Why does the magnetic flux in a CT core decrease with connected load?

[b.shahvir]

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

 

[cabraham]

"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?

Claude

 

[b.shahvir]

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 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?

Kind Regards,
Shahvir

 

[cabraham]

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?

Claude

 

[b.shahvir]

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,
Shahvir

 

[cabraham]

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

 

[b.shahvir]

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.

Sorry! system error

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

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.

 

[cabraham]

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

 

[b.shahvir]

In a CT, if the insulation exhibits zero leakage, then the flux never changes with burden. Claude

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 please simplify or elaborate a bit for me on the insulation leakage part if its not much trouble? Thanx very much.

Kind Regards,
Shahvir

 

[Carl Pugh]

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.

 

[cabraham]

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.

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.

Claude

 

[b.shahvir]

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

Kind Regards,
Shahvir

 

[Phrak]

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.

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?

 

[cabraham]

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?

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.

Claude

 

[Phrak]

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.

 

[Carl Pugh]

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.

 

[Phrak]

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.

Under what conditions?

 

[Averagesupernova]

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.

I haven't had a lot 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.

 

[cabraham]

I haven't had a lot 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.

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.

Claude

 

[Phrak]

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.

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.

 

[cabraham]

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.

Claude

 

[Phrak]

Still, the CT would be destroyed either way.

For that we'd need to look at the hysteresis loop.

 

[Averagesupernova]

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.

 

[Phrak]

Who knows the energy loss in a BH loop? Assigning units is driving me crazy this week...

 

[cabraham]

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.

Claude

 

[cabraham]

In real world terms, considering non-ideal behavior, the CT when terminated in a low-Z has the full 10 amp input, with very low primary voltage Vp. As the secondary terminating resistance increases, the sec & pri voltage both increase. As the secondary resistance increases to very large values, the current in the primary will decrease, since we now add a resistance to the 12 ohms. At Rs open, the maximum value of Vp & Vs takes place, with a value of Ip determined from the B-H curve. In this case the secondary insulation will not likely survive. Once secondary insulation breakdown takes place, the secondary is shorted. Then the Vp & Vs values should decrease.

Off the top of my head, that is what appears to be the case. Imag increasing is not as big a deal as I thought since the one turn primary can handle up to 10 amps easily. But the core flux would increase with an open secondary since the primary current Ip, is not balanced by Is, and all of Ip becomes Imag. The core flux increases substantially.

No matter how we look at it, we should never open secondary a CT.

Claude

 

[Phrak]

90V rms won't appear across the primary of a current transformer, with open secondary, used in series current sensing. To do so, the current transformer will have to both remain inductive, and present an impedence of about 3.5 times the load.

Recall Faraday's law, the voltage per turn is equal the the rate change in flux. Toward saturation the rate change in flux is diminished. The transformer equation holds, to an extend, while the permeability is much greater than air, u>>u_0, but the CT primary presents a diminishing impedence to the circuit. As u --> u_0, the transform becomes more like some looped wires in proximity to an inductor.

Also, one might also wonder about the initial permeability as a function of self heating. Permeability deminished as temperature rises. This will also decrease the inductance of the CT primary.

 

[Averagesupernova]

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.

Claude

Have you read any of my posts? We are still not in agreement of what the primary voltage is doing when a current transformers secondary is open circuited. This has been the whole point of my posts.
-
Do you think taking a current transformer core without the secondary winding and wrapping it around a current carrying conductor will simply block 3/4 to all of the voltage from getting to the load in a 60 Hz circuit? Nothing that you can hold in your hand and simply wrap around a current carrying conductor is able to do what you claim in the real 60 Hz world regardless of whether or not there is a secondary winding.
-
If you use a voltage transformer with multiple primary windings as a makeshift current transformer then yes, I can see what you say could be true but certainly not with a single primary turn.

