How smps starts up (auxiliary winding)

Thread Starter

mz350

Joined Jun 10, 2021
11
Hi

I've been baffled by this, I get that during regular use the auxiliary winding powers the ic.
I mean no power without the ic but the ic still needs power to create power.
Maybe a small capacitor inside the ic that stays charged and sustains it a few cycles before the auxiliary winding kicks in.

Best regards.
 

DickCappels

Joined Aug 21, 2008
7,954
You've got it. In many offline power supplies there is a capacitor and a resistor in the power supply. When the capacitor charges to a sufficiently high voltage the power power supply starts up. In some case a diac or similar switch (that closes when the voltage across it crosses a threshold) is used to connect the capacitor to the power supply.
 

Thread Starter

mz350

Joined Jun 10, 2021
11
You've got it. In many offline power supplies there is a capacitor and a resistor in the power supply. When the capacitor charges to a sufficiently high voltage the power power supply starts up. In some case a diac or similar switch (that closes when the voltage across it crosses a threshold) is used to connect the capacitor to the power supply.
Ok I see. Interesting the diac thingy. I'll search that.
 

Ian0

Joined Aug 7, 2020
3,752
I've not seen a diac in a SMPS, except those self-oscillating halogen-converters that need a pulse to start them up.
In the usual SMPS circuit, there is a capacitor (10uF to 100uF) across the control IC, which is charged up by a large resistor (often a 1W or 2W power resistor) connected to the high voltage rectified mains supply. When the voltage reaches the upper UVLO threshold, the IC starts, and runs off the capacitor, and hopefully, before the voltage has discharged as far as the lower UVLO threshold, the Vaux supply is supplying the IC.
If the voltage discharges as far as the lower UVLO threshold it stops and tries again. (That is why you see failed LED lights producing a brief flash every couple of seconds).
If the capacitor on the mains output is large, but the capacitor on the control IC is small, there may not be enough charge in the IC capacitor to keep it going long enough to charge the output capacitor, (because it can't supply Vaux until the output capacitor is charged) in which case it never gets started.
With the latest efficiency regulations having a large resistor connected across the 340V DC supply dissipating power all the time is a problem, so newer designs include a depletion-mode MOSFET which can switch off the power through the large resistor when Vaux is high enough.
Some more integrated controllers have a separate pin for the HV supply and can switch it off after the Vaux supply is up to voltage.
 

ronsimpson

Joined Oct 7, 2019
1,630
Picture of what Dick said:
Red is operating power. Green is start up power. The IC pulls almost no power until 10Uf uf is charged up. Then the IC can make 10 cycles before the cap runs out of power. Before then the red power is up and running.
1623954566944.png
 

Ian0

Joined Aug 7, 2020
3,752
Thanks @ronsimpson for posting the diagram. I'll just clarify that newer designs, in order to meet energy efficiency regulations, have a method of blocking the current shown by the green arrow, once the IC is supplied via the red arrow from the aux winding. Otherwise the 56k resistor dissipates power all the time. Of course, it's a problem that's twice as bad for 230V supplies.
 
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