ML1565 datasheet(16/21 Pages) MINILOGIC

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P16/21

Rev. C, Sep 2005

ML1565

Design Procedure (4)

Step-Down Component Selection

Step-Down Inductor

The external components required for the step-down are

an inductor, input and output filter capacitors, and

compensation RC network. The ML1565 step-down

converter provides best efficiency with continuous inductor

current. A reasonable inductor value (LIDEAL) can be

derived from:

LIDEAL = 2(VIN) D (1-D) / (IOUT fOSC)

which sets the peak-to-peak inductor current at 1/2 the DC

inductor current. D is the duty cycle:

D = VOUT / VIN

Given LIDEAL, the peak-to-peak inductor current variation is

0.5 IOUT. The absolute peak inductor current is 1.25 IOUT.

Inductance values smaller than LIDEAL can be used to

reduce inductor size. However, if much smaller values are

used,

inductor

current

rises

and

large

output

capacitance may be required to suppress output ripple.

Larger values than LIEDAL can be used to obtain higher

output current, but with typical larger inductor size.

Step-Down Compensation

The relevant characteristics for step-down compensation

are:

Transconductance (from FBSD to COMPSD), gmEA

(135µS)

Step-down slope compensation pole,

PSLOPE = VIN / (π L)

Current-sense amplifier transresistance,Rcs, (0.6V/A)

Feedback regulation voltage, VFB(1.25V)

Step-down output voltage, VSD, in V

Output load equivalent resistance,

RLOAD in Ω = VOUTSD / ILOAD

The key steps for step-down compensation are:

Set the compensation RC zero to cancel the RLOAD

COUT pole.

Set the loop crossover below the lower of 1/5 the

slope compensation pole, or 1/5 the switching

frequency.

If we assume VIN = 3.35V, VOUT = 1.5V, and

IOUT = 350mA, then RLOAD = 4.3Ω

If we select L = 4.7µH and fOSC = 440kHz,

PSLOPE = VIN / (

πL) = 214kHz, so choose fc = 40kHz and

calculate Cc:

Cc = (VFB / VOUT)(RLOAD / RCS) (gm / 2π fc)

=(1.25/1.5)(4.3/0.6) x (135µS/(6.28 x 40kHz))

= 3.2nF

Choose 3.3nF. Now select Rc such that transient droop

requirements are met. For example, if 4% transient droop

is allowed, the input to the error amplifier moves 0.04 x

1.25V, or 50mV. The error amp output drives 50mV x

135µS, or 6.75µA across Rc to provide transient gain.

Since the current-sense transresistance is 0.6V/A, the

value of Rc that allows the required load step swing:

RC=0.6 IIND(PK) / 6.75µA

In a step-down DC-to-DC converter, if LIDEAL is used, output

current relates to inductor current by:

IIND(OK) = 1.25 IOUT

Thus, for a 250mA output load step with VIN = 3.35V and

VOUT = 1.5V:

Rc = (1.25 x 0.6 x 0.25) / 6.75µA = 27.8kΩ

Choose 27kΩ. Note that the inductor does not limit the

response in this case since it can ramp at

(VIN-VOUT)/4.7µH, or (3.35 – 1.5)/4.7µH = 394mA/µs

The output filter capacitor is then chosen so that the

COUTRLOAD pole cancels the RcCc zero:

COUTRLOAD = RcCc

For example: COUT = 27kΩ x 3.3nF / 4.3 = 20.7µF

Choose 22µF. If the output filter capacitor has significant

ESR, a zero occurs at:

ZESR = 1 / (2πCOUTRESR)

If ZESR > fc, it can be ignored, as is typically the case with

ceramic output capacitors. If ZESR is less than fc, it should

be cancelled with a pole set by capacitor Cp connected

from COMPAD to GND.

Cp = COUTRESR / Rc

If Cp is calculated to be < 10pF, it can be omitted.

Auxiliary Controller Component Selection

Diode

For most auxiliary applications, a Schottky diode rectifies

the output voltage. The Schottky diode’s low forward

voltage

and

fast

recovery

time

provide

the

best

performance in most applications. Silicon signal diodes

(such as 1N4148) are sometimes adequate in low-current

(<10mA) high-voltage (>10V) output circuits where the

output voltage is large compared to the diode forward

voltage.

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