AP61200/AP61201/AP61202/AP61203

Document number: DS44251 Rev. 3 - 2

16 of 21

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September 2022

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AP61200/AP61201/AP61202/AP61203

Application Information (continued)

9

Setting the Output Voltage

The AP61200/AP61201/AP61202/AP61203 has adjustable output voltages starting from 0.6V using an external resistive divider. 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. R2 can be determined by the following equation:

������������ =

������. ������ ∙ ������������

������������������������ − ������. ������������

Eq. 5

Table 1 shows a list of recommended component selections for common AP61200/AP61201/AP61202/AP61203 output voltages referencing

Figure 1 and Figure 2.

Table 1. Recommended Component Selections

AP61200/AP61201/AP61202/AP61203

Output Voltage (V)

R1 (kΩ)

R2

(kΩ)

L (µH)

C1 (µF)

C2 (µF)

C3 (pF)

AP61200

AP61201

AP61202

AP61203

1.0

200.0

301.0

1.0

10

10

OPEN

33

1.2

200.0

200.0

1.0

10

10

OPEN

33

1.5

200.0

133.0

1.0

10

10

OPEN

33

1.8

200.0

100.0

1.0

10

10

OPEN

33

2.5

200.0

63.2

1.0

10

10

OPEN

33

3.3

200.0

44.2

1.0

10

10

OPEN

33

10

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. 6

Where:



∆IL is the inductor current ripple



fSW is the buck converter switching frequency

For AP61200/AP61201/AP61202/AP61203

, choose ∆IL to be 30% to 50% of the maximum load current of 2A.

The inductor peak current is calculated by:

������������

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

∆������������

������

Eq. 7

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 1.0µH to 1.5µH

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 50mΩ. Use a larger inductance for improved efficiency under light load conditions.