NCP1422LEDGEVB ON Semiconductor, NCP1422LEDGEVB Datasheet - Page 2
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NCP1422LEDGEVB
Manufacturer Part Number
NCP1422LEDGEVB
Description
EVAL BOARD FOR NCP1422LED
Manufacturer
ON Semiconductor
Datasheets
1.NCP1422MNR2G.pdf
(14 pages)
2.NCP1421LEDEVB.pdf
(6 pages)
3.NCP1422LEDGEVB.pdf
(1 pages)
Specifications of NCP1422LEDGEVB
Design Resources
NCP1422LEDGEVB BOM NCP1422LEDGEVB Schematic NCP1422LEDEVB Gerber Files
Current - Output / Channel
800mA
Outputs And Type
1, Non-Isolated
Voltage - Output
500 ~ 700 mV
Voltage - Input
3.6V
Utilized Ic / Part
NCP1421
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Features
-
Lead Free Status / Rohs Status
Lead free / RoHS Compliant
Other names
NCP1422LEDEVBOS
NCP1422LEDEVBOS
NCP1422LEDGEVBOS
NCP1422LEDEVBOS
NCP1422LEDGEVBOS
Design Steps
components for this circuit. (R2, R3, R4, L1) This shows the
600 mA version as an example:
(Find value of V
equal dependence on V
reference and R4. One could increase the output voltage by
making the voltage across R4 (V
power dissipation in R4 by lowering V
must equal 1.2 V, and that is 0.6 V
to the above calculated value. Choose R2 = 475 kW.
one−cell Li−ion battery pack or a 3−cell NiMH pack so the
1000
The following steps show how to determine the critical
Step 1: Let LED current = I
Step 2: From the LED datasheet, let V
Step 3: Let R3 = 100 kW
Step 4: Let V
Step 5: For I
Step 6: Now, V
Step 7: So, R2 = (V
Step 8: Then choose a standard value of R2 which is close
Step 9: Pick input voltage range. These circuits assume a
(3.5/0.6) * 100 kW − 100 kW = 483 kW
900
800
700
600
500
400
300
200
100
0
3.0
Figure 2. Output Current vs. Input Voltage
3.2
D
R4
600 mA
200 mA
F
800 mA
= 600 mA and V
R4
at 600 mA).
= 0.5 * V
plus the divided voltage off of the LED
F
3.4
INPUT VOLTAGE (V)
/(V
F
ref
variation and tolerance of the
ref
− V
D
100
Figure 4. Electrical to Optical Efficiency vs. Input Voltage
3.6
95
90
85
80
75
70
65
60
55
50
which is 0.6 V. This places
= 600 mA
3.0
R4
R4
R4
)) * R3 − R3 =
200 mA
800 mA
600 mA
V
= 0.6 V, R4 = 1.0 W.
) larger or decrease
F
R4
= 3.5 V @ 600 mA
3.8
F
.
3.2
= 3.5 V
4.0
3.4
INPUT VOLTAGE (V)
http://onsemi.com
AND8171/D
4.2
3.6
2
V
input voltage is assumed to be 3.6 V and has been optimized
around this point.
V
because this circuit decreases LED current as V
from the designed value. This is shown by the following
equation: I
it increases current as V
but then the difference between V
peak current is reduced.
determine the appropriate L1, C1, and C2. For this
application, 6.8
over the load and line range.
this circuit V
(1−V
840 mA
inductor with an I
F
100
F
Step 10: Determine output voltage. Output voltage will be
Step 11: Use the NCP1421 or NCP1422 datasheet to
Step 12: Determine the inductor saturation current. For
Step 13: Add 20% margin to this I
95
90
85
80
75
70
65
60
55
50
= 3.5 V @ 600 mA
3.8
+ V
3.0
in
200 mA
600 mA
/V
Figure 3. Converter Efficiency vs. Input Voltage
R4
800 mA
out
= 4.1 V One can use the 3.6 V as V
D
4.0
). Therefore I
3.2
= 1/R4*(V
in
m
min = 3 V: I
H, 22
sat
4.2
> 1.0 A.
3.4
INPUT VOLTAGE (V)
m
F
ref
F, and 22
Lavg
decreases from the designed value,
− V
Lavg
= 600 mA/(1−(1−V
3.6
F
*(R3/R2 + R3)) Conversely
m
= I
F were found to work well
in
V
and V
out
F
= 3.5 V @ 600 mA
3.8
/ (1−D) where D =
Lavg
out
in
is less, so the
chosen above
and pick an
4.0
F
in
increases
/V
out
)) =
4.2