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参数资料
| 型号： | FAN5061M |
| 厂商： | FAIRCHILD SEMICONDUCTOR CORP |
| 元件分类： | 稳压器 |
| 英文描述： | SWITCHING CONTROLLER, 345 kHz SWITCHING FREQ-MAX, PDSO20 |
| 封装： | SOIC-20 |
| 文件页数： | 8/18页 |
| 文件大小： | 149K |
| 代理商： | FAN5061M |
FAN5061
16
Pr
eliminar
y
Specication
FAN5061 Evaluation Board
Fairchild provides an evaluation board to verify the system
level performance of the FAN5061. It serves as a guide to
performance expectations when using the supplied external
components and PCB layout. Please call the marketing
department at 650-962-7833 for an evaluation board.
Additional Information
For additional information contact Fairchild Semiconductor’s
Analog & Mixed Signal Products Group Marketing Department
at 650-962-7833.
Appendix
Worst-Case Formulae for the Calculation of
Cout, R5, and Cin (Circuit of Figure 1 only)
The following formulae design the FAN5061 for worst-case
operation, including initial tolerance and temperature depen-
dence of all of the IC parameters (initial setpoint, reference
tolerance and tempco, active droop tolerance, current sensor
gain), the initial tolerance and temperature dependence of
the MOSFET, and the ESR of the capacitors. The following
information must be provided:
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);
ESR, the ESR of the ouput caps, per cap (44m
for the
Sanyo parts shown in this datsheet);
RD, the on-resistance of the MOSFET (20m for the
FDB6030);
R
D, the tolerance of the current sensor (usually about 67%
for MOSFET sensing, including temperature).
Irms, the rms current rating of the input caps (2A for the
sanyo parts shown in this datasheet.)
Number of capacitors needed fo Cout = the greater of:
Example: Suppose that the transient limits are ±134mV, cur-
rent I is 14.2A, and the nominal voltage is 2.000V, using
MOSFET current sensing and the usual caps. We have VT+ =
|VT-| = 0.134, IO = 14.2, Vnom = 2.000, and RD = 0.67. We
calculate:
Since X > Y, we choose X, and round up to nd we need 5
capacitors for COUT.
2
Irms
Vin
Vnom
Vin
Vnom
*
IO
Cin
–
=
50 * 10
-6
IO* RD * (1 + RD) * 1.10
R5 =
VT-
ESR * IO
X=
18 * R5 * 1.1
14400 * IO * RD
Y
or
VT+ –0.004 * Vnom +
=
2
5
2.000
5
2.000
*
14.2
Cin
2
3.47
4 caps
=
–
=
10.4K
50 * 10-6
14.2 * 0.020 * (1 + 0.67) * 1.0
R5
4.28
18 * 10400 * 1.1
14400 * 14.2 * 0.020
0.134 – 0.004 * 2.000
0.044 * 14.2
Y
=
+
=
4.66
0.134
0.044 * 14.2
X
=
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