Transformer Primary/Secondary Current Relationships: Split Single Phase Case

[wwhitney]

For the simplest case of an idealized single phase transformer with 2-wire primary and secondary, and linear loads, the primary current will be a real constant times the secondary current, according to the turns ratio. NEC 240.4(F) therefore allows a primary side OCPD to be used for protecting the secondary conductors accordingly.

What happens if the secondary is replaced with a 3-wire center tapped coil? Suppose the turns ratio is 1 for simplicity, since that captures the new behavior. For a given current on the primary coil, I believe the worst case, in terms of large secondary currents, is that all of the load is connected to one-half of the secondary coil, so the secondary current is twice the primary current. Is that correct?

I'm assuming there are no power sources connected directly to the secondary circuit; otherwise, I believe you can generate power source / load pairs that give you zero current on the primary, so the secondary current could be arbitrarily large, given only a cap on primary current.

Then with those assumptions, it seems like a 240.4(F)-type allowance would be that the primary OCPD could protect the secondary conductors, as long as we use twice the primary OCPD value for sizing the secondary conductors. Yes?

Of course, to do this in practice, in addition to needing NEC allowances that are absent, we'd need a special transformer whose secondary coil is sized to handle all of the power on just half of the coil.

Thanks,
Wayne

 

[Carultch]

The issue is that unbalanced distribution of loads on the center-tapped secondary, create a blindspot in the primary OCPD.

Consider a 480V : 120/240V transformer, that is meant for an 80A circuit on the secondary. Suppose we have an overload of 100A on L1 and a much lower load of 20A on L2. The total apparent power will be 100*120 + 20*120 = 14.4 kVA, which will reflect as 30A on the primary. To a 40A primary breaker, this would look like business as usual, and this breaker wouldn't trip. Therefore, the 40A primary breaker doesn't provide the same protection as an 80A secondary breaker, like you'd be expecting it to do.

You can only take credit for primary breakers indirectly protecting the secondary conductors, if overcurrent directly lines up across winding pairs. This is why 240.21(C) limits this practice to qualifying topologies, like 2-wire to 2-wire, and delta-to-delta with 3-wire for both sides. WYE topologies can create the same issues of redistributing unbalanced load currents, for 3-phase loads.

 

[wwhitney]

Consider a 480V : 120/240V transformer

Your example shows that with a 2:1 voltage ratio, you can't say that a 40A primary breaker will prevent secondary currents in excess of 80A.

But I never suggested that it would. Instead, my claim was that the worst that could happen would be an additional factor of 2. So a 40A primary breaker would be overloaded whenever any secondary current is in excess of 160A. Is that correct?

If so, and if the secondary coils, secondary conductors, and equipment were sized for 160A currents, while only requiring 40A*480 = 19.2 kVA of power, then the 40A primary OCPD would prevent any overloads on the primary or secondary side.

This is not meant as a practical example, and there is no NEC allowance for this. But I would like to check that I understand the worst case discrepancy possible. If I've got this right, then I'll think about the 3-phase delta-wye case.

Cheers, Wayne

 

[Carultch]

So a 40A primary breaker would be overloaded whenever any secondary current is in excess of 160A. Is that correct?

If so, and if the secondary coils, secondary conductors, and equipment were sized for 160A currents, while only requiring 40A*480 = 19.2 kVA of power, then the 40A primary OCPD would prevent any overloads on the primary or secondary side.

That's correct. In my example, the worst possible imbalance that would be in the blindspot of the primary breaker, would be 160A on L1 and 0A on L2. In concept, promoting the secondary conductors to a 160A rating, should mitigate this issue, at least when it comes to overload protection.

Since few people would plan on doubling the ampacity to avoid a main breaker, the NEC's CMP probably just decided to keep it simple, and require OCPD on secondaries of non-qualifying topologies.

This is not meant as a practical example, and there is no NEC allowance for this. But I would like to check that I understand the worst case discrepancy possible. If I've got this right, then I'll think about the 3-phase delta-wye case.

For a mixed topology 3-phase transformer, the "blindspot factor" of the primary breaker would be sqrt(3), assuming unity power factor. So, given a 208 Delta to 208/120 wye transformer, you could have up to 173 amps of overload on the secondary, that is in the blindspot of a 100A primary breaker.

 

[wwhitney]

For a mixed topology 3-phase transformer, the "blindspot factor" of the primary breaker would be sqrt(3), assuming unity power factor.

I don't want to assume unity power factor, but I do want to assume linear loads and idealized transformers. Does that change the answer in either case? Since the primary-secondary current relationships are just linear equations with real coefficients, regardless of the voltage-current phase relationship, I don't think it would.

