MAX15021 datasheet(19/24 Pages) MAXIM

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The locations of the zeros and poles should be such

that the phase margin peaks around fCO.

Set the ratios of fCO-to-fZ and fP-to-fCO equal to one anoth-

er, e.g., fCO = fP = 5 is a good number to get approximately

fCO

60° of phase margin at fCO. Whichever technique, it is

important to place the two zeros at or below the double

pole to avoid the conditional stability issue.

The following procedure is recommended:

1) Select a crossover frequency, fCO, at or below one-

tenth the switching frequency (fSW):

2) Calculate the LC double-pole frequency, fLC :

where COUT is the output capacitor of the regulator.

3) Select the feedback resistor, RF, in the range of

3.3k

Ω to 30kΩ.

4) Place the compensator’s first

zero at or below the output filter’s

double-pole, fLC , as follows:

5) The gain of the modulator (GainMOD)—comprised of

the regulator’s pulse-width modulator, LC filter,

feedback divider, and associated circuitry—at the

crossover frequency is:

The gain of the error amplifier (GainE/A) in midband fre-

quencies is:

The total loop gain is the product of the modulator gain

and the error amplifier gain at fCO should be equal to 1,

as follows:

GainMOD x GainE/A = 1

So:

Solving for CI:

6) For those situations where fLC < fCO < fESR < fSW/2,

as with low-ESR tantalum capacitors, the compen-

sator’s second pole (fP2) should be used to cancel

fESR. This provides additional phase margin. On the

system Bode plot, the loop gain maintains its

+20dB/decade slope up to 1/2 of the switching fre-

quency verses flattening out soon after the 0dB

crossover. Then set:

fP2 = fESR

If a ceramic capacitor is used, then the capacitor ESR

zero, fESR, is likely to be located even above one-half of

the switching frequency, that is fLC < fCO < fSW/2 <

fESR. In this case, the frequency of the second pole

(fP2) should be placed high enough not to significantly

erode the phase margin at the crossover frequency.

For example, fP2 can be set at 5 x fCO, so that its con-

tribution to phase loss at the crossover frequency fCO is

only about 11°:

fP2 = 5 x fCO

Once fP2 is known, calculate RI:

7) Place the second zero (fZ2) at 0.2 x fCO or at fLC,

whichever is lower, and calculate R1 using the fol-

lowing equation:

8) Place the third pole (fP3) at 1/2 the switching fre-

quency and calculate CCF from:

9) Calculate R2 as:

where VFB = 0.6V (typ).

R [k] R [k]

[V]

[V] V

[V]

OUT_

ΩΩ

=×

−

C[ F]

0.5 f

[MHz] R [k ]

××

()

πΩ

R[k ]

[kHz] C [ F]

Ω=

××

πμ

R[k ]

[kHz] C [ F]

Ω=

××

πμ

C pF]

[kHz] L[ H] C

[ F]

4 R [k ]

OUT

[

××

()

πμ

[kHz])

[ F] L[ H]

[kHz] C [ F] R [k ]

OUT

××

πμ

Gain

[kHz] C [ F] R [k ]

E/A

=×

πμ

Gain

[MHz])

L[ H] C

[ F]

MOD

OUT

=×

××

πμ

C[ F]

R [k ] 0.5 f

[kHz]

FLC

××

[MHz]

L[ H]

OUT

≈

××

πμ

[

[kHz]

≤

Dual, 4A/2A, 4MHz, Step-Down DC-DC

Regulator with Tracking/Sequencing Capability

MAX15021

Maxim Integrated

××

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