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Current transformer magnetics

  1. Mar 12, 2009 #1
    Hi Guys, :smile:

    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,
  2. jcsd
  3. Mar 12, 2009 #2
    "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 ampere-turns.

    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?

    Last edited: Mar 12, 2009
  4. Mar 12, 2009 #3
    Dear Claude, :smile:

    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 on-load 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 amp-turns 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? :confused:

    Kind Regards,
  5. Mar 12, 2009 #4
    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, amp-turns, 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 amp-turns is counter-mmf. But the CVS (constant voltage source) at the primary generates a counter-counter mmf in amp-turns 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 counter-counter 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 counter-counter mmf.

    Is this better?

  6. Mar 12, 2009 #5
    :rolleyes: 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 in-correct.

    Kind Regards,
    Last edited: Mar 12, 2009
  7. Mar 12, 2009 #6
    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?

  8. Mar 12, 2009 #7
    Sorry! system error :biggrin:

    I think this should be the other way round! The reduction of flux in VT in on-load conditions is less as compared to the reduction in flux in a CT with burden. :rolleyes:
  9. Mar 12, 2009 #8
    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?

  10. Mar 13, 2009 #9

    :smile: Thanx claude you have explained it well as always, but i do not seem to grasp this insulation leakage point you are making! :confused: 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,
    Last edited: Mar 13, 2009
  11. Mar 13, 2009 #10
    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.

    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.
  12. Mar 13, 2009 #11
    Insulation leakage is a slight second order influence, but it does have a slight impact. The operation of a xfmr itself does not depend on leakage, but the amount of current, namely amp-turns, which determines core flux, can be diminished by insulation leakage.

    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 amp-turns 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 amp-turns.

  13. Mar 13, 2009 #12
    :smile: 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,
  14. Mar 14, 2009 #13
    I don't remember this part. As I understand it, the magnetization flux is independent of load to first order. A load on the secondary represented in amp-turns, is equal and opposite to the ampere-turns on the primary, resulting in a net flux change of zero contribution due to load. The main contribution away from this ideal transformer would be due to the winding resistance. Is this what you're talking about?
  15. Mar 14, 2009 #14
    But when the VT is first energized under no load, the primary voltage source value across the primary, its frequency, Np, & Ac (cross sectional area of core) determine the core flux. Along with the ac primary voltage, a magnetizing current is needed per the B-H curve of the ferromagnetic core material. This primary magnetizing current Imag, exists with or without any load on the secondary Is. The amp-turns are not balanced as Imag is needed to establish the flux.

    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 counter-counter 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 volt-seconds 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 amp-turns.

    Last edited: Mar 14, 2009
  16. Mar 14, 2009 #15
    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 over-voltage the primary--if it actually does.
    Last edited: Mar 14, 2009
  17. Mar 15, 2009 #16
    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.
  18. Mar 15, 2009 #17
    Under what conditions?
  19. Mar 15, 2009 #18


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    I haven't had alot to do with current transformers other than installing them. What I want to know is how you can figure that when you unload the secondary that you can get the full source voltage across the primary considering that it is usually a wire passing through a toroid core. You cannot have 120 volts 60 Hz across a wire that is only an inch or two in lenth laid out straight with no bends. You are applying voltage transformer logic to something that is NOT a voltage transformer. Yes I do realize that they are technically the same thing other than a few optimizations. They both obey the same laws. One important optimization is that the primary winding will just about always act as a primary winding on a voltage transformer with the core missing. In other words, treated as a voltage transformer it's magnetizing current would be very high. This is necessary to prevent the primary winding from loading the circuit in which we want to measure the current. (Do we call it loading when inserting an ammeter in series?)
    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.
    Last edited: Mar 15, 2009
  20. Mar 15, 2009 #19
    Sure, you can have 120V rms across one single turn of a CT, but only momentarily. The core flux goes through the roof, saturation occurs, and the current skyrockets.

