Expert analysis: PFC inductor calculation analysis

7. Calculate the number of turns

8. Calculate the wire size (omitted)

Critical Continuous Boost Inductor Design

1. Critical continuous characteristics
Boost power switch zero current is turned on, and the inductor current rises linearly. When the peak current reaches the reference current (sine wave) to be tracked, the switch is turned off, and the inductor current decreases linearly. When the inductor current drops to zero, the switch is turned on again. If the sine wave is completely tracked, according to the law of electromagnetic induction,

or

Where: Ui, Ii are the input voltage and current effective values. When the input voltage and input power are constant, Ton is a constant. When the output power and inductance are constant, the conduction time Ton is inversely proportional to the square of the input voltage Ui.

2. Determine the output voltage

The on-time volt-seconds of the inductor should be equal to the off-time volt-seconds:

but

The switching cycle is

It can be seen that the output voltage Uo must be greater than the input voltage Uip. If the output voltage is close to the input voltage, the cut-off time is much longer than the on-time near the peak of the input voltage, and the switching cycle is very long, that is, the frequency is very low.

If we first determine that the conduction time corresponding to the lowest input voltage (Uimin) is TonL, the conduction time corresponding to the highest input voltage (Uimax) is

According to equations (11) and (12), the relationship between the switching period (frequency) and different voltage ratios can be obtained.

For example, assuming that the on-time is 10μs, 1.414Uimin/Uo=0.65, if the input voltage changes within the range of ±20%, the minimum input voltage is 220×0.8, and the output voltage is Uo =1.414×220×0.8/0.65＝383V. The period is 10/0.35=28.57μs, and the frequency is 35kHz. At 15°, the period is 12μs, which is equivalent to a switching frequency of 83kHz. At the highest input voltage, the highest voltage on-time Tonn＝(0.8/1.2)2×TonL＝4.444μs is obtained from formula (12), and the switching period at the peak is T＝Tonh/(1-1.414×1.2×220/383)＝176μs, which is equivalent to a switching frequency of 5.66kHz.

If we. The output voltage is increased to 410V, the switching period is 25.45μs at the lowest input voltage, and the switching frequency is 39.3kHz. At 15°, it is 11.864μs, and the switching period is 84.5kHz. When the input voltage peak is the highest, the period is 49.2μs, and the switching frequency is 20.3kHz. The frequency variation range is greatly reduced. Even at the zero-crossing of the input voltage, the cut-off time approaches zero, and the switching frequency is about 100kHz. The highest frequency is only 5 times the lowest frequency. However, at an output voltage of 383V, it is 18 times.
From the above calculations, it can be seen that increasing the output voltage will reduce the switching frequency variation range, which is beneficial to output filtering. However, the power tube and rectifier diode require higher voltage ratings, and the conduction loss and switching loss increase. Therefore, for 220V±20% AC input, the output voltage is generally selected to be around 410V. For 110V±20% AC input, the output voltage is selected to be 210V.

3. Maximum peak current

Maximum input current

The maximum peak current in the inductor is 1 times the peak current

4. Determine the inductance

To avoid audio noise, the switching frequency should be above 20kHz within the input voltage range. From the above analysis, it can be seen that the switching frequency is lowest at the highest input voltage peak. Therefore, it is assumed that the switching period at the highest input voltage peak is 50μs. From formula (11), we can get

The minimum input voltage on-time is obtained from formula (12):

According to formula (8), we can get

5. Select the core

Since the on-time is inversely proportional to the square of the input voltage , the core size should be selected at the lowest voltage that avoids saturation at the lowest input voltage peak.

Where: N is the number of turns of the inductor coil; Ae is the effective cross-sectional area of ​​the magnetic core; Bm

The cross-sectional area of ​​the entire window copper

Substituting equation (17) into (16), we can get

Use AP method to select core size.

6. Calculate the number of coil turns

7. Coil wire cross-sectional area

Example: Input 220V ± 20%, output power 200W, critical continuous. Assume the efficiency is 0.95.
Solution: The maximum input current is

Peak current A

Assume the output voltage is 410V, and the minimum frequency is 20kHz when the input voltage is the highest. That is, the period is 50μs, so the conduction time is μs

If magnetic powder core is used, choose Sendust core. LI2=1.48×3.382×10-3=16.9mJ, choose 77439. The effective magnetic permeability is 60, its inductance coefficient AL＝135nH, and the number of turns required for inductance 1.48mH is N＝105 turns

The average magnetic path length of 77439 is l = 10.74 cm, and the magnetic field strength (Oe) is

From the figure, we can see that the magnetic permeability is 60, H=21Oe, and the magnetic permeability drops to 90%. In order to maintain a given inductance at a given peak current, the number of turns needs to be increased to
turns, and 111 turns are selected.

At this time, the magnetic field strength H = 111 × 21/105 = 22.2Oe, μ drops to 0.88, and the inductance

The design requirements are met. The switching frequency is increased by about 1% at the highest voltage. It should be noted that the average current is used here. The actual peak current is twice as large, and the maximum magnetic field strength is twice as large. From the figure, the magnetic permeability drops to 80%, and the magnetic field strength goes from zero to the maximum. The average magnetic permeability is (0.8+1)/2=0.9, which is close to 0.88.

Select the current density j=5A/mm2 and the wire size is

Select d=0.63mm, d'=0.70mm, and cross-sectional area Acu=0.312mm2.

Calculate the window utilization coefficient. Aw = 4.27cm2, then

77439 Sendust powder core has an outer diameter of OD = 47.6 mm and an inner diameter of ID = 23.3 mm. Consider the first layer

Nm1=(π(ID-0.5d'-0.05)/1.05d')-1=96.9, actual 96 turns.

The second layer only needs 15 turns