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Wide-Input, High-Frequency, Triple-Output Supplies
with Voltage Monitor and Power-On Reset
This gain is also set by the ratio of R3/R1 where R1 is
calculated in the OUT1 Voltage Setting section. Thus:
Below is a numerical example to calculate the error-
amplifier compensation values used in the Typical
R 3 =
R 1 × f Z 2
f C × G MOD ( fc )
Applications Circuit of Figure 5:
V IN = 12V (nomimal input voltage)
V RAMP = 1V
Due to the underdamped (Q > 1) nature of the output
LC double pole, the error-amplifier zero frequencies
must be set less than the LC double-pole frequency to
provide adequate phase boost. Set the error-amplifier
first zero, f Z1 , at 1/4th the LC double-pole frequency and
the second zero, f Z2 , at the LC double-pole frequency.
Hence:
V OUT1 = 3.3V
V FB1 = 1.25V
L1A = 1.8μH
C4 = 47μF/ 6.3V ceramic, with R ESR = 0.008 ?
f S = 1.4MHz
The LC double-pole frequency is calculated as:
C 5 =
2
π × R 3 × f PMOD
f PMOD =
1
2 π L 1 A × C 4
=
Set the error-amplifier f P2 at f ZESR , and f P3 to 1/2 the
switching frequency, if f ZESR < 1/2 f S . If f ZESR > 1/2 f S ,
then set f P2 at 1/2 f S and f P3 at f ZESR .
1
2 π 1 . 8 × 10 - 6 × 47 × 10 - 6
= 17 . 3 kHz
The gain of the error amplifier between f P2 and f P3 is
set by the ratio of R3/R I and is equal to:
f ZESR =
1
2 π × R ESR × C 4
=
R 3
R I
= G EA ( fZ 1 - fZ 2 )
f P 2
f PMOD
1
2 π × 0 . 008 × 47 × 10 - 6
= 423 kHz
where R I is the parallel combination of R1 and R4 and
is equal to:
Pick R2 = 8.06k ? .
R 1 = 8 . 06 k ? × ?
- 1 ? = 13 . 3 k ?
R I =
R1 × R4
R 1 + R 4
? 3 . 3 V
? 1 . 25 V
?
?
The modulator gain at DC is:
Therefore:
R I =
R 1 × R I
G MOD ( DC ) =
V IN
G MOD ( fc ) = 12 × ? ? = 0 . 363
R 3 × f PMOD
f P 2 × G EA ( fZ 1 - fZ 2 )
R 4 =
R 1 - R I
C11 can then be calculated as:
and
Pick f C = 100kHz.
= 12
V RAMP
2
? 17 . 4 kHz ?
? 100 kHz ?
G EA ( fZ 1 ? fZ 2 ) =
and C12 as:
C 11 =
1
2 π × R 4 × f P 2
f PMOD
f C G MOD ( fC )
17 . 4 kHz
=
100 kHz × 0 . 363
R 3 = R 1 × G EA ( fZ 1 ? fZ 2 )
= 0 . 479
C 12 =
( 2 π × C 5 × R 3 × f P 3 -1 )
C 5
= 13 . 3 k ? × 0 . 479 = 6 . 37 k ?
24
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