    A CT is placed in series with the network whose current is to be measured. Let's say the source is 120V rms ac, with a 12 ohm load, and a CT in series. With 120V & 12 ohm, the current is 10A rms. If the CT secondary is loaded with 1 ohm, and the CT turns ratio is 50 to 1, then the 1 ohm resistance is reflected back to the primary by the turns ratio squared, or 2500. Hence the primary reflected impedance is 1/2500 or 400 micro-ohm. Let's say the reactance of the primary is 1 milliohm. The combined impedance is a little over 1 milliohm at the primary as the 2 components are in quadrature.

    The voltage divider formed by the load and CT are 12 ohm and 1 milli-ohm. Thus the primary CT voltage is extremely small. But if the CT secondary is open circuited, then this open appears at the primary. The 120V divides between 12 ohms and an open. Thus the one turn CT primary has the full 120V across it.

    Not pretty at all. A CT MUST be terminated in a low impedance keeping the primary reflected impedance and voltage low and safe.

    Last edited: Mar 15, 2009
  21. Mar 15, 2009 #20
    I have 120VAC in series with a 10 Amp load and a current transformer consisting of a primary with one or two loops with an open secondary.

    The load limits the current to 10 amps.

    The core satures at, say, 0.1 amps and the transformer is no longer an inductor. It's a piece of wire with 10 amperes RMS running through it.

    The core will get hot, so the primary will appear nominally resistive. There will be some slight distortion of the waveform at crossover.
  22. Mar 16, 2009 #21
    Maybe we're mixing up real vs. ideal. My example was ideal based. No current exists in the primary due to secondary open except Imag. But a real CT has primary leakage reactance which is likely smaller than the 12 ohm load. But the magnetizing reactance would be larger than 12 ohms. Hence the full 120V would not appear at the primary, but only a part of it. Maybe 90V rms, or even less.

    Still, the CT would be destroyed either way.

  23. Mar 16, 2009 #22
    For that we'd need to look at the hysteresis loop.
  24. Mar 16, 2009 #23


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    Core saturation always causes primary current to INCREASE due to the fact that the self inductance goes away after the point of saturation. The load limits the current. So a saturated core causes the inductive reactance of the primary to drop. Its already low since the idea of a current transformer is to avoid adding any extra impedance in series with the load. Ohms law says the voltage across the primary should drop if there was any voltage there to begin with. I would agree with phrak in that there could be some distortion at crossover since there will be a small part of the cycle around zero current (crossover) that there is enough self inductance to keep things 'normal' at these low currents. However, as soon as the core saturates, the impedance of the primary circuit drops to zero (where it should be) and everything goes on as normal other than a fried current transformer core and possibly secondary winding. I would say if you want to prove this wire a CT in series with a resistor and hook it up to a variac (preferrably isolated). Then using a scope watch the voltage across the resistor as you slowly increase the variac output voltage. You might have to experiment with a slightly loaded secondary in order to prevent the core from saturating in the too-low-to-measure region.
    I wish I had some CTs to experiment with, however, I don't. The best I could probably do is to use a small step down VT and try sneaking a turn of wire around the core and using the primary as the 'secondary' of my newly created CT. My single turn of wire should present a pretty low impedance so it would probably work fine as a CT for experimental purposes.
  25. Mar 17, 2009 #24
    Who knows the energy loss in a BH loop? Assigning units is driving me crazy this week...
  26. Mar 17, 2009 #25
    Also worth noting is that with a CT having its secondary open, the full 120V rms is across the primary, or in the real world case, most of it. Let's say Vp = 90V rms. Vs, for a 50:1 turns ratio, would be 50*90V or 4500V rms. The insulation likely does not withstand this elevated voltage. Even if it does, the shock hazard is enormous.

    A CT should never be open circuited on its secondary. That was my point. If the secondary insulation fails at the elevated voltage, then the CT is for all practical purposes destroyed.

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