CCM Power Factor Correction Inductor Design with Powder

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: • 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. In this case the harmonics are huge, because much of the power is concentrated 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 90~264 Vac input and 5 V 60 A single output power for the design example. • Set the PFC output voltage 380 Vdc, the efficiency of the dc-dc circuit is 90%, and 95% efficiency for PFC circuit, than PFC output power should be 330 W. • Set the operation frequency 70 KHz, then:

Inductor Current calculation • Set the Ripple current to 50% Imax when input is 50% of output voltage, then the delta. I=2. 7 A and: • 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: • So 0. 5 m. H inductance is needed to achieve 2. 7 A ripple current pass through the inductor

Core Selection and Analysis 1. Compute the product of LI 2 where: ) L = inductance required with dc bias ( millihenry I = dc current (amperes) 2. Locate the LI 2 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 -A 7

Core Selection and Analysis • 3. The 77071 core datasheet shows the nominal inductance of this core to be 61 m. H / 1000 turns, ± 8%. Therefore, the minimum inductance of this core is 56. 12 m. H / 1000 turns, and Le is 8. 15 cm. • 4. The number of turns needed to obtain 0. 5 m. H is 94 Turns as per below calculation

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

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:

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.

Core Selection and Analysis • 6. Multiply the minimum AL 56. 12 m. H by 0. 54 yields 30. 3 m. H. • The inductance of this core with 135 turns and 82 oersteds of dc bias will be 0. 55 m. H. • The minimum inductance requirement of 0. 5 m. H 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 cm 2) equals 1. 067 cm 2, which is 36. 4% winding factor on this core (from the core data, the total window area of 2. 93 cm 2). So a 77071 -A 7 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. 7 mm, the length of wire is L=42. 7 mm x 135 turn=5764. 5 mm, and the wire area is 0. 791 mm 2 The resistivity of copper wire at 100 Degree. C would be 2. 3 x 10^-8 ohm-m, so: Than

Thermal Analysis with natural cooling -core loss From the chat of loss, the core loss Pc should be Pc=1000 x 5. 48 = 5. 48 W

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

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