ADDC02808PB

REV. 0

–10–

8.1

7.4

–200

800

–100

0

100

200

300

400

500

600

700

8

7.9

7.8

7.7

7.6

7.5

TIME – µs

Figure 25. Predicted Response for 24 A Step Load Change,

di/dt = 12 A/

µs, with Factory Set Internal Compensation

RESPONSE AT END OF PULSE

The previous section describes how the ADDC02808PB

converter 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

100µs

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

100mV

10

0%

100

90

100µs

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

deviations for these curves are close to each other and

comparable to the negative deviation shown in Figure 6 for

a similarly sized positive step change in load current.

100mV

10

0%

100

90

100µs

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

100mV

10

0%

100

90

1ms

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

What is different about these curves is the settling time. Once

the converter’s output voltage rises above nominal, the

converter 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

capacitance 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

integrator 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 undershoot is.

Even when the load current steps down to 0.1 A, the maximum

deviation of the output voltage is only about 400 mV, or 5%.

However, it is important to realize that if the next power pulse

occurs before this transient is over, then the output voltage will

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