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전자부품 데이터시트 검색엔진 |
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전자부품 데이터시트 검색엔진 |
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MP1527DM 데이터시트(PDF) 10 Page - Monolithic Power Systems |
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MP1527DM 데이터시트(HTML) 10 Page - Monolithic Power Systems |
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10 / 14 page  MP1527 2A, 1.3MHz Step-Up Converter MP1527 Rev 1.8_8/31/05 Monolithic Power Systems, Inc. 10 Monolithic Power Systems at the switching frequency, and so the output ripple is calculated as: IN OUT ESR LOAD SW LOAD OUT IN RIPPLE V V R I f 2 C I ) V V 1 ( V × × + × × − ≈ Where RESR is the equivalent series resistance of the output capacitors. Choose an output capacitor to satisfy the output ripple and load transient requirements of the design. A 4.7µF-22µF ceramic capacitor is suitable for most applications. Selecting the Inductor The inductor is required to force the higher output voltage while being driven by the input voltage. A larger value inductor results in less ripple current that results in lower peak inductor current, reducing stress on the internal n-channel.switch. However, the larger value inductor has a larger physical size, higher series resistance, and/or lower saturation current. A 4.7µH inductor is recommended for most applications. However, a more exact inductance value can be calculated. A good rule of thumb is to allow the peak-to-peak ripple current to be approximately 30-50% of the maximum input current. Make sure that the peak inductor current is below 75% of the current limit at the operating duty cycle to prevent loss of regulation due to the current limit. Also make sure that the inductor does not saturate under the worst-case load transient and startup conditions. Calculate the required inductance value by the equation: I f V ) V - (V V L SW OUT IN OUT IN ∆ × × = × η × × = IN ) MAX ( LOAD OUT ) MAX ( IN V I V I () ) MAX ( IN I % 50 % 30 I − = ∆ Where ILOAD(MAX) is the maximum load current, ∆I is the peak-to-peak inductor ripple current, and η is efficiency. Selecting the Diode The output rectifier diode supplies current to the inductor when the internal MOSFET is off. To reduce losses due to diode forward voltage and recovery time, use a Schottky diode with the MP1527. The diode should be rated for a reverse voltage equal to or greater than the output voltage used. The average current rating must be greater than the maximum load current expected, and the peak current rating must be greater than the peak inductor current. Compensation The output of the transconductance error amplifier (COMP) is used to compensate the regulation control system. The system uses two poles and one zero to stabilize the control loop. The poles are fP1 set by the output capacitor and load resistance and fP2 set by the compensation capacitor C3. The zero fZ1 is set by the compensation capacitor C3 and the compensation resistor R3. These are determined by the equations: fP1 = 1 / (π*C2*RLOAD) fP2 = GEA / (2π*AVEA*C3) fZ1 = 1 / (2π*C3*R3) Where RLOAD is the load resistance, GEA is the error amplifier transconductance, and AVEA is the error amplifier voltage gain. The DC loop gain is: AVDC = AVEA*GCS*(VIN / VOUT)*RLOAD*(VFB / VOUT) or AVDC = AVEA*GCS*VIN*VFB*RLOAD /(VOUT) 2 Where GCS is the current sense gain, VIN is the input voltage, VFB is the feedback regulation threshold, and VOUT is the regulated output voltage. |
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