Keeping core losses low in AC electric motors can certainly be a challenge, but when it’s done, it can wildly increase the efficiency of your motor.

What is core loss, anyways? The total core losses in AC electrical devices consist of hysteresis losses, eddy current losses, and anomalous losses. But, what’s the difference between hysteresis and eddy current loss? Eddy current and anomalous losses?

Let’s break it down:

## AC Electric Motors | Core Losses

### Hysteresis Losses

**Hysteresis losses are associated with the energy required to go through one complete cycle of the BH curve**. This is an important fact to remember. Magnetic parameters such as higher permeability and lower coercive force imply less energy to sweep through the BH curve.

Wrought steel users take advantage of this factor by alloying with silicon and sometimes aluminum to increase the permeability and lower the coercive force.

It’s worth mentioning that alloying also increases the resistivity of the material. However, higher resistivity is more important when considering eddy current losses.

You’ve probably heard that hysteresis losses are proportional to the frequency of operation to the first power (even if you didn’t totally understand it). In pursuit of reducing the hysteresis loss in magnetic materials, manufacturers have started to use materials such as:

- Iron-nickel alloys
- Permendurs
- Amorphous magnetic alloys for transformers

These materials have extremely high permeability and very low coercive force values. As such, they are used in fast-acting circuit breakers (ground fault circuit breakers).

These devices detect the difference in current between the load and ground leg of the electric circuit. If any differences exist, a solenoid is used to deactivate the incoming power. Differences in current as little as 0.03 amps can trip these devices.

**Finding a magnetic material that quickly responds to small amounts of input current is key for these devices.**

### Eddy Current Losses

**Eddy current** losses (also called **Foucault's currents**) are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday's law of induction.

Eddy currents create a magnetic field that resists the change of the magnetic field that created it. Eddy current losses are proportional to the second power to the part’s:

- Frequency
- Induction
- Thickness

Eddy current losses are also dependent on the thickness of the part’s laminations.

As lamination thickness increases, so do eddy current losses. That’s why electrical motor manufacturers are making a push for thinner laminations.

**Higher resistivity reduces eddy current losses** but cannot completely eradicate them. There’s a limit to what changing the magnetic properties (such as resistivity) can accomplish.

### Anomalous Losses

**A****nomalous losses** include the AC heating of the copper wire and potential size effects of the device.

Higher currents will generate more heat, which increases anomalous losses.

**Anomalous losses are usually minimal,** but they can have an effect, so it’s important to include them in core loss equations.

Speaking of which …

## SMC Materials and the Core Loss Equation

Soft magnetic composite materials (SMCs) introduce a new level of efficiency when it comes to electric motor components. These materials consist of ferromagnetic powder particles, ideally coated with a uniform layer of electrical insulating film. When used correctly, SMCs can increase magnetic performance, reduce energy consumption, and much more.

Relative to SMC materials, total core loss can also be calculated as the sum of the three losses described above. The core loss equation looks like this:

** Ptot = Kh * freq * B^1.75 + Ke * freq^2 * B^2 + (B^2* freq^2 * D^2)/(1.8 * density * resistivity * 1,000) **

- Ptot = total core loss in watts/kg
- Kh = hysteresis coefficient = 0.063
- Ke = eddy current coefficient = 0.000027
- B = induction in Tesla
- Density is assumed to be 7.4 g/cm³
- Resistivity = 700

Because SMCs are individual powder particles coated with an electrically insulated material, the eddy current losses will be much smaller than in wrought steels.

Due to compaction, hysteresis losses are the dominant losses in SMC devices up to about 1,000 Hz.

## Hysteresis Loss in SMC Materials

SMC components are compacted at high pressures; this introduces cold work into the iron powder. This cold work effectively reduces the magnetic permeability of the material and increases the coercive forces; both are bad for hysteresis losses.

Total loss calculations of a specific SMC material at 0.5 Tesla induction level and 1,000 Hz frequency equal about 40 watts, of which:

- 25W were hysteresis losses
- 9W were eddy current losses
- About 6W were anomalous losses

These losses were calculated for an SMC material that was “cured” at about 1000 °F.

**Early research showed that the effects of cold work can be partially overcome by annealing the SMC at about 1200 °F**. During curing, the process must be designed so as not to sinter these iron particles (sintering causes significantly higher eddy current losses).

However, annealing at 1200 °F is only a partial fix. To fully anneal the iron, a temperature of 1400 °F to 1500 °F must be used. Only then could annealing almost completely eliminate the detrimental effects of cold working and reduce hysteresis losses.

Assuming that the SMC device can be cured at 1400 °F, **we would expect a 30% increase in permeability and a comparable decrease in coercive force**. The curing then improves the SMC material that would compete with M19 laminations at frequencies above 150 Hz. Compare this to the current state-of-the-art SMCs, where the SMC is competitive with M19 at 400 Hz.

## Avoiding Core Losses With Powder Metal

We know what you’re thinking: This is a lot to take in, and definitely a lot to consider when it comes to optimizing your material to reduce core losses. That’s why partnering with an experienced powder metallurgy supplier can help you cut the stress out of your manufacturing process.

Want to talk to an engineer about how to take pesky hysteresis loss out of the equation? You can do that below: