米6体育平台手机版

ZHCSOO7A August 2021 – November 2021 TPSM8A28 , TPSM8A29

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机械数据 (封装 | 引脚)

RDG|25
- MPQF603

散热焊盘机械数据 (封装 | 引脚)

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8.2.2.4 Choose the Output Capacitor

There are three considerations for selecting the value of the output capacitor.

Stability
Steady state output voltage ripple
Regulator transient response to a change load current

First, the minimum output capacitance must be calculated based on these three requirements. Equation 12 calculates the minimum capacitance to keep the LC double pole below 1/30th the f_SW to meet stability requirements. This requirement helps to keep the LC double pole close to the internal zero.

Equation 12.

C_{O U T_S T A B I L I T Y} > {(\frac{30}{2 π \times f s w})}^{2} \times \frac{1}{0.6 μ H} = {(\frac{30}{2 π \times 600 k H z})}^{2} \times \frac{1}{0.6 μ H} = 105.5 μ F

Equation 13.

C_{O U T_R I P P L E} > \frac{(\frac{V_{I N (M A X)} - V o u t}{V_{I N (M A X)}}) \times (\frac{V o u t}{0.6 μ H \times f s w})}{8 \times V_{r i p p l e} \times f s w} = \frac{(\frac{14.4 V - 1 V}{14.4 V}) \times (\frac{1 V}{0.6 μ H \times 600 k H z})}{8 \times 10 m V \times 600 k H z} = 53.8 μ F

Equation 13 calculates the minimum capacitance to meet the steady state output voltage ripple requirement of 10 mV. This calculation is for FCCM operation and does not include the portion of the output voltage ripple caused by the ESR or ESL of the output capacitors. Equation 14 and Equation 15 calculate the minimum capacitance to meet the transient response requirement of ±50 mV for a 7.5-A step. These equations calculate the necessary output capacitance to hold the output voltage steady while the inductor current ramps up or ramps down after a load step.

Equation 14.

C_{O U T_U N D E R S H O O T} > \frac{0.6 μ H \times {I_{S T E P}}^{2} \times (\frac{V o u t}{V_{I N (M I N)} \times f s w} + t o f f)}{2 \times V_{T R A N S} \times V_{O U T} \times (\frac{V_{I N (M I N)} - V_{O U T}}{V_{I N (M I N)} \times f s w} - t o f f)} = \frac{0.6 μ H \times {7.5}^{2} \times (\frac{1 V}{9.6 V \times 600 k H z} + 220 n s)}{2 \times 50 m V \times 1 V \times (\frac{9.6 V - 1 V}{9.6 V \times 600 k H z} - 220 n s)} = 104.4 μ F

Equation 15.

C_{O U T_O V E R S H O O T} > \frac{0.6 μ H \times {I_{S T E P}}^{2}}{2 \times V_{T R A N S} \times V_{O U T}} = \frac{0.6 μ H \times {7.5}^{2}}{2 \times 50 m V \times 1 V} = 337.5 μ F

The output capacitance needed to meet the overshoot requirement is the highest value, so this sets the required minimum output capacitance for this example. Stability requirements can also limit the maximum output capacitance. Equation 16 calculates the recommended maximum output capacitance. This calculation keeps the LC double pole above 1/100th the f_SW. It can be possible to use more output capacitance but the stability must be checked through a bode plot or transient response measurement. The selected output capacitance is 8 × 47-μF ceramic capacitors. When using ceramic capacitors, the capacitance must be derated due to DC and AC bias effects. The selected capacitors derate to approximately 95% of their nominal value giving an effective total capacitance of approximately 357 μF. This effective capacitance meets the minimum and maximum requirements.

Equation 16.

C_{O U T_S T A B I L I T Y} < {(\frac{50}{π \times f s w})}^{2} \times \frac{1}{0.6 μ H} = {(\frac{50}{π \times 600 k H z})}^{2} \times \frac{1}{0.6 μ H} = 1172.7 μ F

This application uses all ceramic capacitors so the effects of ESR on the ripple and transient were ignored. If using non-ceramic capacitors, as a starting point, the ESR must be below the values calculated in Equation 17 to meet the ripple requirement and Equation 18 to meet the transient requirement. For more accurate calculations or if using mixed output capacitors, the impedance of the output capacitors must be used to determine if the ripple and transient requirements can be met.

Equation 17.

R_{E S R_R I P P L E} < \frac{V_{R I P P L E}}{(\frac{V_{I N (M A X)} - V o u t}{V_{I N (M A X)}}) \times (\frac{V o u t}{0.6 μ H \times f s w})} = \frac{10 m V}{(\frac{14.4 V - 1 V}{14.4 V}) \times (\frac{1 V}{0.6 μ H \times 600 k H z})} = 3.86 m Ω

Equation 18.

R_{E S R_T R A N S} < \frac{V_{T R A N S}}{I_{S T E P}} = \frac{50 m V}{7.5 A} = 6.67 m Ω