Chapter 38: Eddy Current Loss
38.4 Eddy Current Loss
If a coil is wound round a ferromagnetic core, an alternating flux is set up in the core [17661][17663]. An emf is induced in the coil [17671]. Furthermore, a voltage is induced in the core by the alternating flux, which causes eddy currents to circulate [17673][17678][17681].
These eddy currents possess resistance in the core and result in wasting energy [17685][17687]. Laminating the core is the standard way of reducing eddy current loss [17689].
This leads to the equation for eddy power loss [17695][17696]:
Where Pe is the eddy power lost, ke is a constant, Bm is the maximum flux density, f is the frequency, and t is the thickness of the laminations [17697][17699][17701].
The equation shows that core losses are proportional to the square of the thickness of the lamination strips [17709]. Therefore, it is desirable to make the strips as thin as possible [17710]. However, this is not always possible at high frequencies [17713]. For very high frequencies, dust cores consist of carbonyl iron or similar materials [17717].
Practice Exercises: Eddy Current Loss
Problem 6: Given t = 0.5 × 10-3 m, Pe = 100 W/m3, f = 50 Hz [17727].
ke Bm2 = 160,000 [17730]
Finding the new thickness (t2) for a new frequency f2 = 250 Hz and Bm2 = (1/3) Bm [17727]:
t = √(900 / (160,000 × 2502)) = 0.3 mm [17731]
Problem 8: Given f = 50 Hz, P_total = 400 W, Pe = 150 W, V1 = 500 V [17749].
Pe = ke Bm2 f2 t2 [17813]
Total Core Loss Pc = Ph + Pe = 1400 W (after adjustments for new frequency) [17762][17765]
38.5 Separation of Hysteresis and Eddy Current Losses
The total core loss Pc is the sum of hysteresis loss and eddy current loss [17851]:
- Hysteresis Loss (Ph): Ph = kh f (Bm)n [17847]
- Eddy Current Loss (Pe): Pe = ke (Bm)2 f2 t2 [17847]
If the flux density in an inductor or transformer core is maintained constant, then the formulas simplify to Ph = k1 f and Pe = k2 f2 (where k1 and k2 are constants) [17848][17850][17851].
Dividing both sides by the frequency f yields [17853]:
This is of the straight line form y = mx + c [17857]. If Pc / f is plotted vertically against f horizontally, a straight line graph results having a gradient of k2 and a vertical intercept of k1 [17857][17858][17861].
Calculating the Constants
Using values from a given graph or table [17885][17886]:
k1 = 3.389 – (90 × 0.0317) = 0.54 [17886]
Thus, for a given frequency (e.g., f = 50 Hz) [17889]:
Pe = k2 × f2 = 0.032 × 502 = 80 W [17890]
Total Pc = 26 W + 80 W = 106 W [17891]