## 2 Transformer Design Concept  
The design of distribution transformers must have specific characteristics such as power rating, phase voltage, frequency, vector group, impedance percentage, loss power requirements or test standards, etc. The design will consider such conditions so that the results of the method meet the objectives [12]. The procedure begins with Equation (1). \[E_{t}=\frac{V_{p}}{N_{p}}=\frac{V_{S}}{N_{S}} \tag{1}\]  
where the \(E_{t}\) is the transformer ratio; \(V_{p}\) and \(V_{S}\) are the voltage on the primary and secondary side, respectively; \(N_{p}\) and \(N_{S}\) are the number of turns in the primary and secondary coil, respectively. The estimation of the initial value of Et was determined by Equation (2). \[E_{t}=K\sqrt{Q} \tag{2}\]  
The \(K\) is the constant set between 0.3-0.6 and the \(Q\) is the rated power (kVA). The \(A_{g}\) is the core cross-sectional area which can be calculated according to Equation (3). \[E_{t}=4.44 fB_{m}A_{g}.10^{-4} \tag{3}\]  
where \(f\) is the power system frequency (Hz), \(B_{m}\) is the magnetic force line density (T). When the \(A_{g}\) is the net core cross-section area (mm\({}^{2}\)) thus the diameter of the iron core (\(d\)) is determined from Equation (4). \[A_{g}=K_{1}\frac{\pi d^{2}}{4} \tag{4}\]  
when \(K_{1}\) is the core component coefficient, for example, \(K_{1}\) is 0.92, 0.925, 0.93, 0.935, 0.94, 0.945, when the core layer is 6, 7, 8, 9, 10 and 11, respectively. The current density in the windings is determined by the type of conductor, level of loss and the temperature increase of the coil being industry acceptable, generally defined as no more than 3.5 A/mm\({}^{2}\). Insulation considerations shall be set to suit the test voltage rating, where the safety distance used for insulating oil designs is rated as 3 kV/mm for uninsulated oil, paper barrier, and 5 kV/mm for insulating oil, which is a specified period for industrial use. The total core weight consists of the importance of the three axes and the weights of the upper and lower yokes, which can be calculated according to Equation (5). \[W_{Fe}=D_{Fe}A_{Fe}(3 WH+4 CD+2 WD_{1}) \tag{5}\]  
where:  
\(W_{Fe}\): total iron core weight (kg). \(D_{Fe}\): laminate density of 7.65 \(\times\) 10\({}^{-6}\) kg/mm. \(WD_{1}\): widest laminate width (mm). \(WH\): Axial pin height (mm). \(CD\): distance between the center pin (mm). The coil weight (\(M_{cu}\)) can be calculated according to Equation (6). \[M_{cu}=D_{cu}A_{cu}L_{m}N \tag{6}\]  
where:  
\(M_{cu}\): weight of the conductor coil (kg). \(D_{cu}\): conductor density (Cu = 8.9 \(\times\) 10\({}^{-6}\) kg/mm\({}^{3}\)). \(A_{cu}\): net cross-section area of conductor (mm\({}^{2}\)). \(L_{m}\): average cycle length of the coil (mm). \(N\): number of turns of the coil. Equation (7) shows the calculation for the no-load loss (\(P_{NL}\)), which can be calculated from the iron core loss value in terms of watts per kilogram (W/kg) at the magnitude of the current magnetic line density obtained from the manufacturer together [13]. \[P_{NL}=k_{wn}P_{m}W_{Fe} \tag{7}\]where:  
\(P_{NL}\): no-load loss (W). \(k_{wn}\): multiplier coefficient from the operation and quality of the laminate. \(P_{m}\): power dissipation in the iron core (W/kg). The load loss as the sum of the resistance losses in the coil and the stray losses. The power loss from the resistance in the coil can be calculated according to Equation (8). \[R=\frac{\rho L_{m}N}{A_{cu}} \tag{8}\]  
where:  
\(R\): winding resistance (\(\Omega\)). \(\rho\): specific resistance of the conductor winding (\(\Omega\)-mm). The resistance of the winding can be found in Equation (9). \[P_{LL}=k_{cup}I_{P}^{2}R_{P}+k_{cus}I_{S}^{2}R_{S}+lead_{p}+lead_{S}+stray \tag{9}\]  
where:  
\(P_{LL}\): loss at the winding (W). \(k_{cup}\) and \(k_{cus}\): operating coefficient at the primary and secondary windings, respectively. \(I_{p}\) and \(I_{S}\): rated currents at primary and secondary windings (A), respectively.