 

[cabraham]

Have you read any of my posts? We are still not in agreement of what the primary voltage is doing when a current transformers secondary is open circuited. This has been the whole point of my posts.
-
Do you think taking a current transformer core without the secondary winding and wrapping it around a current carrying conductor will simply block 3/4 to all of the voltage from getting to the load in a 60 Hz circuit? Nothing that you can hold in your hand and simply wrap around a current carrying conductor is able to do what you claim in the real 60 Hz world regardless of whether or not there is a secondary winding.
-
If you use a voltage transformer with multiple primary windings as a makeshift current transformer then yes, I can see what you say could be true but certainly not with a single primary turn.

If it is a very small CT, I'd agree. A CT is generaly sized to drive an ammeter. It does not carry anywhere near the full VA of the network whose current is being measured.

But I stand by what I said regarding flux density in the core increasing. The primary voltage will not reach the full 120V, but it will be a portion of it. If it is only 10V of the 120V, with a 50:1 truns ratio, the Vs is still 500V rms, still quite dangerous.

Again, I tended to mix ideal with real when stating my case. A real world CT has a primary reactance due to leakage in series with the magnetizing reactance. For a CT, which is only sized to drive an ammeter or current shunt, the VA is so low that a small core and one turn primary is used. Hence the magnetizing reactance is lower than that for a VT which supplies the entire VA of the network. The full 120V will not appear on Vp in the real world case.

But still, wouldn't you agree that even if 10 out of the 120V appears at Vp, a Vs of 500V rms is still dangerous? Also, if the normal Vp when the CT secondary is low-Z terminated, is 0.1V, then increasing Vp to 10V rms results in a 100-fold increase in core flux, and a greater increase in core losses.

I concede that Vp will not reach anywhere near 120V, but even 10V, or 5V still presents danger of core loss spiking, and shock hazard. You do agree don't you?

Claude

 

[cabraham]

Also, FWIW, the frequency is all important. If the CT one turn primary has a magnetizing reactance of 100 uh, then at 300 kHz, a common freq used in switched-mode power converters, the reactance is 188.5 ohms. With a 12 ohm load, the voltage at the primary after the divider action is 119.76V rms! So , when the frequency is in the neighborhood of 300 kHz, even a measly 100 uh of magnetizing reactance results in practically the whole 120V appearing at the primary. For a 10 uh primary inductance, quite small, the Vp value is 101.23V rms, still quite a bit.

But, at 60 Hz line freq, things are more in line with what you stated. With 100 uh, at 60 Hz, Vp is a mere 0.377V. Still the increase in core flux is substantial. Vs becomes 18.85V rms, not dangerous in most cases.

Maybe I was thinking SMPS with a switching freq in the 100's of kHz, and others were thinking 50/60 Hz. At line freq, it's a different story vs. SMPS freq.

Does this help?

Claude

 

[Averagesupernova]

Why do I get the feeling that you are packpedalling? The frequency definitely makes a difference. That's why my first post specified 120 V 60Hz. You implied it (at least in my mind) when you gave a 120 volt example. But as a rule, if the transformer design is matched to the rest of the circuit it should come out about the same regardless of frequency. That is not to say you can use a 60 Hz transformer on a 100Khz circuit. I'm not saying that. You can say all you want about flux density increasing but when the core saturates the core no longer offers any reactance and will allow an increase in current easily which is reflected in a LOWER primary voltage. Now if you REALLY want to mess your head up, in the ideal world, there is no difference between a current and a voltage transformer. So, in the ideal world the full voltage would definitely appear across the primary with an open secondary.
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Incidentally, I've NEVER said it was a good idea to open the secondary on a current transformer. Unless you are experimenting there is no good or practical reason to do so.

 

[cabraham]

Why do I get the feeling that you are packpedalling? The frequency definitely makes a difference. That's why my first post specified 120 V 60Hz. You implied it (at least in my mind) when you gave a 120 volt example. But as a rule, if the transformer design is matched to the rest of the circuit it should come out about the same regardless of frequency. That is not to say you can use a 60 Hz transformer on a 100Khz circuit. I'm not saying that. You can say all you want about flux density increasing but when the core saturates the core no longer offers any reactance and will allow an increase in current easily which is reflected in a LOWER primary voltage. Now if you REALLY want to mess your head up, in the ideal world, there is no difference between a current and a voltage transformer. So, in the ideal world the full voltage would definitely appear across the primary with an open secondary.
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Incidentally, I've NEVER said it was a good idea to open the secondary on a current transformer. Unless you are experimenting there is no good or practical reason to do so.

