rt9911 Richtek Technology Corporation, rt9911 Datasheet - Page 26

• Manufacturer Part Number: rt9911
• Description: 6 Channel Dc/dc Converters
• Manufacturer: Richtek Technology Corporation

RT9911
The results we will get for CH1 boost compensation are
as follows:
The major steps for getting above results :
1.
2. Set fc (cross over frequency) sufficiently below f
For example :
3.
4.
5.
6. Find ffz, zero and ffp, pole ratio of voltage divider with
7. Get C
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26
V
V
I
f
L, inductance
R
(ceramic output capacitor)
T
R2, the voltage divider resistor in between FB and ground.
C
R
C
connect to ground
C
ignored if C
C
on ceramic output capacitor.
R2 =
R =
dV
Get
ratio
C
OUT(MAX.)
OSC
C
C
DRP
IN
OUT
ESR
F
C
C
P
OUT
C
F
F
, feedforward capacitor in parallel with R1.
C
, connect in between COMP pin and ground. (Can be
, compensation resistor on COMP pin.
, compensation capacitor in series with R
, input voltage.
.
FB
, operating frequency
=
=
(%), Transient droop.
C
, desired output voltage
, ESR (Equivalent Series Resistance) of C
, output capacitance. This compensation is based
R1
=
GM x dV
R
2
= T
OUT
F
R
dI x R
π
ffp
ffz
LOAD
by placing ffp on f
V
CS
, maximum output load
×
DRP
OUT
=
ff
=
P
1
V
Z
R
f
CS
x
V
< 10pF)
R
(%) x V
C
FB
V
×
C
-
FB
OUT
2
LOAD
=
GM
FB
R1
V
x
π
FB
f
f
C
, where dI = transient step,
HP
C
4
,
C
where
x
FB
V
V
OUT
FB
ff
C
Z
and ffz therefore on
=
ratio
f
C
.
ratio
C
OSC
f
C
and
OUT
.
.
8. Evaluate C
Example : Set R1 = 470kΩ, V
V
Results :
1.
2.
3.
Thus,
4.
5.
6.
7.
Choose C
8.
R
FB
ESR
C
R2
C
R
R
C
ratio
C
C
f
ff
(ceramic output capacitor).
Choose 4.7nF.
Half-load transient means load from 0.25A to 0.5A
transient. So, dI = 0.5 − 0.25=0.25A
dV
which is less than 10pF. So, It can be ignored.
C
Z
P
C
LOAD
C
OUT
F
P
= 0.8V, I
=
FB
=
=
=
=
=
=
=
= 5mΩ, and half-load transient droop is 5%.
f
=
R1
HP
dI
C
C
4
R
ratio
=
2
= T
R
=
f
π
OUT
OUT
LOAD
ffp
ffz
C
GM
R
F
CS
=
x
I
R
×
OUT(MAX.)
DRP
C
= 22pF
R
R
f
LOAD
OUT(MAX.)
V
ff
=
V
=
OSC
x
4
R
x
1
x
C
P
C
x
Z
OUT
π
x
OUT
V
R
(%) x V
C
R
CS
2.25
. C
V
50k
×
dV
V
OUT
2
ESR
C
ESR
GM
FB
R1
FB
=
π
-
P
=
FB
f
40kHz
V
C
is for canceling the zero from C
10k
=
.
=
=
=
=
FB
x
C
=
= 0.5A, f
15.2pF,
22.2kHz
3.6
1.8
10
FB
0.8
P
9.4k
V
=
V
can
x
3
OUT
μ
= 5% x 0.8 = 0.04V.
470k
Ω
FB
6 .
3.9nF
=
10k
x
Ω
2.25
be
0.005
=
,
where
IN
OSC
choose
4.25nF,
x
DS9911-04 August 2007
ignore
= 3V, V
1.8
=
= 500kHz, L = 4.7uH,
10.8
=
0.8
-
5pF
0.8
10k
if
μ
where
OUT
C
F.
,
=
P
Ω
Choose
<
376k
.
= 1.8V,
10pF.
Ω
10
OUT
μ
F.