This document provides a 24 step process for designing a 250 VA, 250 Watt isolation transformer with specifications including 230 V input and output voltages, 95% efficiency, and 1.6 T flux density. Key details include: 1) Total power is calculated to be 513.16 Watts accounting for losses. 2) Core geometry is calculated to be 18.04 cm^5 and the closest lamination is EI-150. 3) Primary and secondary winding properties like number of turns and copper losses are calculated based on the specifications. 4) Total copper loss is calculated to be 8.747 Watts and voltage regulation is 3.5%, meeting the specified 5% maximum.

1. 1
Inductor and Transformer Design
Design Problem # 1
Design a (250 VA) 250 Watt isolation transformer with the following specifications
using core geometry 𝑲 𝒈 approach.
Input voltage, 𝑽𝒊 = 230 V
Output voltage, 𝑽 𝒐 = 230 V
Output Power, 𝑷 𝒐 = 250 Watts
Frequency, 𝒇 = 50 Hz
Efficiency, 𝜼 = 95 %
Regulation, 𝜶 = 5 %
Flux density, 𝑩 𝒂𝒄 = 1.6 T
Design Steps: -
Various steps involved in designing this transformer are:
Step # 1: Calculation of total power
Total power,
𝑃𝑡 = 𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 + 𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 =
𝑃𝑂
𝜂
+ 𝑃𝑂 =
250
0.95
+ 250
= 513.16 𝑤𝑎𝑡𝑡𝑠.
Step # 2: Calculation of electrical condition
Electrical conditions,
𝐾𝑒 = 0.145𝐾𝑓
2
𝑓2
𝐵 𝑚
2
× 10−4
= 0.145 × 4.44 × 502
× 1.62
× 10−4
= 1.83
Step # 3: Calculation of core geometry
Core geometry,
𝐾𝑔 =
𝑃𝑡
2𝐾 𝑒 𝛼
=
513.16
2×1.83×5
= 𝟏𝟖. 𝟎𝟒 cm5
Step # 4: Selection of transformer core
For the core geometry calculated in step # 3, the closest lamination number is 𝑬𝑰 −
𝟏𝟓𝟎.
For 𝑬𝑰 − 𝟏𝟓𝟎 lamination,
Magnetic path length (MPL) = 22.9 cm
Core weight =2.334 Kg

2. 2
Copper weight = 853 gm
Mean length turn (MLT) = 22 cm
Iron area, 𝐴 𝑐 = 13.8 cm2
Window area, 𝑊𝑎 = 10.89 cm2
Area product, 𝐴 𝑝 = 𝐴 𝑐 × 𝑊𝑎 = 150 cm2
Core geometry, 𝐾𝑔 = 28.04 cm5
Surface area, 𝐴 𝑡 = 479 cm2
Step # 5: Calculation of primary number of turns
Primary number of turns,
𝑁𝑝 =
𝑉𝑖×104
𝐾 𝑓 𝐵 𝑎𝑐 𝑓𝐴 𝑐
=
513.16×104
4.44×1.6×50×13.8
= 𝟒𝟕𝟎 turns
Step # 6: Calculation of current density
Current density,
𝐽 =
𝑃𝑡×104
𝐾 𝑓×𝐾 𝑢×𝐵 𝑎𝑐×𝑓×𝐴 𝑝
=
513.16×104
4.44×0.4×1.6×50×150
= 240.78 A/cm2
Step # 7: Calculation of input current
Input current,
𝐼𝑖 =
𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
𝑖𝑛𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
=
𝑃𝑜/𝜂
𝑉𝑖
=
(
250
0.95
)
230
= 1.144 A
Step # 8: Calculation of cross-sectional area (bare) of conductor for primary winding
Bare conductor cross-sectional area,
𝐴 𝑤𝑝(𝐵) =
𝐼 𝑖
𝐽
=
1.144
240.78
= 0.00475 cm2
Step # 9: Selection of wire from wire table
The closest Standard Wire Gauge (SWG) corresponding to the bare conductor area
calculated in step # 8 is 21 SWG.
For 21 SWG conductor,
𝐴 𝑤𝑝(𝐵) = 0.00519 cm2
i.e. 0.519 mm2
Resistance for 21 SWG conductor is 33.2
Ω
Km
= 332
𝜇Ω
𝑐𝑚
Step # 10: Calculation of primary winding resistance
Resistance of primary winding,
𝑅 𝑝 = 𝑀𝐿𝑇 × 𝑁𝑝 × 332 × 10−6
= 22 × 470 × 332 × 10−6
= 3.433 Ω

