Efficiency variation with load and maximum efficiency condition

Efficiency of a transformer is the ratio of output power to input power: $\eta = \frac{\text{output}}{\text{output} + \text{losses}}$. Because one loss is constant and the other depends on load, efficiency changes with load.

Losses in a transformer

• Iron (core) loss $P_i$ = hysteresis + eddy-current loss. It depends on voltage and frequency, which are constant, so $P_i$ is constant at all loads.
• Copper loss $P_{cu}$ = $I_1^2R_1 + I_2^2R_2 = I_2^2R_{02}$. It varies with the square of the load current. At a fraction $x$ of full load, $P_{cu} = x^2 P_{cu,FL}$.

Variation of efficiency with load

At fraction $x$ of full load:

$$\eta = \frac{x\,S\cos\phi}{x\,S\cos\phi + P_i + x^2 P_{cu,FL}}$$

• At no load, output is zero, so $\eta = 0$.
• At light load, the fixed iron loss is large compared with the output, so $\eta$ is low but rises quickly.
• As load increases, $\eta$ rises to a maximum where copper loss equals iron loss.
• Beyond that, copper loss (growing as $x^2$) dominates and $\eta$ falls slowly.
• At lower power factor, output is smaller for the same current, so the whole curve is lower.

 eta
  ^        max (Pcu = Pi)
  |         ____
  |      .-'    '--..___   upf
  |    .'  ___...--''--.._  0.8 pf
  |   / .-'
  |  /.'
  | //
  +-------------------------> load (x)
  0      x_max        1.0

Condition for maximum efficiency

Write $\eta$ in terms of secondary current $I_2$ (with $V_2$ and $\cos\phi$ constant):

$$\eta = \frac{V_2 I_2 \cos\phi}{V_2 I_2\cos\phi + P_i + I_2^2 R_{02}}$$

Divide numerator and denominator by $I_2$:

$$\eta = \frac{V_2\cos\phi}{V_2\cos\phi + \dfrac{P_i}{I_2} + I_2R_{02}}$$

$\eta$ is maximum when the denominator is minimum:

$$\begin{aligned} \frac{d}{dI_2}\left(V_2\cos\phi + \frac{P_i}{I_2} + I_2R_{02}\right) &= 0 \\ -\frac{P_i}{I_2^2} + R_{02} &= 0 \\ I_2^2 R_{02} &= P_i \end{aligned}$$

So efficiency is maximum when copper loss equals iron loss (variable loss = constant loss).

The current and load at maximum efficiency are:

$$I_2 = \sqrt{\frac{P_i}{R_{02}}}, \qquad x = \sqrt{\frac{P_i}{P_{cu,FL}}}$$ $$\eta_{max} = \frac{x\,S\cos\phi}{x\,S\cos\phi + 2P_i}$$

Example: if $P_i = 400$ W and $P_{cu,FL} = 1600$ W, maximum efficiency occurs at $x = \sqrt{400/1600} = 0.5$, i.e. at half load. Distribution transformers are designed with low iron loss so that the maximum efficiency occurs at the light loads they carry most of the day.