AP63300Q/AP63301Q

Document number: DS43699 Rev. 1 - 2

17 of 22

www.diodes.com

May 2021

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AP63300Q/AP63301Q

Application Information (continued)

10

Setting the Output Voltage

The AP63300Q/AP63301Q has adjustable output voltages starting from 0.8V using an external resistive divider. An optional external capacitor, C4

in Figure 1, of 10pF to 220pF improves the transient response. The resistor values of the feedback network are selected based on a design trade-

off between efficiency and output voltage accuracy. There is less current consumption in the feedback network for high resistor values, which

improves efficiency at light loads. However, values too high cause the device to be more susceptible to noise affecting its output voltage accuracy.

R1 can be determined by the following equation:

������������ = ������������ ∙ (

������������������������

������. ������������

− ������)

Eq. 6

Table 1 shows a list of recommended component selections for common AP63300Q/AP63301Q output voltages referencing Figure 1.

Table 1. Recommended Component Selections

AP63300Q/AP63301Q

Output Voltage

(V)

R1

(k

Ω)

R2

(k

Ω)

L

(µH)

C1

(µF)

C2

(µF)

C3

(nF)

C4

(pF)

1.2

15.0

30.1

2.2

10

3 x 22

100

OPEN

1.5

26.1

30.1

3.3

10

3 x 22

100

OPEN

1.8

37.4

30.1

3.3

10

2 x 22

100

OPEN

2.5

63.4

30.1

4.7

10

2 x 22

100

56

3.3

93.1

30.1

4.7

10

2 x 22

100

56

5.0

158.0

30.1

6.8

10

2 x 22

100

47

12.0

422.0

30.1

10.0

10

2 x 22

100

12

11

Inductor

Calculating the inductor value is a critical factor in designing a buck converter. For most designs, the following equation can be used to calculate

the inductor value:

������ =

������������������������ ∙ (������������������ − ������������������������)

������������������ ∙ ∆������������ ∙ ������������������

Eq. 7

Where:



∆IL is the inductor current ripple



fSW is the buck converter switching frequency

For AP63300Q/AP63301Q

, choose ∆IL to be 30% to 40% of the maximum load current of 3A.

The inductor peak current is calculated by:

������������

������������������������ = ������������������������������ +

∆������������

������

Eq. 8

Peak current determines the required saturation current rating, which influences the size of the inductor. Saturating the inductor decreases the

converter efficiency while increasing the temperatures of the inductor and the internal power MOSFETs. Therefore, choosing an inductor with the

appropriate saturation current rating is important. For most applications, it is recommended to select an inductor of approximately 2.2µH to 10µH

with a DC current rating of at least 35% higher than the maximum load current. For highest efficiency, the inductor’s DC resistance should be less

than 3

0mΩ. Use a larger inductance for improved efficiency under light load conditions.