June 24, 2016

Common Mode Choke Design

COMMON MODE CHOKE / INDUCTOR DESIGN

 UU Inductor - Common Mode ChokeThe common mode choke is one of the many types of inductors. Common mode choke is usually used in a power supply to filter out noise from the incoming line voltage. Impedance is the most important aspect of a common mode choke.

Impedance of an inductor is usually specified as a minimum (in OHMs) at the desired frequency. The required inductance can be calculated by the following equation:

inductance equation

Where:

Xs = impedance in OHMs (Ω)

F = frequency in Hertz (Hz)

Ls = inductance in Henry (H)

The above formula give you the minimum inductance you need to achieve the required impedance.

Now that you have the inductance, the next steps are:

Step one: determine the wire size.

Wire size can be determined based on the rule of thumb of 500 circular mils per amp:

wiresize equationWhere:

Aw = nominal area for the wire

I = current in amps

Nominal area of wire is listed on wire manufacturer’s data sheets. For example,

http://www.elektrisola.com/enamelled-wire/technical-data-by-size/nema-mw1000c.html

Step two: select a core.

Proper core selection is somewhat trial and error. A ferrite material with 5000 permeability or higher is commonly used. The core must also be large enough to fit the turns of the wire gauge you need. Each winding must fit in a single layer on different sides of the core with enough room to place a spacer (typically 3mm) between them.

Step three: calculate the number of turns.

Number of turns required is given by:Turns EquationWhere:

L = inductance in henries

T = turns

Al = is the Al value of the core (found on core data sheets from core manufacturer)

Design example:

Known requirements:  377 Ω of impedance is required at 50KHz in a circuit carrying 2 Amps of current.

Required minimum inductance:Design Example calculationTo find out wire size:wire sizeFrom the wire chart, the 20AWG has a nominal area of 1024 cmils, therefore, select the 20AWG. Overall maximum diameter for 20AWG is .0666” with a single build insulation.

If we select a core with an Al = 12080nH/T2  and an inside diameter of 0.57”, the required turns are:common choke equations

T=99.3 turns

T=10 turns

Two windings of 10 turns with each using 20AWG size wire will fit on a core with an ID of 0.57” and still leave room for a 3mm spacer.

Then you are done!