NCP1351LEDGEVB ON Semiconductor, NCP1351LEDGEVB Datasheet - Page 4

• Manufacturer Part Number: NCP1351LEDGEVB
• Description: EVAL BOARD FOR NCP1351LEDG
• Manufacturer: ON Semiconductor

Specifications of NCP1351LEDGEVB

• Design Resources: NCP1351 EVB BOM NCP1351LEDGEVB Gerber Files NCP1351LED EVB Schematic
• Current - Output / Channel: 700mA
• Outputs And Type: 1, Isolated
• Voltage - Output: 33V
• Features: Short-Circuit Protection
• Voltage - Input: 85 ~ 265 V
• Utilized Ic / Part: NCP1351
• Core Chip: NCP1351
• Topology: Flyback
• No. Of Outputs: 1
• Output Current: 700mA
• Output Voltage: 33V
• Development Tool Type: Hardware - Eval/Demo Board
• Leaded Process Compatible: Yes
• Rohs Compliant: Yes
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• For Use With/related Products: NCP1351LEDG
• Other names: NCP1351LEDGEVBOS

primary of the inductor (above) if we define a term k equal
to;
assuming operation in boundary conduction mode:
I
VALLEY
September 2008, Rev. 2
I
I
Looking at the waveform of the current flowing in the
And use the equation:
Then we can determine the inductance we require.
If k = 2 then we are in boundary conduction mode as
the ripple current equals twice the average pulse
current, so setting k to 2:
Thus we can now find the primary ripple current
The average input current, I
AVE
I
k
Δ
PK
L
L
1
I
=
I
=
=
L
Δ
100
(
=
=
V
I
I
1
IN
V
283
L
f
(
(min)
×
IN
δT
80
SW
...........................................................(Eq.8)
10
(min)
SW
×
kP
L
δ
×
10
3
80
MAX
T
. 0
IN
×
ON
−
47
×
2
6
)
0 .
×
. 0
2
=
)
100
2
47
×
V
.........................................(Eq.9)
T
25
IN
SW
×
Lf
(min)
10
=
SW
AVE
283
δ
3
max
, is:
=
. 1
μ
32
H
A
............(Eq.10)
........(Eq.11)
∆I
L
t
DN06040/D
www.onsemi.com
peak primary current is 1.32 A.
sense resistor for dissipation purposes. For a stepped-
sawtooth waveform of this type the equation is:
allowing for a drop across the resistor of 0.8 V:
resistors typically cost more.
offset resistor; this has a bias current of 270 µA in it so we
can determine the resistor value:
The average pulse current, I
Demonstrating that ∆I
We can calculate the RMS current in the MOSFET and
Thus:
We can also determine the current sense resistor,
The total power dissipation is:
Two 1.2 Ω resistors in parallel will be used as sub 1 Ω
The threshold voltage for the current sense is set by an
I
I
I
I
......................................................................... (Eq.15)
R
P
R
1
RMS
RMS
AVE
D
OFFSET
SENSE
(
=
sense
δ
I
=
=
=
=
AVE
max
)
V
=
. 0
526
I
=
=
≅
1
IN
P
V
665
170
V
I
=
IN
I
(min)
DROP
δ
I
RMS
SENSE
mA
PK
BIAS
. 0
. 0
×
1
mW
313
2
=
47
R
+
=
. 0
=
SENSE
25
80
1
3
. 1
47
0
=
270
⎛ Δ
⎜ ⎜
⎝
32
8 .
L
=
662
2
×
does equal twice I
=
0
313
I
I
×
=
1
L
8 .
. 0
10
1
. 0
⎞
⎟ ⎟
⎠
mA
1
+
526
2
, is:
mA
61
−
6
1
3
........................ (Eq.14)
≅
...................... (Eq.13)
⎛
⎜
⎝
Ω
2
2
×
................... (Eq.12)
3
×
................. (Eq.16)
0 .
. 0
. 1
. 0
61
32
k
665
Ω
........ (Eq.17)
1
...... (Eq.18)
and that the
⎞
⎟
⎠
2
4