f

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.
rms
rms
cycle
one
over
averaged
apparent
real )
(
I
V
i
v
P
P
PF





3. Power Factor Correction
  
  
1 ) C H 1
2 ) C H 2
0.00%
20.00%
40.00%
60.00%
80.00%
100.00%
1 3 5 7 9 11 13 15 17 19 21
Harmonic Number
• 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:
A
Vin
P
I out
46
.
5
2
95
.
0
90
330
2
min
max 








7. A
I
i
I
pp
rms
94
.
3
)
2
46
.
5
(
)
3
1
2
7
.
2
(
)
2
(
)
3
1
2
(
2
2
2
m
ax
2







• The RMS value of two signals is the root
sum of the squares of the RMS values of
each of the two signals.
• Set the Ripple current to 50% Imax when
input is 50% of output voltage, then the
deltaI=2.7A and:
Inductor Current calculation

8. • Calculate the inductance required:
mH
i
t
V
L
pp
5
.
0
7
.
2
000
,
70
2
1
190






Inductance Calculation
• So 0.5mH inductance is needed to achieve
2.7A ripple current pass through the inductor

9. 1. Compute the product of LI2 where:
L = inductance required with dc bias ( millihenry )
I = dc current (amperes)
Core Selection and Analysis
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 77071 core line. The part number for a
60µ core of this size is 77071-A7
A
mH
LIrms 


 76
.
7
94
.
3
5
.
0 2
2

10. • 3. The 77071 core datasheet shows the nominal
inductance of this core to be 61 mH / 1000 turns,
±8%. Therefore, the minimum inductance of this
core is 56.12 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
Core Selection and Analysis
Turns
AL
L
N 94
10
12
.
56
10
5
.
0
9
3




 


11. • we calculate the magnetic force as
Core Selection and Analysis
Oersteds
8
.
56
15
.
8
94
.
3
94
4
.
0
4
.
0








e
L
NI
H
• The magnetizing force (dc bias) is 56.8
oersteds, yielding around 70% of initial
permeability. DC BIAS

12. • 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:
Turns
AL
N
N
load
noload
need 135
7
.
0
94



Core Selection and Analysis

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.
Oersteds
82
15
.
8
94
.
3
135
4
.
0
4
.
0








e
L
NI
H

14. Core Selection and Analysis
• 6. Multiply the minimum AL 56.12 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.
mH
N
AL
L
55
.
0
135
10
3
.
30 2
3
2








15. • 7. The wire table indicates that #19 wire is needed for
4.0 amperes. Therefore, 135 turns of #19 wire
(0.00791 cm2) equals 1.067 cm2, which is 36.4%
winding factor on this core (from the core data, the
total window area of 2.93 cm2).
So a 77071-A7 core with 135 turns of #19 wire will
meet the requirements.
Core Selection and Analysis

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:







 


3
6
8
10
2
.
11
10
791
.
0
616
.
0
10
3
.
2
S
l
Rdc
 Than
mW
R
I
P dc
rms
w 174
10
2
.
11
94
.
3 3
2
2





 

17. mT
f
A
N
D
V
B
s
e
on
in
150
70000
2
.
67
135
5
.
0
190










From the chat of loss, the core loss Pc should be
Pc=1000 x 5.48 = 5.48W
Thermal Analysis with natural cooling
-core loss

18. Total inductor loss:
mW
mW
mW
P
P
P c
w
total 5654
5480
174 




Temperature rise approximated:
   
 
833
.
0
2 








cm
Area
Surface
milliwats
loss
power
Total
C
Rise
e
Temperatur o
C
cm
mW
T o
53
48
5654
833
.
0
2










 Design passed
Thermal Analysis with natural cooling
-total loss and temperature rise

19. The End

20. Kool Mµ® Core Selector Chart
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