I apparently overlooked your mention of 60 Hz as the frequency. You did say "120V 60 Hz", but I was thinking in terms of a CT used in an SMPS, which I'm familiar with. At this low a freq, 50/60 Hz, the CT presents a very low reactance. So we agree regarding line freq. As far as a CT & VT having no difference in the ideal world, I've always stated such. The laws of Ampere & Faraday describe both VT & CT operation.

As far as core saturation goes, sure the reactance plummets when viewed in steady state frequency domain terms, but it's the single ac cycle we're dealing with & the hysteresis & eddy current losses increase. The core is driven into saturation in 1 direction, then the polarity of the ac reverses, driving the core back into its normal region, then into saturation the other way. The Vp value is larger than normal while the core is in the non-saturated region of its B-H curve. In saturation, the reactance is very low but why would the Vp value plummet?

When a core is subjected to an amp-turns value resulting in a specific H value, then the core saturates, H keeps increasing with increasing NpIp, whereas B flattens out due to saturation. The volt-second product flattens out, so with increasing time, Vp decreases gradually. But core losses are great since the B-H curve is traversing a high flux swing. I would expect a distorted Vp waveform, which someone else already mentioned. When the core enters saturation, the Vp does not sharply plummet, because B does not sharply plummet. Vp/Np is proportional to B where NpIp is proportional to H. The flux has to increase despite saturation. Why would we think otherwise?

But I don't understand why you say that when the core saturates, the voltage drops to near zero, so that flux is low? The flux is not related to volts, but volt-seconds. The current in the primary Ip, is limited by the power source & the load resisance. So we can model this scenario as a forced value of H, and compute B based on core B-H properties. The above paragraph suggests that core flux does not drop but, rather, it increases slightly, at the onset of saturation and continues to increase, albeit slightly. Whereas Vp will gradually decrease since the volt-second product, or B, is converging upon a limit, & as time goes forward, Vp must gradually decrease. Vp does not, however, plummet at the onset of saturation. Have I missed something?

These are good mental excercises. They keep me thinking. BR.

 

[b.shahvir]

Assume ideal conditions i.e. zero winding impedances in pri and sec. Also assume load or burden impedance to be zero. The sec will draw infinite amps, ideally! So, what will happen to core flux in this case? Will it increase, decrease, remain constant or disappear (become zero) altogether? The reply would be interesting!

 

[Averagesupernova]

I'll post more later, not sure if I'll have time tonight but I'll try and for sure won't have time tomorrow until late tomorrow night. Not a cop out, seriously.

 

[Averagesupernova]

...but I was thinking in terms of a CT used in an SMPS, which I'm familiar with. At this low a freq, 50/60 Hz, the CT presents a very low reactance. So we agree regarding line freq.

I would think a properly designed circuit would have a current transformer that is matched to the circuit concerning inductive reactance. A CT designed for a 60 Hz circuit would be a poor choice for a 100 Khz circuit. It would probably have enough inductance to insert too much impedance in series with the load which of course would be reflected as a voltage across the primary. So, when the higher frequency CT is properly sized down inductance-wise it should behave the same as the 60 Hz version when the secondary is opened. For the sake of this discussion though, let's stick to 60 Hz at least until we get that part ironed out.

But I don't understand why you say that when the core saturates, the voltage drops to near zero, so that flux is low?

