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Usually the Auxiliary supply is added for off-line converters, in order to provide power for the dc-to-dc control IC. It is generally needed when creating a lower, isolated voltage from a higher voltage. It increases efficiency (the control IC draws its power from a stepped-down voltage) and improves reliability (the alternative, using a resistor from the input voltage to supply the capacitor, might not provide enough current at peak demand).
I am trying to put the UC3845 working without being mounted on a full circuit configuration. I've a doubt here regarding that dc-to-dc control IC. Please, see the figure below (taken from the datasheet from ON Semiconductor):
That auxiliary bobbin in the primary side is used instead of using the optcoupler to sample from the secondary side?
Can someone tell how can I simulate that without having the full circuit mounted? (I am using a breadboard) Is it sufficient to put 2.5V on that pin? Since the other pin of the Error Amplifier is 2.5V. BTW, if this was in a circuit, like the figure above, we should select the pair of resistor in order to put 2.5V in that pin? Is that it?
dear ZekeR:
can you give an example for this paragraph"It is generally needed when creating a lower, isolated voltage from a higher voltage."？
thanks !
Usually the Auxiliary supply is added for off-line converters, in order to provide power for the dc-to-dc control IC. It is generally needed when creating a lower, isolated voltage from a higher voltage. It increases efficiency (the control IC draws its power from a stepped-down voltage) and improves reliability (the alternative, using a resistor from the input voltage to supply the capacitor, might not provide enough current at peak demand).
This is often done simply for efficiency's sake. To boot
and run the PWM controller you need (say) 100mA.
This would come from the input supply, and at 40V in
you'd be talking 4W draw with most of it dumped in a
limiting resistor. That's 4% off the top in a 100W brick.
Now if you either-ORed the feed with a winding derived
10V source, that same 100mA is 1W and only a 1%
inefficiency term. Given how hard you struggle to squeeze
percentage points out of the powertrain, getting 3%
back for a few turns of wire and a cheap diode or two
is a bargain.
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