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Current transformer magnetics 
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#1
Mar1209, 02:04 AM

P: 190

Hi Guys,
I am grappling with a peculier (but somewhat common) problem with Current Transformer (CT) magnetic circuit. I will be very grateful if someone could help me out with my problem. 1)In an ideal Voltage Transformer (VT), the magnetic flux developed in the core remains constant at all loads. But the magnetic flux in the core of a CT reduces/neutralizes as soon as a 'Burden' (load) is connected to the secondary winding of a CT! Hence, why is the magnetic flux in a CT core not constant (with connected load, also) as is the case with a VT ? Why does this reduction/dampening in magnetic flux occur in case of a CT, since a CT basically resembles an ideal VT as far as the core magnetic circuit is concerned ? 2) I am looking for a phasor diagram for an UNLOADED CT (open circuited secondary) but i am unable to find one yet. I will be extremely grateful if someone could provide me with one or guide me to a suitable link. Thanks & Regards, Shahvir 


#2
Mar1209, 08:29 AM

P: 1,036

"Unloaded CT" or "open secondary CT" is dangerous. So is a shorted VT. With a CT the primary is placed in series with the circuit whose current is to be measured. The flux in the core would be dangerously high as would the voltage on the secondary if said secondary was open. All primary current would become magnetizing current. A CT secondary must always be terminated in a low impedance, either a short (shunt or switch). or an ammeter. The flux in the core is determined by the difference between primary and secondary ampereturns.
An *unloaded* CT has its secondary shorted. If a low valued resistor is placed across the secondary and the short removed, then the primary current times the reciprocal turns ratio equals the secondary current. Multiplying by secondary resistance gives secondary voltage. The primary voltage is then determined by the turns ratio. Does this help? Claude 


#3
Mar1209, 12:17 PM

P: 190

Dear Claude,
Thanks for the reply. Though your reply is technically correct, but it does not cover the point i am after! i need to understand what is that factor/s which causes the difference between a VT and a CT ?.....as far as the magnetic circuit is concerned. So, in brief why does the magnetic flux in a VT core remain constant even at onload conditions, whilst the flux in the core of a CT reduces/dampens as soon as it is loaded by a burden ? I am aware of the neutralizing effects of secondary ampturns on the core flux of a loaded CT, but this effect is common for a loaded VT too! Then what factor/s causes the flux in a CT to behave differently (reduce) from a VT? Kind Regards, Shahvir 


#4
Mar1209, 12:49 PM

P: 1,036

Current transformer magnetics
Ok, I understand your question. The "difference" between a CT & VT is only in physical construction & optimization. Both types operate under 2 laws, Ampere's law AL, & Faraday's law, FL.
With a VT, the flux is determined by the amplitude of the constant voltage source inputted to the primary. When a load is added, the secondary current, ampturns, tends to reduce the core flux, since Lenz' law states that induced flux/current/voltage always are oriented in a direction so as to *oppose* the original quantities. The mmf due to secondary ampturns is countermmf. But the CVS (constant voltage source) at the primary generates a countercounter mmf in ampturns and restores the flux to the original value. With a CT, the secondary is shorted as its *no load* condition. The primary presents a small leakage reactance and a large magnetizing reactance in parallel with the secondary small leakage reactance. Hence the primary flux is established by the primary *current*, as opposed to primary voltage for a VT. When the short on the secondary is increased to a small resistance, like a current shunt, or an ammeter, the flux will decrease due to the couner emf. Increasing the secondary resistance from 0.01 to let's say 0.1 ohm results in an increase in secondary voltage per Ohm. By Lenz law, the core flux would decrease. But a countercounter emf occurs at the primary since the secondary resistance reflected over to the primary increases. A larger voltage is present on the primary. The volts per turn on both sides balance & the original core flux is retained. The CT & the VT both operate under AL & FL. The main difference is that a CT's core flux is determined by primary current, and the secondary loading will vary the counter emf which reduces flux. But the primary emf increases to oppose this property. A VT's core flux is established by primary voltage. Loading the secondary results in current/mmf which counters the flux. The primary current increases providing countercounter mmf. Is this better? Claude 


#5
Mar1209, 01:11 PM

P: 190

Could the flux dampening be due to the voltage drop across the load in series with the CT when the burden is connected? this volt drop will reduce the voltage across the magnetizing current branch in parallel with the burden (if one refers the equivalent ckt. of a Xmer) reducing the flux in CT core since the magnetizing current now drops due to the reason mentioned above.
If so, then there will be a small drop (though practically negligible) in the magnetic flux in the core of a VT too, due to voltage drop across the winding impedance which appears in series with the primary wdg. of a VT (as per equivalent ckt.). Again, the voltage across magnetizing branch will drop (though negligibly) which in turn will reduce the magnetizing current by a small amount.....decreasing the magnetic flux in core of a VT too! In brief, i think the flux in the core of a VT will also reduce (though practically by a very small amount) due to the above mentioned reasons. Please correct me if my understanding is incorrect. Kind Regards, Shahvir 


#6
Mar1209, 01:43 PM

P: 1,036

You understanding is correct except for the "burden" concept you presented. For a VT, no burden is an *open* secondary. But for a CT, no burden is a *shorted* secondary.
With a VT, with no load, the primary voltage & magnetizing current establish the core flux. When the secondary is loaded, there is a voltage drop in both windings due to IR (resistance & increased current). So, the actual voltage impressed across the core is slightly reduced, but it is a small difference. A similar analogy holds for a CT. Unburdened, the secondary is shorted. When a resistance is added to the secondary, the voltage on the secondary, & primary as well increases. A small leakage current results from the increased voltage. So, the actual current in the primary which is driving the core flux gets reduced ever so slightly, so that the core flux gets reduced. With a CT, the reduction of flux under burden is less than that for the VT because insulators are better than conductors. Now do I make sense? Claude 


