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NCP1028
Figure 24. A typical startup sequence showing the V CC capacitor
voltage evolution versus time.
Suppose our power supply takes 10 ms (t startup ) to bring
the output voltage to its target value. We know that the
switcher consumption is around 2.0 mA (I CC1 ). Therefore,
we can calculate the amount of capacitance we need, to
hold V CC above 7.5 V at least for 10 ms while delivering
2.0 mA:
C w
ICC1tstartup
D VCC
or, by replacing with the above values,
C w 2m · 10 m w 20 m F then select a 33 m F for the V CC
1
capacitor.
Fault Condition – Short-Circuit on V CC
In some fault situations, a short-circuit can purposely
occur between V CC and GND. In high line conditions
(V HV = 370 V DC ) the current delivered by the startup
device will seriously increase the junction temperature. For
instance, since IC1 equals 3.0 mA (the min corresponds to
Figure 25. The startup source now features a
dual-level startup current.
the highest T J ), the device would dissipate 370 3m=
1.1 W. To avoid this situation, the controller includes a
novel circuitry made of two startup levels, IC1 and IC2. At
C
The first startup period is calculated by the formula
V=I t, which implies a 33 m 1.3/650 m = 66 ms
powerup, as long as V CC is below a 1.3 V level, the source
delivers IC1 (around 650 m A typical), then, when V CC
reaches 1.3 V, the source smoothly transitions to IC2 and
delivers its nominal value. As a result, in case of
short-circuit between V CC and GND, the power dissipation
will drop to 370 650 m = 240 mW. Figure 25 portrays
this particular behavior.
startup time for the first sequence (t 1 ). The second
sequence (t 2 ) is obtained by toggling the source to 4.0 mA
with a delta V of VCC ON – VCCth = 8.5 – 1.5 = 7.0 V,
which finally leads to a second startup time of
7 33 m /6.0 m = 39 ms. The total startup time becomes
66 m + 39 m = 105 ms as a typical value. Please note that
this calculation is approximated by the presence of the knee
in the vicinity of the transition.
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