參數(shù)資料
型號: MAX1566-MAX1567
廠商: Maxim Integrated Products, Inc.
英文描述: Six-Channel, High-Efficiency, Digital Camera Power Supplies
中文描述: 六通道,高效率,數(shù)碼相機(jī)電源
文件頁數(shù): 29/35頁
文件大?。?/td> 1134K
代理商: MAX1566-MAX1567
M
Six-Channel, High-Efficiency, Digital
Camera Power Supplies
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29
For the example:
C
OUT
= 68k
x 6.8nF / 10
= 46μF
Choose 47μF for C
OUT
. If the available C
OUT
is sub-
stantially different from the calculated value, insert the
available C
OUT
value into the above equation and
recalculate R
C
. Higher substituted C
OUT
values allow a
higher R
C
, which provides higher transient gain and
consequently less transient droop.
If the output filter capacitor has significant ESR, a zero
occurs at the following:
Z
ESR
= 1 / (2
π
x C
OUT
x R
ESR
)
If Z
ESR
> f
C
, it can be ignored, as is typically the case
with ceramic output capacitors. If Z
ESR
is less than f
C
,
it should be cancelled with a pole set by capacitor C
P
connected from CC to GND:
C
P
= C
OUT
x R
ESR
/ R
C
If C
P
is calculated to be <10pF, it can be omitted.
Step-Down Component Selection
This section describes component selection for the
step-down converter, and for the main converter if used
in step-down mode (SUSD = GND).
Step-Down Inductor
The external components required for the step-down
are an inductor, input and output filter capacitors, and
compensation RC network.
The MAX1566/MAX1567 step-down converter provides
best efficiency with continuous inductor current. A rea-
sonable inductor value (L
IDEAL
) can be derived from
the following:
L
IDEAL
= [2(V
IN
) x D(1 - D)] / I
OUT
x f
OSC
This sets the peak-to-peak inductor current at 1/2 the
DC inductor current. D is the duty cycle:
D = V
OUT
/ V
IN
Given L
IDEAL
, the peak-to-peak inductor current is 0.5
I
OUT
. The absolute-peak inductor current is 1.25 I
OUT
.
Inductance values smaller than L
IDEAL
can be used to
reduce inductor size; however, if much smaller values are
used, inductor current rises, and a larger output capaci-
tance may be required to suppress output ripple. Larger
values than L
IDEAL
can be used to obtain higher output
current, but typically with larger inductor size.
Step-Down Compensation
The relevant characteristics for step-down compensa-
tion are as follows:
Transconductance (from FB to C
C
), gm
EA
(135μS)
Step-down slope-compensation pole, P
SLOPE
= V
IN
/
(
π
L)
Current-sense amplifier transresistance, R
CS
(0.6V/A)
Feedback-regulation voltage, V
FB
(1.25V)
Step-down output voltage, V
SD
, in V
Output-load equivalent resistance, R
LOAD
, in
=
V
OUT
/ I
LOAD
The key steps for step-down compensation are
as follows:
1) Set the compensation RC to zero to cancel the
R
LOAD
C
OUT
pole.
2) Set the loop crossover below the lower of 1/5 the
slope compensation pole or 1/5 the switching
frequency.
If we assume V
IN
= 2.5V, V
OUT
= 1.8V, and I
OUT
=
350mA, then R
LOAD
= 5.14
.
If we select f
OSC
= 500kHz and L = 5.6μH.
P
SLOPE
= V
IN
/ (
π
L) = 142kHz, so choose f
C
= 24kHz
and calculate C
C
:
C
C
= (V
FB
/ V
OUT
)(R
LOAD
/ R
CS
)(gm / 2
π
x f
C
)
= (1.25 / 1.8)(5.14 / 0.6) x [135μS / (6.28 x 24kHz)]
= 6.4nF
Choose 6.8nF.
Now select R
C
so transient-droop requirements are
met. As an example, if 4% transient droop is allowed,
the input to the error amplifier moves 0.04 x 1.25V, or
50mV. The error-amp output drives 50mV x 135μS, or
6.75μA across R
C
to provide transient gain. Since the
current-sense transresistance is 0.6V/A, the value of R
C
that allows the required load-step swing is as follows:
R
C
= 0.6 I
IND(PK)
/ 6.75μA
In a step-down DC-to-DC converter, if L
IDEAL
is used,
output current relates to inductor current by the following:
I
IND(PK)
= 1.25 I
OUT
So for a 250mA output load step with V
IN
= 2.5V and
V
OUT
= 1.8V:
R
C
= (1.25 x 0.6 x 0.25) / 6.75μA = 27.8k
Choose 27k
.
Note that the inductor does somewhat limit the response
in this case since it ramps at (V
IN
- V
OUT
) / 5.6μH, or
(2.5 - 1.8) / 5.6μH = 125mA/μs.
The output filter capacitor is then chosen so the C
OUT
R
LOAD
pole cancels the R
C
C
C
zero:
C
OUT
x R
LOAD
= R
C
x C
C
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