I think you are overcomplicating things. Consider this scenario: Suppose we have a laminated iron core with lots of windings on it which gives us many henries of inductance. Now, suppose we hook a battery to this coil with a resistor that limits current to 50 amperes. The full battery voltage will appear across the coil the moment the switch is closed. Slowly the current ramps up and is limited to 50 amperes by the resistor. At this point we will say that the core has not saturated. But suppose that by design it does saturate at 75 amperes. Now we will cut the value of the current limiting resistor in half and do the same thing over. Current is now limited to 100 amperes. We connect the battery up and the current ramps up and goes past 50 amperes and continues on towards 75 amperes. It ramps up as expected, but when it reaches 75 amperes the core becomes saturated and there is no longer a CMF to counter the current. Now at 75 amperes the circuit is only limited in current by the resistor and the current slams up from 75 amperes to 100 amperes almost instantly. THIS is what happens when a core saturates. In current transformers that have the secondary winding properly shorted, there is virtually no chance of core saturation to begin with no matter how small the core is (within reason) as long as it is run within design limits. I believe you have shown this in previous posts. So, the core by design is shrunk down as small as possible which puts it somewhat close to becoming saturated if the secondary is terminated into something other than a short or at least something other than what it was designed for. One reason for this is cost reduction by reducing materials, another is that less core means less series impedance. The idea is to keep the primary voltage as low as possible in normal operation so all the voltage ends up across the load instead of part of it across the transformer. So now we have a transformer core that easily goes into saturation if the secondary is opened or partially opened. The hypothetical case I gave above with the 50/75/100 amp will apply here too. Since the CT core will saturate easily when run this way, it saturates virtually right away in the AC cycle. A saturated core will offer little impedance to current. I don't know how many different ways I can cover this to try to get you to understand why I say what I do. I guess it all depends on how far into the AC cycle that the core actually does saturate, but it is my belief that by design they would be made to saturate fairly quickly for reasons I gave above.

 

[Averagesupernova]

Assume ideal conditions i.e. zero winding impedances in pri and sec. Also assume load or burden impedance to be zero. The sec will draw infinite amps, ideally! So, what will happen to core flux in this case? Will it increase, decrease, remain constant or disappear (become zero) altogether? The reply would be interesting!

Well, I suppose it depends on what ideal is, but with 100% coupling and no leakage, infinite current in the both coils will result in no flux in the core. Doesn't transformer design theory tell us this? The only way flux actually builds up in the core is when a voltage starts to appear across the coil right? Since the flux in the core is what indirectly causes the reactance then if we say infinite current this implies zero reactance. I know it may sound like I am contradicting myself but it really isn't a contradiction.

 

[cabraham]

I would think a properly designed circuit would have a current transformer that is matched to the circuit concerning inductive reactance. A CT designed for a 60 Hz circuit would be a poor choice for a 100 Khz circuit. It would probably have enough inductance to insert too much impedance in series with the load which of course would be reflected as a voltage across the primary. So, when the higher frequency CT is properly sized down inductance-wise it should behave the same as the 60 Hz version when the secondary is opened. For the sake of this discussion though, let's stick to 60 Hz at least until we get that part ironed out.



I think you are overcomplicating things. Consider this scenario: Suppose we have a laminated iron core with lots of windings on it which gives us many henries of inductance. Now, suppose we hook a battery to this coil with a resistor that limits current to 50 amperes. The full battery voltage will appear across the coil the moment the switch is closed. Slowly the current ramps up and is limited to 50 amperes by the resistor. At this point we will say that the core has not saturated. But suppose that by design it does saturate at 75 amperes. Now we will cut the value of the current limiting resistor in half and do the same thing over. Current is now limited to 100 amperes. We connect the battery up and the current ramps up and goes past 50 amperes and continues on towards 75 amperes. It ramps up as expected, but when it reaches 75 amperes the core becomes saturated and there is no longer a CMF to counter the current. Now at 75 amperes the circuit is only limited in current by the resistor and the current slams up from 75 amperes to 100 amperes almost instantly. THIS is what happens when a core saturates. In current transformers that have the secondary winding properly shorted, there is virtually no chance of core saturation to begin with no matter how small the core is (within reason) as long as it is run within design limits. I believe you have shown this in previous posts. So, the core by design is shrunk down as small as possible which puts it somewhat close to becoming saturated if the secondary is terminated into something other than a short or at least something other than what it was designed for. One reason for this is cost reduction by reducing materials, another is that less core means less series impedance. The idea is to keep the primary voltage as low as possible in normal operation so all the voltage ends up across the load instead of part of it across the transformer. So now we have a transformer core that easily goes into saturation if the secondary is opened or partially opened. The hypothetical case I gave above with the 50/75/100 amp will apply here too. Since the CT core will saturate easily when run this way, it saturates virtually right away in the AC cycle. A saturated core will offer little impedance to current. I don't know how many different ways I can cover this to try to get you to understand why I say what I do. I guess it all depends on how far into the AC cycle that the core actually does saturate, but it is my belief that by design they would be made to saturate fairly quickly for reasons I gave above.

