MAX1566-MAX1567 Datasheet, PDF(31/35 Page) Maxim Integrated Products – Six-Channel, High-Efficiency, Digital Camera Power Supplies

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MAX1566-MAX1567 Datasheet, PDF (31/35 Pages) Maxim Integrated Products – Six-Channel, High-Efficiency, Digital Camera Power Supplies

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Six-Channel, High-Efficiency, Digital

Camera Power Supplies

The CC RC zero is then used to cancel the fP pole, so:

RC = RLOAD x COUT x VOUT / [(2VOUT - VIN) x CC]

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,

ZRHP, at the following:

ZRHP = (1 - D)2 x RLOAD / (2Ï x L)

where (1 - D) = VIN / VOUT (in a boost converter).

There is a complex pole pair at the following:

f0 = VOUT / [2Ï x VIN (L x COUT)1/2]

If the zero due to the output capacitance and ESR is

less than 1/10 the right-half-plane zero:

ZCOUT = 1 / (2Ï 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) (VFB / VOUT) [gM / (2Ï x ZCOUT)]

Choose RC to place the integrator zero, 1 / (2Ï x RC x

CC), at f0 to cancel one of the pole pairs:

RC = VIN(L x COUT)1/2 / (VOUT 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) (VFB / VOUT) (gM / (2Ï x fC))

Place:

1 / (2Ï x RC x CC) = 1 / (2Ï x RLOAD x COUT), so that

RC = RLOAD x COUT / CC

Or, reduce the inductor value for discontinuous operation.

MAX1567 AUX2 Inverter Compensation,

Discontinuous Inductor Current

If the load current is very low (â¤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 < [VIN / (|VOUT| + VIN)]2 RLOAD / (2fOSC)

A discontinuous current inverter has a single pole at the

following:

fP = 2 / (2Ï 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 /

(2Ï x fC)]

where K = 2L x fOSC / RLOAD, and VRAMP is the internal

slope-compensation voltage ramp of 1.25V.

The CC RC zero is then used to cancel the fP pole, so:

RC = (RLOAD x COUT) / (2CC)

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,

ZRHP, at:

ZRHP = [(1 - D)2 / D] x RLOAD / (2Ï x L)

where D = |VOUT| / (|VOUT| + VIN) (in an inverter).

There is a complex pole pair at:

f0 = (1 - D) / (2Ï(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 / (2Ï x COUT x RESR) < ZRHP / 10

Then choose CC such that the crossover frequency fC

occurs at ZCOUT. The ESR zero provides a phase boost

at crossover:

CC = (VIN / VRAMP) [VREF / (VREF + |VOUT|)] [gM /

(2Ï x ZCOUT)]

Choose RC to place the integrator zero, 1 / (2Ï 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

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