"Franc Zabkar" wrote in message
...
On Sun, 25 Sep 2011 01:29:05 +0100, "Arfa Daily"
put finger to keyboard and composed:

Typically, the startup supply doesn't come *directly* from the high
voltage
rail, It is normally fed via a diode, and the self-powering rail is
arranged
to be a higher voltage than the startup supply. This results in the diode
feeding the startup supply, to become reverse biased once the supply fully
starts, which then results in the current draw from that supply reducing
to
as good as zero, and thus likewise dropping the power dissipation in the
startup resistor to virtually zero.

Arfa

Your comment took me by surprise. Just to make sure I wasn't having a
brain fart, I consulted an application note:
http://www.nalanda.nitc.ac.in/indust...de/slua143.pdf

I confess that I haven't really thought about this subject before, but
according to page 4 of the PDF, there is no back-biased diode. AISI, a
diode that is fed from a 170VDC supply cannot become reverse biased by
a much lower bootstrap voltage.

The app note suggests that the UC3842 will remain in the off state
until the capacitor on its Vcc pin charges up to the UVLO (under
voltage lockout) turn-on voltage of 16V. During this time the IC draws
only 1mA.

After the UVLO turn-on threshold is reached, the IC turns on and
pulses the chopper. The bootstrap winding then generates the Vcc
supply for the IC and prevents the capacitor from discharging below
the IC's UVLO turn-off threshold of 10V.

Fig 31 on page 13 of the same document has an application circuit for
an isolated +/-12V supply. The 56K resistor (R2) can pass about 3mA.
This means that 2mA goes toward charging the capacitor. The UC3842
datasheet specifies a typical operating current of 14mA.

http://www.elektronik.sk/datasheet/UC3842.pdf

If you were to disconnect the bootstrap winding, then the output
voltage will only remain alive for as long as it takes the Vcc
capacitor to discharge from 16V to 10V.

I = C . dV/dt

so ...

dt = C x dV / I

= 100uF x (16V - 10V) / (14 - 3)mA
= 55 msec


- Franc Zabkar

Hmmm. OK. I'll buy that. I must admit that I haven't taken too much notice
of how that bit of the circuitry works in recent years. Perhaps I'm just
going back to the early days when I was first taught about these things on
manufacturers' service courses. Maybe there used to be a zener on the end of
the startup resistor, and then a diode. The diode could then be reverse
biased by the self-powering supply, and the current in the startup resistor
would decrease to that of the zener's draw only. I'll have to have a look
back on some of the older schematics for VCRs and DVDs etc

Arfa