| Determine minimum number of turns for a 120V60Hz transformer primary |
Not
really, but everyone always talks about turns ratios, but forgets to
mention the minimum number
of turns for the primary. Well, the iron/silicon/metal core of the
transformer can accept only so
much magnetic flux before it saturates and can't take more. If you go
beyond this, the inductance
drops a lot and you end up drawing a lot more current off the powerline
and it will get really hot,
not good. A conservative rule of thumb, for transformer laminations you
may salvage from a junked 60Hz transformer:
Number of turns needed for the 120V60Hz primary = 800/(area of the core
in square inches). You
measure the height (in the below diagram "L") of the pile of
laminations, and the width ("W") of the center leg of the E lamination.
In other words, imagine a single turn of wire tight on the center core,
the area of
the loop this single turn forms is the area. In the below diagram, it's L
times W. Do not include
the outer legs of the E, or the I. A bigger area will make for a lower
number of turns. Once you have
the number of primary turns, then you can do the turns ratio to get the
number of turns for the
secondary. Add a few more turns to make up for resistance of the wire
voltage losses. If your
powerline frequency is 50Hz, you need 60/50 times the above 120V60Hz
result for your primary for 120V, and
twice that for 240V. Which works out to be 2.4 times the above 120V60Hz
result above
Of course, you also need to select magnet wire thick enough to handle the current the load will impose,
but thin enough so you can get all the turns around the core without getting too fat that the outside
parts of the E lams won't fit over. And also allow for insulation to insulate the core lams from
the bottom of the windings, and insulation between the primary and the secondary (safety regs want 1400VDC
of insulation) and the sides of the windings too. Winding the turns neatly will allow tighter packing of the
turns). If the wire works out to be too thin for the currents, you will need a bigger core. This equation should get
you in the ballpark:

A is the core cross section in sq inches and W is the VoltAmp output.
A more general equation:

E is the winding voltage, ie, primary input voltage, or a secondary output voltage.
F is the powerline (mains) frequency
H is the number of lines of magnetic flux per square inch of iron, for most power transformers
it's 56300 RMS. The iron won't accept much more.
N is the number of turns for the above mentioned E primary or secondary.
A is the cross section area in sq inches of the core
If E is taken as 1 it gives the turns per volt for any winding on the core.
Working this equation to check if it matches the simple rule of thumb:
Do some algebra, set A to 1 square inch, F to 60Hz, E to 120V, and H to 56300, and we have
N=(108 E)/(A H F (4.44)) = (108 *120)/(1 * 56300 * 60 * (4.44)) = 800. But this is circular fudging, as I got the 56300 working the equation the other way.
High permeability iron alloys used in transformers reach magnetic saturation at 1.6 to 2.2 teslas (T).
Okay, a telsa is the number of lines of magnetic flux per square meter. There's 1553 square inches in a
square meter. Taking the above 56300 * 1552 = 87377600. But the winding voltage E is an RMS value, and
the magnetic saturation mentioned would be a limit, and you have to take into account of the peak
value of E. Recalculating the H fudge variable with the peak value of E instead of its RMS value
gets us 78400 per square inch for H. Which gets us 123000000 for square meters. Combining it with the 108
in the denominator in the above equation gets us 1.23, which looks like it may be telsas. If so, this looks like
a conservitive value to stay well away from saturation. All this handwaving may or may not be valid,
as I'm no physicist.
But if my handwaving is good, it looks to validate the
rule of thumb "Number of turns needed for the 120V60Hz primary = 800/(area of the core in square inches)".