Download Power Factor Correction Design Consideration

CCM Power Factor
Correction Inductor
Design with Powder
Core
By Jacki_wang
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:
Preal
(v  i)averaged over one cycle
PF 

Papparent
Vrms  Irms
• 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.
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.


100.00%
80.00%
60.00%
40.00%
20.00%
0.00%

1
3
5
7 9 11 13 15 17 19 21
Harmonic Number

1
2
)
)

In this case the harmonics are huge,
C
H
1
because
much
of
the
power
is
concentrated
C
H
2

in a short period of time in each cycle.
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.
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.
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:
I max
Pout
330

 2
 2  5.46 A
Vinmin 
90  0.95
Inductor Current calculation
•
Set the Ripple current to 50% Imax when
input is 50% of output voltage, then the
deltaI=2.7A and:
I rms 
i pp
I max 2
1 2
(

) (
)
2
3
2
2.7
1 2
5.46 2
(

) (
)  3.94 A
2
3
2

•
The RMS value of two signals is the root
sum of the squares of the RMS values of
each of the two signals.
Inductance Calculation
• Calculate the inductance required:
1
190 
V t
2  70,000
L

 0.5 mH
i pp
2.7
• So 0.5mH inductance is needed to achieve
2.7A ripple current pass through the inductor
Core Selection and Analysis
1. Compute the product of LI2 where:
L = inductance required with dc bias ( millihenry )
I = dc current (amperes)
LI rms  0.5  3.94  7.76 mH  A
2
2
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
Core Selection and Analysis
• 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
N
L
0.5 10 3

 94Turns
9
AL
56.12 10
Core Selection and Analysis
• we calculate the magnetic force as
0.4NI 0.4    94  3.94
H

 56.8Oersteds
Le
8.15
• The magnetizing force (dc bias) is 56.8
oersteds, yielding around 70% of initial
permeability.
DC BIAS
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:
N need
N noload 94


 135Turns
ALload 0.7
Core Selection and Analysis
• 5. An recalculate of the preceding result yields the
following:
1. Calculate the dc bias level in oersteds:
0.4NI 0.4   135  3.94
H

 82Oersteds
Le
8.15
The permeability versus DC Bias curve shows a
54% initial permeability at 82 oersteds for 60µ
material.
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.
L  AL  N 2
 30.3 10 3 1352
 0.55mH
• The minimum inductance requirement of 0.5 mH
has been achieved with the dc bias.
Core Selection and Analysis
• 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.
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:
l 2.3 108  0.616
3
Rdc  

11
.
2

10

6
S
0.79110
Than
3
Pw  I rms  Rdc  3.94 11.2 10  174mW
2
2
Thermal Analysis with natural cooling
-core loss
B 
Vin  Don
190  0.5

 150mT
N  Ae  f s 135  67.2  70000
From the chat of loss, the core loss Pc should be
Pc=1000 x 5.48 = 5.48W
Thermal Analysis with natural cooling
-total loss and temperature rise
Total inductor loss:
Ptotal  Pw  Pc  174mW  5480mW  5654mW
Temperature rise approximated:
 Total power loss milliwats  

Temperature Rise C  
2
Surface Area cm


 
o
 5654 mW 

T  
2 
 48 cm 
 
0.833
 53 C
o
Design passed
0.833
The End
Kool Mµ® Core Selector Chart
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