參數(shù)資料
型號(hào): MAX1565
廠商: Maxim Integrated Products, Inc.
英文描述: Small, High-Efficiency, Five-Channel Digital Still Camera Power Supply
中文描述: 小型、高效率、5路輸出數(shù)碼相機(jī)電源
文件頁數(shù): 21/26頁
文件大?。?/td> 512K
代理商: MAX1565
Discontinuous Inductor Current
When the inductor current falls to zero on each switching
cycle, it is described as discontinuous. The inductor is
not utilized as efficiently as with continuous current. This
often has little negative impact in light-load applications
since the coil losses may already be low compared to
other losses. A benefit of discontinuous inductor cur-
rent is more flexible loop compensation and no maxi-
mum duty-cycle restriction on boost ratio.
To ensure discontinuous operation, the inductor must
have a sufficiently low inductance to fully discharge on
each cycle. This occurs when:
L < [V
IN
2
(V
OUT
- V
IN
)/V
OUT3
] [R
LOAD
/(2 f
OSC
)]
A discontinuous current boost has a single pole at:
f
P
= (2V
OUT
- V
IN
)/(2
π
R
LOAD
C
OUT
V
OUT
)
Choose the integrator capacitor such that the unity-gain
crossover (f
C
) occurs at f
OSC
/10 or lower. Note that for
many auxiliary circuits, such as those powering motors,
LEDs, or other loads that do not require fast transient
response, it is often acceptable to overcompensate by
setting f
C
at f
OSC
/20 or lower. C
C
is then determined by:
C
C
= [2V
OUT
V
IN
/((2V
OUT
- V
IN
)V
RAMP
)]
[V
OUT
/(K(V
OUT
- V
IN
))]
1/2
[(V
FB
/V
OUT
)
(g
M
/(2
π
f
C
))]
where K = 2 L f
OSC
/R
LOAD,
and V
RAMP
is the internal
slope compensation voltage ramp of 1.25V. The C
C
R
C
zero is then used to cancel the f
P
pole, so:
R
C
= R
LOAD
C
OUT
V
OUT
/[(2V
OUT
- V
IN
) C
C
]
Continuous Inductor Current
Continuous inductor current can sometimes improve
boost efficiency by lowering the ratio between peak
inductor current and output current. It does this at the
expense of a larger inductance value that requires larger
size for a given current rating. With continuous inductor
current boost operation, there is a right-plane zero at:
f
RHPZ
= (1 - D)
2
R
LOAD
/(2
π
L)
where (1 - D) = V
IN
/V
OUT
(in a boost converter). A com-
plex pole pair is located at:
f
0
= V
OUT
/[2
π
V
IN
(L C
OUT
)
1/2
]
If the zero due to the output capacitor capacitance and
ESR is less than 1/10 the right-plane zero:
Z
COUT
= 1/(2
π
C
OUT
R
ESR
) < f
RHPZ
/10
Choose C
C
such that the crossover frequency f
C
occurs at Z
COUT
. The ESR zero provides a phase boost
at crossover.
C
C
= (V
IN
/V
RAMP
)(V
FB
/V
OUT
)(g
M
/(2
π
Z
COUT
))
Choose R
C
to place the integrator zero, 1/(2
π
R
C
C
C
), at
f
0
to cancel one of the pole pairs:
R
C
= V
IN
(L C
OUT
)
1/2
/(V
OUT
C
C
)
If Z
COUT
is not less than f
RHPZ
/10 (as is typical with
ceramic output capacitors) and continuous conduction is
required, then cross the loop over before f
RHPZ
and f
0
:
f
C
< f
0
/10, and f
C
< f
RHPZ
/10
In that case:
C
C
= (V
IN
/V
RAMP
)(V
FB
/V
OUT
)(g
M
/(2
π
f
C
))
Place 1/(2
π
R
C
C
C
) = 1/(2
π
R
LOAD
C
OUT
), so that R
C
=
R
LOAD
C
OUT
/C
C
or reduce the inductor value for dis-
continuous operation.
Applications Information
LED, LCD, and Other Boost Applications
Any auxiliary channel can be used for a wide variety of
step-up applications. These include generating 5V or
some other voltage for motor or actuator drive, generating
15V or a similar voltage for LCD bias, or generating a
step-up current source to efficiently drive a series array
of white LEDs for display backlighting. Figures 5 and 6
show examples of these applications.
M
Small, High-Efficiency, Five-Channel
Digital Still Camera Power Supply
______________________________________________________________________________________
21
MAX1565
(PARTIAL)
AUX_
PWM
OUTSU
DL_
FB_
4.7
μ
H
TO V
BATT
4.7
μ
F
22
μ
F
15V
100mA
1.1M
100k
Figure 5. Using an AUX_ Controller Channel to Generate LCD
Bias
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