Ferrite Core Inductor Design Calculation

In previous article we discussed about constraints for ferrite core inductor design for converter circuit. This ferrite core inductor serves as filter for ripple currents and designed according to specific power electronic converter circuit. Now, by using those constraints we are going to design a practical ferrite core filter inductor as an example.

Mobile users please use landscape mode for better view of formulas in this articles.

Practical Filter Inductor Design Calculations

Let’s take an example for designing inductor according to discussed constraints in previous article. (An unsolved problem from a book of Fundamentals of Power Electronics by Erickson has been taken as an design problem)

Problem Statement for Ferrite Core Inductor Design:

Problem is designing a ferrite core inductor for ripple current filter in a boost converter. This converter converts 28 V DC to 48V for 150W of maximum load. Converter operates at frequency of 100 kHz. Peak flux density is 0.225 T and fill factor is 0.5. Copper losses of of 0.5% of load power are allowed. Ripples are 10% of DC inductor current.

1. Calculating Required Inductance:

For calculating ferrite core filter inductor, first of all we will calculate required inductance according to requirement of circuit.

Calculating Duty Cycle for Boost Converter

Therefore, we will start from calculating Duty Cycle D for this converter circuit:
we know that $$ D = 1 – \frac{V_{in} \times eff}{V_{out}}$$
So, D = 0.533 with 80% efficiency.

Sampling Time Calculation

Now, sampling time $T_s$ will be inverse of frequency
$$T_s = \frac{1}{f_s} $$

$$T_s= 1 \times 10^{-5}$$

Ripple Current Calculation

Now, calculating ripple current for which filter inductor is to be designed. As given in design requirements, that current ripple is 10% of maximum current. Also 0.5% of copper losses are there. So,
$$\Delta i_m = 0.1 \times [I_{max}+0.5 \% I_{m}]$$

$$\Delta i_m =0.1 \times \frac{P}{V_{out}} \times (1+0.005) $$

$$\Delta i_m= 0.3140 A$$


Finally, gathering all the values and putting in formula for required inductance of ferrite core inductor of boost converter circuit. $$L= \frac{V \times D \times T_s}{2 \Delta i_m}$$
So, Putting Values we got
$$L= \frac{ 28 \times 0.533 \times 1 \times 10^{-5}}{2 \times 0.3140}$$

$$L=0.2376 mH$$
Hence, 0.2376mH of inductance should be placed for required boost converter.

Geometric Constraints for Ferrite Core:

Now, we will find ‘Geometry constant’ using $K_g$ method whose value combines all the constraints required for core geometry.
$$ K_g \geq \frac{\rho L^2 I^2_{m} }{B^2_{m}R K_u} \times 10^8 (cm^5) $$
We have:

  • $ \rho$ is resistivity of copper and it is $1.72 \times 10^{-6} \Omega cm$.
  • L is inductance (calculated earlier)
  • $I_{max}$ is max current.
  • $B_{max}$ is maximum operating flux density
  • R is winding resistance
  • $K_u$ is winding fill factor

Putting values, (mobile phone users !! please switch to landscape mode for better view)
$$K_g \geq \frac{1.72 \times 10^{-6} \times (0.2388\times10^{-3})^2 \times (3.125)^2}{(0.225)^2 \times R \times 0.5 } \times 10^8 (cm^5)$$

Here, R is unknown. We can find by $P_{loss}=I_{max}^2\times R$. i.e. we get
$$R = \frac{P_{loss}}{I_{max}^2} =\frac{0.5 \% \times P_{max}}{I_{max}^2} $$

$$R= \frac{0.5 \times 150}{100 \times (3.140)^2}$$
So, Putting all the values in inequality of Geometric constant $K_g$.

$$K_g \geq \frac{1.72 \times 10^{-6} \times (0.2388 \times 10^{-3})^2 \times (3.140)^4}{(0.225)^2 \times 0.75 \times 0.5 } \times 10^8 (cm^5)$$

$$K_g \geq 0.050 (cm^5) = 50 \times 10^{-3}$$

Finally, we have put all the constraints in geometric constant and obtained a value which is threshold for minimum geometric parameters for ferrite core.

Now, we have checked standard tables for the geometric values of the core. According to tables, pot type core is selected (value of $K_g \geq 0.050$) with following values.

  • Core type : 2616 (mm)
  • Cross sectional Area ($A_c$) : 0.948 ($cm^2$)
  • Bobbin Winding Area ($W_A$) : 0.406 ($cm^2$)
  • Mean length per turn (MLT) : 5.28 (cm)
  • Magnetic Path length ($l_m$) : 3.75 (cm)
  • Terminal Resistance ($R_{th}$) : $30 ^0 C / W$
  • Core Weight : 20 (g)

Therefore, by using the core of this specification, core will neither be very large nor too short to saturate.
Furthermore, You can see similarity with Transformer design calculation that we calculate core size according to magnetic as well as winding size.

Finally, more articles, or projects from experienced engineer, stay tuned. Or contact us or share your experiences through us.

What do you think?

This site uses Akismet to reduce spam. Learn how your comment data is processed.