Wide-Input, High-Frequency, Triple-Output Supplies
with Voltage Monitor and Power-On Reset
This gain between f Z1 and f Z2 is also set by the ratio of
R3/R1, where R1 is selected in the OUT1 Voltage
Setting section. Therefore:
f PMOD =
1
2 π L 1 A × C 4
R 3 =
R 1 × f PMOD
f ZESR × G MOD ( fc )
=
1
2 π 6 . 2 μ H × 560 μ F
= 2 . 7 kHz
And similar to Case 1, C5 can be calculated as:
f ZESR =
1
2 π R ESR × C 4
C 5 =
2
π × R 3 × f PMOD
=
1
2 π × 0 . 015 ? × 560 μ F
= 18 . 95 kHz
Pick R2 = 8.06k ? . Then:
Set the error-amplifier third pole, f P3 , at approximately
1/2 the switching frequency. The gain of the error
amplifier at f C (between f P2 and f P3 ) is set by the ratio
R 1 = 8 . 06 k ? ×
3 . 3 V
1 . 25 V
- 1 = 13 . 3 k ?
G MOD ( DC ) = = 12
of R3/R I and is also equal to:
1
G EA ( fc ) =
G MOD ( fc )
V IN
V RAMP
Pick f C = 50kHz, which is less than f S / 5.
G EA ( fZ 1 ? fZ 2 ) =
where R I is:
R I =
R1 × R4
R 1 + R 4
2 . 7 kHz 2
G MOD ( DC ) = 12 ×
18 . 95 kHz × 50 kHz
f PMOD
f ZESR G MOD ( fc )
= 0 . 0923
Therefore:
R I = R 3 × G MOD ( fc ) × G D
2 . 7 kHz
= = 1 . 543
18 . 95 kHz × 0 . 0923
R 3 = R 1 × G EA ( fZ 1 - fZ 2 ) = 13 . 3 k ? × 1 . 543 = 20 . 48 k ?
Similar to Case 1, R4, C11, and C12 can be calculated as:
Use 20k ? .
R 4 =
R 1 × R I
R 1 - R I
C 5 =
2
π × R 3 × f PMOD
=
2
π × 20 k ? × 2 . 7 kHz
= 11 . 8 nF
R 1 × R I 13 . 3 k ? × 1 . 846 k ?
1
C 11 =
2 π × R 4 × f ZESR
C 5
C 12 =
2 π × C 5 × R 3 × f P 3 - 1
Below is a numerical example to calculate the error-
amplifier compensation values for Case 2:
Use 12nF.
R I = R 3 × G MOD ( fc ) = 20 k ? × 0 . 0923 = 1 . 846 k ?
R 4 = = = 2 . 14 k ?
R 1- R I 13 . 3 k ? - 1 . 846 k ?
Use 2.2k ? .
V IN = 12V (nomimal input voltage)
V RAMP = 1V
V OUT1 = 3.3V
C 11 =
1
2 π × R 4 × f ZESR
=
1
2 π × 2 . 2 k ? × 18 . 95 kHz
= 3 . 82 nF
V FB1 = 1.25V
L1A = 6.2μH
C4 = 560μF/ 10V OS-Con capacitor, with ESR = 0.015 ?
f S = 300kHz
Use 3.9nF.
Pick f P 3 = f S / 2 = 150kHz.
26
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