LT1248IS Linear Technology, LT1248IS Datasheet - Page 10

IC PFC CTRLR AVERAGE CURR 16SOIC

LT1248IS

Manufacturer Part Number
LT1248IS
Description
IC PFC CTRLR AVERAGE CURR 16SOIC
Manufacturer
Linear Technology
Datasheet

Specifications of LT1248IS

Mode
Average Current
Frequency - Switching
300kHz
Current - Startup
250µA
Voltage - Supply
15.5 V ~ 27 V
Operating Temperature
-40°C ~ 125°C
Mounting Type
Surface Mount
Package / Case
16-SOIC (3.9mm Width)
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
LT1248IS
Manufacturer:
LINEAR/凌特
Quantity:
20 000
Part Number:
LT1248IS#PBF
Manufacturer:
LINEAR/凌特
Quantity:
20 000
Part Number:
LT1248IS#TRPBF
Manufacturer:
LINEAR/凌特
Quantity:
20 000
LT1248
A
For 382V V
In Figure 6, a new technique for supply voltage eliminates
the need for an extra inductor winding. It uses capacitor
charge transfer to generate a constant current source
which feeds a Zener diode. Current to the Zener is equal
to (V
switching frequency. For V
1000pF, and f = 100kHz, Zener current will be 36mA. This
is enough to operate the LT1248, including the FET gate
drive. Normally soft-start is not needed because the
LT1248 has overcurrent limit and overvoltage protection.
If soft-start is used with a 0.01 F capacitor on SS pin,
V
hold V
C4 to 100 F ensures start-up, but start-up time will be
extended if the same 90k trickle charge resistor is used.
Output Capacitor
The peak-to-peak 120Hz output ripple is determined by:
For 180 F at 300W load, I
V
higher capacitance should be used. The selection of the
output capacitor should also be based on the operating
ripple current through the capacitor. The ripple current
can be divided into three major components. The first is at
120Hz; it’s RMS value is related to the DC load current as
follows:
The second component contains the PF switching fre-
quency ripple current and its harmonics. Analysis of the
ripple is complicated because it is modulated with a 120Hz
signal. However computer numerical integration and Fou-
rier analysis approximate the RMS value reasonably close
to the bench measurements. The RMS value is about 0.82A
at a typical condition of 120VAC, 200W load. This ripple is
line-voltage dependent, and the worst case is at low line.
10
OUT
P-P
PPLICATI
V
V
where I
I
I
1RMS
2RMS
OUT
P-P
OUT
= 2 • 0.78A • 7.4 = 11.5V. If less ripple is desired,
ramps up slower during start-up. Then C4 has to
CC
/(V
= (2) (I
= 0.82A at 120VAC, 200W
longer, and the circuit may not start. Increasing
– V
CC
LOAD(DC)
OUT
0.71
– 2V) = N
Z
)(C)(f), where V
LOAD(DC)
and 18V V
I
O
LOAD(DC)
Z: capacitor impedance at 120Hz.
: DC load current.
U
P
)(Z)
S
/N
CC
S
LOAD(DC)
.
I FOR ATIO
, Np/Ns 19.
U
OUT
Z
is Zener voltage and f is
= 382V, V
= 300W/385V = 0.78A,
W
Z
= 18V, C =
U
The third component is the switching ripple from the load,
if the load is a switching regulator.
For the United Chemicon KMH 400V capacitor series,
ripple current multiplier for currents at 100kHz is 1.43. The
equivalent 120Hz ripple current can be then found:
For a typical system that runs at an average load of 200W
and 385V output:
The 120Hz ripple current rating at 105 C ambient is 0.95A
for the 180 F KMH 400V capacitor. The expected life of the
output capacitor may be calculated from the thermal
stress analysis:
where:
In our example L
0.95A. T
Assuming the operating ambient temperature is 60 C, the
approximate life time is:
For longer life, a capacitor with a higher ripple current
rating or parallel capacitors should be used.
I
I
I
I
I
I
I
L = L
L:
L
T
L
3RMS
RMS
LOAD(DC)
1RMS
2RMS
3RMS
RMS
O
A
O
T
T
T
: Operating ambient temperature.
: hours of load life at rated ripple current and rated
K
K
O
: Capacitor internal temperature rise at rated condi-
: Capacitor internal temperature rise at operating
= (I
O
= (0.37A)
2000
= (I
expected life time
ambient temperature.
tion. T
R is capacitor ESR, and KA is a volume constant.
condition.
RMS
O
2
I
0.71
0.82A at 120VAC
I
LOAD(DC)
LOAD(DC)
= 0.52A
can then be calculated from:
(105 C
1RMS
/0.95A)
2
K
(105 C 10 C) – (60 6.6 C)
0.52A = 0.37A
O
)
= (I
2
2
= 2000 hours and T
= 0.52A
T
(0.82A/1.43)
2
2
K
(I
R)/(KA). Where I is the rated current,
) – (T
10
2RMS
T
10
A
K
/1.43)
= (0.77A/0.95A)
T
O
)
2
2
(0.52A/1.43)
(I
3RMS
57,000 hours
K
= 10 C at rated
2
/1.43)
10 C = 6.6 C
2
= 0.77A
2

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