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ML4812CP Datasheet(PDF) 9 Page - Fairchild Semiconductor

Part # ML4812CP
Description  Power Factor Controller
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Manufacturer  FAIRCHILD [Fairchild Semiconductor]
Direct Link  http://www.fairchildsemi.com
Logo FAIRCHILD - Fairchild Semiconductor

ML4812CP Datasheet(HTML) 9 Page - Fairchild Semiconductor

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ML4812
CURRENT SENSE AND SLOPE (RAMP) COMPENSATION
COMPONENT SELECTION
Slope compensation in the ML4812 is provided internally.
Rather than adding slope to the noninverting input of the
PWM comparator, it is actually subtracted from the
voltage present at the inverting input of the PWM
comparator. The amount of slope compensation should be
at least 50% of the downslope of the inductor current
during the off time, as reflected to the inverting input of
the PWM comparator. Note that slope compensation is
required only when the inductor current is continuous
and the duty cycle is more than 50%. The downslope of
the inductor current at the verge of discontinuity can be
found using the expression given below:
di
dt
VV
L
VV
mH
As
L
OUT
INDRY
=
-
=
-
=m
380
20
2
018
./
(9)
The downslope as reflected to the input of the PWM
comparator is given by:
S
VV
L
R
N
PWM
OUT
INDRY
S
C
=
-
´
(10)
S
V
mH
Vs
PWM =
-
´=
m
380
20
2
100
80
0 225
./
Where RS is the current sense resistor and NC is the turns
ratio of the current transformer (T1) used. In general,
current transformers simplify the sensing of switch
currents (especially at high power levels where the use of
sense resistors is complicated by the amount of power
they have to dissipate). Normally the primary side of the
transformer consists of a single turn and the secondary
consists of several turns of either enameled magnet wire
or insulated wire. The diameter of the ferrite core used in
this example is 0.5" (SPANG/Magnetics F41206-TC). The
rectifying diode at the output of the current transformer
can be a 1N4148 for secondary currents up to 75mA
average.
Sense FETs or resistive sensing can also be used to sense
the switch current. The sensed signal has to be amplified
to the proper level before it is applied to the ML4812.
The value of the ramp compensation (SCPWM) as seen at
the inverting terminal of the PWM comparator is:
SC
R
RC
R
PWM
M
TT
SC
=
´
´´
25
.
(11)
The required value for RSC can therefore be found by
equating: SCPWM = ASC × SPWM, where ASC is the amount
of slope compensation and solving for RSC. The value of
GM OUT depends on the selection of RAMP COMP.
R
V
ImA
k
P
IN
PEAK
SINE PEAK
==
´
=
(max)
.
.
()
260 1414
05
750 Ω
(12)
R
VR
V
k
k
M
CLAMP
P
IN PEAK
=
´
=
´
´
=
()
.
.
.
49 750
90 1414
288
(13)
The peak of the inductor current can be found
approximately by:
I
P
V
A
LPEAK
OUT
IN RMS
=
´
=
´
=
1414
1414 200
90
314
.
.
.
()
(14)
Selection of NC which depends on the maximum switch
current, assume 4A for this example is 80 turns.
R
VN
I
S
CLAMP
C
LPEAK
=
´
=
´
=W
49 80
4
100
.
(15)
Where RS is the sense resistor, and VCLAMP is the current
clamp at the inverting input of the PWM comparator. This
clamp is internally set to 5V. In actual application it is a
good idea to assume a value less than 5V to avoid
unwanted current limiting action due to component
tolerances. In this application, VCLAMP was chosen as
4.9V.
Having calculated RS, the value SPWM and of RSC can
now be calculated:
R
R
AS
R
C
R
k
KnF
k
SC
M
SC
PWM
T
T
SC
=
´
´´
´
=
´
´´
´
´
=
25
25 288
07 0 225 10
14
1
33
6
.
..
.( .
)
(16)
The following values were used in the calculation:
RM = 28.8kΩ
ASC = 0.7
RT = 14kΩ
CT = 1nF
VOLTAGE REGULATION COMPONENTS
The values of the voltage regulation loop components are
calculated based on the operating output voltage. Note
that voltage safety regulations require the use of sense
resistors that have adequate voltage rating. As a rule of
thumb if 1/4W resistors are chosen, two of them should
be used in series. The input bias current of the error
amplifier is approximately 0.5µA, therefore the current
available from the voltage sense resistors should be
significantly higher than this value. Since two 1/4W
resistors have to be used the total power rating is 1/2W.
The operating power is set to be 0.4W then with 380V
output voltage the value can be calculated as follows:
RV
W
k
1
2
380
0 4
360
==
() / .
(17)
Choose two 178k
Ω, 1% connected in series. Then R2 can
be calculated using the formula below:
R
VR
VV
Vk
VV
k
REF
OUT
REF
2
1
5356
380
5
4747
=
´
-
=
´
-
=
.
(18)
TYPICAL APPLICATIONS (Continued)
REV. 1.0 10/10/2000
9


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