3412EFE 데이터 시트보기 (PDF) - Linear Technology

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3412EFE

2.5A, 4MHz, Monolithic Synchronous Step-Down Regulator

Linear Technology 

LTC3412

APPLICATIO S I FOR ATIO

The junction temperature, TJ, is given by:

TJ = TA + TR

where TA is the ambient temperature.

As an example, consider the LTC3412 in dropout at an

input voltage of 3.3V, a load current of 2.5A and an

ambient temperature of 70°C. From the typical perfor-

mance graph of switch resistance, the RDS(ON) of the P-

channel switch at 70°C is approximately 97mΩ. There-

fore, power dissipated by the part is:

PD = (ILOAD2)(RDS(ON)) = (2.5A)2(97mΩ) = 0.61W

For the TSSOP package, the θJA is 37.6°C/W. Thus the

junction temperature of the regulator is:

TJ = 70°C + (0.61W)(37.6°C/W) = 93°C

which is below the maximum junction temperature of

125°C.

Note that at higher supply voltages, the junction tempera-

ture is lower due to reduced switch resistance (RDS(ON)).

Checking Transient Response

The regulator loop response can be checked by looking at

the load transient response. Switching regulators take

several cycles to respond to a step in load current. When

a load step occurs, VOUT immediately shifts by an amount

equal to ΔILOAD(ESR), where ESR is the effective series

resistance of COUT. ΔILOAD also begins to charge or

discharge COUT generating a feedback error signal used by

the regulator to return VOUT to its steady-state value.

During this recovery time, VOUT can be monitored for

overshoot or ringing that would indicate a stability prob-

lem. The ITH pin external components and output capaci-

tor shown in Figure 1 will provide adequate compensation

for most applications.

Design Example

As a design example, consider using the LTC3412 in an

application with the following specifications: VIN = 2.7V to

4.2V, VOUT = 2.5V, IOUT(MAX) = 2.5A, IOUT(MIN) = 10mA, f

= 1MHz. Because efficiency is important at both high and

low load current, Burst Mode operation will be utilized.

First, calculate the timing resistor:

ROSC

=

3.23 •1011

1• 106

−

10k

=

313k

Use a standard value of 309k. Next, calculate the inductor

value for about 40% ripple current at maximum VIN:

L

=

⎛ 2.5V ⎞

⎝⎜ (1MHz)(1A)⎠⎟

⎛⎝⎜1−

2.5V

4.2V

⎞⎠⎟

=

1.01μH

Using a 1μH inductor, results in a maximum ripple current

of:

ΔIL

=

⎛

⎝⎜

2.5V ⎞

(1MHz)(1μH)⎠⎟

⎛⎝⎜1−

2.5V

4.2V

⎞⎠⎟

=

1.01A

COUT will be selected based on the ESR that is required to

satisfy the output voltage ripple requirement and the bulk

capacitance needed for loop stability. In this application,

two tantalum capacitors will be used to provide the bulk

capacitance and a ceramic capacitor in parallel to lower the

total effective ESR. For this design, two 100μF tantalum

capacitors in parallel with a 10μF ceramic capacitor will be

used. CIN should be sized for a maximum current rating of:

IRMS=

(2.5A)

⎛⎝⎜

2.5V

4.2V

⎞⎠⎟

4.2V

2.5V

−

1

=

1.23ARMS

Decoupling the PVIN and SVIN pins with a 22μF ceramic

capacitor and a 220μF tantalum capacitor is adequate for

most applications.

The burst clamp and output voltage can now be pro-

grammed by choosing the values of R1, R2 and R3. The

voltage on the MODE pin will be set to 0.32V by the resistor

divider consisting of R2 and R3. A burst clamp voltage of

0.32V will set the minimum inductor current, IBURST, as

follows:

IBURST

=

(0.32V

−

0.2V)⎛⎝⎜

3.75V

0.8V

⎞⎠⎟

=

563mA

3412fb

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