ADP2302 datasheet(37/44 Pages) AD

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Data Sheet

AD5737

Rev. B | Page 37 of 44

AICC SUPPLY REQUIREMENTS—SLEWING

The AICC current requirement while slewing is greater than in

static operation because the output power increases to charge

the output capacitance of the dc-to-dc converter. This transient

current can be quite large (see Figure 57), although the methods

described in the Reducing AICC Current Requirements section

can reduce the requirements on the AVCC supply.

If not enough AICC current can be provided, the AVCC voltage

drops. Due to this AVCC drop, the AICC current required for

slewing increases further, causing the voltage at AVCC to drop

further (see Equation 3). In this case, the VBOOST_x voltage and,

therefore, the output voltage, may never reach their intended

values. Because the AVCC voltage is common to all channels,

this voltage drop may also affect other channels.

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.5

1.0

1.5

2.0

2.5

TIME (ms)

AICC

IOUT

VBOOST

0mA TO 24mA RANGE

1kΩ LOAD

fSW = 410kHz

INDUCTOR = 10µH (XAL4040-103)

TA = 25°C

Figure 57. AICC Current vs. Time for 24 mA Step Through 1 kΩ Load

with Internal Compensation Resistor

Reducing AICC Current Requirements

Two main methods can be used to reduce the AICC current

requirements. One method is to add an external compensation

resistor, and the other is to use slew rate control. These methods

can be used together.

Adding an External Compensation Resistor

A compensation resistor can be placed at the COMPDCDC_x pin

in series with the 10 nF compensation capacitor. A 51 kΩ exter-

nal compensation resistor is recommended. This compensation

increases the slew time of the current output but reduces the AICC

transient current requirements. Figure 58 shows a plot of AICC

current for a 24 mA step through a 1 kΩ load when using a 51 kΩ

compensation resistor. The compensation resistor reduces the

current requirements through smaller loads even further, as

shown in Figure 59.

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0mA TO 24mA RANGE

1kΩ LOAD

fSW = 410kHz

INDUCTOR = 10µH (XAL4040-103)

TA = 25°C

0.5

1.0

1.5

2.0

2.5

TIME (ms)

AICC

IOUT

VBOOST

Figure 58. AICC Current vs. Time for 24 mA Step Through 1 kΩ Load

with External 51 kΩ Compensation Resistor

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0mA TO 24mA RANGE

500Ω LOAD

fSW = 410kHz

INDUCTOR = 10µH (XAL4040-103)

TA = 25°C

0.5

1.0

1.5

2.0

2.5

TIME (ms)

AICC

IOUT

VBOOST

Figure 59. AICC Current vs. Time for 24 mA Step Through 500 Ω Load

with External 51 kΩ Compensation Resistor

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