NCP1253_12 datasheet(11/15 Pages) ONSEMI

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NCP1253

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200k

100k

4.7uF

1N4007

Cbulk

22uF

input

mains

1N4935

47uF

1N4148

Vcc

aux.

Figure 25. The Startup Resistor Can Be Connected to the Input Mains for Further Power Dissipation Reduction

The first step starts with the calculation of the needed VCC

capacitor which will supply the controller until the auxiliary

winding takes over. Experience shows that this time t1 can

be between 5 and 20 ms. Considering that 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

10m

w 3.3 mF

(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

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

2.5

4.7m

2.5

w 34 mA

(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:

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,rms 2

* VCCon

Rstart−up

(eq. 3)

To make sure this current is always greater than 49

mA, the

maximum value for Rstart−up can be extracted:

Rstart−up v

Vac,rms 2

* VCCon

ICVCC,min

85 1.414

* 18

49m

v 413 kW

(eq. 4)

This calculation is purely theoretical, considering 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. This brings a decrease in the charging current

and an increase of the start−up resistor, for the benefit of

standby power. Laboratory experiments on the prototype are

thus mandatory to fine tune the converter. If we chose the

400k resistor as suggested by Equation 4, the dissipated

power at high line amounts to:

PRstart,max +

Vac,peak 2

4Rstart−up

320

400k

105k

1.6Meg

+ 66 mW

(eq. 5)

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 25 elegantly

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