Also, I will try working through the math in the 3 phase delta-wye case, but both answers come out to be the ratio of the primary line-line voltage to the smallest secondary voltage present. Is there some easy argument that says that has to be the factor?

Cheers, Wayne

 

[Carultch]

I don't want to assume unity power factor, but I do want to assume linear loads and idealized transformers. Does that change the answer in either case? Since the primary-secondary current relationships are just linear equations with real coefficients, regardless of the voltage-current phase relationship, I don't think it would.

Also, I will try working through the math in the 3 phase delta-wye case, but both answers come out to be the ratio of the primary line-line voltage to the smallest secondary voltage present. Is there some easy argument that says that has to be the factor?

Cheers, Wayne

I haven't explored what the possibilities are for phase-shifted loads. My expectation is that unity power factor would be the worst-case scenario for the blindspot factor. It would be sqrt(3) for 3-phase mixed-topology, and 2 for single phase 2-wire to 3-wire.

The way I calculated it for 3-phase:
1. Start with 100A on phase A, and 0A on the other two phases.
2. Assign primary winding AB to be paired with secondary winding A. Correspondingly, primary BC and secondary B, as well as primary CA and secondary C.
3. Take the V_LN for the secondary wye system, and V_LL for the primary delta system. This forms the turns ratio of the individual winding pair, and calculate the corresponding inter-phase currents on the primary.
4. Use Ia = sqrt(Iab^2 + Ica^2 + Iab*Ica), and other applicable formulas, to translate the inter-phase currents on the primary to the individual phase conductor currents.
5. This calculates for us, the primary current that corresponds to the 100A secondary current, for which I got 57.7A on primary phases A and B (and zero on primary phase C). If we consider a 100A OCPD on the primary, and want the corresponding maximum secondary overload, then we'd take the reciprocal of this ratio, and multiply by 100A.

If it were a 480V delta to 120/208V wye, then a 100A maximum unbalanced secondary current, would correspond a 25A primary current on A and B. This would mean a 100A primary OCPD would enable up to a 400A secondary overload, in a setup where you'd likely have a 225A circuit on the secondary.

 

[wwhitney]

The way I calculated it for 3-phase:
1. Start with 100A on phase A, and 0A on the other two phases.
2. Assign primary winding AB to be paired with secondary winding A. Correspondingly, primary BC and secondary B, as well as primary CA and secondary C.
3. Take the V_LN for the secondary wye system, and V_LL for the primary delta system. This forms the turns ratio of the individual winding pair, and calculate the corresponding inter-phase currents on the primary.

For a transformer that is, say, 480V delta to 480Y/277V wye, the turns ratio is 1/sqrt(3) (or its reciprocal, I don't know if there's a standard presentation). If I want to stick with a 1:1 turns ratio, I'd need to consider say a 240V delta to 416Y/240V transformer.

4. Use Ia = sqrt(Iab^2 + Ica^2 + Iab*Ica), and other applicable formulas, to translate the inter-phase currents on the primary to the individual phase conductor currents.

Seems like for the case of secondary currents (100,0,0,-100) (A,B,C,N), the primary coil currents are just (100/sqrt(3),0,0) for your example. And then the line currents are just the pairwise differences, so for appropriate sign and ordering choices, (100/sqrt(3), -100/sqrt(3), 0). Which agrees with what you got, but doesn't require any formulas.

So to me that seems like the "blind spot" factor for this example is 1, in that the current ratio is just the turns ratio. I.e. I'd say we need to compare the voltages across the coils as our starting point, not compare the L-L voltages.

Cheers, Wayne

 

[wwhitney]

Rather than start a new thread, some general comments on delta-wye currents. If I'm painfully reworking some standard treatment, please point me to a good reference to review. : - )

So for a delta-wye transformer (idealized transformer and AC power source with all linear circuit elements, so all waveforms are sinewaves of the fundamental frequency, and representable as complex numbers), we have 4 possible sets of currents. [A diagram would be good here, but hopefully the text is clear enough.]

1) Secondary line currents, which I'll call (I_a, I_b, I_c, I_n). If we define the positive sense as, say, towards the transformer in all case, these sum to zero as complex numbers. So there are 3 independent variables here.

2) Secondary coil currents. They just coincide with (I_a, I_b, I_c), since each coil is L-N.

3) Primary coil currents, which we can call (I_AB, I_BC, I_CA). But if those coils (AB, BC, CA) correspond with the secondary coils (a,b,c), then (I_AB, I_BC, I_CA) = (I_a, I_b, I_c) for one choice of sign convention on the coil currents per choice of the wye side configuration. [Whether the wye side of the transformer is connected with the dotted ends toward the neutral or away. Hopefully that's right, I haven't fully dived into the dot notation.]

