ML4819
The value of R (pin 3) depends on the selection of R
VOLTAGE REGULATION COMPONENTS
9
2
(pin 6).
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 through-hole resistors are used, two of
them should be put 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:
V
IN(MAX)PEAK
260 ×1.414
0.72mA
R =
=
= 510kΩ
= 22kΩ
(12)
2
I
SINE (PEAK)
VCLAMP ×R2
4.8 × 510kΩ
80 ×1.414
R9 >
=
(13)
V
IN(MIN)PEAK
Choose R9 = 27kW
The peak of the inductor current can be found
approximately by:
R5 = (380V)2 /0.4W = 360kΩ
(17)
Choose two 178kW, 1% connected in series.
1.414 × POUT
1.414 × 200
ILPEAK
=
=
= 3.14A
(14)
V
90
IN(MIN)RMS
Then R6 can be calculated using the formula below:
Next select N , which depends on the maximum switch
C
V
×R
5V × 356kΩ
380V − 5V
REF 5
R =
=
= 4.747kΩ
current. Assume 4A for this example. N is 80 turns.
6
(18)
C
V − V
B
REF
V
CLAMP ×NC
ILPEAK
Choose 4.75kW, 1%. One more critical component in the
voltage regulation loop is the feedback capacitor for the
error amplifier. The voltage loop bandwidth should be set
such that it rejects the 120Hz ripple which is present at
the output. If this ripple is not adequately attenuated it
will cause distortion on the input current waveform.
Typical bandwidths range anywhere from a few Hertz to
15Hz. The main compromise is between transient
response and distortion. The feedback capacitor can be
calculated using the following formula:
4.9× 80
R11
=
=
= 100Ω
(15)
4
Where R is the sense resistor, and V
is the current
11
CLAMP
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 V
was chosen as 4.8V.
CLAMP
Having calculated R the value S
and of R can
18
11
PWM
1
now be calculated:
C =
8
3.142×R ×BW
5
(19)
1
380V − 20 100
C =
= 0.44µF
S
=
×
= 0.225V /µs
8
PWM
3.142×356kΩ × 2Hz
2mH
80
2.5×R
9
R
=
=
18
18
A
× S
×R × C
SC
PWM T T
(16)
2.5× 28.8k
R
30kΩ
6
0.7 × (0.225×10 ) ×14kΩ ×1nF
Choose R18 = 33kW
The following values were used in the calculation:
R = 27kW
T
A
T
= 0.7
9
SC
R = 14kW
C = 1nF
11