I perfectly understand what you're saying about the current spiking upward at the onset of saturation. I've been a practicing EE 31 yrs. and am currently a doctoral candidate. After grad school (MS) I worked at an aerospace & defense corporation, fortune 50, in the magnetics design group. Among my products was CTs. Here is where you and I differ.

"Reactance" is a small signal linear concept. Inductance at the onset of core saturation is ambiguous. There is a "normal" inductance, as well as an "incremental" inductance. Picture a B-H curve where we are right at the onset of saturation. What is the inductance value? By definition, N*phi = L*I, or L = N*phi / I. But phi = Ac * B, where Ac = core cross sectional area. Hence L = Np*Ac*B / I. But B flattens out once saturation is reached. I, meanwhile, keeps increasing. A line drawn from the origin to the operating point on the B-H curve gives the normal value of L, or Lnorm. This value gradually decreases as I increases and the core goes deeper into saturation. But, if we draw a line tangent to the operating point, and take the slope Np*Ac*delta B / delta I, that is the incremental inductance, Lincr. At the onset of saturation it is this value, Lincr, that plummets.

Nonetheless, if we force a constant B, i.e. as in a VT, then one saturation is reached, the magnetizing current will indeed skyrocket. Since B flattens out, while H continues rightward (or leftward for negative half of cycle), forcing an increase in B incurs a huge increase in H.

But what happens when we force H, i.e. current mode of operation where the current is fixed by an independent power source and a load resistor? Picture the B-H curve. We increase H up to the onset of saturation. When we continue to increase H. remember that increasing H results in only a small increase in B, by definition of saturation. B is related to volt-seconds, not volts. Thus the volt-second product keeps increasing, but converges upon an asymptotic limit. As time progresses, the volts will decrease. The saturated core does not provide the near zero reactance you're saying it does.

Every 60 Hz cycle results in the B-H loop being traversed. The area inside the loop is the hysteresis loss (non-conservative), and more importantly, the area between the loop and the B axis is the stored and recoverable per cycle energy (conservative). Thus a core overdriven into saturation on both halves of the ac cycle stores much more energy and flux than an inductor with an incremental inductance value equal to that of the CT in saturation.

Your position is that since the CT core is saturated, its reactance is tiny and thus the Vp value is tiny. But you are not considering Lnorm, but rather Lincr. The B-H loop clearly demonstrates that the flux, energy and Vp value are determined by Lnorm, and not Lincr. Big difference. Is this helpful? BR.

 

[b.shahvir]

Hi, can someone please explain to me the logic behind the representation of 'Magnetic Hysteresis loss' as a resistance in electrical equivalent circuits?... will be extremely grateful.

I have studied some info on this subject on the net. Even though the physics of Hysteresis Loss has been explained appropriately, I wish to dwell deeper into the phenomenon to get to the source of my doubt. In spite of the relevant theories, it becomes difficult to visualize a magnetic phenomenon into an electrical parameter as mentioned by me earlier. When we talk of energy expended by the power source to ‘pump up’ the magnetic field in the core, sounds a bit abstract!

I want to understand what phenomenon in the magnetic hysteresis causes the machine (Xmer or motor) to suck an additional losses (wattful) current from the power source.
Is it a result of the delay caused by the magnetic dipole friction in the magnetic core... which prevents the magnetic field from building up quickly? I would be grateful if someone could elaborate on this part. Thanx.

Kind Regards,
Shahvir

 

[Phrak]

Thus the volt-second product keeps increasing, but converges upon an asymptotic limit. As time progresses, the volts will decrease. The saturated core does not provide the near zero reactance you're saying it does.

