LX1562IDM Microsemi Analog Mixed Signal Group, LX1562IDM Datasheet - Page 18

• Manufacturer Part Number: LX1562IDM
• Description: IC CONTROLLER PFC 13.1V 8SOIC
• Manufacturer: Microsemi Analog Mixed Signal Group

Specifications of LX1562IDM

• Mode: Discontinuous Conduction (DCM)
• Frequency - Switching: 1.7MHz
• Current - Startup: 200µA
• Voltage - Supply: 11 V ~ 25 V
• Operating Temperature: 0°C ~ 100°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

LX1562/1563
18
OUTPUT VOLTAGE REQUIREMENT
INDUCTOR PEAK CURRENT
It can be shown by referring to Figure 33 that the inductor peak
current is always twice the average input current.
FIGURE 32 — NORMALIZED OPERATING FREQUENCY vs.
0.15
0.05
I
I
I
I
0.2
0.1
IN(t)
IN
INpeak
LP
0.3
=
=
= I
= Inductor peak current at peak input voltage.
FIGURE 33 — INDUCTOR CURRENT
1
T
P
0.4
AVE [ I
=
(I
I
OFF-TIME DUTY CYCLE
2
LP
(D') Off Time Duty Cycle
L
) (T)
2
0.5
L
(t) ]
D' =
f =
0.6
=
P R O D U C T D A T A B O O K 1 9 9 6 / 1 9 9 7
4 LP
Inductor Peak
Current Envelope
2 V
S
V
V
Average
AC Input Current
O
(continued)
I
E C O N D
2
O
O
²
L
AC
f
n
f
= (1 - D') D'²
n
0.7
P
A P P L I C A T I O N I N F O R M A T I O N
-G
R O D U C T I O N
0.8
E N E R AT I O N
T
ON
0.9
I
L
T
OFF
1.0
P
O W E R
D
Maximum peak input current can be calculated using:
INDUCTOR DESIGN
The inductor value is calculated assuming a 50KHz operating
frequency at the nominal AC voltage using the following equation:
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.
product method (AP) or the K
for selecting proper core size. In this example, we apply the K
approach using the following steps:
Step 1: Calculate K
A T A
F
Once the inductance is known, we can either use the area
A C T O R
L
L
choose T = 20µsec (50kHz)
K
1
1
g
=
I
where:
assuming:
I
I
=
=
P
P
LP/min AC
K
where:
Assume: P
S
.95 (
1.724 * 10
g
=
=
C
H E E T
V
=
(.95)(141)
O N T R O L L E R
O
V
P
1.6
- V
4 P
2 x 80
O
230 - 120 2
V
CU
= 2 * 1.2 = 2.4A
P
P
2P
230
O
T V
V
(
L
B
I
P
O
LP
g
-8
P
1
CU
CU
= 95%, P
L
Converter efficiency
Peak AC input voltage
using
P
1
B
2
I
= 1.6W (2% of total output)
LP
) 20 * 10
= 1.2A
4 * 80
450 * 10
2
Required inductance
1.724 * 10
Maximum flux density
Maximum peak inductor current
Maximum copper dissipation
where:
)
2
g
O
(based on copper losses method),
0.15
= 80W, V
-6
-6
* (120 2)
* (2.4)
V
V
T
P
-8
O
O
P
m
Efficiency
Output DC voltage
Peak AC input voltage
Switching period
Output Power
2
Pmin
2
2
= 100 2 = 141
= 3.21 * 10
= 448µH
Copyright © 1996
Rev. 1.3a
-12
m
5
8/30
g