4) Primary line currents, which we can call (I_A, I_B, I_C), choosing the positive sense again to be towards the transformer. Then for one choice of wye side configuration we have (I_A, I_B, I_C) = (I_AB - I_BC, I_BC - I_CA, I_CA-I_AC). [Hopefully I got my sign/configuration choices correct, i.e. the above is possible and also fully specifies the choices.]

Then 3 and 4 together give (I_A, I_B, I_C) = (I_a - I_b, I_b - I_c, I_c - I_a). This is a degenerate transformation from 3 independent variables to 3 variables with one relationship, namely I_A + I_B + I_C = 0 Which relationship is evident from the previous equation, and is also evident from the conservation of charge going through the boundary of the transformer.

So what is the kernel (preimage of zero) of this mapping from (I_a, I_b, I_c) to I_A, I_B, I_C)? It is just (I,I,I) for any complex current I, i.e. all secondary non-neutral currents have the same magnitude and phase.

But for this to happen with I nonzero, one of the secondary coil voltages must be more than 90 degrees out of phase from the current I. Can this happen with linear loads, or does it imply that there is a power source connected to the transformer secondary?

Since the math says the primary currents would be zero, I think conservation of energy says there's no power coming into the transformer on the primary side, so I assume that means there has to be a power source on the secondary side.

Thanks, Wayne

 

[Carultch]

But for this to happen with I nonzero, one of the secondary coil voltages must be more than 90 degrees out of phase from the current I. Can this happen with linear loads, or does it imply that there is a power source connected to the transformer secondary?

It can happen just with loads, but in a practical sense, it will most likely happen with a power source instead of a load. Because we generally try to avoid building loads that do this, and most loads will usually have a dominant type of phase shift.

Resonance in LC circuits is one way that a 180 degree phase shift can be produced. An LC circuit with no resistance produces a 180 degree phase shift at all frequencies, and produces an infinite steady state amplitude when driven at exactly the resonant frequency. Including a resistance in the circuit will mitigate the resonance behavior, and the phase shift won't be 180 degrees everywhere, but only close to the resonant frequency.

How can this be, with conservation of energy? What ends up happening is that the energy cannot exit an ideal LC circuit. An ideal LC circuit that is driven at its resonant frequency will start at zero and ramp up in amplitude indefinitely.

 

[Carultch]

For a transformer that is, say, 480V delta to 480Y/277V wye, the turns ratio is 1/sqrt(3) (or its reciprocal, I don't know if there's a standard presentation). If I want to stick with a 1:1 turns ratio, I'd need to consider say a 240V delta to 416Y/240V transformer.


Seems like for the case of secondary currents (100,0,0,-100) (A,B,C,N), the primary coil currents are just (100/sqrt(3),0,0) for your example. And then the line currents are just the pairwise differences, so for appropriate sign and ordering choices, (100/sqrt(3), -100/sqrt(3), 0). Which agrees with what you got, but doesn't require any formulas.

So to me that seems like the "blind spot" factor for this example is 1, in that the current ratio is just the turns ratio. I.e. I'd say we need to compare the voltages across the coils as our starting point, not compare the L-L voltages.

What I call the "blindspot factor", is the following:
Blindspot factor = M*Vs/(B*Vp)

Primary breaker rating: B
Maximum possible secondary current in an unbalanced case: M
Primary voltage (phase-to-phase): Vp
Secondary voltage (phase-to-phase): Vs

Since the primary current is 1/sqrt(3) times the secondary current in this example, and Vp/Vs = 1, this means that the blindspot factor is sqrt(3).

 

[Carultch]

An LC circuit with no resistance produces a 180 degree phase shift at all frequencies, and produces an infinite steady state amplitude when driven at exactly the resonant frequency. Including a resistance in the circuit will mitigate the resonance behavior, and the phase shift won't be 180 degrees everywhere, but only close to the resonant frequency.

I revisited this, and realized I needed to correct it. I misinterpreted this yesterday.

The ideal LC circuit either has a phase shift of +90 deg or -90 deg, depending on whether L dominates or C dominates. At resonance exactly, you can't really define the phase angle.
The LRC circuit will have 90 degrees of phase shift at resonance, and an approach to a 180 degree phase shift in the stopband, so that would be where it meets the intent of your question. Granted, it doesn't draw a lot of current there anyway, since it is the stopband.

 

[wwhitney]

What I call the "blindspot factor", is the following:
Blindspot factor = M*Vs/(B*Vp)

Yes, I understand and I agree now that makes sense. I was thinking that there was something qualitatively different between the 3 phase and single phase cases, but now I'm thinking not.

