MAX1586A Datasheet, PDF(25/32 Page) Maxim Integrated Products – High-Efficiency, Low-IQ PMICs with Dynamic Core for PDAs and Smart Phones

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MAX1586A Datasheet, PDF (25/32 Pages) Maxim Integrated Products – High-Efficiency, Low-IQ PMICs with Dynamic Core for PDAs and Smart Phones

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High-Efficiency, Low-IQ PMICs with

Dynamic Core for PDAs and Smart Phones

must also have low impedance at the switching fre-

quency. Ceramic, polymer, and tantalum capacitors

are suitable, with ceramic exhibiting the lowest ESR

and lowest high-frequency impedance.

Output ripple with a ceramic output capacitor is

approximately:

VRIPPLE = IL(PEAK) [1 / (2Ï x fOSC x COUT)]

If the capacitor has significant ESR, the output ripple

component due to capacitor ESR is:

VRIPPLE(ESR) = IL(PEAK) x ESR

Output capacitor specifics are also discussed in the

Compensation and Stability section.

Compensation and Stability

The relevant characteristics for REG1, REG2, and

REG3 compensation are:

1) Transconductance (from FB_ to CC_), gmEA

2) Current-sense amplifier transresistance, RCS

3) Feedback regulation voltage, VFB (1.25V)

4) Step-down output voltage, VOUT, in V

5) Output load equivalent resistance, RLOAD = VOUT /

ILOAD

The key steps for step-down compensation are:

1) Set the compensation RC zero to cancel the RLOAD

COUT pole.

2) Set the loop crossover at or below approximately

1/10th the switching frequency.

For example, with VIN(MAX) = 5V, VOUT = 2.5V for

REG2, and IOUT = 800mA, then RLOAD = 3.125â¦. For

REG2, RCS = 0.75V/A and gmEA = 87ÂµS.

Choose the crossover frequency, fC â¤ fOSC / 10.

Choose 100kHz. Then calculate the value of the com-

pensation capacitor, CC:

CC = (VFB / VOUT) x (RLOAD / RCS) x (gm / (2Ï x fC))

= (1.25 / 2.5) x (3.125 / 0.75) x (87 x 10-6 / (6.28

x 100,000)) = 289pF

Choose 330pF, the next highest standard value.

Now select the compensation resistor, RC, so transient-

droop requirements are met. As an example, if 3% tran-

sient droop is allowed for the desired load step, the

input to the error amplifier moves 0.03 x 1.25V, or

37.5mV. The error-amplifier output drives 37.5mV x

gmEA, or IEAO = 37.5mV x 87ÂµS = 3.26ÂµA across RC to

provide transient gain. Find the value of RC that allows

the required load-step swing from:

RC = RCS x IIND(PK) / IEAO

Table 4. Compensation Parameters

PARAMETER

REG1 REG2 REG3

Error-Amplifier

Transconductance, gmEA

Current-Sense Amp

Transresistance, RCS

87ÂµS 87ÂµS 68ÂµS

0.5V/A 0.75V/A 1.25V/A

Table 5. Typical Compensation Values

COMPONENT OR

PARAMETER

VOUT

Output Current

Inductor

Load-Step Droop

Loop Crossover Freq (fC)

COUT

REG1

3.3V

1300mA

3.3ÂµH

100kHz

330pF

240kâ¦

22ÂµF

REG2

2.5V

900mA

6.8ÂµH

100kHz

270pF

240kâ¦

22ÂµF

REG3

1.3V

500mA

10ÂµH

100kHz

330pF

240kâ¦

22ÂµF

where IIND(PK) is the peak inductor current. In a step-

down DC-DC converter, if LIDEAL is used, output cur-

rent relates to inductor current by:

IIND(PK) = 1.25 x IOUT

So for an 800mA output load step with VIN = 3.6V and

VOUT = 2.5V:

RC = RCS x IIND(PK) / IEAO = (0.75V/A) x

(1.25 x 0.8A) / 3.26ÂµA = 230kâ¦

We choose 240kâ¦. Note that the inductor does not limit

the response in this case since it can ramp at (VIN -

VOUT) / L, or (3.6 - 2.5) / 3.3ÂµH = 242mA/Âµs.

The output filter capacitor is then selected so that the

COUT RLOAD pole cancels the RC CC zero:

COUT x RLOAD = RC x CC

For the example:

RLOAD = VOUTx ILOAD = 2.5V / 0.8A =

3.125â¦

COUT = RC x CC / RLOAD = 240kâ¦ x 330pF /

3.125â¦ = 25ÂµF

We choose 22ÂµF.

Recalculate RC using the selected COUT.

RC = COUT x RLOAD / CC = 208kâ¦

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