|
TRANSFORMER DESIGN AND SIMULATION
Narrative:
A transformer design was performed with the following parameters:
Input Voltage or Primary Voltage (RMS) is Vp= 220V at 60Hz. Desires to have Output voltage of 12v on the Secondary.
The Primary winding calculated to be 1.0mm dia wire and requires 369 Turns with measured resistance of 2.26 Ohms.
Rp = 2.26 Ohms.
The designed Current for primary is based on 250W rating, hence 250W/220v
Ip = 1.13Amp.
Let's find Inductance of the Primary coil on following steps:
First, Power the Transformer without load on Secondary and measure the no-load current.
No load Current=0.0971 Amp
1. Calculate Impedance of primary coil:
Impedance Z = Vp/Ip
Impedance Z = 220/0.097 = 2,265.70 Ohm
2. Calculate Inductive Reactance (Xi):
Impedance: Z² = Xi ² + (Xr - Xc) ² , Xc=0:
Xi = √ ( Z² - R³)
Xi = √ (2265.70² - 2.26²) = 2265.70 Ohm
Inductance of Primary Coil
XL = 2π x f x Lp
So...
Lp = XL / 2 π f
Lp = 2265.70/ (2 x π x 60)
Lp = 6.03 Henry (Same as Simulated Value)
Relation of Turn ratio to coil Turns, Current, Voltage, and Inductors.
Relationship Equation:
T.R. = Vs/Vp = N's/Np = Is/Ip = sqrt (Ls/Lp)
Thus, using this relationship, we can then determine the Inductor on the Secondary coil.
Find first the coil Turn Ratio.
T.R = 12/220 = 0.05454
Given Inductor value of Primary, Lp = 6.03
and the Turn ratio..
TR² = Ls/Lp
Ls = Lp x T.R.² = 6.06 x (0.05454)²
Ls = 0.018 Henry
Given the Primary windings, We can also determine Secondary winding..just for demonstration...😊
0.05454 = Ts/Tp
Ts = 0.05454 x 365 turn
Ts = 19.90 Turns
Notes:
The software model parameters has the Users option available to enter the Inductor value of the Primary coil and the Turn ratio. There is no input available for the user to enter the Inductor value for the Secondary coil.
Maybe because given the input for Primary Inductor and Turn ratio, the Secondary Inductor value can be calculated behind the scene (ín the algorithm) as we have shown here.
|