3. 3
Step # 11: Calculation of copper loss in primary winding
Primary winding copper loss
𝑃𝑝 = 𝐼 𝑝
2
× 𝑅 𝑝 = 1.1442
× 3.433 = 4.493 Watts
Step # 12: Calculation of secondary winding turns
Number of turns in the secondary winding
𝑁𝑠 =
𝑁 𝑝×𝑉𝑠
𝑉𝑖
[1 +
𝛼
100
] =
469×230
230
[1 +
5
100
] = 493
Step # 13: Calculation of bare conductor area for secondary winding
Cross-sectional area of bare conductor for secondary winding
𝐴 𝑤𝑠(𝐵) =
𝐼 𝑖
𝐽
=
1.087
240.78
= 0.00451 cm2
= 0.451 mm2
Step # 14: Selection of conductor size required for secondary winding
From the wire table, the closest cross-sectional area (i.e. next to) is found by choosing
the conductor size as 21 SWG.
For 21 SWG wire, bare conductor area is 0.519 mm2
, for which resistance/cm is
332 𝜇Ω/𝑐𝑚.
Step # 15: Calculation of secondary winding resistance
Secondary winding resistance
𝑅 𝑠 = 𝑀𝐿𝑇 × 𝑁𝑠 × 332 × 10−6
= 22 × 493 × 332 × 10−6
= 3.601 Ω
Step # 16: Calculation of copper in secondary winding
Copper loss in secondary winding,
𝑃𝑠 = 𝐼 𝑜
2
× 𝑅 𝑠 = 1.0872
× 3.601 = 4.255 Watts
Step # 17: Calculation of total copper loss
Total copper loss,
𝑃𝑐𝑢 = 𝑃𝑝 + 𝑃𝑠 = 4.493 + 4.255 = 8.747 Watts
Step # 18: Calculation of voltage regulation
Voltage regulation,
𝛼 =
𝑃𝑐𝑢
𝑃𝑜
=
8.747
250
= 0.035 = 3.5 %
Step # 19: Calculation of Watts per Kg (W/K)
Watts/Kg,
𝑊
𝐾
= 0.000557𝑓1.68
𝐵𝑎𝑐
1.86
= 0.000557 × 501.68
× 1.61.86
= 0.9545

4. 4
Step # 20: Calculation of core loss
Core loss,
𝑃𝑓𝑒 =
𝑊
𝐾
× 𝑊𝑡𝑓𝑒 × 10−3
= 0.9545 × 2334 × 10−3
= 2.23 Watts
Step # 21: Calculation of total loss
Total loss in the transformer,
𝑃Σ = 𝑃𝑐𝑢 + 𝑃𝑓𝑒 = 8.748 + 2.23 = 10.978 Watts
Step # 22: Calculation of Watts/unit area
Watts/unit area,
𝜓 =
𝑃Σ
𝐴 𝑡
=
10.978
479
= 0.023 Watts/cm2
Step # 23: Calculation of temperature rise
Temperature rise,
𝑇𝑟 = 450𝜓0.826
= 450 × 0.0230.826
= 19.95 0
C
Step # 24: Calculation of window utilization factor
Window utilization factor,
𝐾 𝑢 = 𝐾 𝑢𝑝 + 𝐾 𝑢𝑠 =
𝑁 𝑝×𝐴 𝑤𝑝(𝐵)
𝑊𝑎
+
𝑁 𝑠×𝐴 𝑤𝑠(𝐵)
𝑊𝑎
=
470×0.00519+493×0.00519
10.89
= 0.46