This transformer equation is known as the “transformer design equation”. This transformer equation is applicable for inputs that are of most any voltage. The usual voltage forms are sinusoids as seen in lower frequency silicon steel type transformers such as UL Class 2 recognized power transformers. The MDT software allows application-related parameters to be calculated for all available EPCOS ferrite cores and / or materials. It provides access to their digitized material data including their graphical representations.

In this post we comprehensively discuss how to design and calculate your own ferrite transformer by suitably calculating the various necessary parameters such as primary turns, Bmax of the ferrite core, the secondary turns, the core dimensions, auxiliary winding and other related variables.

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. SMPS Transformer Coil Transformer Calculation Programs. Before “Advanced SMPS Transformer design program” ExcellentIT and “Ir2153 SG3525 Frequency Calculation Programs for Switch mode power supplies with raisers” shared a very useful programs. Programs prepared by Russian-born Vladimir Denisenko sharing a person known as a Russian electronic. Electronics Projects, SMPS Transformer Coil Transformer Calculation Programs 'electronics software tools, ' Date 2019/08/04. SMPS transformer design tool. Here's a full and complex design tool for calculating the SMPS transformer based on a push-pull, half-bridge and full-bridge topology. It has many pop-up tips to help you to use it and correctly paste the values. I think that it'll be very useful for many of you, as for the beginners and so as for the guru of DIY.

For transformer model, the distinction among a push-pull and a full-bridge transformer meant for similar voltage and power will likely be that the push-pull transformer will demand a center tap, which means it will need two times the amount of primary turns as the full-bridge transformer.

Calculation of required turns is in fact pretty basic I’ll clarify this here.

For description, I’ll work with an illustration and deal with calculation process.

Excellent It Smps Transformer Calculation Tool Supply

For example let's imagine the ferrite transformer is intended for a 250W inverter. The chosen topology is push-pull. The power supply is a 12V battery. Output voltage of the DC-DC converter stage is going to be 310V.

Switching frequency is 50kHz. The chosen core is ETD39. Keep in mind that the output of the transformer will probably be high frequency AC (50kHz square wave in this instance). As i make reference to an output of high voltage DC (eg 310VDC mentioned previously), this is actually the DC output attained following rectification (making use of ultrafast recovery diodes set up as bridge rectifier) and filtering (using LC filter).

In the course of operation, the battery voltage is not going to be fixed at 12V. With higher loads, the voltage is going to be under 12V. With reduced loads and around fully charged battery, the voltage could possibly be greater than 13V. Therefore, it should be considered that the input voltage is not really consistent, rather is varying. In inverters, the battery low-cut is commonly established at 10.5V. Thus, we’ll work with this as our most reasonable input voltage.

Excellent It Smps Transformer Calculation Tool

Vin(min) = 10.5V

Calculating Primary Turns

The formula for determining the quantity of necessary primary turns is:

For the push-pull transformer, this can be one-half the required range of turns.

N(pri) indicates amount of primary turns; N(sec) signifies amount of secondary turns; N(aux) suggests amount of auxiliary turns and so forth. However just N (with no subscript) describes turn or winding ratio.

For calculating the desired volume of primary turns utilizing the formula, the variables or specifics that must be deemed are:

Vin(nom) - Moderate Input Voltage. We will consider this as 12V. Therefore, Vin(nom) = 12.

f = The working switching frequency in Hertz. Considering that our switching frequency is 50kHz, f = 50000.

B(max) = Maximum flux density in Gauss. Should you be accustomed to applying Tesla or milliTesla (T or mT) for flux density, remember 1T = 104Gauss.

B(max) actually is determined by the design and the transformer cores used. In our designs, we'll typically consider Bmax to stay within the range 1300G to 2000G.

This can be suitable for the majority of transformer cores. In this particular illustration, we may focus on 1500G. Hence Bmax = 1500. Keep in mind that too big a B(max) can cause the transformer to saturate. Far too lower B(max) will likely be not optimally utilizing the core

Ac = Effective Cross-Sectional Area in cm2. You will find these details through the datasheets of the ferrite cores. Ac can also be occasionally known as Ae.

For ETD39, the productive cross-sectional area offered in the datasheet/specification sheet we are talking about is TDK E141.

You may download this from here: Ferrite Core Dimension, the productive cross-sectional area (in the specification pdf, it’s called Ae however as mentioned, it’s one and the same as Ac) which is presented as 125mm^2. Which is equal to 1.25cm^2. So, Ac = 1.25 for ETD39.

Now, we have access to the values of all necessary variables for calculation N(pri) - the amount of necessary primary turns.

Vin(nom) = 12, f = 50000, Bmax = 1500, Ac = 1.25

Inserting these quantities into the formula:

N(pri) = 3.2

We won’t be applying fragmentary winding, therefore we are going to round off N(pri) for the closest whole number, in this particular scenario, let's round it down to 3 turns.

At this point, before we complete this and choose N(pri) = 3, we better make certain that B(max) continues to be within tolerable range. As we’ve lowered the quantity of turns from the calculated number (down to 3.0 from 3.2), B(max) will certainly go higher.

Calculating Maximum Flux Density of Ferrite Core

We at this point have to determine exactly how much B(max) has increased and if that continue to be an appropriate value.

Vin(nom) = 12, f = 50000, N(pri) = 3, Ac = 1.25

B(max) = 1600

The new figure of B(max) is actually within tolerable range and thus we could continue with N(pri) = 3.

Therefore, we now recognize that for the primary, our transformer will have to have 3 turns + 3 turns.

In just about any design, if you want to modify the values, it is simple to achieve this task. However bear in mind to evaluate that B(max) is appropriately selected.

As an example, suppose due to structure complications, winding 3 turns + 3 turns gets challenging, you could employ 2 turns + 2 turns or 4 turns + 4 turns.

Maximizing number of turns won’t harm - you will merely would be under utilizing the core. However, reducing quantity of turns may raise B(max), therefore just reconfirm to make certain B(max) is appropriately selected. The range we are using for B(max) (1300G to 2000G) is simply an estimation. It will eventually be suitable for most of the cores. Nonetheless, with several cores, it is possible to go bigger in order to decrease the amount of turns.

Planning lower may simply be not utilizing the core optimally, yet may at times be necessary in case number of turns is far too low.

We can start using a fixed B(max) and go on to estimate N(pri) following that. You can even designate a value of N(pri) and subsequently find out if B(max) is acceptable. Otherwise, you can easily maximize or minimize N(pri) as necessary and next see if B(max) was suitable or not, keep repeating this right up until you obtain an acceptable end result.

For instance, you might have fixed N(pri) = 2 and determined B(max) and determined that this had been way too high. Thus, you place N(pri) = 3 and determined B(max) and realized that it was alright. Or perhaps you could have began with N(pri) = 4 and worked out B(max) and made a decision that it was way too small. Therefore, you fixed N(pri) = 3 and computed B(max) and opted it absolutely was good.

At this point it’s time to proceed on to the seconda