5
\$\begingroup\$

I often read mention of a transformer core's "cross-sectional area", of how it affects the transformers VA rating (although I can't find any formulas for that.)

I haven't found anything mentioning the length that the magnetic flux must travel.

Does this mean that both of the following cores are equivalent, other than heat dissipation?

enter image description here

How would the inductance, VA rating, and other properties compare between these two toroidal cores?

\$\endgroup\$

4 Answers 4

7
\$\begingroup\$

Some aspects are equivalent, some are different.

Let's assume they are the same cross sectional area, and you want to make a transformer, perhaps a power transformer, operating from a fixed mains input voltage and frequency, with a sine wave waveform.

The first, most important thing, before doing anything else, is to make sure that the core operates at the highest field practical, but still short of saturation.

Neglecting copper losses, all of the input voltage is balanced by the voltage induced in the primary by the changing field in the core. If we allow losses, we find things are not much different, and in fact due to the voltage drop on the copper giving a lower voltage, a little better for saturation, so we'll stick with the ideal case for simplicity.

As the flux changes, it induces a voltage in each turn of the primary, that can be given by \$V = area\frac{dB}{dt}\$, when all quantities are measured in SI units. For an input sinewave, as in a mains transformer, this is \$V_{peak} = area.2{\pi}fB_{peak}\$. For an input square wave, as in an inverter, the constant is slightly different, and \$V_{peak} = area.4fB_{peak}\$

To find the minimum number of turns to put onto the primary, we take the peak input voltage, and divide by the volts per turn. As both cores have the same area, both cores will require the same number of turns on their primaries to operate below saturation You will notice that this expression did not involve the core length or the core permeability, or BH curves.

While it's tempting to think that the primary inductance might be a way to arrive at the number of turns here, anybody that asks you to go that route is signing you up for a lot of trouble. The inductance depends on the core permeability, but we don't know the permeability until we know the field (is it above or below saturation level, below, maybe 2000, above, maybe 10), we don't know the field until we know the current ... you see where this is headed. The only way to determine the field is through the applied volts, and the core area. And I've put the formula for that above.

The longer core will draw a higher magnetising current. This is the off-load current, which creates the flux in the core. There are two ways to understand this increase. One is that as we need the same H field to drive the B field, whose changes are balancing our input voltage, and as the H field is Ampere.Turns/length, and the length has doubled, we need twice as many AT. The other is to say that the primary inductance has halved, as it involves a 1/length, so the input voltage will drive twice as much current through it. The two approaches are equivalent.

The longer core allows a larger hole in the middle, the 'copper window', which means we can include many more turns of wire, or wire that is thicker. This means higher voltage or current output respectively. As the area goes up as the square of the length, the VA of the transformer is increased four fold.

If these were inductors (and you wouldn't make inductors on an un-gapped high mu material core (the why is for a different question)) with the same number of turns, then if you passed the same current through both, then the H field in the small core would be greater, and so this one would saturate first. We already know that fed from the same voltage, they will run at the same field.

So in summary, comparing the two cores...

the number of primary turns is the same
the primary inductance is half
so the magnetising current is doubled
the copper window area has increased fourfold
which has increased the VA about 4 times

\$\endgroup\$
4
\$\begingroup\$

The length of the core is used for determining the magnetic field strength, H (measured in ampere turns per metre). The "per metre" part is the mean length (\$l_e\$) of the core shown below: -

enter image description here

So, if you know the coil current and the number of turns, you can put a value to H. Then, if you have the B-H curve for the core you can predict the flux density: -

enter image description here

Keeping below about 400 mT largely avoids core saturation but it varies a bit between this or that core. The above picture is for Ferroxcube 3F3 material. For iron or silicon steel the saturation levels are significantly higher than most ferrites.

Regarding inductance, the generally accepted formula is this: -

enter image description here

As you can see, both \$A_e\$ and \$l_e\$ (top picture) play a part in defining inductance and, with a larger toroid it can be seen that the inductance is proportionally lower than for a smaller toroid of the same cross sectional area and material.

If the ratio of mean lengths between the toroids is 2:1, the inductances of both can be made the same by increasing the number of turns by \$\sqrt2\$ for the larger toroid. So turns have increased by \$\sqrt2\$ to equalize inductance values but the bigger toroid will saturate less at a given current because although ampere turns have increased, the length \$l_e\$ is still twice that of the smaller toroid.

\$\endgroup\$
1
  • \$\begingroup\$ I see that the flat-earth society has been around to down-vote this answer. \$\endgroup\$
    – Andy aka
    Commented Feb 7, 2017 at 9:07
0
\$\begingroup\$

The length of the core increases its reluctance in a completely analogous way to how the length of a wire increases its resistance.

The consequence of higher reluctance are (for the same crosssection and number of turns): lower inductance, higher saturation current

\$\endgroup\$
0
\$\begingroup\$

VA rating

How do we increase the VA rating of a transformer? We either need to increase the maximum allowed current, or we need to increase the maximum allowed voltage (or both). The maximum allowed current is in general increased by making the wire in the winding thicker to reduce wire resistance. If the core cross sectional area and material remain the same, the maximum allowed voltage (to avoid core saturation - see below) can in general only be increased by increasing the number of turns.

The longer/bigger core simply has a bigger hole and therefore more space for either:

  • thicker wire (higher maximum current)
  • more turns (higher maximum voltage)
  • or a combination of the two

which again enables a higher VA rating than the smaller core allows. Note that if the winding remains the same (same wire thickness and number of turns), the VA rating will in general be the same for two transformers made from the two cores.

Inductance

A transformer's inductance is in general proportional to its core cross sectional area and inversely proportional to its core length. It also increases with the square of the number of turns:

Inductance

(where N is the number of turns, A is the cross sectional area and l is the length of the core)

For an AC input to a transformer with constant voltage, the inductance decides the magnetisation current of the transformer. (It is important to distinguish between the magnetisation current and the current discussed in the previous section which is used to transfer power to the secondary) This means that the longer core has a higher magnetisation current in the winding if the number of turns remains unchanged..

Core saturation

It is crucially important to operate below heavy saturation when designing a transformer. The two cores will saturate at the same voltage if the winding remains unchanged. Even though the magnetisation current has increased because of lower inductance, a longer core will allow a higher magnetisation current before saturation:

enter image description here

, so the length of the core cancels out when finding the expression for the saturation voltage:

enter image description here

enter image description here

This is where there is a major difference between the cross sectional area and the length. The area is not cancelled out when deriving an expression for the saturation voltage.

Core loss

The core loss consists of two main parts: hysteresis loss and eddy current loss. If the material and lamination thickness of the cores are the same, and the voltage and frequency are the same, these losses are mainly dependent on the flux density and the core volume. If also the windings are the same, the flux density B will be the same in the two cores, and the core loss is proportional to the core volume only. Since the volume is bigger in the longer core, the core loss will be higher.

Eddy current loss:

Eddy current loss

where B is the flux density.

Hysteresis loss:

enter image description here

(where x is the Steinmetz constant which is between 1.5 and 2.5 - depending on the material)

Since the magnetisation current and core losses become higher and the VA rating does not improve, one could say that it is wasteful to use a longer core if you do not also increase the size of the winding.

(Note that in this answer I have ignored some non-ideal effects like flux leakage and that the mean distance of the wire from the core typically increases if the number of turns is increased etc.)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.