ML1565 datasheet(17/21 Pages) MINILOGIC

17 / 21 page

P17/21

Rev. C, Sep 2005

ML1565

Design Procedure (5)

Auxiliary controller Component Selection

External MOSFET

All ML1565 auxiliary controllers drive external logic-level

N-channel MOSFETs. Significant MOSFET selection

parameters are:

On-resistance (RDS(ON))

Maximum drain-to-source voltage (VDS(MAX))

Total gate charge (QG)

Reverse transfer capacitance (CRSS)

DL_ swings between OUTSU and GND. Use a MOSFET

with on-resistance specified at or below the main output

voltage. The gate charge, QG, includes all capacitance

associated with charging the gate and helps to predict

MOSFET transition time between on and off states.

MOSFET

power

dissipation

combination

on-resistance and transition losses. The on-resistance

loss is:

PRDSON = D IL

2RDS(ON)

where D is the duty cycle, IL is the average inductor

current, and RDS(ON) is MOSFET on-resistance. The

transition loss is approximately:

PTRANS = (VOUT IL fOSC tT) / 3

were VOUT is the output voltage, IL is the average inductor

current, fOSC is the switching frequency, and tT is the

transition time. The transition time is approximately QG / IG,

where QG is the total gate charge, and IG is the gate drive

current (typically 0.5A). the total power dissipation in the

MOSFET is:

PMOSFET = PRDSON + PTRANS

Auxiliary Compensation

The auxiliary controllers employ voltage-mode control to

regulate their output voltage. Optimum compensation

somewhat

depends

whether

the

design

uses

continuous or discontinuous inductor current.

Discontinuous Inductor Current

When the inductor current falls to zero on each switching

cycle, it is described as discontinuous. The inductor is not

utilized as efficiently as with continuous current. This often

has little negative impact in light-load applications since

the coil losses may already be low compared to other

losses. A benefit of discontinuous inductor current is more

flexible loop compensation and no maximum duty-cycle

restriction on boost ratio.

To ensure discontinuous operation, the inductor must have

a sufficiently low inductance to fully discharge on each

cycle. The occurs when:

L < [ VIN

2 (VOUT – VIN) / VOUT3 ] [ RLOAD / (2fOSC) ]

A discontinuous current boost has a single pole at:

fP = (2VOUT – VIN) / (2π RLOAD COUT VOUT)

Choose the integrator capacitor such that the unity-gain

Crossover (fC) occurs at fOSC / 10 or lower. Note that for

many auxiliary circuits, such as those powering motors,

LEDs, or other loads that do not require fast transient

response, it is often acceptable to over compensate by

setting fC at fOSC / 20 or lower. Cc is then determined by:

Cc = [2VOUTVIN / (2VOUT – VIN) VRAMP]]

[VOUT / (K(VOUT – VIN))

1/2 [VFB / VOUT]

(gM / 2π fC))]

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

compensation voltage ramp of 1.25V. The CcRc zero is

then used to cancel the fP pole, so:

Rc = RLOAD COUT VOUT / [ (2VOUT – VIN) Cc]

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-plane zero at:

fRHPZ = (1-D)

2 RLOAD /(2πL)

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

pole pair is located at:

f0 = VOUT /[2π VIN (L COUT)

1/2]

If the zero due to the output capacitor capacitance and

ESR is less than 1/10 the righe-plane zero:

ZCOUT = 1 / (2πCOUT RESR) < fRHPZ / 10

Choose Cc such that the crossover frequency fC occurs at

ZCOUT. The ESR zero provides a phase boost at crossover.

Cc = (VIN / VRAMP)(VFB / VOUT)(gM / (2πZCOUT))

Choose Rc to place the integrator zero, 1/(2πRcCc), at f0 to

cancel one of the pole pairs:

Rc = VIN (L COUT)

1/2 / (VOUT Cc)

If ZCOUT is not less than fRHPZ / 10 (as is typical with ceramic

output capacitors) and continuous conduction is required,

then cross the loop over before fRHPZ and f0:

fC < f0/10, and fC < fRHPZ / 10

In that case:

Cc = (VIN/ VRAMP) (VFB/ VOUT) (gM / 2π fC)

Place 1/ (2πRcCc) = 1 / (2π RLOADCOUT), so that Rc =

RLOADCOUT / Cc or reduce the inductor value for

discontinuous operation.

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