What’s about Common Mode Choke

Common modechoketypically consist of a magnetic core and two sets of winding. Visually, they appear as two sets of copper wires wound around a ring core, and they also come in different structural designs such as UU-type, and chip-type. Common-modechoke are usually connected in series on the L line and N line of the power supply. Their main function is to allow normal operating current to pass through while suppressing common-mode high-frequency interference.

Figure 1 | Schematic of the appearance structure of a common-mode inductor

Figure 2 | Schematic of common-mode inductor connected in series on the L/N line

For Example: What’s 6.8mH meaning?

mH Is it the unit of inductance?. for example: L=6.8mH

Conversion relationship: 1H = 1000mH. When engineers say 'this common-mode inductor is 6.8mH', they are typically referring to the common-mode inductance value of the component. Inductance is one of the important parameters that describe this common-mode inductor.

For Example: What’s 1A, 2A refer to?

It’s rated current, take 6.8mH/2A for example.

6.8mH : Inductance   2A: Allowable working current related indicators

The larger the current, the higher the requirements are typically for the copper wire gauge, DCR (Direct Current Resistance), and temperature rise. Therefore, you cannot simply say, 'I need a 6.8 mH inductor.' You also need to know: what actual current will flow through it.

What’s AL Value?

Take“AL=5000nH/N2” for example, we can treat it like: Request wind how many pcs to received how much inductance value?

Common Formulas: L = AL×N²( L: Inductance Value; AL: The core’s AL Value; N: Winding set’s Turns)

So, once we get the target inductance value & AL value, we can calculate out based on: N =√(L / AL)

How to do based on the formulas?    

For example, what needs to be done now is:  6.8 mH       

Provided by the magnetic core manufacturer:  AL = 5000 nH/N²     

First, unify the units:6.8 mH = 6,800,000 nH    

Then substitute the formula directly: N = √(L / AL)       

In other words:  N = √(6,800,000 ÷ 5000)      

The calculation yields: N ≈ 37

What does it mean? It's simple: theoretically, each winding has approximately 37 turns. Therefore, this common-mode inductor can be initially designed as: 37 turns + 37 turns

This is what people refer to as: 'How many turns should this core be wound with?' Here, T (Turns) refers to the number of turns.      

What’s DCR meaning?

DCR stands for Direct Current Resistance.

Copper wires themselves are not completely resistance-free. Generally, the longer and thinner the wire, the larger the DCR. The basic formula is: R = ρL / S (R is resistance;ρis the resistivity of copper; L is the length of the copper wire; S is the cross-sectional area of the copper wire. )

Therefore, when engineers say, 'This common-mode DCR is too high,' they usually mean that the DC resistance of the winding is too large, which may result in greater losses and temperature rise.

Why is it still necessary to confirm 'what wire gauge'?"

Because copper wires need to carry the actual working current. Take 1A for example.

An initial estimate can be made based on current density first: S = I / J(S - Cross-sectional area of copper wire; I - Current; J - Design current density)

Take example as:  J = 3 A/mm² 

Thus, S = 1÷3≈0.33 mm²

Convert to the diameter of round copper wire:  d =√(4S/π)   about   0.65 mm

So engineers might discuss: 'Should we use a 0.6 or 0.7 gauge wire for 1 A?' Essentially, this is about balancing current, DCR, temperature rise, and winding space.

Conclusion

If you read request like“6.8mH, 2A, AL5000, 37T, 0.65mm copper wire, DCR....”, please relax, because it is easy to understand like:

Thus, that means: Select a core -----> Calculate the number of turns based on AL----> choose the copper wire based on Current ----> Wound it ---> Test current, DCR, Temperature rise and high-frequency impedance

Is it easy to understand? For more discuss, please contact us freely.

Original information fromhttps://mp.weixin.qq.com/s/SYkG4MdSEdYUkaMDMJTCpA