SLUU195 − June 2004 |
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where |
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• D is the duty cycle for a single phase |
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• NPH is the number of active phases |
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• K (NPH) is the intermediate function for calculation |
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In this case, NPH=4 and Dmin=0.107 which yields k=0.573. |
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The actual output ripple is calculated in equation (7) |
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| VOUT |
| KNPH, D + | 1.5 V |
| 0.573 + |
| (7) | ||||
IRIPPLE + L | f |
| 0.6m H | 420 kHz | 3.41 A |
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| 1.0 |
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A |
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− |
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| NPH = 4 |
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InputRMS | 0.8 |
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Normalized− |
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0.6 |
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| NPH = 3 |
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| NPH = 1 |
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RMSCout(nom) | 0.4 |
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| NPH = 2 |
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0.2 |
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I |
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| 0 | NPH = 6 |
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| 0 |
| 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
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| Duty Cycle − % |
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| Figure 5. Output Ripple Current Cancellation |
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Selection of the output capacitor is based on many application variables, including function, cost, size, and availability. There are three ways to calculate the output capacitance.
1.The minimum allowable output capacitance is determined by the amount of inductor ripple current and the allowable output ripple, as given in equation (8).
COUT(min) | + | IRIPPLE | + | 3.41 A |
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| + 101 mF | (8) |
8 f VRIPPLE | 8 420 kHz | 10 mV |
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In this design, COUT(min) is
influenced by ESR and transient considerations.
2.ESR limitation. (To limit the ripple voltage to 10 mV, the capacitor ESR should be less than the value calculated in equation (9)).
RC t+ | VRIPPLE | + | 10 mV | + 2.93 mW | (9) |
IRIPPLE | 3.41 A |
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10TPS40090