AP5101
1.5A Step-Down Converter with 1.4MHz Switching
Frequency
AP5101
Document number: DS32258 Rev. 1 - 2
9 of 15
www.diodes.com
July 2010
© Diodes Incorporated
NEW PRODUCT
Applications Information (Continued)
Component Selection
The output voltage can be adjusted from 0.81V to 15V using an external resistor divider. Table 1 shows a list of resistor
selection for common output voltages. Resistor R1 is selected based on a design tradeoff between efficiency and output
voltage accuracy. For high values of R1 there is less current consumption in the feedback network. However the trade off is
output voltage accuracy due to the bias current in the error amplifier. R2 can be determined by the following equation:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−×= 1
0.81
OUT
V
2
R
1
R
V
OUT
(V) R
1
(kΩ) R
2
(kΩ)
1.8 80.6 (1%) 64.9 (1%)
2.5 49.9 (1%) 23.7 (1%)
3.3 49.9 (1%) 16.2 (1%)
5.0 49.9 (1%) 9.53 (1%)
Table 1. Resistor Selection for Common Output Voltage
Compensation Components
The AP5101 has an external COMP pin through which system stability and transient response can be controlled. COMP pin is
the output of the internal trans-conductance error amplifier. A series capacitor-resistor combination sets a pole-zero
combination to control the characteristics of the control system. The DC gain of the voltage feedback loop is given by:
OUT
V
FB
V
VEA
A
CS
G
LOAD
R
VDC
A ×××=
Where V
FB
is the feedback voltage (0.810V), R
LOAD
is the load resistor value, G
CS
is the current sense trans-conductance and
A
VEA
is the error amplifier voltage gain.
The control loop transfer function incorporates two poles. One is due to the compensation capacitor (C3) and the output
resistor of error amplifier, and the other is due to the output capacitor and the load resistor. These poles are located at:
VEA
A3C2
EA
G
P1
f
××π
=
LOAD
R2C2
1
P2
f
××π
=
Where G
EA
is the error amplifier trans-conductance.
One zero is present due to the compensation capacitor (C3) and the compensation resistor (R3). This zero is located at:
3R3C2
1
Z1
f
××π
=
