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DRV590GQCR Datasheet(PDF) 10 Page - Texas Instruments |
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DRV590GQCR Datasheet(HTML) 10 Page - Texas Instruments |
10 / 16 page DRV590 SLOS365A – AUGUST 2001 – REVISED AUGUST 2002 10 www.ti.com APPLICATION INFORMATION LC filter in the frequency domain (continued) If L = 10 µH and C = 10 µF, the resonant frequency is 9.2 kHz, which corresponds to –69 dB of attenuation at the 500-kHz switching frequency. For VDD = 5 V, the amount of ripple voltage at the TEC element will be approximately 1.7 mV. The average TEC element has a resistance of 1.5 Ω, so the ripple current through the TEC is approximately 1.13 mA. At the 1-A maximum output current of the DRV590, this 1.13 mA corresponds to 0.113% ripple current, causing less than 0.0001% reduction of the maximum temperature differential of the TEC element (see equation 1). LC filter in the time domain The ripple current of an inductor can be calculated using equation 4. DI L + V DD * V TEC DTs L D = duty cycle (0.5 worst case) Ts = 1/fs = 1/500 kHz For VDD = 5 V, VTEC = 2.5 V, and L = 10 µH, the inductor ripple current is 250 mA. To calculate how much of that ripple current will flow through the TEC element, however, the properties of the filter capacitor must be considered. For relatively small capacitors (less than 10 µF) with very low equivalent series resistance (ESR, less than 10 m Ω), such as ceramic capacitors, equation 5 may be used to estimate the ripple voltage on the capacitor due to the change in charge. DV C + p2 2 (1–D) fo fs 2 V TEC fo + 1 2 p L 3C D = duty cycle fs = 500 kHz For L = 10 µH and C = 10 µF, the cutoff frequency fo = 9.2 kHz. For a worst case duty cycle of 0.5 and VTEC = 2.5, the ripple voltage on the capacitors is 2 mV. The ripple current may be simply calculated by dividing the ripple voltage by the TEC resistance of 1.5 Ω, resulting in a ripple current through the TEC element of 1.33 mA. Note that this is similar to the value calculated using the frequency domain approach. For larger capacitors (greater than 10 µF) with relatively high ESR (greater than 100 mΩ), such as electrolytic capacitors, the ESR drop dominates over the charging-discharging of the capacitor. Equation 6 can be used to estimate the ripple voltage. DV C + D I L R ESR ∆L = inductor ripple current RESR = filter capacitor ESR For a 100- µF electrolytic capacitor, an ESR of 0.1 Ω is common. If the 10-µH inductor is used, delivering 250 mA of ripple current to the capacitor (as calculated above), then the ripple voltage is 25 mV. This is over ten times that of the 10- µF ceramic capacitor, as ceramic capacitors typically have negligible ESR. (4) (5) (6) |
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