Transformer Losses: Copper and Hysteresis

Transformer Losses

• Practical power transformers are highly efficient, but not perfect devices.
• Small power transformers have an efficiency range of 80% to 90%.
• Large utility distribution transformers may exceed 98% efficiency.
• Total power loss is a combination of copper loss (I²R loss) and losses due to eddy currents and hysteresis in the core.
• Copper loss, eddy-current loss, and hysteresis loss convert electrical energy into heat.

Copper Losses

• Copper loss, or I²R loss, results from the resistance of wire in the primary and secondary coils.
• These losses are emitted as heat.
• Minimizing resistance in primary and secondary windings reduces copper losses.
• Increase wire diameter to lower resistance.
• Weight and physical dimensions must be factored in when reducing copper losses.

Hysteresis Losses

• Reversing magnetism in the core releases energy as heat, known as hysteresis loss.
• Hysteresis loss is proportional to the core's area, materials, and the frequency of applied current.
• Air-core transformers do not experience hysteresis losses.
• Minimizing core area and using supermalloy and permalloy materials (silicon-iron alloys) reduces hysteresis.

Eddy Current Losses

• A transformer's changing magnetic field induces an EMF in the core, creating circulating eddy currents.
• If the core was solid iron, the large eddy currents create large heat losses and opposing flux.
• This results in more current flowing in the primary to maintain the core's magnetic field, further increasing losses.
• Transformer cores are laminated to minimize eddy current losses.
• Thin, insulated laminations impede current flow, reducing losses.

Transformer Efficiency Calculations

• Transformer efficiency requires knowing input and output power.
• Input power equals primary voltage times current.
• Output power is secondary voltage times current.
• The difference between input and output power represents power loss.
• The formula for percentage efficiency is: Efficiency (%) = (Pout / Pin) × 100

Maximum Power Transfer

• Impedance matching is studied by looking at transfer between parts of a circuit between amplifier stages or amplifier to a loudspeaker.
• Every energy source has internal resistance/impedance making 100% power use by the load impossible to achieve in ideal conditions.
• The goal is to make maximum power is available to the load.

Impedance Matching Example 1

• With a 24 V source and 6 Ω internal impedance connected to a 2 Ω load:
• Total circuit resistance is 8 Ω.
• Circuit current is 3 A.
• Power in the source: P = 54 W
• Power in the load: P = 18 W
• The source is consuming there times the power of the load, meaning it becomes really hot and is probably damaged.

Impedance Matching Example 2

• With a 24 V source and 6 Ω internal impedance connected to a 6 Ω load:
• Total circuit resistance is 12 Ω.
• Circuit current is 2 A.
• Power in the source: P = 24 W
• Power in the load: P = 24 W
• The source is consuming there times the power of the load, meaning it becomes really hot and is probably damaged.

Impedance Matching Example 3

• With a 24 V source and 6 Ω internal impedance connected to an 18 Ω load:
• Total circuit resistance is 24 Ω.
• Circuit current is 1 A.
• Power in the source: P = 6 W
• Power in the load: P = 18 W
• The load is consuming the amount of power as in the first example, and is three times that of the source.

Power Transfer Summary

• Region 3 heat generated in the source (battery), where Rload is less than Rsource.
• Region 2 provide maximum power transfer (antenna to a radio), where Rload equals Rsource.
• Area 1 (Rload is far less than Rsource) should never be never equal to each other to prevent source damage.
• Power supplies need a way to connect sources and loads of different impedance.
• The connection matches both impedances without power loss.

Turns Ratio

• Connecting dissimilar impedances uses a transformer to calculate the turns ratio: (NP/NS)² = ZP/ZS

Power Supply Example

• Power supply delivers power to a load:
• V = 200 V
• Load draws 8 A
• Then the resistance is: R = 25 Ω
• AC power delivered through a transformer and load now uses the same power.

Transformer Calculations

• Transformer steps voltage down from 200 V to 50 V.
• Current steps up by a factor of 4.
• Turns ratio: T = VP/VS = 4:1
• Secondary current: ISec = 32 A
• Voltage of 50V and a current of 32 A: RSec = 1.5625 Ω
• With the First circuit, V = 200 V, R = 25 Q, and second produced V = 50 V produce a voltage transformation ratio of 4:1 .
• The impedance transformation ratio impedance values of each side of the transformer is compared in a ratio.
• We assume that the total impedance equal to the resistance values calculated earlier then R Pri = Z pri and R Sec = Z Sec.
• Impedance Ratio = Zpri : ZSec = 16:1
• The impedance transformation ratio (16) is the turns ratio squared (4²).
• These calculations ignore transformer losses assuming 100% efficiency.

Transformer Ratings

• When a transformer is used in a circuit, turns ratio must be considered.

• The voltage, current and power-handling capabilities of the primary and secondary windings must be considered.

• Maximum safe voltage is determined by insulation type/thickness.

• Better insulation allows a higher maximum voltage.

• Maximum current is determined by the diameter of the wire.

• Excessive current causes higher power dissipation as heat, which may degrade insulation.

• The transformer may be permanently damaged.

• Power-handling depends on the transformer's ability to dissipate heat.

• Improved heat removal increases capacity achieved by immersing in oil or using cooling fins.

• Power-handling is measured in volt-amperes or watts.

• Two common generator frequencies (60 Hz and 400 Hz).

• Increased frequency increases inductive winding reactance with a greater voltage drop across the windings and a smaller drop across the load.

• Increased frequency won't damage the transformer.

• Decreased frequency decreases winding reactance.

• This will increase the current through the winding.

• Excessive current may damage the transformer.

• Use transformers at or above their normal frequency.

Auto-Transformers Configuration

• In an auto-transformer, primary and secondary aren't separate.
• A single wire coil is "tapped" to create primary and secondary windings.
• The secondary voltage relates to the primary voltage as if there were two distinct windings.
• Adjustable taps provide variable output voltage.
• These are often referred to by brand name, like Variac.
• Movable tap selects output voltage, higher or lower than Ep.
• At point A, Eş < Ep; and at point B, Eş > Ep.

Auto-Transformer Example

• 200 V is applied between points B and C.
• Secondary voltage of 100 V is available from A and B.
• 300 V is available from points A and C.
• Current supplied through B and C produces a back EMF of nearly 200 V between B and C
• Also producing an EMF of 100 V between A and B.
• Two EMFs are in phase, being induced by the same field.
• Combined, they give a secondary voltage of 300 V between A and C
• 100 V induced between A and B can be taken as a secondary.