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MAX1584 Datasheet, PDF (24/29 Pages) Maxim Integrated Products – 5-Channel Slim DSC Power Supplies
5-Channel Slim DSC Power Supplies
MAX1585 AUX2 Inverter Compensation,
Discontinuous Inductor Current
If the load current is very low (40mA or less), discontin-
uous 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 < [VIN / (|VOUT| + VIN)]2 RLOAD / (2fOSC)
A discontinuous current inverter has a single pole at:
fP = 2 / (2π x RLOAD x COUT)
Choose the integrator cap so the unity-gain crossover,
fC, occurs at fOSC / 10 or lower. Note that for many AUX
circuits that do not require fast transient response, it is
often acceptable to overcompensate by setting fC at
fOSC / 20 or lower.
CC is then determined by the following:
CC = [VIN / (K1/2 x VRAMP][VREF / (VOUT + VREF)] [gM /
(2π x fC)]
where:
K = 2L x fOSC / RLOAD, and VRAMP is the internal volt-
age ramp of 1.25V.
The CC RC zero then is used to cancel the fP pole, so:
RC = (RLOAD x COUT) / (2 CC)
MAX1585 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,
ZRHP, at:
ZRHP = [(1 - D)2 / D] x RLOAD / (2π x L)
where D = |VOUT| / (|VOUT| + VIN) (in an inverter).
There is a complex pole pair at:
f0 = (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:
ZCOUT = 1 / (2π x COUT x RESR) < ZRHP / 10
Then choose CC so the crossover frequency, fC, occurs
at ZCOUT. The ESR zero provides a phase boost at
crossover.
CC = (VIN / VRAMP)[VREF / (VREF + |VOUT|)][gM / (2π x
ZCOUT)]
Choose RC to place the integrator zero, 1 / (2π x RC x
CC), at f0 to cancel one of the pole pairs:
RC = (L x COUT)1/2 / [(1 - D) x CC]
If ZCOUT is not less than ZRHP / 10 (as is typical with
ceramic output capacitors) and continuous conduction
is required, then cross the loop over before ZRHP and f0:
fC < f0 / 10, and fC < ZRHP / 10
In that case:
CC = (VIN / VRAMP)[VREF / (VREF + |VOUT|)][gM / (2π x fC)]
Place:
1 / (2π x RC x CC) = 1 / (2π x RLOAD x COUT), so that
RC = RLOAD x COUT / CC
Or, reduce the inductor value for discontinuous operation.
Applications Information
LED, LCD, and Other Boost Applications
Any AUX 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, generat-
ing 15V or a similar voltage for LCD bias, or generating
a step-up current source to efficiently drive a series
array of white LEDs to display backlighting. Figures 5
and 6 show examples of these applications.
Multiple-Output Flyback Circuits
Some applications require multiple voltages from a sin-
gle converter channel. This is often the case when gen-
erating voltages for CCD bias or LCD power. Figure 7
shows a two-output flyback configuration with AUX_.
The controller drives an external MOSFET that switches
the transformer primary. Two transformer secondaries
generate the output voltages. Only one positive output
voltage can be fed back, so the other voltages are set
by the turns ratio of the transformer secondaries. The
load stability of the other secondary voltages depends
on transformer leakage, inductance, and winding resis-
tance. Voltage regulation is best when the load on the
secondary that is not fed back is small compared to the
load on the one that is fed back. Regulation also
improves if the load current range is limited. Consult
the transformer manufacturer for the proper design for
a given application.
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