Rainbow Electronics MAX1567 User Manual
Page 31
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
/ (2
π 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
/ [2
π 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 / (2
π 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
/ (2
π x Z
COUT
)]
Choose R
C
to place the integrator zero, 1 / (2
π 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
/ (2
π x f
C
))
Place:
1 / (2
π x R
C
x C
C
) = 1 / (2
π 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 / (2
π 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
/
(2
π 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
/ (2
π x L)
where D = |V
OUT
| / (|V
OUT
| + V
IN
) (in an inverter).
There is a complex pole pair at:
f
0
= (1 - D) / (2
π(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 / (2
π 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
/
(2
π x Z
COUT
)]
Choose R
C
to place the integrator zero, 1 / (2
π 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