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ADP5053

Data Sheet

Rev. B | Page 28 of 36

DESIGN EXAMPLE

This section provides an example of the step-by-step design

procedures and the external components required for Channel 1.

Table 13 lists the design requirements for this specific example.

Table 13. Example Design Requirements for Channel 1

Parameter

Specification

Input Voltage

VPVIN1 = 12 V ± 5%

Output Voltage

VOUT1 = 1.2 V

Output Current

IOUT1 = 4 A

Output Ripple

ΔVOUT1_RIPPLE = 12 mV in CCM mode

Load Transient

±5% at 20% to 80% load transient, 1 A/µs

Although this example shows step-by-step design procedures

for Channel 1, the procedures apply to all other buck regulator

channels (Channel 2 to Channel 4).

SETTING THESWITCHINGFREQUENCY

The first step is to determine the switching frequency for the

ADP5053 design. In general, higher switching frequencies

produce a smaller solution size due to the lower component

values required, whereas lower switching frequencies result in

higher conversion efficiency due to lower switching losses.

The switching frequency of the ADP5053 can be set to a value

from 250 kHz to 1.4 MHz by connecting a resistor from the RT

pin to ground. The selected resistor allows the user to make

decisions based on the trade-off between efficiency and solution

size. (For more information,see the Oscillatorsection.) However,

the highest supported switching frequency must be assessed by

checking the voltage conversion limitations enforced by the

minimum on time and the minimum off time (see the Voltage

Conversion Limitations section).

In this design example, a switching frequency of 600 kHz

achieves a good combination of small solution size and high

conversion efficiency. To set the switching frequency to

600 kHz, use the following equation to calculate the resistor

value, RRT:

RRT (kΩ) = [14,822/fSW (kHz)]1.081

Therefore, select standard resistor RRT = 31.6 kΩ.

SETTING THEOUTPUTVOLTAGE

Select a 10 kΩ bottom resistor (RBOT) and then calculate the top

feedback resistor using the following equation:

RBOT = RTOP × (VREF/(VOUT − VREF))

where:

VREF is 0.8 V for Channel 1.

VOUT is the output voltage.

To set the output voltage to 1.2 V, choose the following resistor

values: RTOP = 4.99 kΩ and RBOT = 10 kΩ.

SETTING THECURRENT LIMIT

For 4 A output current operation, the typical peak current limit

is 6.44 A. For this example, choose RILIM1 = 22 kΩ (see Table 10).

For more information,see the Current-Limit Protectionsection.

SELECTING THEINDUCTOR

The peak-to-peak inductor ripple current, ΔIL, is set to 35% of

the maximum output current. Use the following equation to

estimate the value of the inductor:

L = [(VIN − VOUT) × D]/(ΔIL × fSW)

where:

VIN = 12 V.

VOUT = 1.2 V.

D is the duty cycle (D = VOUT/VIN = 0.1).

ΔIL = 35% × 4 A = 1.4 A.

fSW = 600 kHz.

The resulting value for L is 1.28 µH. The closest standard

inductor value is 1.5 µH; therefore, the inductor ripple current,

ΔIL, is 1.2 A.

The peak inductor current is calculated using the following

equation:

IPEAK = IOUT + (ΔIL/2)

The calculated peak current for the inductor is 4.6 A.

Calculate the rms current of the inductor using the following

equation:

OUT

RMS

∆

The rms current of the inductor is approximately 4.02 A.

Therefore, an inductor with a minimum rms current rating of

4.02 A and a minimum saturation current rating of 4.6 A is

required. However, to prevent the inductor from reaching its

saturation point in current-limit conditions, it is recommended

that the inductor saturation current be higher than the maximum

peak current limit, typically 7.48 A, for reliable operation.

Based on these requirements and recommendations, the TOKO

FDV0530-1R5inductor,with a DCR of 13.5mΩ,is selected for this

design.

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