The output filter inductor of the switching power supply is also called a smoothing choke. Its main function is to flatten the rectified current to obtain a more stable output and a smoother waveform. In order to achieve better results, can a larger inductance value be used? The answer is not so. If the value is too large, it will cause adverse effects in other aspects.

　　In reference [1], the author encountered such a situation when he was trying to produce a 500W half-bridge switching power supply.

　　Since the output current is relatively large, a current splitting method is adopted, using two MPP magnetic powder cores with an outer diameter of φ40 mm. (Reference [1], page 161)

　　At the beginning, 24 turns were wound, that is, N=24, L=58μH. When Io=15~30A (average IL=7.5~15A per piece), the high-voltage switch pulse waveform had serious self-excited jitter, the high-frequency oscillation was significantly aggravated, and the strong spike interference was reflected from the secondary side to the primary side circuit. Even on the grid input line and the +20V auxiliary power line, high-frequency noise interference with an amplitude of up to 5~6V was superimposed, and the pulses were clearly visible at the two output ends of the control module SG3525A and the midpoint of the high-voltage switch tube.

　　Then reduce the number of turns by 10 turns, that is, N=14, L=20.6μH. Io=20～25A (average IL=10～12.5A per cell), it starts to stabilize. When Io=30A (average IL=15A per cell), the trailing edge of the high voltage pulse waveform still jitters.

　　Finally, the number of turns is reduced to only 8-10 turns, L = 10.1μH, and it can work stably even when Io = 30A (average IL = 15A per piece).

　　Now, let’s make a brief analysis of the above situation.

　　According to the calculation formula for output filter inductance given by the author at the back of the book (page 234 of document [1]):

L=(Vi－Vo)ton/(2Iomin) (1)

　　Iomin is generally taken as (5-10)% of Io. For a single magnetic core, IL=10% of 15A is 1.5A.

　　Switching frequency fsw=80kHz. That is, T=12.5μs. Vi=18V, Vo=15V.

ton=(Vo/Vi)×(T/2) (2)

ton=(15/18)×(12.5/2)=5.2μs

L=(18－15)×5.2/(2×1.5)=5.2μH.

　　This tells us that the minimum inductance is 5.2μH, or the critical inductance is 5.2μH. The following is an analysis of the above situation based on the principle of volt-ampere (micro)second balance.

　　The magnetic energy W is

W=(1/2)LI2(VAs)or(VAμs)(3)

　　The electrical power P is

P = (1/2) LI2fsw (VA) (4)

　　The formula is converted to

P/fsw=(1/2)LI2

Figure 1 Electrical schematic diagram

　　Substituting VI for P and T for 1/fsw, we get the following relationship:

VIT=LI2/2 (5)

　　According to Figure 1, V=Vi=18V, I=IL=15A.

VIT=18×15×12.5=270×12.5=3375VAμs.

　　When N=24, Lo=58μH, 15A, the actual inductance value is 60%, L15=58×0.6=35μH.

　　When N=14, Lo=20.6μH, 15A, the actual inductance value is 80%, L15=20.6×0.8=16.5μH.

　　When N=10, Lo=10.6μH, and 15A, the actual inductance value is 95%, L15=10.6×0.95=10.1μH.

　　The magnetic energy of each group is obtained as

WN=24=0.5×35×152=3937.5VAμs (6)

WN=14=0.5×16.5×152=1856.25VAμs (7)

WN=10=0.5×10.1×152=1136.25VAμs (8)

　　The magnetic energy calculated above is 3375VAμs. In fact, when the duty cycle is equal to 0.5, it is halved to only 1687.5VAμs.

　　Obviously, this magnetic energy - 1687.5VAμs cannot satisfy the two conditions of equations (6) and (7). Only when the magnetic energy fully meets the requirements in equation (8), can it work stably.

　　Ideally, the inductance value can change with the output current. The starting inductance value should be determined according to the core saturation curve and should be 1.5 to 3 times the critical inductance value. It should not be too large. If the above analysis is correct, please correct me.