ML4812
TYPICAL APPLICATIONS (Continued)
The recommended maximum duty cycle is 95% at
100KHz to allow time for the input inductor to dump its
The inductor can be allowed to decrease in value when
the current sweeps from minimum to maximum value.
This allows the use of smaller core sizes. The only
requirement is that the ramp compensation must be
adequate for the lower inductance value of the core so
that there is adequate compensation at high current.
energy to the output capacitors. For example, if: V
=
OUT
380V and D
(max) = 0.95, then substituting in (3)
= 20V. The effect of drying out is an
ON
yields V
INDRY
increase in distortion at low voltages.
Step 4: The presence of the ramp compensation will
change the dry out point, but the value found
above can be considered a good starting point.
Based on the amount of power factor correction
the above value of L1 can be optimized after a
few iterations.
For a given output power, the instantaneous value of the
input current is a function of the input sinusoidal voltage
waveform, i.e. as the input voltage sweeps from zero volts
to a maximum value equal to its peak so does the current.
The load of the power factor regulator is usually a
switching power supply which is essentially a constant
power load. As a result, an increase in the input voltage
will be offset by a decrease in the input current.
Gapped Ferrites, Molypermalloy, and Powdered Iron
cores are typical choices for core material. The core
material selected should have a high saturation point and
acceptable losses at the operating frequency.
By combining the ideas set forth above, some ground
rules can be obtained for the selection and design of the
input inductor:
One ferrite core that is suitable at around 200W is the
#4119PL00-3C8 made by Philips Components
(Ferroxcube). This ungapped core will require a total gap
of 0.180" for this application.
Step 1: Find minimum operating current.
1.414 ´P (min)
IN
IIN(min)PEAK
=
OSCILLATOR COMPONENT SELECTION
(4)
V (max)
IN
The oscillator timing components can be calculated by
using the following expression:
V (max) = 260V
IN
P (min) = 50W
IN
1.36
fOSC
=
(6)
RT ´ CT
then:
I (min)
= 0.272A
PEAK
IN
For example:
Step 2: Choose a minimum current at which point the
inductor current will be on the verge of drying
out. For this example 40% of the peak current
found in step 1 was chosen.
Step 1: At 100kHz with 95% duty cycle T
= 500ns
OFF
calculate C using the following formula:
T
TOFF ´ IDIS
CT
=
= 1000pF
(7)
VOSC
then:
I
= 100mA
LDRY
Step 2: Calculate the required value of the timing
resistor.
Step 3: The value of the inductance can now be found
using previously calculated data.
1.36
fOSC ´ CT 100KHz ´ 1000pF
= 13.6kW chooseRT = 14kW
1.36
RT
=
=
V
´ DON(max)
INDRY
(8)
L1=
ILDRY ´ fOSC
20V ´ 0.95
100mA ´ 100KHz
(5)
=
= 2mH
8
MICRO-LINEAR [ MICRO LINEAR CORPORATION ]
