A resistive load on the secondary, up to and including a short, will not increase the flux in the core. In fact, drawing current from the secondary will slightly decrease the core flux due to the IR drop in the primary.

Wouldn't a high load cause a large decrease in the core flux, since the secondary's current creates an opposing magnetic field that cancels out the primary's?

Presumably power supply authorities do their best to remove any DC offset from the mains supply. Considering that most consumers are connected to sub-stations with transformers it's unlikely under normal circumstances that a DC offset in the mains supply would be observed or if it did exist, it could not propagate very far down the distribution system. The next transformer lower down in the system network would not pass the DC onwards.

Wikipedia: "On transformers connected to long, overhead power transmission lines, induced currents due to geomagnetic disturbances during solar storms can cause saturation of the core and operation of transformer protection devices."

 

This thread is very old, even though some members are still around. If you want to start a new conversation, I suggest you open a thread of your own. If you just want to comment, even though it's not exactly fitting, this one can be extended.

 

This thread is very old

I knew someone would do this.

 

Wouldn't a high load cause a large decrease in the core flux, since the secondary's current creates an opposing magnetic field that cancels out the primary's?

The secondary current doesn't completely cancel the primary's field. The OP stated the true facts of the situation in the first post:

"When you attach a resistive load to the secondary then the secondary load current is in phase with the secondary voltage. This causes additional current to flow in the primary, which must also be in phase with the primary voltage.

However, the secondary current does not cause additional flux because the secondary magnetomotive force (MMF) is cancelled by an opposing MMF in the primary, so the net flux remains unchanged by loading."

The flux in the core is set by the voltage applied to the primary winding and a little current (the magnetizing current) in the primary is related to this flux. A secondary load causes secondary current, and also some primary current. The increment of primary current is "canceled" by the secondary current, but the secondary current doesn't cancel the magnetizing current.

The primary current causes a small loss of the "effective" voltage applied to the core due to the IR drop in the resistance of the primary winding. This in turn causes a small drop in the core flux, so a secondary load actually causes a slight decrease in core flux.

 

The increment of primary current is "canceled" by the secondary current, but the secondary current doesn't cancel the magnetizing current.

Ah! Thank you. So the flux that is found in the core of the transformer is actually unrelated to the function of the transformer, and is just the result of the non-ideal inductive component of the primary? It's just the flux that would always be there if you completely remove the secondary coils and turn it into a plain inductor (same as open-circuited secondary)?

Like the case of a plain inductor, if you apply AC to the primary, energy is being stored in this magnetic field, but then it's returned to the primary circuit on the next half-cycle, and has nothing to do with the power transfer to the secondary?

 

It's just the flux that would always be there if you completely remove the secondary coils and turn it into a plain inductor (same as open-circuited secondary)?

Yes, except for the slight reduction in flux that is caused by the voltage drop in the resistance of the primary when the additional current flows in the primary to supply the loaded secondary current.

If you could wind the coils of a transformer with superconducting wire and completely remove the core, the transformer would still work as a transformer, except that the magnetizing current would be very large. But, if the voltage source driving the primary could supply that large magnetizing current without any problem, a load on the secondary would still cause an incremental increase in the primary current and the transformer would operate normally. The secondary would never know that the large magnetizing current was present.

In an ideal transformer, the primary inductance with nothing connected to the secondary would be infinite, and the magnetizing current would be zero. The main purpose of an iron core in a transformer is to cause the magnetizing current to be as small as possible.

Another way to cause the magnetizing current to be small is to increase the frequency of operation. Notice that radio frequency transformers can operate without a core (sometimes called an air core). In this case, even though the primary inductance is not very high, the frequency of operation is high enough to cause the magnetizing current to be reasonably small.

 

I knew someone would do this.

Then why did you do it? The thread is two years old, and hijacking is a rule here. If we see a question brought up (instead of a comment) it will be split.

 

Thanks. Everything makes much more sense now. So the difference between a flyback converter and a forward converter is that the flyback stores energy in the magnetic field during half the cycle, and then transfers it to the secondary during the second half, while a forward converter acts like a transformer and passes the energy directly through on the first cycle. The magnetic field in the forward converter's core is only incidental, from the magnetizing inductance, and the energy stored there gets passed back to the primary during the second half of the cycle. The flyback transformer needs to store lots of energy without saturating, while the forward transformer can be optimized to reduce the inductance and not saturate as easily.

In an ideal transformer, the primary inductance with nothing connected to the secondary would be infinite, and the magnetizing current would be zero.

I was trying to think about what would happen in this case. Seems like this would mean that all superconducting coils have infinite inductance and can't be energized, though. That doesn't seem right.

 

Superconductors don't have inductance per sea, just no resistance. They have a perfect Q, but they still follow the same rules about Henry's and windings. They would make a really tight filter if the cap was of the same quality.

 

I was trying to think about what would happen in this case. Seems like this would mean that all superconducting coils have infinite inductance and can't be energized, though. That doesn't seem right.

The inductance of superconducting coils is not much different from the inductance of a coil with the same geometry made with copper. Inductance is a function of geometry and the permeability of the core (if there is one) material.

However, the "apparent" inductance of a transformer primary depends on what may be connected to the secondary. A transformer wound with superconducting wire and with an "unobtanium" core of infinite permeability would have a primary with infinite inductance and would draw no current (magnetizing or otherwise) IF nothing were connected to the secondary (I'm ignoring parasitic capacitance). But if you measured the primary inductance with an LCR meter when a load was connected to the secondary, the measured primary inductance would no longer be infinite.

This means that an ideal transformer draws no magnetizing current and the only primary current is determined by the load on the secondary, but even with no load on the secondary there will still be flux in the core; the transformer can be "energized" by applying a voltage to the primary even if no magnetizing current flows. Infinite inductance simply means that it takes no current to produce a flux.