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機(jī)械數(shù)據(jù) (封裝 | 引腳)

MOW|59

散熱焊盤機(jī)械數(shù)據(jù) (封裝 | 引腳)

訂購信息

7.2.2.5.1 Output Voltage Ripple

The output-voltage ripple is the second criterion for output capacitor selection. Use Equation 15 to calculate the minimum output capacitance required to meet the output-voltage ripple specification.

Equation 15.

C_{O U T (m i n)} = \frac{I_{R I P P L E}}{8 \times f_{s w} \times V_{O U T (R I P P L E)}} = \frac{9.62 A}{8 \times 550 k H z \times 20 m V} = 110 μ F

In this case, the target maximum output-voltage ripple is 20 mV. Under this requirement, the minimum output capacitance for ripple is 110 μF. This capacitance value is smaller than the output capacitance required for the transient response, so select the output capacitance value based on the transient requirement. Considering the variation and derating of capacitance, in this design, two 470-μF low-ESR tantalum polymer bulk capacitors and four 47-μF ceramic capacitors were selected to meet the transient specification with sufficient margin. Therefore, the selected nominal C_OUT is equal to 1128 μF.

With the output capacitance value selected the ESR must be considered. This is an important consideration in this example because it uses mixed output capacitor types. First use Equation 16 to calculate the maximum allowable impedance for the output capacitor bank at the switching frequency to meet the output voltage ripple specification. Equation 16 indicates the output capacitor bank impedance should be less than 2.1 mΩ. The impedance of the ceramic capacitors is calculated with Equation 17 and the impedance of the bulk capacitor is calculated with Equation 18. The result from Equation 18 shows the impedance of the bulk capacitor at the switching frequency is dominated by its ESR. Equation 19 calculates the total output impedance of the output capacitor bank at the switching frequency to be 1.2 mΩ, which meets the 2.1-mΩ requirement.

Equation 16.

Equation 17.

Z_{C E R_f_{s w}} = \frac{1}{2 π \times f_{s w} \times C_{C E R}} = \frac{1}{2 π \times 550 k H z \times (4 \times 47 μ F)} = 1.5 m Ω

Equation 18.

Z_{B U L K_f_{s w}} = \sqrt{{E S R}_{B U L K}^{2} + {(\frac{1}{2 π \times f_{S W} \times C_{B U L K}})}^{2}} = \sqrt{{(\frac{10 m Ω}{2})}^{2} + {(\frac{1}{2 π \times 550 k H z \times (2 \times 470 μ F)})}^{2}} = 5.3 m Ω

Equation 19.

Z_{C O U T_f_{s w}} = \frac{Z_{C E R_f_{s w}} \times Z_{B U L K_f_{s w}}}{Z_{C E R_f_{s w}} + Z_{B U L K_f_{s w}}} = \frac{1.5 m Ω \times 5.3 m Ω}{1.5 m Ω + 5.3 m Ω} = 1.2 m Ω