MAX1567ETL+T Maxim Integrated Products, MAX1567ETL+T Datasheet - Page 31

• Manufacturer Part Number: MAX1567ETL+T
• Description: IC DGTL CAM PWR-SUP 6CH 40TQFN
• Manufacturer: Maxim Integrated Products

Specifications of MAX1567ETL+T

• Applications: Controller, Digital Camera
• Voltage - Input: 0.7 ~ 5.5 V
• Number Of Outputs: 7
• Voltage - Output: 1.25 ~ 5 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 40-TQFN Exposed Pad
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

The C
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-half-plane zero,
Z
where (1 - D) = V
There is a complex pole pair at the following:
If the zero due to the output capacitance and ESR is
less than 1/10 the right-half-plane zero:
Then choose C
at Z
crossover:
Choose R
C
If Z
ceramic output capacitors) and continuous conduction
is required, then cross the loop over before Z
In that case:
Place:
1 / (2π x R
Or, reduce the inductor value for discontinuous operation.
If the load current is very low (≤40mA), discontinuous
current is preferred for simple loop compensation and
freedom from duty-cycle restrictions on the inverter
input-output ratio. To ensure discontinuous operation,
the inductor must have a sufficiently low inductance to
fully discharge on each cycle. This occurs when:
RHP
C
C
R
), at f
C
COUT
C
COUT
C
, at the following:
= (V
Z
C
C
= R
L < [V
COUT
R
= (V
0
C
IN
AUX Step-Up, Continuous Inductor Current
LOAD
to cancel one of the pole pairs:
R
is not less than Z
. The ESR zero provides a phase boost at
C
C
f
Z
0
C
zero is then used to cancel the f
RHP
/ V
IN
x C
to place the integrator zero, 1 / (2π x R
f
= V
IN
= 1 / (2π x C
C
= V
MAX1567 AUX2 Inverter Compensation,
/ V
RAMP
< f
R
/ (|V
C
C
x C
OUT
= (1 - D)
IN
C
) = 1 / (2π x R
IN
RAMP
so the crossover frequency f
0
______________________________________________________________________________________
(L x C
= R
OUT
OUT
/ 10, and f
/ V
) (V
/ [2π x V
Discontinuous Inductor Current
OUT
LOAD
) (V
| + V
FB
x V
OUT
2
OUT
FB
/ V
OUT
(in a boost converter).
x R
IN
RHP
x C
)
IN
/ V
OUT
Six-Channel, High-Efficiency, Digital
1/2
C
)]
x R
LOAD
LOAD
2
/ [(2V
(L x C
< Z
OUT
OUT
/ (V
/ 10 (as is typical with
R
) [g
ESR
LOAD
RHP
) (g
OUT
/ C
/ (2π x L)
x C
M
OUT
) < Z
OUT
/ (2π x Z
M
C
/ 10
OUT
/ (2f
x C
/ (2π x f
- V
)
RHP
1/2
P
C
), so that
OSC
IN
RHP
]
pole, so:
)
) x C
/ 10
C
COUT
)
C
occurs
and f
))
C
]
)]
C
0
x
:
Camera Power Supplies
A discontinuous current inverter has a single pole at the
following:
Choose the integrator cap so the unity-gain crossover,
f
circuits that do not require fast transient response, it is
often acceptable to overcompensate by setting f
f
C
C
where K = 2L x f
slope-compensation voltage ramp of 1.25V.
The C
Continuous inductor current may be more suitable for
larger load currents (50mA or more). It improves effi-
ciency by lowering the ratio between peak inductor cur-
rent 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
inverter operation, there is a right-half-plane zero,
Z
where D = |V
There is a complex pole pair at:
If the zero due to the output-capacitor capacitance and
ESR is less than 1/10 the right-half-plane zero:
Then choose C
occurs at Z
at crossover:
Choose R
C
If Z
ceramic output capacitors) and continuous conduction
is required, then cross the loop over before Z
C
OSC
RHP
C
C
C
, occurs at f
), at f
is then determined by the following:
= [V
COUT
C
, at:
/ 20 or lower.
C
Z
(2π x f
C
COUT
= (V
IN
R
0
Z
C
(2π x Z
to cancel one of the pole pairs:
is not less than Z
RHP
/ (K
C
R
COUT
zero is then used to cancel the f
IN
C
to place the integrator zero, 1 / (2π x R
C
f
OUT
R
= 1 / (2π x C
1/2
)]
C
f
MAX1567 AUX2 Inverter Compensation,
OSC
/ V
P
= [(1 - D)
= (L x C
C
f
< f
0
C
= 2 / (2π x R
x V
OSC
COUT
= (R
RAMP
. The ESR zero provides a phase boost
| / (|V
= (1 - D) / (2π(L x C)
such that the crossover frequency f
0
/ 10 or lower. Note that for many AUX
RAMP
/10, and f
LOAD
/ R
OUT
)]
) [V
OUT
2
LOAD
)] [V
Continuous Inductor Current
/ D] x R
OUT
REF
| + V
)
x C
1/2
RHP
LOAD
REF
C
, and V
/ (V
x R
OUT
/ [(1 - D) x C
IN
< Z
LOAD
) (in an inverter).
/ (V
/ 10 (as is typical with
REF
ESR
x C
) / (2C
RHP
OUT
RAMP
) < Z
+ |V
OUT
1/2
/ (2π x L)
/ 10
)
+ V
C
OUT
)
)
RHP
is the internal
C
P
REF
]
RHP
|)] [g
pole, so:
/ 10
)] [g
and f
M
M
C
/
C
/
31
at
0
C
x
: