Exciting current waveform with non-linear B-H core

For a sinusoidal supply voltage the core flux is forced to be sinusoidal; because the B–H (φ–i) curve is non-linear (saturation and hysteresis), the exciting current is non-sinusoidal. It is peaky and contains odd harmonics, mainly the 3rd.

Why the flux is sinusoidal

Neglecting winding resistance and leakage,

$$v = V_m \cos\omega t = N\frac{d\phi}{dt} \Rightarrow \phi = \frac{V_m}{N\omega}\sin\omega t = \phi_m \sin\omega t$$

So the flux follows the voltage waveform, whatever the core is like. The current needed for each value of flux is then read from the φ–i curve.

Shape of the exciting current (graphical method)

 phi-i curve            waveforms vs time
   phi                  phi (sine)   i0 (peaky)
    |    ___              |   _          ^
    |  /                  | /   \       / \
    | /                   |/     \  __./   \.__
 ---+------ i0         ---+-------\----------- t
   /|                             \_/
 _/ |

Steps: for each instant take φ from the sine wave, find the matching i on the φ–i curve, and plot i against time.

1. Saturation only (no hysteresis): near the peak of φ the core is saturated, so a small extra flux needs a large current. The current is peaked, symmetric about the flux peak, and in phase with φ. It contains the fundamental plus large 3rd, 5th, … harmonics.
2. With hysteresis: the rising and falling branches differ, so the current wave is also unsymmetrical and its zero crossings come before those of the flux. The current leads the flux slightly.

Expression and components

The exciting current can be written as a Fourier series of odd harmonics:

$$i_0(t) = I_1\sin(\omega t + \theta_1) + I_3\sin(3\omega t + \theta_3) + I_5\sin(5\omega t + \theta_5) + \dots$$

The fundamental component is split into:

• Magnetising current $I_m = I_0 \sin\phi_0$, in phase with φ (lags V by 90°); it sets up the flux.
• Core-loss current $I_c = I_0\cos\phi_0$, in phase with V; it supplies hysteresis and eddy losses, $P_c = V I_c$.

$$I_0 = \sqrt{I_m^2 + I_c^2}$$

Consequences

• The 3rd harmonic is typically 30–40% of the fundamental in a transformer worked near the knee.
• In three-phase banks, a delta winding or a neutral path is provided for these 3rd-harmonic currents; otherwise the flux and voltages become distorted.
• Because $I_0$ is only 2–5% of rated current, the distortion is usually neglected and an equivalent sine wave is used in the equivalent circuit.