constant-current (series-loaded) device. That is, the input VA is insignificant, having no influence upon the primary circuit. Its inter-related variables like flux, volts, current, and burden are not constant but more likely non-linear.

If you need more technical detail than given above, contact me!

Regards, Phil Corso, PE {Boca Raton, FL, USA} [[email protected]] ([email protected])

Curt Wuollet

Simply transformer action. The turns ratios of most power transformers are low so the voltages are low also. In a CT the primary is 1 turn so the turns ratio is high. Normally the primary voltage drop is low because of the reflected impedance of the load which is usually a very low resistance like a current shunt. When you remove the load, the primary drops appreciable voltage according to it's now higher inductive reactance and the voltage is multiplied by the high turns ratio and you have sometimes very high voltages. The biggest danger is when you open the load under power as the voltage rises quickly enough to maintain an arc. If the transformer or something connected to it is not insulated for high voltages you can get a fire. If Jumper Joe is pulling the wire off, he can become the easiest path to ground.

Basic electrical theory (which I am not going to go into) - an open CT present itself as a large inductance to the primary current source and develops a large voltage on it primary winding. The secondary winding will then develop a voltage that is N times larger than the primary (because the transformer ratio is 1:N - Primary:Secondary).
Responding to Curt's Mar12, 12:00pm and Meir's Mar 12, 12:02pm comments:

Tsk, tsk... I too appreciate not-to-technical answers... but, they should be technically correct:

A) Turns-ratio is not responsible for overvoltage. Even CTs with lower ratios, like 5:1 or 1:1, have the cautionary notice (at least in the USA!) Furthermore primary voltage is a fraction of Volts. So, using turns-ratio as the primary to secondary multiplier, secondary volts would be in the order of volts to tens of volts.

B) Overvoltage occurs when the secondary open-circuits and the demagnetizing effect of the secondary emf is lost. Flux density quickly increases, limited only by core saturation.

C) As core saturation progresses, waveshape changes. The usual sinewave changes to a peaked wave. Such a waveshape has an extremely high dV/dT characteristic. It is very likely that insulation flashover and/or fire results from dV/dT... not from a high amplitude sine wave!

Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

Gary Wiggins

The CT is designed to lower current to a safe and measureable level. To accomplish this, the voltage level on the secondary is raised. P=IE, if P is constant, I and E are then inversely proportional. The windings ratio determines the secondary voltage levels. While the secondary is connected to a measuring device, it is essentially shorted. Opening the secondary circuit immediately raises the voltage to levels pre-determined by the primary voltage and the windings ratio. There are shunted CT's available that greatly reduce potential hazards typically associated with standard design CT's. Always remove power and install shorting bars prior to working on CT circuits.

Curt Wuollet

Yes, but it is technically correct. You are also correct in that there is di/dt action going on. The theoretically instantanious collapse of the field is going to cause theoretically infinite voltage. And that field is generated by primary current. You will find that within physical limits, the change in flux density affects both windings. Actually a better model than "normal" transformer action might be the common Kettering ignition coil. And I don't recall stating that the voltage would be sinusoidal although you might get lucky and interrupt the sinusoidal current at a zero crossing. In any case, it's a bad idea. The non-technical explanation is you are going to get a big spike to light an arc and greater than normal voltage to help keep it going.


Shailesh Patel

Dear josh Case : Power through aprox. transformer is constant

when you think about power transformer then voltage is aprox. constant and current in primary is depend on secondary load. so if you short power transformer then current in primary become very huge and may (must) be damage to winding and insulation.

But about currant transformer current passing in primary is depend (=same) on busbas or cable where current transformer mount (which is not dependent on load of secondary of CT) & voltage of primary is depend on secondary load. if your secondary of CT is open than as per magnetic coupling very high voltage generated (because of secondary current = 0 ) at secondary. And insulation will be fail.

Shailesh Patel
A few off-list questions have prompted revisiting this topic, as well as two related ones, namely, "Current Transformer" (#1026172605) and "Current Transformer Failures" (#1026113301). From the questions asked it is obvious the topic is still in a state of flux (pun intended!)

Some commenters expressed a 'very strong' belief that overvoltage is due to CT turns-ratio. Others, myself included, believe that while ratio is present, it is not the reason for the high voltage. A new approach is presented here to resolve the issue.

Consider the example submitted by Pagnair... a CT whose current ratings are is 2,000/1/5. But, ratio alone does not complete the specification. So, for calculation purposes, I will assume it's capacity is 10 VA (volt-amperes) for the 1-Amp tap.

Tagging primary and secondary parameters with subscripts p and s, respectively, letting the winding turns-ratio = Np / Ns, then:

A) Primary Parameters. The pri / sec current-ratio (Ap / As) is 2,000:1. Primary voltage, Vp, is determined by the simple calculation VA / Ap, or 10 / 2,000 = 5.0 mV. Primary ampere-turns, ATp, are 2,000 x 1.0 = 2,000 AT.

B) Secondary Parameters. Secondary voltage, Vs, is VA / As = 10 Volts. Secondary ampere-turns, ATs, are As x Ns / Np = 2,000 AT!

C) Ampere-Turns Impact on Core-Flux. Note, from above, the primary and secondary ampere-turns are equal. The direction of their flux flows are in opposition. But, a small core-flux is still required to cover both iron-core and winding losses. This net core-flux, ATe, is quite small, as low as 1/2% to 2% of the rated ampere-turns. (An aside, it is this value that establishes the CT's accuracy!)

