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MIC2155YML Datasheet(PDF) 22 Page - Micrel Semiconductor |
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MIC2155YML Datasheet(HTML) 22 Page - Micrel Semiconductor |
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22 / 35 page ![]() Micrel, Inc. MIC2155/2156 November 2009 22 M9999-111209-B For this example, assume the output transient loading is small and the filter design is based on output ripple voltage requirement. The inductance value is calculated by the equation below: OUT S ) MAX ( IN OUT ) MAX ( IN OUT I 2 . 0 f V ) V V ( V L × × × × η − × η × = Where: fS is the switching frequency 0.2 is the ratio of AC ripple current to DC output current VIN(MAX) is the maximum input voltage IOUT is output current of the each channel or ½ of the total output current η is the converters efficiency For this example: H 1 A 15 2 . 0 kHz 500 V 12 88 . 0 ) V 8 . 1 V 12 88 . 0 ( V 8 . 1 L μ = × × × × − × × = If another inductor value is used, the ripple current for each channel is calculated from the formula below: A 3 H 1 kHz 500 V 12 88 . 0 ) V 8 . 1 V 12 88 . 0 ( V 8 . 1 I L f V ) V V ( V I PP S ) MAX ( IN OUT ) MAX ( IN OUT PP = μ × × × − × × = × × × η − × η × = The output capacitors see less ripple current than each channel because they are out of phase. The normalizing factor is: 6 . 3 H 1 kHz 500 V 8 . 1 L f V OUT S OUT = μ × = × The output ripple current in the two-phase configuration is approximately: A 3 . 2 H 1 kHz 500 V 8 . 1 65 . 0 L f V 65 . 0 OUT S OUT = μ × × = × × For the input and output voltage in this application, going to a two-phase design decreased the total output ripple current from 3APP to 2.3APP. The peak inductor current in each channel is equal to the average output current plus one half of the peak to peak inductor ripple current: A 5 . 16 A 3 5 . 0 A 15 I 5 . 0 I I PP OUT PK = × + = × + = The RMS inductor current is used to calculate the I 2 × R losses in the inductor: A 1 . 15 15 A 3 12 1 1 A 15 I I I 12 1 1 I I 2 ) RMS ( INDUCTOR 2 OUT PP OUT ) RMS ( INDUCTOR = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + × = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + × = Maximizing efficiency requires the proper selection of core material and minimizing the winding resistance. The high frequency operation of the MIC2155 requires the use of ferrite materials for all but the most cost sensitive applications. Lower cost iron powder cores may be used but the increase in core loss will reduce the efficiency of the power supply. This is especially noticeable at low output power. The inductor winding resistance decreases efficiency at the higher output current levels. The winding resistance must be minimized although this usually comes at the expense of a larger inductor. The power dissipated in the inductor is equal to the sum of the core and copper losses. At higher output loads, the core losses are usually insignificant and can be ignored. At lower output currents the core losses can be a significant contributor. Core loss information is usually available from the magnetics vendor. For this example a Cooper HCF1305-1R0 inductor was chosen. Core loss for this application was taken from the data sheet and is 15mW. Winding resistance is 1.9mΩs Copper loss in the inductor is calculated by the equation below: PINDUCTOR(COPPER) = (IINDUCTOR(RMS)) 2 x R WINDING = 15.12 x 1.9mΩ = 0.43W |
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