Rainbow-electronics MAX1567 Manual de usuario Pagina 31

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MAX1566/MAX1567
Six-Channel, High-Efficiency, Digital
Camera Power Supplies
______________________________________________________________________________________ 31
The C
C
R
C
zero is then used to cancel the f
P
pole, so:
R
C
= R
LOAD
x C
OUT
x V
OUT
/ [(2V
OUT
- V
IN
) x C
C
]
AUX Step-Up, 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-half-plane zero,
Z
RHP
, at the following:
Z
RHP
= (1 - D)
2
x R
LOAD
/ (2π x L)
where (1 - D) = V
IN
/ V
OUT
(in a boost converter).
There is a complex pole pair at the following:
f
0
= V
OUT
/ [2π x V
IN
(L x C
OUT
)
1/2
]
If the zero due to the output capacitance and ESR is
less than 1/10 the right-half-plane zero:
Z
COUT
= 1 / (2π x C
OUT
x R
ESR
) < Z
RHP
/ 10
Then choose C
C
so 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
/ (2π x Z
COUT
)]
Choose R
C
to place the integrator zero, 1 / (2π x R
C
x
C
C
), at f
0
to cancel one of the pole pairs:
R
C
= V
IN
(L x C
OUT
)
1/2
/ (V
OUT
x C
C
)
If Z
COUT
is not less than Z
RHP
/ 10 (as is typical with
ceramic output capacitors) and continuous conduction
is required, then cross the loop over before Z
RHP
and f
0
:
f
C
< f
0
/ 10, and f
C
< Z
RHP
/ 10
In that case:
C
C
= (V
IN
/ V
RAMP
) (V
FB
/ V
OUT
) (g
M
/ (2π x f
C
))
Place:
1 / (2π x R
C
x C
C
) = 1 / (2π x R
LOAD
x C
OUT
), so that
R
C
= R
LOAD
x C
OUT
/ C
C
Or, reduce the inductor value for discontinuous operation.
MAX1567 AUX2 Inverter Compensation,
Discontinuous Inductor Current
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:
L < [V
IN
/ (|V
OUT
| + V
IN
)]
2
R
LOAD
/ (2f
OSC
)
A discontinuous current inverter has a single pole at the
following:
f
P
= 2 / (2π x R
LOAD
x C
OUT
)
Choose the integrator cap so the unity-gain crossover,
f
C
, occurs at f
OSC
/ 10 or lower. Note that for many AUX
circuits 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 the following:
C
C
= [V
IN
/ (K
1/2
x V
RAMP
] [V
REF
/ (V
OUT
+ V
REF
)] [g
M
/
(2π x f
C
)]
where K = 2L x 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
x C
OUT
) / (2C
C
)
MAX1567 AUX2 Inverter Compensation,
Continuous Inductor Current
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
RHP
, at:
Z
RHP
= [(1 - D)
2
/ D] x R
LOAD
/ (2π x L)
where D = |V
OUT
| / (|V
OUT
| + V
IN
) (in an inverter).
There is a complex pole pair at:
f
0
= (1 - D) / (2π(L x C)
1/2
)
If the zero due to the output-capacitor capacitance and
ESR is less than 1/10 the right-half-plane zero:
Z
COUT
= 1 / (2π x C
OUT
x R
ESR
) < Z
RHP
/ 10
Then 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
REF
/ (V
REF
+ |V
OUT
|)] [g
M
/
(2π x Z
COUT
)]
Choose R
C
to place the integrator zero, 1 / (2π x R
C
x
C
C
), at f
0
to cancel one of the pole pairs:
R
C
= (L x C
OUT
)
1/2
/ [(1 - D) x C
C
]
If Z
COUT
is not less than Z
RHP
/ 10 (as is typical with
ceramic output capacitors) and continuous conduction
is required, then cross the loop over before Z
RHP
and f
0
:
f
C
< f
0
/10, and f
C
< Z
RHP
/ 10
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