7.4 Transformers and Losses

Key Takeaways

  • Aircraft AC systems use a 400 Hz frequency, enabling much lighter and smaller transformer cores compared to standard 50/60 Hz ground systems.
  • The turns ratio determines voltage, current, and impedance transformation: Vp/Vs = Np/Ns = Is/Ip.
  • Core losses consist of eddy currents (mitigated by laminations) and hysteresis (mitigated by using magnetically soft silicon steel).
  • Copper losses are I²R heating losses in the windings, reduced by using thicker conductors, while flux leakage is minimized by interleaving windings.
  • Autotransformers save weight by using a single tapped winding but do not provide electrical isolation between primary and secondary circuits.
Last updated: July 2026

Why This Matters for the Exam

In aviation maintenance engineering, electrical power management is critical. Aircraft alternators typically generate 115V AC, but various onboard systems require different voltage levels. For instance, cabin lighting, instrumentation, and avionics may require 28V AC, 5V AC, or even higher voltages for specific radar equipment. Transformers are the solid-state electromagnetic devices that perform these conversions with high efficiency.

For the EASA Part-66 Module 3 exam, a technician must master not only the mathematical relationships governing turns ratios, voltage, and current, but also the physical construction and loss mechanisms that affect transformer efficiency. In aviation, efficiency translates directly to weight. The use of a 400 Hz AC frequency on aircraft (compared to 50/60 Hz on the ground) is a deliberate engineering choice: because the rate of change of magnetic flux is much higher at 400 Hz, the core volume and weight of transformers and motors can be reduced by up to 75% for the same power rating. However, high frequency increases core losses, making the selection of core materials and lamination techniques critical.


Transformer Construction and Operating Principles

A basic transformer consists of two or more coils of insulated wire wound around a common ferromagnetic core. The winding connected to the AC source is the primary winding, and the winding connected to the load is the secondary winding.

Operating Principle

Transformers operate on the principle of mutual induction (Faraday's Law of Electromagnetic Induction). When an alternating current flows through the primary winding, it creates a continuously varying magnetic flux in the core. This changing flux travels through the core and cuts across the turns of the secondary winding, inducing an alternating electromotive force (EMF) or voltage in the secondary winding. Transformers cannot operate on steady DC because a changing magnetic field is required to induce voltage.

Core Construction

To guide the magnetic flux efficiently and minimize losses, transformer cores are constructed using high-permeability materials. There are two primary configurations:

  1. Core-Type: The windings surround the core legs. This construction is simpler and easier to insulate, making it common in high-voltage applications.
  2. Shell-Type: The core surrounds the windings. The windings are wound on a central leg, and the flux splits into two external paths. This provides excellent mechanical support and minimizes flux leakage, making it highly common in aviation equipment.

Core materials are typically made of silicon steel (an alloy of steel with 3% to 4.5% silicon). The silicon increases the electrical resistivity of the steel, which suppresses eddy currents, while maintaining high magnetic permeability. To further combat eddy currents, the core is not a solid block of metal. Instead, it is constructed from thin sheet metal plates called laminations (typically 0.35 mm to 0.5 mm thick), which are coated with an insulating varnish. This breaks up the electrical path for circulating currents while leaving the magnetic path uninterrupted.


The Turns Ratio and Transformer Mathematics

The voltage induced in the secondary winding is directly proportional to the ratio of the number of turns in the primary winding (Np) to the number of turns in the secondary winding (Ns).

The Transformer Equation

The fundamental mathematical relationship for an ideal transformer is:

Vp / Vs = Np / Ns = Is / Ip

Where:

  • Vp = Primary Voltage (Volts)
  • Vs = Secondary Voltage (Volts)
  • Np = Number of turns in the primary winding
  • Ns = Number of turns in the secondary winding
  • Ip = Primary Current (Amperes)
  • Is = Secondary Current (Amperes)

Key Rules:

  • Step-Up Transformer: The secondary winding has more turns than the primary (Ns > Np). Consequently, the secondary voltage is higher than the primary voltage (Vs > Vp), but the secondary current is proportionally lower (Is < Ip).
  • Step-Down Transformer: The secondary winding has fewer turns than the primary (Ns < Np). The secondary voltage is lower than the primary voltage (Vs < Vp), but the secondary current is higher (Is > Ip).
  • Conservation of Power: In an ideal transformer (100% efficient), the power input to the primary equals the power output from the secondary: Pin = Pout, or Vp x Ip = Vs x Is.
  • Impedance Matching: The impedance ratio is the square of the turns ratio: Zp / Zs = (Np / Ns)^2. This is crucial for audio and RF coupling circuits.

Transformer Efficiency and Loss Mechanisms

No physical transformer is 100% efficient. The efficiency (eta) is calculated as:

Efficiency (eta) = (Pout / Pin) x 100% = (Pout / (Pout + Losses)) x 100%

In modern aircraft transformers, efficiency typically exceeds 95% under optimal loads. The remaining percentage is dissipated as heat due to four main losses:

1. Copper (I²R) Losses

Also known as winding resistance losses, copper losses are caused by the electrical resistance of the copper wire used in the windings. As current flows through the windings, it produces heat according to P = I^2 x R.

  • Mitigation: Use high-purity copper conductors with a larger cross-sectional area (thicker wire) for the windings. Since the secondary winding of a step-down transformer carries high current, it requires much thicker wire than the high-voltage primary winding.

2. Eddy Current Losses

Eddy currents are small loop currents induced inside the conductive iron core itself by the alternating magnetic flux. Because the core is a conductor, these currents circulate and generate heat (I^2 x R loss in the core).

  • Mitigation: Construct the core from thin, varnished laminations rather than solid iron. The laminations are oriented parallel to the magnetic flux lines. This cuts off the path of the eddy currents (acting as open circuits) while allowing the magnetic flux to pass freely. The addition of silicon to the steel also raises the electrical resistance of the core material, suppressing eddy currents further.

3. Hysteresis Losses

Hysteresis is the magnetic friction within the core material. As the alternating AC current reverses direction, the magnetic domains in the iron core must continuously rotate and realign. The core material resists this alignment, and the energy required to overcome this resistance is dissipated as heat. Hysteresis loss is proportional to the AC frequency.

  • Mitigation: Construct the core using magnetically "soft" materials, such as silicon-alloy sheet steel, which have a narrow hysteresis loop. This means the material has low coercivity and can be magnetized and demagnetized easily with minimal energy loss.

4. Flux Leakage (Copper-to-Iron Coupling Loss)

In an ideal transformer, all magnetic flux generated by the primary winding passes through the core to link with the secondary winding. In reality, some flux lines leak into the surrounding air and do not cut across the secondary winding. This results in inductive voltage drops and reduces the effectiveness of the transformer.

  • Mitigation: Use shell-type core construction, place the windings as close together as possible, or wind the secondary directly on top of the primary (interleaved windings).

Autotransformers

An autotransformer is a special type of transformer that has only a single continuous winding wound on a laminated core. The primary and secondary circuits share a common section of this winding. The output voltage is tapped from various points along the winding.

Advantages:

  • Weight and Space Saving: Since there is only one winding, it requires less copper and a smaller core, resulting in a lighter and cheaper design. This is highly advantageous in weight-sensitive aircraft systems.
  • Higher Efficiency: Because there is less winding resistance, copper losses are lower.

Disadvantages:

  • No Electrical Isolation: Since the primary and secondary share a physical conductor, there is no electrical separation between the input and output circuits. If the winding breaks (opens) on the common side, the full primary voltage can appear across the secondary terminals, risking catastrophic damage to sensitive low-voltage avionics. For this reason, autotransformers are restricted to non-critical applications on aircraft.

Worked Exam Calculation Scenarios

Scenario 1: Step-Down Voltage and Winding Current

An aircraft instrument transformer steps down 115V AC from the primary bus to 28V AC for cabin panel illumination. The primary winding has 960 turns.

  1. Calculate the number of turns required in the secondary winding: Using Vp / Vs = Np / Ns: Ns = Np x (Vs / Vp) Ns = 960 x (28 / 115) = 233.74 turns Rounding to the nearest whole turn yields 234 turns.

  2. If the 28V cabin lighting load draws a current of 12 A, calculate the current drawn by the primary winding (assuming an ideal transformer): Using Is / Ip = Vp / Vs: Ip = Is x (Vs / Vp) Ip = 12 A x (28 V / 115 V) = 2.92 A

Scenario 2: Efficiency and Total Losses

An aircraft transformer is rated at 2.5 kVA. When operating at its full rated load with a unity power factor (output power of 2500 W), the transformer experiences 45 W of copper losses and 35 W of core losses (combined eddy current and hysteresis losses).

  1. Calculate the total loss power: Plosses = Pcopper + Pcore Plosses = 45 W + 35 W = 80 W

  2. Calculate the input power: Pin = Pout + Plosses Pin = 2500 W + 80 W = 2580 W

  3. Calculate the efficiency (eta): eta = (Pout / Pin) x 100% eta = (2500 / 2580) x 100% = 96.9%

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Transformer Voltage and Flux Flow
Test Your Knowledge

Why are the cores of aircraft transformers constructed using thin laminations insulated from one another?

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Test Your Knowledge

An aircraft transformer has a turns ratio of 5:1 (Np:Ns). If a primary voltage of 115V AC is applied and the secondary load draws 10A, what are the secondary voltage and primary current? (Assume an ideal transformer).

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Test Your Knowledge

What is a major safety disadvantage of using an autotransformer in an aircraft's electrical system?

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