0.20 mm vs 0.23 mm GOES

Engineering Question

When does a thinner lamination justify its material and processing cost?

Key Data

• Compare specific loss at the same flux density, frequency and material test method.
• JFE patent Table 1: at 1.7 T, 50 Hz and a 2 mm lap, 0.20 mm material gave 0.69 W/kg sheet loss and 0.86 W/kg core loss; 0.23 mm gave 0.73 and 0.92 W/kg respectively.

Technical Explanation

Thinner steel can reduce the classical eddy-current component of loss. Total core loss also depends on material grade, domain treatment, joint geometry and the mechanical condition after processing. Thickness alone therefore cannot establish the finished-core result.

The cited patent comparison used three-phase laboratory cores with five-step laps, a 15 mm stack and equal B8 of 1.91 T. It demonstrates a result under defined conditions, not a production guarantee or a ranking that applies to every pair of commercial grades.

For a given net steel build, thinner material requires more laminations. Handling, alignment and cutting workload change. The quotation should account for these operations and any change in effective steel area, rather than treating gauge as a simple substitution.

Engineering Implications

• Compare suitable grades in both gauges against the actual loss target and operating frequency.
• Review joint geometry and the expected stack build before freezing the material choice.

Procurement Implications

• Evaluate the additional material and processing cost against the value of the expected finished-core loss reduction.
• Request comparable core evidence where loss requirements are critical; a lower certificate value alone is insufficient.

What to Confirm Before Ordering

• Grade, nominal thickness, coating and domain treatment.
• Drawing, net steel area, excitation conditions and acceptance limits.
• Test scope, reporting basis and any agreed material substitution.