FAN5070 Datasheet, PDF (12/14 Pages) Fairchild Semiconductor – High Performance Programmable Synchronous DC-DC Controller

FAN5070
Appendix
Worst-Case Formulae for the Calculation of
C_in, C_out, R5, R7 and R_offset (Circuits similar
to Figure 1 only)
The following formulae design the FAN5070 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:
V_S+, the value of the positive static voltage limit;
|V_S-|, the absolute value of the negative static voltage limit;
V_T+, the value of the positive transient voltage limit;
|V_T-|, the absolute value of the negative transient voltage
limit;
I_O, the maximum output current;
V_nom, the nominal output voltage;
V_in, the input voltage (typically 5V);
I_rms, the ripple current rating of the input capacitors, per cap
(2A for the Sanyo parts shown in this data sheet);
R_D, 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 *
C_in =
2
V_nom
V_nom
–
V_in
V_in
Irms
R_offset = ^{VS+ – .014 * Vnom – .029 * 1KΩ}
.029 * Vnom
R7 = ^{IO* RD * (1 + ∆RD)}
45 * 10^-6
Number of capacitors needed for C_OUT = the greater of:
ESR * I_O
X=
V_T- + V_S+ – .024 * V_nom
or
Y=
V_T+ – V_S+ +
ESR * I_O
14400 * I_O * R_D
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 V_S+ = 0.089, |V_S-| = 0.079, V_T+ = |V_T-| = 0.134, I_O =
14.2, V_nom = 2.000, and ∆R_D = 1.67. We calculate:
14.2 *
C_in =
2
2.000 2.000
–
5
5
= 3.47 ⇒ 4 caps
2
R_offset = ^{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 C_OUT.
A detailed explanation of this calculation may be found in
Applications Bulletin AB-24.
14400 * I_O* R_D * (1 + ∆R_D) *1.1
R5 =
18 * (V_S+ + V_S- – .024 * Vnom)
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REV. 1.0.1 8/7/01