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參數(shù)資料
| 型號(hào)： | NCP9003SNT1G |
| 廠商： | ON SEMICONDUCTOR |
| 元件分類： | 顯示驅(qū)動(dòng)器 |
| 英文描述： | Compact Backlight LED Boost Driver |
| 中文描述： | LED DISPLAY DRIVER, PDSO5 |
| 封裝： | LEAD FREE, TSOP-5 |
| 文件頁(yè)數(shù)： | 9/22頁(yè) |
| 文件大?。?/td> | 294K |
| 代理商： | NCP9003SNT1G |
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NCP9003
http://onsemi.com
9
Output Load Drive
In order to optimize the built
in Boost capabilities, one
shall operate the NCP9003 in the continuous output current
mode. Such a mode is achieved by using and external
reservoir capacitor (see Table 1) across the LED.
At this point, the peak current flowing into the LED
diodes shall be within the maximum ratings specified for
these devices. Of course, pulsed operation can be achieved,
due to the EN signal Pin 4, to force high current into the
LED when necessary.
The Schottky diode D1, associated with capacitor C2
(see Figure 9), provides a rectification and filtering
function.
When a pulse
operating mode is acceptable:
A PWM mode control can be used to adjust the output
current range by means of a resistor and a capacitor
connected across FB pin. On the other hand, the
Schottky diode can be removed and replaced by at
least one LED diode, keeping in mind such LED shall
sustain the large pulsed peak current during the
operation.
TYPICAL OPERATING CHARACTERISTICS
0
2.50
10
20
30
40
50
60
70
80
90
100
3.00
3.50
4.00
V
bat
(V)
4.50
5.00
5.50
Y
5 LED/10 mA
3 LED/10 mA
4 LED/10 mA
2 LED/10 mA
0
2.50
10
20
30
40
50
60
70
80
90
100
3.00
3.50
4.00
V
bat
(V)
4.50
5.00
5.50
Y
5 LED/4 mA
3 LED/4 mA
4 LED/4 mA
2 LED/4 mA
Figure 10. Overall Efficiency vs. Power Supply @
I
out
= 4.0 mA, L = 22 H
Figure 11. Overall Efficiency vs. Power Supply @
I
out
= 10 mA, L = 22 H
0
2.50
10
20
30
40
50
60
70
80
90
100
3.00
3.50
4.00
Vbat (V)
4.50
5.00
5.50
Y
5 LED/20 mA
3 LED/20 mA
4 LED/20 mA
2 LED/20 mA
0
2.50
10
20
30
40
50
60
70
80
90
100
3.00
3.50
4.00
Vbat (V)
4.50
5.00
5.50
Y
5 LED/15 mA
3 LED/15 mA
4 LED/15 mA
2 LED/15 mA
Figure 12. Overall Efficiency vs. Power Supply @
I
out
= 15 mA, L = 22 H
Figure 13. Overall Efficiency vs. Power Supply @
I
out
= 20 mA, L = 22 H
Yield = f(V
bat
) @ I
out
= 4.0 mA/L
out
= 22 H
Yield = f(V
bat
) @ I
out
= 10 mA/L
out
= 22 H
Yield = f(V
bat
) @ I
out
= 15 mA/L
out
= 22 H
Yield = f(V
bat
) @ I
out
= 20 mA/L
out
= 22 H
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