The asymtotic limit, B= B_n + H \mu_{i} is obtained from the incremental permeability, which is, of course the permeability of air. It is one of a family of curves having constant slope. We should shift our perspective, with the permeability of air set to unity, to scales where \Delta B = \Delta H.

In best case, the incremental inductance approches a lower limit as \mu \rightarrow \mu_{0}[/tex]; the incremental inductance is that of an inductor with the core removed.<br /> <br /> L=N^2 \mu_{0}A_e / l_e<br /> <br /> In reality it&#039;s less, as both A_e and l_e vary, taking up all space; the mutual inductance between windings is decreased.

 

[Averagesupernova]

cabraham, can you explain how an SCR works?

 

[Phrak]

Hi, can someone please explain to me the logic behind the representation of 'Magnetic Hysteresis loss' as a resistance in electrical equivalent circuits?... will be extremely grateful..

A piece of fermomagnetic material is full of domains. A domain is a region of uniform magnetic dipole moment. In a piece of deGaussed material the domains point in random directions constrained by the crystaline latice to discrete orientations. Magnetizing the material means you are making more domains point in a preferred direction. The more domains that align, the further up (or down) the B-H curve you are. It takes energy to reorientate that domains. I don't know why. Ask the guys proficient in solid state physics. It could be bulk lattice distoration, or energy state changes of the atoms at grain boundries--any number of things...

 

[cabraham]

cabraham, can you explain how an SCR works?

Yes, I can, but the OP is about CT magnetics. For an SCR treatise, just visit a semiconductor OEM site. Since I do not produce SCRs, my explanation would just be a verbatim repeat of an OEM explanation. I'd recommend you go straight to the horse's mouth.

If you have a specific question about an SCR, that is not covered in the OEM treatise, I will gladly help you, if I can. BR.

Claude

 

[Averagesupernova]

Yes, I can, but the OP is about CT magnetics. For an SCR treatise, just visit a semiconductor OEM site. Since I do not produce SCRs, my explanation would just be a verbatim repeat of an OEM explanation. I'd recommend you go straight to the horse's mouth.

If you have a specific question about an SCR, that is not covered in the OEM treatise, I will gladly help you, if I can. BR.

Claude

Saturable Core Reactor. Power magnetics should be right up your alley I'd say. I might add it is closely related to the discussion at hand. Thought you'd see the close relationship. Sorry for the assumption.

 

[cabraham]

Saturable Core Reactor. Power magnetics should be right up your alley I'd say. I might add it is closely related to the discussion at hand. Thought you'd see the close relationship. Sorry for the assumption.

Well, I've used semiconductor SCRs in my career, more so than saturable reactors, so that is what immediately came to mind.

SCR will now refer to sat core reactor. I was going to mention the SCR concept when replying to you, but for brevity I opted not to. It's worth discussing.

Examine a B-H loop. If we bias the inductor with dc, so that it has a quiescent point at a flux well beneath saturation, it has a permeability to an ac signal with a substantial slope. It's value is mu_inc, the incremental slope of the B-H curve at the dc bias point. The inductance, and thus the reactance, to a small ac signal swing, is determined by mu_inc which sets Linc.

If we increase the dc bias to a point beyond the onset of saturation, the slope to the tangent of the B-H curve is much less than before. Hence the mu value plummets, as does the inductance. Linc & mu_inc are eventually that of air for hard saturation.

That is the point you were making. But there is a difference when dealing with an overdriven CT. We are not biasing the core into saturation, and examining the small-signal inductance and mu. Rather, we are traversing through the entire B-H curve. The inductance is by definition, the flux linkage per unit current, or L = N*phi/I. Although the core saturates, the effective mu & L are much greater than the SCR case. If I get time, I'll make a sketch and attach it. For now, draw the B-H curve, and examine what happens for large signal ac stimulus. The energy stored and recovered is in the area between the B-H curve and the B axis (outside the loop). Of course, the area inside the loop is lost as hysteresis. But both ares increase greatly.

Did I help?

