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ADDC02808PBTV Scheda tecnica(PDF) 10 Page - Analog Devices |
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ADDC02808PBTV Scheda tecnica(HTML) 10 Page - Analog Devices |
10 / 20 page ADDC02808PB REV. A –10– 8.1 7.4 –200 800 –100 0 100 200 300 400 500 600 700 8.0 7.9 7.8 7.7 7.6 7.5 TIME – s Figure 25. Predicted Response for 24 A Step Load Change, di/dt = 12 A/ µs, with Factory Set Internal Compensation Optimized for CLOAD = 4,000 µF and RESR = 2.5 mΩ RESPONSE AT END OF PULSE The previous section describes how the ADDC02808PB con- verter responds to the positive step change in load current that occurs at the beginning of a power pulse. This section will discuss the converter’s response at the end of the power pulse when the load current is abruptly returned to a small value. Figures 26-29 show the converter’s measured output voltage as the load current is stepped from 25 A down to 4 A, 2 A, 1 A, and 0.1 A, respectively. The load capacitance is 1,000 µF with 100mV 10 0% 100 90 100 s VO Figure 26. Output Voltage Transient Response to a 25 A to 4 A Step Change in Load, di/dt/ = 12 A/ µs, with 1,000 µF Load Capacitance (RESR = 10 mΩ) 100mV 10 0% 100 90 100 s VO Figure 27. Output Voltage Transient Response to a 25 A to 2 A Step Change in Load, di/dt/ = 12 A/ µs, with 1,000 µF Load Capacitance (RESR = 10 mΩ) RESR = 10 m Ω. The di/dt is 12 A/µs. As can be seen, the peak deviations for these curves are close to each other and com- parable to the negative deviation shown in Figure 6 for a simi- larly sized positive step change in load current. 100mV 10 0% 100 90 100 s VO Figure 28. Output Voltage Transient Response to a 25 A to 1 A Step Change in Load, di/dt/ = 12 A/ µs, with 1,000 µF Load Capacitance (R ESR = 10 mΩ) 100mV 10 0% 100 90 1ms VO Figure 29. Output Voltage Transient Response to a 25 A to 0.1 A Step Change in Load, di/dt/ = 12 A/ µs, with 1,000 µF Load Capacitance (R ESR = 10 mΩ) What is different about these curves is the settling time. Once the converter’s output voltage rises above nominal, the con- verter cannot help to discharge the load capacitor. It can only reduce its output current to zero; it cannot draw a negative current. As such, the time it takes to bring the output voltage back down to its nominal value depends on the load current during the low load portion of the cycle. The rate at which the output voltage falls to its nominal value is the load current divided by the load capacitance (including the 150 µF capaci- tance that is inside the converter). The smaller the load current, the longer it takes to get the output voltage back to its nominal value. During the time that the output voltage is too high, the integra- tor in the converter’s feedback circuitry is continuing to ramp out of range. As the output voltage then falls below its nominal value, it must have an undershoot error to bring the integrator back into range. As can be seen from these figures, the lower the load current, the longer the output voltage remains too high, and the longer and the greater the output voltage under- shoot is. |
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