RT9911PQV RICHTEK [Richtek Technology Corporation], RT9911PQV Datasheet - Page 31
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RT9911PQV
Manufacturer Part Number
RT9911PQV
Description
6 Channel DC/DC Converters
Manufacturer
RICHTEK [Richtek Technology Corporation]
Datasheet
1.RT9911PQV.pdf
(32 pages)
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Company:
Part Number:
RT9911PQV
Manufacturer:
NICHICON
Quantity:
40 000
The results we will get for CH5 asyn-inverter in CCM
compensation are as follows :
The major steps for getting above results :
1.
2. Select suitable inductor to ensure I
3. Find RHPZ(Right Hand Plan Zero) location.
4. Set f
For example:
5.
6.
DS9911-04 August 2007
I
f
L, inductance
C
on ceramic output capacitor.
R
(ceramic output capacitor)
R2, the voltage divider resistor in between FB and V
C
R
C
connect to ground
C
ignored if C
R2
L <
RHPZ(Boost
R
Find
Get
D = duty cycle =
OUT(MIN.)
OSC
then
CCM,
where
which is duty to V
OUT
ESR
F
C
P
C
LOAD
, feedforward capacitor in parallel with R1.
, connect in between COMP pin and ground. (Can be
, compensation resistor on COMP pin.
, compensation capacitor in series with R
=
, operating frequency
2
R
, output capacitance. This compensation is based
, ESR (Equivalent Series Resistance) of C
C
the
R2
R1
I x
C
(cross over frequency) sufficiently below RHPZ.
=
R
V
=
OUT(MIN.)
, minimum output laod
x
=
I
LOAD
load
IN
OUT(MAX)
GM
1M
V
V
V
f
P
x
f
LP
f
FB
OUT
C
REF
C
< 10pF)
(1
)
pole
=
x
x
=
=
−
x V
-
abs(V
G
0
-
R
RHPZ
I
D)
V
f x
V + abs(V
OUT(MAX.)
- 1
V
doc
-
LOAD
,
OUT
P
f :
OUT
IN
(-8)
OSC
D
6
FB
abs(V
0
LP
,
=
OUT
where
If .
transfer function.
=
duty
=
(1
or
2
R1
125k
2
)
-
OUT
π
D
lower
π
D)
.
L
cycle
x
=
OUT
G
2
R
)
1M
Ω
doc
,
LOAD
where
)
Ω
2
=
V
=
and
OUT
V
(1
x
OUT(MIN.)
IN
V
C
-
abs(V
D)
+
V
IN
OUT
OUT
abs(V
2
,
,
OUT
=
works in
(-8)V
OUT
)
C
REF
and
)
OUT
.
7. Find
8.
9. Find C
10.Evaluate C
PCB Layout Considerations
The feedback netwok should be very close to the FB
pin.
The compensation network should be very close to the
COMP pin and avoid through VIA.
For CH3 current sense, CS should be close to the drain
site of external NMOS.
Keep high current path as short as possible.
C
C
which is the conjugate double pole from LC filter.
C
double pole.
(ceramic output capacitor).
C
F
P
=
=
=
C
2
f
2
cdp
π
OUT
π
f
by placing its zero on fcdp to cancel another
×
f x
=
f
x
cdp
cdp
2
1
1
P
π
R
. C
R
ESR
x
×
- 1
x
C
R
R1
P
(LC)
D
C
is for canceling the zero from C
.
.
C
to
P
2
cancel
can
,
be
one
ignore
of
the
if
RT9911
www.richtek.com
C
double
P
<
10pF.
pole.
OUT
31