I'm trying to understand what happens to a distribution transformer in relation to overvoltage conditions (in terms of core saturation and current flow; not insulation failure). I understand that once the voltage has risen high enough to drive the magnetic core past the saturation point, that the magnetizing current will rise dramatically in relation to the increase in the flux field.
1. Is the current increase (on the primary) at this point analogous to a motor at locked rotor, where the resistance of the windings themselves are the limiting factor?
2. Is there a substantial difference in the amount of this current in relation to full load vs. no load on the secondary (other than the load itself)?
I find it easier to think of things from the opposite direction: what is it that _limits_ the current flow in a transformer, both during normal and saturated conditions.
Start by considering an 'air core' inductor, simply a coil of wire. This has essentially no saturation issues. I'll also start by talking about applied DC voltage, not AC.
The voltage applied to this _inductor_ will act to change the current flowing through the inductor. But when you have a changing current through the coil, you get a corresponding changing magnetic field developed by the coil. This changing magnetic field induces a voltage in the coil, and if we ignore resistance, we will find that the magnetic field is changing at just the right rate to induce a voltage which balances out the applied voltage.
Put another way: the rate change of current through an inductor is proportional to the applied voltage and inversely proportional to the inductance.
When you put an iron core in the inductor, you change the relationship between current flowing through the coil and magnetic flux developed. But you don't change the relation between change in magnetic flux and induced voltage. What happens is that you need less current to develop the same magnetic flux, and thus less _change in current_ to develop the same voltage. Net result: with the same coil and the same applied voltage, the current flow increases more slowly.
Put another way: the addition of the iron core increases the inductance of the coil.
But the core has an issue: saturation. Above a certain magnetic flux density, the boost in magnetic flux caused by the core starts to diminish rapidly. This means that once you reach saturation, the inductor will act more like an air core inductor. With the same applied voltage, the current flowing through the coil increases rapildy.
Put another way, when the core saturates, the inductance of the coil goes down.
Looking at applied AC voltages: At any _instant_ in time, the rate change of current in the coil depends upon the applied voltage and the inductance of the coil. But with AC, the applied voltage is constantly changing, which means that the rate change of current flow will be constantly changing. The current flow at any instant in time is proportional to the _integral_ of the applied voltage, and inversely proportional to the inductance.
Saturation makes the inductance not constant. What happens is that at the peak of the current flow curve in the inductor, the inductance goes down, and the current flow rises more rapidly, so you get pointy tips on the current flow curve.
This saturation current flow is still lagging the applied voltage, and is still reactive power; energy is still being stored in the inductor magnetic field. However this increased current flow is accompanied by increased resistance losses, and thus increased real power consumption by the inductor.
A transformer primary is just an inductor, but with a secondary coil magnetically coupled to the circuit. Any current flowing in the secondary is in some sense subtracting from the magnetic field, thus reducing the 'back emf' in the primary and increasing the primary current flow to compensate. The additional primary current flow means more voltage drop in the primary versus any resistance, thus reducing the voltage that the magnetic flux needs to act against. So as you increase secondary current, if there is resistance in the primary circuit, the peak magnetic flux and magnetizing current will go down. The significance of this will depend upon circumstances.
Note that when a motor is in locked rotor, you essentially have a transformer with a _shorted_ secondary. Lots of current will flow, but it is in phase with the supply voltage, delivering real, rather than reactive power.
Re: Transformer core saturation from overvoltage#130156 02/19/0603:54 PM02/19/0603:54 PM
As background for this question (or followup, if you will), it arose around the placement of surge protection on the transformer.
Popular application for arresters utilizes combination cutouts (ie-an arrester mounted on the fuse cutout that is in parallel with the fuse/transformer). A new engineer was promoting going back to tank mounted arresters that would put the arrester in series with the fuse.
The argument against this was that, even though the closer proximity of the arrester to the transformer provided better surge protection, its discharge current would flow through the fuse causing more blown fuses.
The argument against the first argument was that the excess current from core saturation was much more likely to cause the fuse to blow, and that the additional discharge current would be comparatively insignificant.
Re: Transformer core saturation from overvoltage#130157 02/20/0608:57 AM02/20/0608:57 AM
Hmmm. I am pretty much guessing here, since I don't really know much about surge protection devices at line voltage levels.
I don't understand what you mean by 'in parallel with the fuse/transformer' versus 'in series with the fuse'. My understanding is that the surge protection is some form of 'shunt' protection, where any excessive voltage causes current flow from phase to phase, or phase to ground. In this case the 'surge protection' current would always go through the fuses unless the surge protection devices are upstream of the fuses.
But the other issue is that I don't know what these surge protection devices are there to protect against. If you are talking about devices which protect against very short duration high voltage spikes, then the inductance of the transformer is a pretty effective 'barrier' to having much current flow into the transformer during the event, so I wouldn't expect much in the way of 'transient' current flow into the transformer. Instead I would expect the extremely high voltage to stress the transformer insulation. Add the surge protection device, and I expect the high voltage to show up as a current spike through the protector.
But we are not talking about many 'amp seconds' here, so I wouldn't expect much dissipation in the fuse.
In that web site you can find everything about the Arresters.
Concerning to fuses (overcurrent protection of power transformer) Firstly this fuse is working by formule: I^2 x t where I is the rated current of Power transformer. Some time the power transformer should support the load exceeding with 1.6 the rated load, and formul become: 1.6 x I^2 x t
Don't worry about burning your fuses from overvoltage from lightening: million Volts and kiloAmpers for microseconds. The secon type Arrester for Switching Overvoltages is more for consideration when the duration of the event is in seconds , but the level of over voltage is sqrt(3) Vmax=Vrated x sqrt(3),and current that is not verry high.