rt9911 Richtek Technology Corporation, rt9911 Datasheet - Page 25

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

The major steps for getting above results :
1.
2. Find RHPZ(Right Hand Plan Zero) location.
3. Set f
4. Get
5. Select Rc based on the allowed transient droop.
6. Get
7. Find ffz, zero and ffp, pole ratio of voltage divider with
8. Get C
9. Evaluate C
Example : Set R1 = 470kΩ, V
V
R
Results:
1.
2.
3.
DS9911-04 August 2007
FB
ESR
C
For example : f
, where dI = transient step, dV
C
(ceramic output capacitor).
R2
RHPZ(Boost
R
R = dI x (
C
f
R2
RHPZ(Boost
R
C
F
P
= 0.8V, I
LOAD
C
f
LOAD
.
=
= 5mΩ, and half-load transient droop is 5%.
=
=
=
=
ratio
C
RHPZ
C
C
C
2
R1
R1
F
(cross over frequency) sufficiently below RHPZ.
OUT
C
=
OUT
x
=
6
by placing ffp on f
V
π
I
x
=
I
=
OUT(MAX)
OUT(MAX.)
OUT
OUT(MAX.)
⎛
⎜ ⎜
⎝
V
⎛
⎜
⎝
(1- D)
x
R
=
V
ffp
ffz
P
=
(V
1
R
V
R
OUT
ESR
ffz
OUT
. C
R
1
)
11kHz
R
)
C
FB
R
OUT
LOAD
-
=
C
=
=
CS
V
LOAD
x
P
V
C
R
R
x
.
) x
V
FB
R1
C
FB
V
= RHPZ/6
is for canceling the zero from C
LOAD
-
C
=
LOAD
,
OUT
P
FB
= 0.5A, f
V
D
⎞
⎟
⎠
GM x dV
C
=
6.6
,
can
FB
x
where
=
470k
)
R
(1
x
Duty
2
Ω
GM
⎞
⎟ ⎟
⎠
π
2
CS
be
(1
-
,
C
π
f
D)
(1
2
C
L
-
and ffz therefore on
3.3
π
ignore
IN
OSC
FB
D)
ffz
-
Cycle
x
2
L
D)
0.8
= 1.8V, V
=
2
FB
V
-
=
V
= 500kHz, L = 4.7uH,
,
0.8
66.3kHz,
=
OUT
ratio
Where
FB
= T
V
f
if
=
C
V
OUT
- 1
=
C
DRP
IN
x
150k
P
V
(1
<
OUT
V
(%) x V
=
OUT
10pF.
IN
-
where
0.54
D)
Ω
= 3.3V,
FB
ratio
f
C
OUT
.
4.
Thus,
5.
6.
7.
8.
Choose C
9.
which is less than 10pF. So, It can be ignored.
CH1 Sync-Buck (Select Pin = Low Logic) and CH2
Sync-Buck :
CH1 sync-buck (select pin=low logic) and CH2 sync-buck
are converters employ current-mode control to simplify
the control loop compensation.
Hand Plan Zero) in the buck topology but there is a high
frequency pole f
frequency) is chosen sufficient less than f
The fixed parameters for CH1 and CH2 buck compensation
are as follows:
The input parameters for CH1 and CH2 buck compensation
are as follows:
Transconductance (from FB to COMP), GM = 200us
Current sense transresistance, R
Feedback voltage, V
R1, the voltage divider resistor in between V
FB.
R
C
ratio =
C
C =
ff
Half-load transient means load from 0.25A to 0.5A
transient. So, dI=0.5 − 0.25=0.25A
dV
C
Choose
Z
C
OUT
F
P
C
FB
=
=
=
=
C
dI
ratio
=
2
= T
R
f
π
ffp
ffz
OUT
⎛
⎜ ⎜
⎝
C
R
R
LOAD
GM
F
(1
R
×
CS
2
6.8nF.
C
DRP
= 150pF
LOAD
=
ff
R
π
-
1
=
x
1
x R
f
D)
Z
C
x
V
C
11k
(%) x V
C
GM
V
4.1
×
OUT
dV
C
FB
⎞
⎟ ⎟
⎠
ESR
R1
x
HP
=
FB
=
x
R
=
=
23k
⎛
⎜
⎝
2.68kHz
CS
=
>= f
126pF,
FB
3.3
0.8
V
FB
V
22μF x 0.005
OUT
6
=
FB
x
= 5% x 0.8 = 0.04V.
= FB = 0.8V
6 .
OSC
23k
= 4.1
6.8n
23k
⎞
⎟
⎠
x
where
Ω
/π . The f
(1
There is no RHPZ (Right
=
−
22
D)
CS
μ
= 4.8pF ,
. F
=
= 0.3V/A
RT9911
6.3nF.
www.richtek.com
HP
C
.
(cross over
OUT
and
25