D2
1N4007
12
5
D1
1N4007
input
mains
D4
1N4007
D3
1N4007
NCP1251
R3
200k
3
10
R2
200k
Cbulk
22uF
VCC
11
R1
200k
1
D6
1N4148
D5
1N4935
2
4
C1
4.7uF
C3
47uF
aux.
Figure 40. The Startup Resistor Can Be Connected to the Input Mains for Further Power Dissipation Reduction
The first step starts with the calculation of the VCC
capacitor which will supply the controller when it operates
until the auxiliary winding takes over. Experience shows
that this time t1 can be between 5 ms and 20 ms. If we
consider we need at least an energy reservoir for a t1 time of
10 ms, the VCC capacitor must be larger than:
CVCC
w
ICCt1
VCCon * VCCmin
w
3m
10m w 3.3 mF
9
(eq. 1)
Let us select a 4.7 mF capacitor at first and experiments in
the laboratory will let us know if we were too optimistic for
the time t1. The VCC capacitor being known, we can now
evaluate the charging current we need to bring the VCC
voltage from 0 to the VCCon of the IC, 18 V typical. This
current has to be selected to ensure a start−up at the lowest
mains (85 V rms) to be less than 3 s (2.5 s for design margin):
Icharge w
VCConCVCC w 18
2.5
4.7m
w 34 mA
2.5
(eq. 2)
If we account for the 15 mA that will flow inside the
controller, then the total charging current delivered by the
start−up resistor must be 49 mA. If we connect the start−up
network to the mains (half−wave connection then), we know
that the average current flowing into this start−up resistor
will be the smallest when VCC reaches the VCCon of the
controller:
ICVCC,min
+
Vac,rmsǸ2
p
*
VCCon
Rstart*up
(eq. 3)
To make sure this current is always greater than 49 mA,
then the minimum value for Rstart−up can be extracted:
Rstart*up
v
Vac,rmsǸ2
p
*
VCCon
ICVCC,min
v
85
1.414
p
*
18
v
413.5
kW
49m
(eq. 4)
This calculation is purely theoretical, and assumes a
constant charging current. In reality, the take over time can
be shorter (or longer!) and it can lead to a reduction of the
VCC capacitor. Hence, a decrease in charging current and an
increase of the start−up resistor, thus reducing the standby
power. Laboratory experiments on the prototype are thus
mandatory to fine tune the converter. If we chose the 413 kW
resistor as suggested by Equation 4, the dissipated power at
high line amounts to:
ǒ Ǔ Vac,peak 2
230 Ǹ2 2
PRstart*up + 4Rstart*up + 4 413k
(eq. 5)
+ 2302 + 64 mW
0.827Meg
Now that the first VCC capacitor has been selected, we
must ensure that the self−supply does not disappear when in
no−load conditions. In this mode, the skip−cycle can be so
deep that refreshing pulses are likely to be widely spaced,
inducing a large ripple on the VCC capacitor. If this ripple is
too large, chances exist to touch the VCCmin and reset the
controller into a new start−up sequence. A solution is to
grow this capacitor but it will obviously be detrimental to the
start−up time. The option offered in Figure 40 elegantly
solves this potential issue by adding an extra capacitor on the
auxiliary winding. However, this component is separated
from the VCC pin via a simple diode. You therefore have the
ability to grow this capacitor as you need to ensure the
self−supply of the controller without jeopardizing the
start−up time and standby power. A capacitor ranging from
22 to 47 mF is the typical value for this device.
One note on the start-up current. If reducing it helps to
improve the standby power, its value cannot fall below a
certain level at the minimum input voltage. Failure to inject
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