FAN5071 데이터 시트보기 (PDF) - Fairchild Semiconductor

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FAN5071

High Performance Programmable Synchronous DC-DC Controller for Multi-Voltage Platforms

Fairchild Semiconductor 

PRODUCT SPECIFICATION

FAN5071

Appendix

Worst-Case Formulae for the Calculation of

Cin, Cout, R5, R7 and Roffset (Circuits similar

to Figure 1 only)

The following formulae design the FAN5071 for worst-case

operation, including initial tolerance and temperature depen-

dence of all of the IC parameters (initial setpoint, reference

tolerance and tempco, internal droop impedance, current

sensor gain), the initial tolerance and temperature depen-

dence of the MOSFET, and the ESR of the capacitors. The

following information must be provided:

VS+, the value of the positive static voltage limit;

|VS-|, the absolute value of the negative static voltage limit;

VT+, the value of the positive transient voltage limit;

|VT-|, the absolute value of the negative transient voltage

limit;

IO, the maximum output current;

Vnom, the nominal output voltage;

Vin, the input voltage (typically 5V);

Irms, the ripple current rating of the input capacitors, per cap

(2A for the Sanyo parts shown in this data sheet);

RD, the resistance of the current sensor (usually the MOSFET);

∆RD, the tolerance of the current sensor (usually about 67%

for MOSFET sensing, including temperature); and

ESR, the ESR of the output capacitors, per cap (44mΩ for

the Sanyo parts shown in this data sheet).

IO *

Cin =

2

Vnom

Vnom

–

Vin

Vin

Irms

Roffset = VS+ – .014 * Vnom – .029 * 1KΩ

.029 * Vnom

R7 = IO* RD * (1 + ∆RD)

45 * 10-6

Number of capacitors needed for COUT = the greater of:

ESR * IO

X=

VT- + VS+ – .024 * Vnom

or

Y=

VT+ – VS+ +

ESR * IO

14400 * IO * RD

18 * R5 * 1.1

Example: Suppose that the static limits are +89mV/-79mV,

transient limits are ±134mV, current I is 14.2A, and the nom-

inal voltage is 2.000V, using MOSFET current sensing. We

have VS+ = 0.089, |VS-| = 0.079, VT+ = |VT-| = 0.134, IO =

14.2, Vnom = 2.000, and ∆RD = 1.67. We calculate:

14.2 *

Cin =

2

2.000 2.000

–

5

5

= 3.47 ⇒ 4 caps

2

Roffset = 0.089 – .014 * 2.000 – .029 *1000 = 15.8Ω

0.29 + 2.000

R7 = 14.2 * 0.020 * (1 + 0.67) = 10.5KΩ

45 * 10-6

R5 = 14400 * 14.2 * 0.020 * (1 + 0.67) * 1.1 = 3.48KΩ

18 * (0.089 + 0.079 – .024 * 2.000)

0.044 * 14.2

X=

= 3.57

0.134 + 0.089 – .024 * 2.00

Y=

0.044 * 14.2

14400 * 14.2 * 0.020

0.134 – 0.089 +

18 * 3640 * 1.1

= 6.14

Since Y > X, we choose Y, and round up to find we need 7

capacitors for COUT.

A detailed explanation of this calculation may be found in

Applications Bulletin AB-24.

14400 * IO* RD * (1 + ∆RD) *1.1

R5 =

18 * (VS+ + VS- – .024 * Vnom)

REV. 1.0.4 1/29/02

13

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