Hi,
I have attached a application note from ST Microelectronics. I have a doubt in this circuit. am not able to understand the output voltage fast discharge circuit. can anybody explain me this?
If the D17 is not conducting (less than 17V across it). then Q10 does not switch on. If Q10 is not on, then Q11 is switched on - base current supplied by R77. The resistor R78 (560R) is connected across the capacitors, discharging them fairly quickly.
I notice that on a lot of laptop power supplies, the power LED stays on for a long time after the power is disconnected due to energy storage in the capacitors. This circuit is designed to reduce this.
BTW There is a typo in the text: "D17, R75+R78, Q10 and Q15" should read "D17, R75+R78, Q10 and Q11"