参数资料
| 型号： | NCP1421DMR2 |
| 厂商： | ON Semiconductor |
| 文件页数： | 11/14页 |
| 文件大小： | 0K |
| 描述： | IC REG BST SYNC ADJ 0.6A 8MICRO |
| 产品变化通告： | LTB Notification 03/Jan/2008 |
| 标准包装： | 1 |
| 类型： | 升压（升压） |
| 输出类型： | 可调式 |
| 输出数： | 1 |
| 输出电压： | 1.5 V ~ 5 V |
| 输入电压： | 1 V ~ 5 V |
| 频率 - 开关： | 1.2MHz |
| 电流 - 输出： | 600mA |
| 同步整流器： | 是 |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 8-TSSOP，8-MSOP（0.118"，3.00mm 宽） |
| 包装： | 剪切带 (CT) |
| 供应商设备封装： | Micro8? |
| 其它名称： | NCP1421DMR2OSCT |
NCP1421
TYPICAL APPLICATION CIRCUIT
V IN
C1
22 m F
L
6.5 m H
Shutdown
Open Drain
Input
R2 200 k
C4
10 p*
R3
220 k
R1
350 k
1
2
3
NCP1421
FB
LBI/EN
LBO
OUT
LX
GND
8
7
6
C2
22 m F
+
V OUT =3.3 V
500 mA
R4
330 k
4
REF
BAT
5
C3
Low Battery
200 nF
Open Drain
Output
*Optional
Figure 26. Typical Application Schematic for 2 Alkaline Cells Supply
GENERAL DESIGN PROCEDURES
Switching mode converter design is considered a
complicated process. Selecting the right inductor and
Determine the Steady State Duty Ratio, D, for typical
V IN . The operation is optimized around this point:
capacitor values can allow the converter to provide
optimum performance. The following is a simple method
based on the basic first?order equations to estimate the
VOUT
VIN
+
1
1 * D
+ 1 * 2.4 V + 0.273
inductor and capacitor values for NCP1421 to operate in
Continuous Conduction Mode (CCM). The set component
D + 1 *
VIN
VOUT
3.3 V
ILAVG + OUT + 500 mA + 688 mA
R1 + R2
* 1
L +
+
+ 6.5 m H
2 IRIPPLE * P
R1 + 200 k 3.3 V * 1 + 350 k
COUT u
values can be used as a starting point to fine tune the
application circuit performance. Detailed bench testing is
still necessary to get the best performance out of the circuit.
Design Parameters:
V IN = 1.8 V to 3.0 V, Typical 2.4 V
V OUT = 3.3 V
I OUT = 500 mA (600 mA max)
V LB = 2.0 V
V OUT?RIPPLE = 45 mV p?p at I OUT = 500 mA
Calculate the feedback network:
Select R2 = 200 k
VOUT
VREF
1.20 V
Calculate the Low Battery Detect divider:
Determine the average inductor current, I LAVG, at
maximum I OUT :
I
1 * D 1 * 0.273
Determine the peak inductor ripple current, I RIPPLE?P,
and calculate the inductor value:
Assume I RIPPLE?P is 20% of I LAVG . The inductance of the
power inductor can be calculated as follows:
VIN tON 2.4 V 0.75 m S
2 (137.6 mA)
A standard value of 6.5 m H is selected for initial trial.
Determine the output voltage ripple, V OUT?RIPPLE, and
calculate the output capacitor value:
V OUT?RIPPLE = 40 mV P?P at I OUT = 500 mA
IOUT tON
VOUT * RIPPLE * IOUT ESRCOUT
R3 + R4
COUT u
+ 18.75 m F
V LB = 2.0 V
Select R4 = 330 k
VLB
VREF
* 1
where t ON = 0.75 uS and ESR COUT = 0.05 W ,
500 mA 0.75 m S
45 mV * 500 mA 0.05 W
2.0 V * 1 + 220 k
R3 + 300 k
1.20 V
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