ADP5003ACPZ-R7 datasheet(24/31 Pages) AD

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ADP5003

Data Sheet

Rev. A | Page 24 of 31

ADAPTIVE HEADROOM CONTROL DESIGN EXAMPLE

This section provides an example of the step by step design

procedures and the external components required for the buck

regulator using adaptive headroom control. Table 11 lists the

design requirements for this example.

Table 11. Example Design Requirements for the Buck

Regulator Using Adaptive Headroom Control

Parameter

Specification

Input Voltage

VPVIN1 = 12 V

Output Voltage

VPVOUT2 = 1.3 V

Output Current

ILOAD1 = ILOAD2 = 3 A

Buck Load Transient

±100 mV at 20% to 80% load transient

SETTING THE SWITCHING FREQUENCY FOR THE

BUCK REGULATOR USING ADAPTIVE HEADROOM

CONTROL

Similar to the buck design example, a switching frequency of

600 kHz is used to achieve a good combination of small solution

size and high conversion efficiency. To set the switching frequency

to 600 kHz, use Equation 1 to calculate the resistor value, RRT:

RRT (kΩ) = 1.78 × 1011/fSW (kHz)

Therefore, select standard resistor RT = 294 kΩ.

SETTING THE OUTPUT VOLTAGE FOR THE LDO

REGULATOR USING ADAPTIVE HEADROOM

CONTROL

Select a value for the top feedback resistor (RTOP2) and then

calculate the bottom resistor (RBOT2) by using the following

equation:

RBOT2 = (RTOP2 × VPVOUT2)/((VREFOUT × ALDO) – VPVOUT2)

(19)

where:

VPVOUT2 is the LDO output voltage.

VREFOUT is 2 V.

ALDO is the LDO regulator gain.

To set the output voltage to 1.3 V, RTOP2 is set to 100 kΩ giving

an RBOT2 value of 65 kΩ.

SELECTING THE INDUCTOR FOR THE BUCK

REGULATOR USING ADAPTIVE HEADROOM

CONTROL

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

the maximum output current. Use the Equation 8 to estimate

the value of the inductor:

L = ((VPVIN1 – VPVOUT1) × D)/(ΔIL × fSW)

where:

VPVIN1 = 12 V.

VPVOUT1 = VPVOUT2 + VHR = 1.7 V.

VPVOUT2 = 1.3 V.

VHR is the adaptive headroom voltage at steady state current. See

Table 4 and Figure 25 for the approximate values of VHR vs. load

current. For this example, use VHR equal to 0.4 V for steady state

load current of 3 A.

D is the duty cycle (D = VVOUT1/VPVIN1).

ΔIL = 35% × 3 A = 1.05 A.

fSW = 600 kHz.

The resulting value for L is 2.32 μH. The selected standard

inductor value is 2.2 μH; therefore, ΔIL is 1.1 A.

To calculate the peak inductor current (IPEAK), use Equation 9:

IPEAK = ILOAD1 + (ΔIL/2)

The calculated peak current for the inductor is 3.55 A.

SELECTING THE OUTPUT CAPACITORS FOR THE

BUCK REGULATOR USING ADAPTIVE HEADROOM

CONTROL

To ensure that the LDO regulator does not track the buck

output, the undershoot voltage must be set to a value less than

the minimum adaptive headroom voltage. Use Equation 17 and

Equation 18 to calculate the capacitance.

For estimation purposes, use KOV = KUV = 2; therefore,

COUT_OV = 40.7 μF and COUT_UV = 6.92 μF.

It is recommended to use a single 47 μF ceramic capacitor for the

output of the buck and a single 10 μF for the output of the LDO.

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