1. CCM Power Factor Correction Inductor Design with Powder Core
By Jacki_wang

2. Power Factor Definition
Power Factor (PF) is a term describing the input characteristic of an electrical appliance that is powered by alternating current (ac).
It is the ratio of “real power” to “apparent power” or:
Where v and i are instantaneous values of voltage and current.
RMS indicates the root-mean-squared value of the voltage or current.
The apparent power (Vrms x Irms), in effect, limits the available output power.

3. Power Factor Correction
Here’s the input current of a power supply without PFC. The current is concentrated at the peak of the voltage waveform, where the input rectifier conducts to charge the input energy-storage capacitor.
In this case the harmonics are huge,
because much of the power is concentrated in a short period of time in each cycle.

4. Why Choose Powder Core
Normally, because of the low loss coefficient, we use the ferrite core for the PFC inductor.
However, the space for PFC components is smaller and smaller due to the slim requirement of power supply.
The powder core have higher saturate flux, can conduct the same energy with smaller size core vs ferrite.

5. CCM Inductor in PFC Circuit
Normally, a boost circuit will be used for the power factor correction, inductor in active PFC circuit is a really choke, and it is very significant because the energy is carry by the choke from input to output circuit. The key point of designing PFC choke is: 1. Will not saturate at maximum peak current. 2. The loss can be accepted accordance to the temperature rise.

6. Inductor Current calculation
We use a 90~264Vac input and 5V 60A single output power for the design example.
Set the PFC output voltage 380Vdc, the efficiency of the dc-dc circuit is 90%, and 95% efficiency for PFC circuit, than PFC output power should be 330W .
Set the operation frequency 70KHz, then:

7. Inductor Current calculation
Set the Ripple current to 50% Imax when input is 50% of output voltage, then the deltaI=2.7A and:
The RMS value of two signals is the root sum of the squares of the RMS values of each of the two signals.

8. Inductance Calculation
Calculate the inductance required:
So 0.5mH inductance is needed to achieve 2.7A ripple current pass through the inductor

9. Core Selection and Analysis
1. Compute the product of LI2 where:
L = inductance required with dc bias ( millihenry )
I = dc current (amperes)
2. Locate the LI2 value on the core selector chart, this coordinate passes through the 60µ section of the permeability line and, proceeding upwards, intersects the horizontal core line. The part number for a 60µ core of this size is A7

10. Core Selection and Analysis
3. The core datasheet shows the nominal
inductance of this core to be 61 mH / 1000 turns,
±8%. Therefore, the minimum inductance of this
core is mH / 1000 turns, and Le is 8.15cm.
4.The number of turns needed to obtain 0.5 mH is
94Turns as per below calculation

11. Core Selection and Analysis
we calculate the magnetic force as
The magnetizing force (dc bias) is 56.8 oersteds, yielding around 70% of initial permeability.
DC BIAS

12. Core Selection and Analysis
The turns with DC bias should be calculate by divide the turns of no load by the percentage of DC bias，then adjusted turns are as below calculation:

13. Core Selection and Analysis
5. An recalculate of the preceding result yields the following:
1. Calculate the dc bias level in oersteds:
The permeability versus DC Bias curve shows a 54% initial permeability at 82 oersteds for 60µ material.

14. Core Selection and Analysis
6. Multiply the minimum AL mH by 0.54 yields 30.3 mH.
The inductance of this core with 135 turns and 82 oersteds of dc bias will be 0.55 mH.
The minimum inductance requirement of 0.5 mH has been achieved with the dc bias.

15. Core Selection and Analysis
7. The wire table indicates that #19 wire is needed for 4.0 amperes. Therefore, 135 turns of #19 wire ( cm2) equals cm2, which is 36.4% winding factor on this core (from the core data, the total window area of 2.93 cm2).
So a A7 core with 135 turns of #19 wire will meet the requirements.

16. Thermal Analysis with natural cooling
-Wire loss
From the core datasheet, the MLT with 40% wound would be 42.7mm, the length of wire is L=42.7mm x 135turn=5764.5mm, and the wire area is 0.791mm2
The resistivity of copper wire at 100DegreeC would
be 2.3 x 10^-8 ohm-m, so:
Than

17. Thermal Analysis with natural cooling
-core loss
From the chat of loss, the core loss Pc should be
Pc=1000 x 5.48 = 5.48W

18. Thermal Analysis with natural cooling -total loss and temperature rise
Total inductor loss:
Temperature rise approximated:
Design passed

19. The End

20. Kool Mµ® Core Selector Chart
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