What is the formula for Transformer Design?
Answer:
There are two approaches used in designing transformers. One uses the long formulas, and the other uses the Wa product. The Wa product is simply the cores window area multiplied by the cores area. Some say it simplifies the design, especially in C-core (cut core) construction. Most manufacturers of C-cores have the Wa product added into the tables used in their selection. The designer takes the area used by a coil and finds a C-core with a similar window area. The Wa product is then divided by the window area to find the area of the core. Either way will bring the same result.
For a transformer designed for use with a sine wave, the universal voltage formula is:
E={\frac {2 \pi f N a B} {\sqrt{2}}} \!=4.44 f N a B
thus,
E={4.44 f N a B} \!
where,
* E is the sinusoidal rms or root mean square voltage of the winding,
* f is the frequency in hertz,
* N is the number of turns of wire on the winding,
* a is the cross-sectional area of the core in square centimeters or inches,
* B is the peak magnetic flux density in teslas per square centimeter, gausses per square centimeter, or lines (maxwells) per square inch.
* P is the power in volt amperes or watts,
* W is the window area in square centimeters or inches and,
* J is the current density.
* Note: 10 kilogauss = 1 tesla.
This gives way to the following other transformer equations for cores in square centimeters:
N={\frac {E 10^8} {{4.44 f B a}}} \!
B={\frac {E 10^8} {{4.44 f N a}}} \!
a={\frac {E 10^8} {{4.44 f B N}}} \!
P={0.707 J f W a B} \!
The derivation of the above formula is actually quite simple. The maximum induced voltage, e, is the result of N times the time-varying flux:
e = N dφ/dt
If we are using RMS voltage values and E equal the rms value of voltage then
e = E{\sqrt{2}}
and
E = dφ/dt{\frac {N} {\sqrt{2}}}
Since the flux is created by a sinusoidal voltage, it too varies sinusoidally:
φ(t) = Φmaxsinwt = ABmaxsinwt, where A = area of the core
Taking the derivative we have:
dφ(t)/dt = wABmaxcoswt
Substituting into the above equation and using
w = 2 πf and the fact that we are only concerned with the maximum value yields
E = \frac {{2\pi}fNAB} {\sqrt{2}}
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