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参数资料
型号： LX1562IDM-TR
厂商： MICROSEMI CORP-ANALOG MIXED SIGNAL GROUP
元件分类： 稳压器
英文描述： 0.5 A POWER FACTOR CONTROLLER, PDSO8
封装： ROHS COMPLIANT, PLASTIC, SOIC-8
文件页数： 10/25页
文件大小： 430K
代理商： LX1562IDM-TR
SECOND-GENERATION POWER FACTOR C ONTROLLER
LX1562/1563
PRODUCT DA T ABOOK 1996/1997
Copyright 1996
Rev. 1.3a
8/30
18
P RODUCTION D ATA
S HEET
APPLICA TION INFORMA TION
Maximum peak input current can be calculated using:
I
P
=
where:
η≡ Converter efficiency
V
P
≡ Peak AC input voltage
assuming:
η = 95%, P
O = 80W, VPmin = 100 √2 = 141
I
P
=
= 1.2A
I
LP/min AC
= 2 * 1.2 = 2.4A
2P
O
ηV
P
OUTPUT VOLTAGE REQUIREMENT (continued)
0.05
(D') Off Time Duty Cycle
0.2
(f
n)
Normalized
Operating
Frequency
0.1
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.15
f
n = (1 - D') D'
D' =
2 V
AC
V
O
η V
O
4 LP
O
f =
f
n
INDUCTOR PEAK CURRENT
It can be shown by referring to Figure 33 that the inductor peak
current is always twice the average input current.
Inductor Peak
Current Envelope
Average
AC Input Current
I
L
T
ON
T
OFF
FIGURE 33 — INDUCTOR CURRENT
I
IN(t)
=
AVE [ I
L (t) ]
I
IN
==
I
INpeak =
I
P
=
I
LP
= Inductor peak current at peak input voltage.
I
L
2
(I
L) (T)
2
1
T
I
LP
2
Σ
FIGURE 32 — NORMALIZED OPERATING FREQUENCY vs.
OFF-TIME DUTY CYCLE
INDUCTOR DESIGN
The inductor value is calculated assuming a 50KHz operating
frequency at the nominal AC voltage using the following equation:
L
1
=
where:
η≡ Efficiency
V
O ≡ Output DC voltage
V
P ≡ Peak AC input voltage
T
≡ Switching period
P
O ≡ Output Power
L
1ILP
2
B
P
CU
2
1.724 * 10
-8
1.6
450 * 10
-6 * (2.4)2
0.15
2 x 80
(.95)(141)
η
T V
P
2
4 P
O
V
O - VP
V
O
L
1
=
= 448H
choose T = 20sec (50kHz)
Figure 32 shows that at nominal AC line (D' = 0.74) the normalized
frequency is 0.142 and dropping to 0.13 at maximum line
condition. This translates to a 10% drop in operating frequency
which is still well above the audible range.
Once the inductance is known, we can either use the area
product method (AP) or the K
g (based on copper losses method),
for selecting proper core size. In this example, we apply the K
g
approach using the following steps:
Step 1: Calculate K
g using
K
g =(
)2
where:
L
1
≡ Required inductance
≡ 1.724 * 10-8 m
B
≡ Maximum flux density
I
LP
≡ Maximum peak inductor current
P
CU ≡ Maximum copper dissipation
Assume: P
CU = 1.6W (2% of total output)
K
g =
= 3.21 * 10
-12 m5
.95 (
) 20 * 10
-6 * (120√2)2
4 * 80
230 - 120
√2
230
O
B
SO
LETE
PRO
D
UCT
NOT
RECOMMENDED
FOR
NEW
DESIGNS
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