On different forums, I often find people asking for help in
calculating the required turns for a ferrite transformer they are going to use
in high-frequency/SMPS inverters. In a high-frequency/SMPS inverter, the
ferrite transformer is used in the step-up/boost stage where the low voltage DC
from the battery is stepped up to high voltage DC. In this situation, there are
really only two choices when selecting topology – push-pull and full-bridge.
For transformer design, the difference between a push-pull and a full-bridge
transformer for same voltage and power will be that the push-pull transformer
will require a center tap, meaning it will require twice the number of primary
turns as the full-bridge transformer.

Calculation of required turns is actually quite simple and I’ll
explain this here.

For explanation, I’ll use an example and go through the
calculation process.

Let’s say the ferrite transformer will be used in a 250W
inverter. The selected topology is push-pull. The power source is a 12V
battery. Output voltage of the DC-DC converter stage will be 310V. Switching
frequency is 50kHz. The selected core is ETD39. Remember that the output of the transformer will be high frequency AC (50kHz square wave in this case). When I refer to an output of high voltage DC (eg 310VDC mentioned above), this is the DC output obtained after rectification (using ultrafast recovery diodes configured as bridge rectifier) and filtration (using LC filter).

During operation, the battery voltage does not stay fixed at
12V. With high loads, the voltage will be less than 12V. With low loads and
near-fully charged battery, the voltage may be higher than 13V. So, it must be
kept in mind that the input voltage is not constant, but is variable. In
inverters, the battery low-cut is usually set at 10.5V. So, we’ll take this as
our lowest possible input voltage.

Vinmin = 10.5V

The formula for calculating the number of required primary
turns is:

For our push-pull transformer, this will be one-half the
required number of turns.

N

_{pri}means number of primary turns; N_{sec}means number of secondary turns; N_{aux}means number of auxiliary turns and so on. But just N (with no subscript) refers to turns ratio.
For calculating the required number of primary turns using
the formula, the parameters or variables that need to be considered are:

- Vin
_{(nom) }– Nominal Input Voltage. We’ll take this as 12V. So, Vin_{(nom)}= 12. - f – The operating switching frequency in Hertz. Since our switching frequency is 50kHz, f = 50000.
- B
_{max}– Maximum flux density in Gauss. If you’re used to using Tesla or milliTesla (T or mT) for flux density, just remember that 1T = 10^{4}Gauss. B_{max}really depends on the design and the transformer cores being used. In my designs, I usually take B_{max}to be in the range 1300G to 2000G. This will be acceptable for most transformer cores. In this example, let’s start with 1500G. So B_{max}= 1500. Remember that too high a B_{max}will cause the transformer to saturate. Too low a B_{max}will be under utilizing the core. - A
_{c}– Effective Cross-Sectional Area in cm^{2}. You will get this information from the datasheets of the ferrite cores. A_{c}is also sometimes referred to as A_{e}. For ETD39, the effective cross-sectional area given in the datasheet/specification sheet (I’m referring to TDK E141. You can download it from here: www.tdk.co.jp/tefe02/e141.pdf ), the effective cross-sectional area (in the specification sheet, it’s referred to as A_{e}but as I’ve said, it’s the same thing as A_{c}) is given as 125mm^{2}. That is equal to 1.25cm^{2}. So, A_{c}= 1.25 for ETD39.

So now, we’ve obtained the values of all required parameters
for calculation Npri – the number of required primary turns.

Vin

_{(nom)}= 12 f = 50000 B_{max}= 1500 A_{c}= 1.25
Plugging these values into the formula:

N

_{pri}= 3.2
We won’t be using fractional windings, so we’ll round off N

_{pri}to the nearest whole number, in this case, rounded down to 3 turns. Now, before we finalize this and select N_{pri}= 3, we better make sure that B_{max}is still within acceptable bounds. As we’ve decreased the number of turns from the calculated figure (down to 3.0 from 3.2), B_{max}will increase. We now need to figure out just how much B_{max}has increased and if that is still an acceptable value.
Vin

_{(nom)}= 12 f = 50000 N_{pri}= 3 A_{c}= 1.25
B

_{max}= 1600
The new value of B

_{max}is well within acceptable bounds and so we can proceed with N_{pri}= 3.
So, we now know that for the primary, our transformer will
require 3 turns + 3 turns.

In any design, if you need to adjust the values, you can
easily do so. But always remember to check that B

_{max}is acceptable.- For example, if for construction difficulties, winding
3 turns + 3 turns becomes difficult, you may use 2 turns + 2 turns or 4 turns +
4 turns. Increasing number of turns won’t hurt – you’ll just be under utilizing
the core. However, decreasing number of turns increases B
_{max}, so just recheck to make sure B_{max}is okay. The range I’ve stated for B_{max}(1300G to 2000G) is just an estimate. It will work for most cores. However, with many cores, you can go higher to decrease the number of turns. Going lower will just be under utilizing the core, but may sometimes be required if number of turns is too low.

