CALCULATING THE PRIMARY WINDING AND AIR GAP FOR A SWITCHMODE TRANSFORMER

In the previous blog we showed how to calculate the inductance needed for an example power supply with a max input of 1000 volts and a power rating of 150 watts. Now lets see how to calculate the number of turns required. This is the standard formula for a square wave voltage so it approximates the switchmode condition. This formula will give us a starting point and it also verifies that we have at least the number of turns to withstand the applied voltage without magnetic saturation:

N = Vpk * 10^8 / (4 * Bmax * Ac * F)

where N is the number of turns, Bmax is the maximum allowed flux density without saturation for the material to be used ( some type of ferrite ), Ac is the core area, and the frequency F. We will work in CGS units and the flux density will be in Gauss units.

For a 150 watt transformer, from experience I will estimate the size of the transformer as a cube approximately 5 cm on a side ( about 2 inches.) This will also be the approximate size of a magnetic structure of say an EI or EE structure. Assume that the structure will be an EE form with a single air gap in the center of the structure. With this size and form I will further estimate (based on experience that the core area will be a square cross-section 1 cm on a side or 1 sq. cm. Assume further that the operating frequency will be 100KHZ. Now we can calculate our initial number of turns from our equation above using Bmax as 4000 (you will need to check what Bmax is for the ferrite material you plan to use) as

N = 1000 *10^8 / (4 * 4000 * 1 * 100000) = 62.5 round off to 63

Now we will need to know what is the magnetic path length of the core structure we have selected. You will most likely use the number you find in the manufactures catalog. Here we will estimate it to be 10 cm based on the size of the magnetic structure and its form. Also we will use 1/2 of the Bmax value as our max operating flux level, or 2000 Gauss. Then we will use our data so far and the following formula to calculate the effective permeability of the core as

Ue = B * Pl / ( 1.257 * N * I )

We will use 1 amp as our max current I, flux density B of 2000 Gauss and for the magnetic path length Pl our estimate of 10 cm:

Ue = 2000 * 10 / ( 1.257 * 63 * 1 ) = 252.55 round off to 253

From the catalog we need to obtain the permeability, U, of the ferrite material we are going to use. Lets assume that the basic permeability is 2500. Now we can make a first calculation of the air gap from the formula

Lg = Pl * ( 1 / Ue ) - Pl * ( 1 / U)

Lg = 10 * ( 1 / 253 ) - 10 * ( 1 / 2500 ) = 0.0355 cm

The above would be enough calculations to build an experimental transformer, but it might be wise to do some more calculations to verify the previous results. I will not do all of the numerical examples here but I will show the equations you can use , still working in CGS units.

Re-calculate the number of turns based on the effective permeability, Ue, and the required current I :

N = B * Pl / ( 1.257 * Ue * I ) = 62.89

This value of 62.89 checks well with our original calculation of 62.5, so we will stay with 63 turns for the primary of the transformer.

Next we can calculate the actual inductance of our transformer primary for our design to this point:

L = 1.257 * Ue * N^2 * Ac * 10^-8 / Pl = 0.00126 Henry

This value is less than what we originally decided we needed (from the previous post) of 0.005 H. So we will need to increase the area of the core by a factor of approximately 4 increasing Ac to 4 cm sq.

Now we need to check on the core area, Ac, that we need so we will compute it from the desired inductance value and effective permeability, assuming we did not change it:

Ac = L * Pl * 10^8 / ( 1.257 * Ue * N^2 ) = 3.96 cm^2

The result of the calculation is obvious from looking at the equation. We could have also increased Ue or N to achieve the same result, but it is better to increase Ac so that the maximum operating magnetic flux is not increased. Now it would be good to run back through our above calculations with our new value for L and Ac to get a design closer to what is needed (several iterations may be needed.)

But lets look at a little more design work. It may be better to compromise between the original estimate of Ac and the new, so that you might want to try some new values for Ac and N. One thing to note in the inductance formula is that L is proportional to the square of N so we can just do a minor increase in N along with a more practical value for Ac to get the 5 mH inductance we want. To do this we will "push" the design a little making it a little more risky but easier to make the transformer. For example, as a more practical value lets try 2 cm sq.for Ac. Also lets increase the turns N by a factor of about 1.4 or to 90 turns. We can wind the primary as two layers of 45 turns each. After re-doing the inductance and operating flux density calculations we have the following (the reader should be able to do these calculations now):

V = 1000 volts (spec.)
I = 1.0 amps (spec.)
Ac = 2 cm sq. (changed)
N= 90 turns (changed)
Ue = 253 (did not change this)

L = 0.00515 mH
B = 2862 Gauss

So now the changes yield the inductance you want at the price of increasing the operating flux density but still staying considerably less than the maximum of 4000 Gauss. This would be a possible starting point to for your design, but you will still need to determine how to design the secondary windings. We will cover this in another post.