Calculation of coupled inductors to achieve zero current ripple

LeP →∞, LeS = L _{2, the effect of which is to make the primary} side ripple current zero and the secondary side ripple unchanged.

This can be explained as follows: achieving zero ripple is simply to push the ripples in the two mutually coupled coils to (or concentrate) one coil, and the current flowing through the remaining coil is a DC current.

2.2 Explaining the principle of ripple reduction from the perspective of magnetic circuit theory ^[2][3]

The winding of the two inductors is shown in Figure 2.

The equivalent magnetic circuit model of the two coupled inductors is the same as the leakage inductance model of the transformer (Figure 3). The coupling coefficient of winding 1 can be defined as

k1 = (9 ₎

Where: φ _m and φ _l1 are shown in Figure 2.

Figure 2 Inductor UI winding structure

Since v = N dφ/d t , the above formula can also be written as

k1 = (10 ₎

According to the equation N φ = Li, we can get

k ₁ = (11)

Therefore, _k1 can be regarded as the voltage division coefficient of the inductor voltage in the model shown in Figure 3. The ideal transformer primary voltage _vip in the figure has the same shape as the input voltage, but the amplitude is reduced by k1 _{times . The} transformer ratio N1 / N2 _is selected so that the transformer secondary voltage is equal to the primary input voltage v , and the same voltage is applied to both ends of the inductor _Ll2 at the same time , so the current ripple on Ll2 will be zero (d _i / _dt = vLl2 / Ll2 = 0 ) . Therefore, the condition for zero current ripple on the inductor _{Ll2 is}

k ₁ = N ₁ / N ₂ (12)

Figure 3 Equivalent circuit model of coupled inductor

This condition can be understood as follows: the turns ratio of the two windings must completely compensate for the leakage flux of the primary winding, so that the voltage induced by the primary winding on the secondary side is equal to the given voltage of the primary side.

2.3 Derivation of the calculation formula of coupled inductance using the equivalent magnetic resistance model

Figure 4 shows the UI winding structure of the coupled inductor and its T-type magnetic resistance equivalent model. Because a DC current flows through one of the windings, an air gap is required to prevent the core from saturating. And from the following analysis, it can be seen that the two inductor windings wound on the same core can achieve zero ripple by adjusting the air gap size.

In Figure 4 (b), Rx1 and Rx2 are the magnetic resistances of the two air gaps, and _{R1 is} the magnetic resistance of the magnetic core. The calculation formula is as follows _:

R _xi = x _i /μ ₀ S _e

R _l = l _e /μ ₀ S _e

Where: Se and Le are _{the equivalent cross- sectional} area and equivalent magnetic path length of the core respectively ^[4] .

Reference [4] introduces a method for determining the equivalent magnetic path length le of a magnetic core, and explains that for a given magnetic core, its equivalent magnetic path length is fixed.

From Figure 4(b), combined with the zero ripple condition k = N ₁ / N ₂ obtained above, the zero ripple magnetic resistance expression can be obtained as

k ₁ ==(13)

(a) UI winding structure diagram of coupled inductor

(b) T-type magnetoresistance equivalent model

Figure 4 Coupled inductor UI winding structure and magnetic resistance model

From the model shown in FIG4(b), if it is assumed that the output inductor current has achieved zero ripple, that is, di2 _/ dt = 0 , then the calculation formula of the primary inductance (i.e., input inductance) can be obtained from Kirchhoff's second law of magnetic circuit:

L ₁ = N ₁ ² /( R _x1 + R _l ∥ R _x2 ) (14)

When considering the magnetic saturation limit, the following equation holds true:

φ _1max =( I _1max + I _2max )≤ B _M S _e

So there is

N ₁ ≥( I _1max + I _2max )(15)

Combining formula (13), we have

N ₂ = N ₁ (16)

According to the formula obtained above, by selecting the value of L1 (Note: if the input inductor current is to be zero ripple, the value of L2 should be selected ₎ , the number of turns and air gap _value required to achieve zero ripple can be calculated.

3 Simulation Results

To verify the obtained results, the Cuk circuit (Figure 1) was simulated using Pspice to make the output inductor current zero ripple, with the following parameters:

V _i =60V, V _o =50V, f _s =40kHz, P =100W.

In order to prevent the input current ripple from being too large, L ₁ = 200μH is taken, and the EE55 type magnetic core is used. The two inductor windings are wound on the middle column and one of the side columns respectively. The number of turns of the input inductor is calculated to be 19 by formula (15), and the air gap size is calculated to be about 0.34mm by formula (14). The number of turns of the output inductor is 53 by formula (16), and its inductance is measured to be about 814μH. Based on the above parameters, the simulation waveforms are shown in Figures 5 and 6. According to the simulation results, the ripple factor of the output current is calculated to be about 1.3%.

Figure 5: Inductor current waveform without coupling

Figure 6 Current waveform when the output inductor current ripple is zero

The simulation results show that by using the data calculated by the proposed calculation formula for simulation, the current ripple of the inductor at one end can be approximately zero.

4 Conclusion

This paper analyzes and studies the coupled inductor. Through the magnetic resistance equivalent model of the coupled magnetic circuit, a calculation formula for the coupled inductor to achieve zero current ripple is given, and the correctness of the calculation formula is verified through simulation.