#7
Mar1209, 01:58 PM

P: 190




#8
Mar1209, 03:39 PM

P: 1,036

Ideally, for a perfect CT and/or VT, the change in flux due to burdening would be zero. If the windings in a VT exhibit zero resistance, then the flux never changes with burden. In a CT, if the insulation exhibits zero leakage, then the flux never changes with burden.
In the real world, resistance & leakage are present. But we can approach an ideal insulator much more than an ideal conductor, at room temperature anyway. If the VT was built with superconducting windings, then I agree with you that the VT exhibits less flux change when burdened. Otherwise not. Have I explained it well? Claude 


#9
Mar1309, 02:17 AM

P: 190

Thanx claude you have explained it well as always, but i do not seem to grasp this insulation leakage point you are making! Doesn't the CT flux reduce/change due to the series impedance presented by the 'system load' in series with the primary winding of the CT (or VT for that matter) ?........could you plz simplify or elaborate a bit for me on the insulation leakage part if its not much trouble? Thanx very much. Kind Regards, Shahvir 


#10
Mar1309, 12:33 PM

P: 384

For an ideal transformer, insulation leakage has nothing to do with the operation of the transformer.
There is no difference in flux between a current transformer and a potential transformer. If the voltage to a potential transformer goes to zero, the flux goes to zero. If the voltage to a potential transformer goes to a maximum, the flux goes to a maximum. (Assuming the correct load {burden} on the secondary of a current transformer) If the current in a current transformer goes to zero, the flux goes to zero. If the current in a current transformer goes to a maximum, the flux goes to a maximum. The following equations can be used to calculate the voltages and currents in either a ideal potential or current transformer. Epri/Esec=Npri/Nsec=Isec/Ipri Isec=Esec/Rload For an ideal current transformer, you start with the load resistance and voltage(burden) and just plug the values in the formulas given. This will give you the the secondary current, the primary voltage and current and the turns ratio. 


#11
Mar1309, 01:01 PM

P: 1,036

For an ideal current xfmr, you do NOT start with the load resistance and voltage burden. Rather, you start with the primary current. All of this, of course, is based on the presumption that the load resistance in the CT primary is much greater than the reflected secondary impedance. The primary current is defined by the external network whose current is to be measured. The secondary current is then defined by the turns ratio, Np*Ip = Ns*Is. The secondary voltage is defined per Ohm's law, Vs = Is*Rs, where Rs is the resistance across the secondary terminals. Finally, Vp is defined by Vs per Vp/Np = Vs/Ns. Regarding the insulation, I explained it above. The primary current is the network current to be measured. Ideally all of the primary ampturns link the CT core and determine the flux. If the insulation is damp, leaky, operating at high temp, a slight leakage current exists. The current leaking through the insulation does not couple the secondary if it does not link the core. Hence this small fraction of current does not contribute to the core flux. Again, insulation is so darn good, that this effect is negligible. For all practical purposes it can be ignored. But if the secondary resistance, Rs, is increased, then both Vs & Vp increase. The insulation leakage increases with higher voltage, and the core flux undergoes a slight decrease. Again, it's too small to even worry about. Insulation leakage slightly influences core flux in an indirect fashion. By shunting a little of the primary current, the core is linked by slightly less ampturns. Claude 


#12
Mar1309, 01:22 PM

P: 190

Thanx guys for solving my doubt (especially u claude!) thanx very much for ur time. Additional inputs/suggestions regarding the same are welcome.
Kind Regards, Shahvir 


#13
Mar1409, 04:51 PM

P: 4,513




#14
Mar1409, 05:01 PM

P: 1,036

When the secondary is loaded, Is tends to produce a counter mmf which reduces the core flux. But the primary constant voltage source immediately increases the primary current Ip. So now the primary current consists of Imag plus Ip. Ip produces a countercounter mmf which opposes the counter mmf of Is, and the core flux is about the same as the no load case. In real world xfmrs, the voltage drop in the windings slightly reduce the core flux as less voltseconds per turn is reaching the core. Does this make sense? Thus Np*Ip balances Ns*Is. But Np*Imag is never balanced. Imag is necessary to have core flux, so Imag does not participate in the balance of ampturns. Claude 


#15
Mar1409, 08:43 PM

P: 4,513

That all sounds good to me, Claude.
I haven't had anything to do with current mode transformers. The only use for them, that I know of, is for line monitoring instrumentation where the transformer is connected in series with a load on the AC line. I can't quite see how, in this application, unloading the secondary of the currrent transformer would overvoltage the primaryif it actually does. 


#16
Mar1509, 01:20 PM

P: 384

Opening the secondary of a current transformer overvoltages both the primary and the secondary. Since the primary usualy has only one or a few turns and the secondary has many turns, the problem/danger is usually the secondary.
With an ideal current transformer: When secondary of the current transformer is opened, the full line voltage is applied to the primary. If the current transformer turns ratio is 1/1000 and the primary line voltage is 120 volt, there will be 120, 000 volt on the secondary. In practice, the core usually saturates and there are high voltage pulses. 


#17
Mar1509, 01:34 PM

P: 4,513




#18
Mar1509, 08:20 PM

P: 2,499

 Not sure, but we are probably actually in agreement considering what you actually posted concerned an IDEAL current transformer. Big difference in this case I suppose. 


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