ACT6390 データシートの表示（PDF） - Active-Semi, Inc
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ACT6390 Datasheet PDF : 12 Pages
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ACT6390/ACT6391
Rev 1, 22-Aug-13
the cross-over frequency (unity gain-frequency)
should be less than one-fifth of the right-half-plane
zero fZ(RHP), and lower than one-fifteenth of switch-
ing frequency fsw.
fZ (RHP )
=
VIN 2 × RLOAD
2VOUT 2 × π × L
(16)
Choose fC
=
1
5
fZ (RHP ) , then calculate CCOMP:
( ) CCOMP
= VFB
VOUT
× RLOAD
RCS
× GM
2πfC
1−D
=
VIN ×VFB
V2
OUT
×
RLOAD × GM
RCS × 2πfC
(17)
Select RCOMP to meet the transient-droop require-
ments.
α
×VFB
×GM
× RCOMP
=
RCS
× VOUT × IOUT
VIN × η
× ⎜⎛1 +
⎝
K
2
⎟⎞
⎠
RCOMP
=
RCS
× VOUT
× IOUT
⎜⎛1 +
⎝
K
2
α ×VFB × GM ×VIN × η
⎟⎞
⎠
(18)
(19)
Where:
α is the transient droop percentage which can be
accepted, calculated by:
α = ΔVOUT
VOUT
(20)
K: is defined in equation (4)
η: is the typical efficiency.
VFB: is the feedback voltage, 1.24V
GM: is the trans-conductance of the error amplifier.
The output capacitor is chosen to set the output
pole for canceling the RCOMP, CCOMP zero.
COUT
= RCOMP × CCOMP
RLOAD
(21)
CCOMP2 is optional and can be used when the output
capacitor has significant ESR. The ESR will form a
zero as follows:
fZ (ESR )
=
2π
1
× RESR
× COUT
(22)
If this zero occurs at a higher frequency than the
cross-over frequency, it can be ignored. Otherwise,
it should be canceled with the pole set by capacitor
CCOMP2,
CCOMP2
=
COUT × RESR
RCOMP
(23)
If the value of CCOMP2 calculated by (23) is smaller
than 10pF, CCOMP2 can be omitted.
For example:
(3.3V )2 × ⎜⎛ 12V ⎟⎞
fZ (RHP )
=
2 × (12V )2
⎝ 250mA⎠
×π ×10μH
≈
57.8kHz
(24)
Choose
fC
=
1
5
fZ (RHP )
= 11.56kHz
CCOMP
=
3.3V ×1.24V
(12V)2
×
48Ω
0.45Ω
×
2π
150μS
×11.56kHz
=
6.26nF
(25)
Choose CCOMP = 6.8nF
Assume that 200mV of transient droop can be
accepted:
α = 200mV = 1
12V 60
(26)
0.45Ω×12V ×250mA⎜⎛1 + 0.4 ⎟⎞
RCOMP =
1
⎝ 2 ⎠ =186.3kΩ
×1.24V ×150μS ×3.3V ×0.85
(27)
60
Choose RCOMP = 180kΩ
COUT
=
RCOMP ×CCOMP
RLOAD
= 180kΩ ×6.8nF
⎜⎛ 12V ⎟⎞
= 25.5μF
(28)
⎝ 0.25A ⎠
COUT can be chosen to be either 22µF or 33µF,
choose 33µF to reduce droop.
RCOMP
= RLOAD ×COUT
CCOMP
= 48Ω × 33μF
6.8nF
= 233kΩ
(29)
If a ceramic capacitor is used with an assumed
ESR of 20mΩ,
fZ (ESR )
=
2π
1
× 33 μF
× 20mΩ
=
241kHz
(30)
fZ(ESR) > fC
Since the zero frequency is greater than the pole
frequency ,CCOMP2 can be omitted.
If a tantalum capacitor is used, whose ESR is about
0.5Ω,
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