D) Net Core-Flux Produces Output. Using a 1% value, the exciting ATe is, in this case, 20-AT. This inturn (pun not intended) results in an output voltage of 0.5 Volt per ampere-turn.

E) Effect of Opening Secondary. Opening the secondary results in the full primary ampere-turns increasing the core-flux, i.e., from 20 to 2,000 ampere-turns. This of course produces 2,000 AT x 0.5 V / AT volts, or 1,000 Volts at the secondary.

F) Conclusion. CT-ratio is not the cause of overvoltage. Instead it is the substitution of an overwhelming primary ampere-turns for the very small excitation ampere-turns! Note, the calculations above are based solely on an rms-treatment of both flux and voltage. Core-saturation effects were not considered!

Regards, Phil Corso, PE {Boca Raton, FL, USA} [[email protected]] ([email protected])

Phil Corso, PE

You know I'm a stickler (some say obsessed) for Technical Correctness. Several contributors to this and related current transformer topics made
several errors:

1) When transferring the secondary impedance, Zs, into the primary as Zs', the correct expression is,

Zs' = Zs x (Np/Ns)^2, where,

Np and Ns are the primary and secondary winding turns, respectively.

For the CT example cited the secondary impedance reflected into the primary is,

Zs' = 10 x (1/2,000)^2 = 2.5E-6.

2) This low value illustrates that secondary impedance has no influence on the primary circuit... contrary to what some stated.

3) Some presented the formula for the development of an extreme secondary voltage as Es = Ep x current-ratio! Again using the CT cited, for an open circuit condition the secondary voltage would have been 5 mV x 2,000 or 10 Volts. Clearly not the extreme voltage observed!

If any of you disagree, then present a point of view free of obfuscation!

Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])
I have read with great interest the various comments expressed and I feel that a number of questionable interpretations have been presented.
Under open circuit conditions the total primary current will be equal to the total excitation
force of Ampereturns/metre, thus inducing maximum
induction (B of 1.9-2.0 tesla for normal materials).

The voltage generated in the secondary winding under open circuit will be directly proportional to:
Number of secondary turns
Cross sectional area core

Extremely large voltages can be generated in Cts that have large turn ratios, i.e. 5000/1, and have
large CSA as typically used in protection class cts for differential protection.

My experience in laboratory tests shows that in CT designs with AT/m up to 1500, the relationship of generated voltage is linear and proportional to the items listed above. I have developed curves to estimate open circuit voltages and would greatly appreciate to have some comparisons if anyone has similar information.

Sten Maraldo
[email protected]

Darryl Muetzel

Mr Corso,
I have a problem - maybe you can point me in the right direction.

I have an application where my customer is requiring FS(instrument security factor)>=5.

We were unable to meet this with the BCT design - so we suggested to use a secondary protector (current shunt) that would short circuit when the voltage across the burden would reach a certain voltage (due to over current during a short circuit event).

This was agreed to - but when I went to the vendor (ITI/GE) to see if they could make up a secondary protector (similar to their OCP line of open circuit protectors) with a lower activation voltage than their current designs- they said that they are unable to design any new devices for several reasons - which aren't really important.

SOOO - for my question - do you have any information on who could manufacture such a secondary protector (activation voltage is 60V)?

Any help you (or anyone else) could give would be very much appreciated.

Thanks - Darryl Muetzel ([email protected])

Phil Corso, PE

Darryl, cts have been used in protective relaying applications for about a century. Your attempt to use a new technology has me puzzled. Please provide additional information:

1) What are details (primary kV, primary Amp, ratio, SC level, burden impedance, etc)?
2) Why do you believe an overvoltage will occur?
3) Are you concerned about saturation effects?
4) Why the intent to use a new technology?
5) What is the customer's reason for Inst FS?
6) What, if any, is the cost-to-benefit ratio?

Phil Corso, PE ([email protected])
The primary current in the ct is not controlled.
This is determined by the load on the power side.
When a primary current flows, a secondary current is indused as per the ct ratio. If we keep the ct secondary open, it tries to develop a voltage to pass that secondary current. This cause insulation to break down in ct.

Manu Thomas

Q1) What will happen when CT secondary is kept open and energize the primary?

Q2) Is it only dv/dt or di/dt relation for overvoltage across CT sry winding?
I think that when we open ct and there is current running in the primary coil this cause high voltage in secondary coil but pwr tr when is open no current running in primary coil
hozgab [at]
we made an experiment.
A 3000/1 Current transformer was gradually energized in steps with secondary open right from beginning.

We observed that at 10% rated current in primary. after about 10 seconds, the CT failed. The secondary when tested was, found to have shorted, also confirmed by significant reduction of winding resistance indicating interlayer shorts. Since we did not open secondary while primary was energised, it is not a Ldi/dt voltage failure.

Also, when tested earlier, this CT showed a knee point voltage of 750 volts at the secondary, which is well below the insulation level of the enamel wire used for secondary, i.e. 2 kV.
Thus even when open, the voltage developed may have exceeded the knee voltage to a saturation level of the order of at most 1 kV.

Thus I am not clear. what c`d be the cause of insulation failure?
Dhamdev... good work!

Regarding your question about not reaching the varnish breakdown voltage when you tested for the kneepoint, the CT had not yet been saturated.

Regards, Phil Corso