What is Transformer Efficiency? – A Complete Explanation

The power that enters through the primary of a power transformer does not remain the same at the output circuit. In simple words, a transformer cannot transfer the entire input power to the load.

A transformer is an electrical machine. So, it has its own efficiency level. No machine in the universe can have 100% efficiency. In this article, we will focus on transformer efficiency. We will learn what efficiency means and how to derive its equation. We will also discuss the loading condition at which transformer efficiency becomes maximum.

Losses in a Transformer

Since a transformer is not an ideal machine, it has its own losses. These losses are the only reasons for power reduction at the transformer output. We have already discussed transformer losses in detail in a different article. In short, there are mainly two types of loss in a transformer.

1. No-load loss (Constant loss)
2. Load loss (Variable loss)

The first one is no-load loss. No-Load Loss in a transformer consists of hysteresis loss and eddy current loss. The magnetic flux in the transformer core remains almost constant at all loading conditions, even at no-load. So, hysteresis loss and eddy current loss also remain constant at all loading conditions. That is why we also refer to these losses as constant losses.

We express hysteresis loss as$$W_H = K_H \, f \, B_m^{1.6} \; \text{watts}$$We express eddy current loss as$$W_E = K_E \, f^2 \, K_S^2 \, B_m^2 \; \text{watts}$$Where, K_H represents the hysteresis constant, and K_E represents the eddy current constant. $K_S$ is the form factor. f and B_m represent the power frequency and maximum flux density, respectively.

The second one is load loss. The load loss occurs due to the ohmic resistance (I²R) loss in the transformer windings. In practical transformers, load loss also includes stray losses. So, total load loss equals Ohmic (Copper) loss + Stray loss.

Transformer Efficiency

The output power of a transformer is:$$\text{Output power} = \text{Input power} – \text{Losses}$$Efficiency $(\eta)$ is defined as:$$\eta = \frac{\text{Output power}}{\text{Input power}} \times 100\%$$So, we can write:$$\eta = \frac{\text{Input} – \text{Losses}}{\text{Input}} \times 100\%$$$$\eta = \left(1 – \frac{\text{Losses}}{\text{Input}}\right) \times 100\%$$

Copper Loss and Iron Loss

Copper loss of the transformer is:$$W_{cu} = I_1^2R_{01} \quad \text{or} \quad I_2^2R_{02}$$

Where, $R_{01}$ represents equivalent resistance referred to the primary. $R_{02}$ represents equivalent resistance referred to the secondary.

Iron loss (no-load loss) is:$$W_i = W_H + W_E$$

Input Power

Primary input power is:$$P_{in} = V_1 I_1 \cos\theta_1$$Here, V₁ and I₁ represent input voltage and current. \(\theta_1\) is the phase angle between the input voltage and current. So, efficiency becomes:$$\eta = 1 – \frac{I_1^2R_{01} + W_i}{V_1 I_1 \cos\theta_1}$$

This can also be written as:$$\eta = 1 – \frac{I_1R_{01}}{V_1 \cos\theta_1} – \frac{W_i}{V_1 I_1 \cos\theta_1}$$

Condition for Maximum Efficiency

A transformer has a unique loading condition at which efficiency becomes maximum. To find it, we differentiate efficiency with respect to current $I_1$ and equate it to zero$$\frac{d\eta}{dI_1} = 0$$After differentiation,$$-\frac{R_{01}}{V_1\cos\theta_1} + \frac{W_i}{V_1 I_1^2\cos\theta_1} = 0$$So we get,$$\frac{R_{01}}{V_1\cos\theta_1} = \frac{W_i}{V_1 I_1^2\cos\theta_1}$$$$W_i = I_1^2 R_{01}$$​

Or equivalently,$$W_i = I_2^2 R_{02}$$

This proves that the maximum efficiency occurs when the copper loss​ becomes equal to the core loss.