- I’ve started off with a set B
_{max}and gone on to calculate N_{pri}from there. You can also assign a value of N_{pri}and then check if B_{max}is okay. If not, you can then increase or decrease N_{pri}as required and then check if B_{max}is okay, and repeat this process until you get a satisfactory result. For example, you may have set Npri = 2 and calculated Bmax and decided that this was too high. So, you set Npri = 3 and calculated Bmax and decided it was okay. Or you may have started with Npri = 4 and calculated Bmax and decided that it was too low. So, you set Npri = 3 and calculated Bmax and decided it was okay.

Now it’s time to move on to the secondary. The output of our DC-DC
converter is 310V. So, the transformer output must be 310V at all input voltages,
from all the way up from 13.5V to all the way down to 10.5V. Naturally,
feedback will be implemented to keep the output voltage fixed even with line
and load variations – changes due to battery voltage change and also due to
load change. So, some headroom must be left for feedback to work. So, we’ll
design the transformer with secondary rated at 330V. Feedback will just adjust
the voltage required by changing the duty cycle of the PWM control signals.
Besides feedback, the headroom also compensates for some of the losses in the
converter and thus compensates for the voltage drops at different stages – for example,
in the MOSFETs, in the transformer itself, in the output rectifiers, output
inductor, etc.

This means that the output must be capable of supplying 330V
with input voltage equal to 10.5V and also input voltage equal to 13.5V. For
the PWM controller, we’ll take maximum duty cycle to be 98%. The gap allows for
dead-time.

At minimum input voltage (when Vin = Vinmin), duty cycle will
be maximum. Thus duty cycle will be 98% when Vin = 10.5 = Vinmin. At maximum
duty cycle = 98%, voltage to transformer = 0.98 * 10.5V = 10.29V.

So, voltage ratio (secondary : primary) = 330V : 10.29V =
32.1

Since voltage ratio (secondary : primary) = 32.1, turns ratio
(secondary : primary) must also be 32.1 as turns ratio (secondary : primary) =
voltage ratio (secondary : primary). Turns ratio is designated by N. So, in our
case, N = 32.1 (I’ve taken N as the ratio secondary : primary).

N

_{pri}= 3
N

_{sec}= N * N_{pri}= 32.1 * 3 = 96.3
Round off to the nearest whole number. N

_{sec}= 96.
Thus 96 turns are required for the secondary. With proper
implementation of feedback, a constant 310VDC output will be obtained
throughout the entire input voltage range of 10.5V to 13.5V.

Here, one thing to note is that even though I took 98% as the
maximum duty cycle, maximum duty cycle in practice will be smaller since our
transformer was calculated to provide 330V output. In the circuit, the output will
be 310V, so the duty cycle will be even lower. However, the advantage here is
that you can be certain that the output will not drop below 330V even with
heavy loads since a large enough headroom is provided for feedback to kick in and
maintain the output voltage even at high loads.

If any auxiliary windings are required, the required turns can
be easily calculated. Let me show with an example. Let’s say we need an
auxiliary winding to provide 19V. I know that the output 310V will be regulated,
whatever the input voltage may be, within the range initially specified (Vinmin
to Vinmax – 10.5V to 13.5V). So, the turns ratio for the auxiliary winding can
be calculated with respect to the secondary winding. Let’s call this turns
ratio (secondary : auxiliary) N

_{A}.
N

_{A}= N_{sec}/ N_{aux}= V_{sec}/ (V_{aux}+ V_{d}). V_{d}is the output diode forward drop. Let’s assume that in our application, a schottky rectifier with a V_{d}= 0.5V is used.
So, N

_{A}= 310V / 19.5V =15.9
N

_{sec}/ N_{aux}= N_{A}
N

_{aux}= N_{sec}/ N_{A}= 96 / 15.9 = 5.96
Let’s round off N

_{aux}to 6 and see what the output voltage is.
V

_{sec}/ (V_{aux}+ V_{d}) = N_{A}= N_{sec}/ N_{aux}= 96 / 6 =16.0
(V

_{aux}+ V_{d}) = V_{sec}/ N_{A}= 310V / 16.0 = 19.375V
V

_{aux}= 19.375V – 0.5V = 18.875V (rounded off)
I would say that’s great for an auxiliary supply. If in your calculations you come to a voltage that is too far off the required target voltage and thus greater
accuracy is required, take V

For example, if in our previous example, instead of18.875V we had gotten 19.8V but needed more accuracy, we could've used 24V or thereabouts and used a voltage regulator to give 19V output.

_{aux}as something higher and use a voltage regulator.For example, if in our previous example, instead of18.875V we had gotten 19.8V but needed more accuracy, we could've used 24V or thereabouts and used a voltage regulator to give 19V output.

So, there we have it. Our transformer has 3 turns + 3 turns
for primary, 96 turns for secondary and 6 turns for auxiliary.

Here’s our transformer:

Calculating required number of turns for a transformer is
actually a simple task and I hope that I could help you understand how to do
this. I hope this tutorial helps you in your ferrite transformer designs. Do
let me know your comments and feedback.