Getting back to the single phase case in the OP, and sticking with turns ratio 1, suppose we construct our transformer with 120V/240V secondary out of two individual 2-wire primary, 2-wire secondary transformers, as in the diagram below. Then the primary side OCPD protects the secondary conductors: as phasors, current in X2 = current in coil 4 = current in coil 2 = current in L2; similarly for X1 and L1. Yes?

Now if we (a) combine N1 = N2, then (b) put the transformers on a common core, and then (c) delete N1/N2, we are back to the usual case in the OP where the blind spot factor is 2. [Steps (a) and (b) could also be done in either order.] I'm having trouble seeing which of the steps increases the blind spot factor from 1 to 2. Presumably step (c), but how does that change the primary circuit to cause the currents to be rebalanced to create the blind spot factor of 2?

Thanks,
Wayne

 

[synchro]

The ampere turns (in phasor form) of all the windings on a common core have to sum to zero. This is assuming the magnetic flux flowing through all coils is the same, i.e. there's negligible leakage flux. Therefore, if all the windings shown above are on a common core, N1 AND N2 are connected but not driven, and say we put a load drawing 10A across X2 and X0, then 5A will flow between L1 and L2. The primary will have twice as many turns utilized as the secondary, and so it can create the same value of ampere turns with half of the current drawn on the secondary.

 

[wwhitney]

The ampere turns (in phasor form) of all the windings on a common core have to sum to zero.

Thanks for your response. In the notion of ampere-turns, is turns a scalar, so ampere-turns has the same phase as the current?

Therefore, if all the windings shown above are on a common core, N1 AND N2 are connected but not driven, and say we put a load drawing 10A across X2 and X0, then 5A will flow between L1 and L2.

OK, that's for the simple case of a single primary core, the terminals N1 and N2 connected to each other, but N1/N2 not connected to any primary circuit conductor. What, if anything, changes if that N1/N2 terminal is connected to the primary neutral conductor?

Cheers, Wayne

 

[synchro]

In the notion of ampere-turns, is turns a scalar, so ampere-turns has the same phase as the current?

Yes. In common usage ampere turns is also a scalar, but just like when applying Kirchoff's Laws with AC voltages and currents, a phasor/vector representation may be necessary.

OK, that's for the simple case of a single primary core, the terminals N1 and N2 connected to each other, but N1/N2 not connected to any primary circuit conductor. What, if anything, changes if that N1/N2 terminal is connected to the primary neutral conductor?

Unless the L1-N and L2-N voltages are well balanced, there may be a high current that flows. This would be similar to applying voltage sources to both a primary and secondary winding.

 

[jim dungar]

Thanks for your response. In the notion of ampere-turns, is turns a scalar, so ampere-turns has the same phase as the current?


OK, that's for the simple case of a single primary core, the terminals N1 and N2 connected to each other, but N1/N2 not connected to any primary circuit conductor. What, if anything, changes if that N1/N2 terminal is connected to the primary neutral conductor?

Cheers, Wayne

If the N connection is made to the source neutral it will become fixed in relation to L1 and L2. The fixed relationship will mean it will be harder to tolerate imbalance and it is likely some circulating currents will form.

 

[wwhitney]

Unless the L1-N and L2-N voltages are well balanced, there may be a high current that flows.

Assuming the transformers coils are perfectly balanced, if the primary voltages are not, I take it the current that flows is determined by the requirement to balance the voltages seen by the transformer coils. E.g. if relative to some 0V reference, L1 = 120V, N = 1V, and L2 = -121V, if the N and L2 conductors (and upstream source) are purely resistive and of equal resistance, enough current would flow L2-N to shift the voltages to L1 = 120V, N = 0V, and L2 = -120V? And this depends on the common core criterion; if the cores are separate as in my drawing, then there is nothing that requires the volts/turn to be the same on each coil.

Now for the common core, primary neutral connected, primary side perfectly balanced case, how are the primary currents determined from the secondary currents? If the primary side has no impedance, it seems like the primary currents are under constrained--i.e. 10A on the secondary X0-X2 could be 5A on the primary L1-L2, or it could 10A on the primary N-L2. But then in reality with the primary having impedance, the current division would be determined by the requirement to maintain the balance of the primary voltages?

Thanks,
Wayne

 

[gar]

230702-2144 EDT

Going back to the first post I am going to make the following assumptions based on how a residential single phase transformer would be arranged.

The primary is a single two terminal coil. The secondary is an identical winding with the total secondary winding the same number of turns as the primary, but also has a center tap brought out. Thus, both windings have the same size wire, and number of turns. We assume the transformer is wound so that there is about equal internal power loss at full load in each winding.

This means, if there is no load on 1/2 of the secondary, that the approximate maximum load power on the transformer secondary can be only 1/2 of the transformer rating.

.