Claude

 

[Averagesupernova]

I can't seem to find something I've read on the web before about saturable core reactors. You didn't mention it in your last post. At one point it was explained that an SCR is driven into saturation by the bias winding but it is actually 2 windings. Don't take this word for word because I can't remember it completely and can't find it through google. What was explained, and it made sense at the time, was that one coil saturates for one half of the AC cycle and the other coil for the other half. I recall having to read it twice at the time before it made sense. Right now I can't think of how that would work.
-
From what I read on wiki in the last couple of hours, the BH curve of the typical SCR is sharper than other types of cores. This is something I did not know. So it goes into saturation quickly compared to a slower transition of typical transformers. It has occurred to me that when an AC current runs through a CT that it is in fact running the core through the entire BH curve with every cycle. In fact, I realized this early on before this discussion got off the ground very far at all. But, it is still my belief that the amount of inductance the relatively small core of a CT offers is not enough to impede the current. If it is, then I have a lot of practical joking to do by ordering up a bunch of surplus toroids (no windings) and install them in some unsuspecting souls breaker box. (Kidding of course) However, something just occurred to me. A while back I was searching on forums.mikeholt.com. Not until now did I recall this:
-
http://forums.mikeholt.com/showthread.php?t=111133&highlight=phase+short

Read the thread and tell me what you think. Obviously current transformers get disconnected and left in place from time to time. The pic in the above link shows a problem that wasn't really related to current transformers and apparently the people residing in this house didn't have any complaints. So I would assume that these particular CTs didn't drop any noticable voltage on the 'one turn' with opened secondarys. Unrelated to CTs, the pics in that mikeholt thread just make me cringe. I've run into some things that have had melted and burned up wires before, but the near phase to phase short that was about to occur in those pics could have really caused quite the fireball/explosion inside that box. WOW!

 

[Phrak]

"The inductance is by definition, the flux linkage per unit current, or L = N*phi/I."

--except that it's not a relevant equation. This equation is applicable when coupling between each loop is unity.

 

[cabraham]

"The inductance is by definition, the flux linkage per unit current, or L = N*phi/I."

--except that it's not a relevant equation. This equation is applicable when coupling between each loop is unity.

Ahh, I don't think so, please recheck your reference. The equation I gave has nothing to do with coupling between 2 loops. It holds for a simple inductor. Self-inductance is what it refers to. That is the most basic definition of inductance. Any EE, e/m fields, or physics reference will affirm it. If not, then just what is the basic definition of L?

Claude

 

[Phrak]

Ahh, I don't think so, please recheck your reference. The equation I gave has nothing to do with coupling between 2 loops. It holds for a simple inductor. Self-inductance is what it refers to. That is the most basic definition of inductance. Any EE, e/m fields, or physics reference will affirm it. If not, then just what is the basic definition of L?

Claude

Maybe... How do you define \Phi ?

edit: I should add an example. You have an inductor consisting of a long wire with a single loop and a distance away two loops that are coincident. Obviously the flux through the two loops will be geater. Which flux is called phi?

 

[m.s.j]

In transformer regardless of VT or CT, the linkage flux between primary and secondary is obtained by difference of primary and secondary's ampere-turn. In voltage and current transformers, the linkage flux is established by linkage EMF. The linkage EMF in VT is made by primary voltage (neglect from primary drop voltage) and in CT is made by secondary drop voltage on CT internal impedance and burden resistance. Indeed, the flux curve work point of CT is changed by load variation but it is fixing in voltage and power transformers if we can neglect the primary voltage drops.

REGARDS
M.S.J

------------------------------
http://electrical-riddles.com

 

[Phrak]

In transformer regardless of VT or CT, the linkage flux between primary and secondary is obtained by difference of primary and secondary's ampere-turn. In voltage and current transformers, the linkage flux is established by linkage EMF. The linkage EMF in VT is made by primary voltage (neglect from primary drop voltage) and in CT is made by secondary drop voltage on CT internal impedance and burden resistance. Indeed, the flux curve work point of CT is changed by load variation but it is fixing in voltage and power transformers if we can neglect the primary voltage drops.

REGARDS
M.S.J

------------------------------
http://electrical-riddles.com

Not to confuse mmf with emf.