LT1508 LINER [Linear Technology], LT1508 Datasheet - Page 8

• Manufacturer Part Number: LT1508
• Description: Power Factor and PWM Controller(Voltage Mode)
• Manufacturer: LINER [Linear Technology]

APPLICATIONS
8
LT1508
Current Amplifier (PFC Section)
The current amplifier has a 110dB DC gain, 3MHz unity-
gain frequency and a 2V/ s slew rate. It is internally
clamped at 8.5V. Note that in the current averaging opera-
tion, high gain at twice the line frequency is necessary to
minimize line current distortion. Because CA
to swing 5V over one line cycle at high line condition,
20mV AC will be needed at the inputs of the current
amplifier for a gain of 260 at 120Hz. Especially at light load
when the current loop reference signal is small, lower gain
will distort the reference signal and line current. But, if
signal gain at switching frequency is too high, the system
behaves more like a current mode system and can cause
subharmonic oscillation.
V
IN
V
IN
R
1M
IAC
–20
–40
80
60
40
20
0
0.1
–
+
I
AC
V
I
I
IN
M
IN
U
1
FREQUENCY (Hz)
M
R
4k
0.15
REF
OUT
INFORMATION
Figure 2
Figure 3
R
L
U
S
10
I
R5
4k
SENSE
20k
R6
100
W
–
+
300pF
C4
CA
1508 • F03
1000
D1
0.001 F
OUT
60
45
30
15
0
–15
–30
C3
V
U
may need
OUT
LT1508
CA
C
470 F
OUT
1508 • F02
OUT
To avoid subharmonic oscillations, the amplified downslope
of the inductor current must be less than the slope of the
oscillator ramp.
If the current amplifier gain at 100kHz is less than 4.4,
there will be no subharmonic oscillation. The open-loop
gain of the current loop is given by:
The current error amp, with R5 = 4k, R6 = 20k, C3 =
0.001 F and C4 = 300pF, provides zero pole compensa-
tion resulting in 16kHz loop crossover frequency. The
current amp gain at 100kHz is 1.7. The resulting current
loop gain and phase margin is shown in Figure 4.
Multiplier
The multiplier has high noise immunity and superior
linearity over its full operating range. The current gain is
I
error amplifier output voltage required at the input to the
multiplier is:
M
=
= (I
V
=
V
CA(OUT)
(5V)(500 H)(100k)
CA(OUT)
(j)(2 f)(500 H)(5V)
V
V
AC
(382V)(0.15 )
RS
RS
(382V)(0.15 )
I
EA
–20
–40
80
60
40
20
2
0
)/(200 A
0.1
=
(V
(j)(2 f)(L)(V
(V
OSC
(V
OUT
1
2
OUT
)(L)(f
) with I
FREQUENCY (kHz)
)(R
Figure 4
)(R
= 4.4
=
S
SW
10
3648
(j)(f)
)
S
OSC
EA
)
)
= (VA
)
100
OUT
1508 • F04
– 2V)/ 25k. The
1000
60
45
30
15
0
–15
–30