Choose f
C
= 20kHz. Calculate C
C
:
C
C
= (V
FB
/V
OUT
)(R
LOAD
/R
CS
)(gm/2
π
f
C
)(1 - D)
= (1.25/3.35)(6.7/0.3) x (135μS/(6.28 x 20kHz)
(2/3.35) = 5.35nF
Choose 6.8nF. Now select R
C
such that transient droop
requirements are met. For 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 tran-
sient gain. Since the current-sense transresistance is
0.3V/A, the value of R
C
that allows the required load
step swing:
R
C
= 0.3 I
IND(PK)/
6.75μA
In a step-up DC-to-DC converter, if L
IDEAL
is used, out-
put current relates to inductor current by:
I
IND(PK)
= 1.25 I
OUT
/(1 - D) = 1.25 I
OUT
V
OUT
/V
IN
Thus, for a 400mA output load step with V
IN
= 2V and
V
OUT
= 3.35V:
R
C
= [1.25(0.3 x 0.4 x 3.35)/2)]/6.75μA = 37k
Note that the inductor does not limit the response in this
case since it can ramp at 2V/3.3μH, or 606mA/μs. The
output filter capacitor is then chosen so that the C
OUT
R
LOAD
pole cancels the R
C
C
C
zero:
C
OUT
R
LOAD
= R
C
C
C
For example:
C
OUT
= 37k
x 6.8nF/6.7 = 37.5μF
Since a reasonable value for C
OUT
is 47μF rather than
37.5, choose 47μF and rescale RC:
R
C
= 47μF x 6.7/6.8nF = 46.3k
which provides a slightly higher transient gain and con-
sequently less transient droop than previously selected.
If the output filter capacitor has significant ESR, a zero
occurs at:
Z
ESR
= 1/(2
π
C
OUT
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 COMPSU to GND:
C
P
= C
OUT
R
ESR
/R
C
If C
P
is calculated to be < 10pF, it can be omitted.
Step-Down Component Selection
Step-Down Inductor
The external components required for the step-down
are an inductor, input and output filter capacitors, and
compensation RC network. The MAX1565 step-down
converter provides best efficiency with continuous
inductor current. A reasonable inductor value (LIDEAL)
can be derived from:
L
IDEAL
= 2 (V
IN
) D (1 - D)/(I
OUT
f
OSC
)
which 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 varia-
tion 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 capacitance may be required to suppress
output ripple.
Larger values than L
IDEAL
can be used to obtain higher
output current, but with typically larger inductor size.
Step-Down Compensation
The relevant characteristics for step-down compensa-
tion are:
1)
Transconductance (from FBSD to COMPSD), gm
EA
(135μS)
2)
Step-down slope compensation pole, P
SLOPE
=
V
IN
/ (
π
L)
3)
Current-sense amplifier transresistance, R
CS
,
(0.6V/A)
4)
Feedback regulation voltage, V
FB
(1.25V)
5)
Step-down output voltage, V
SD
, in V
6)
Output load equivalent resistance, R
LOAD
,
in
= V
OUTSD
/I
LOAD
The key steps for step-down compensation are:
1)
Set the compensation RC 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 fre-
quency.
If we assume V
IN
= 3.35V, V
OUT
= 1.5V, and I
OUT
=
350mA, then R
LOAD
= 4.3
.
M
Small, High-Efficiency, Five-Channel
Digital Still Camera Power Supply
______________________________________________________________________________________
19