patrick01

You were right I was wrong Seems I got Ohm's Law all wrong but nothing better to get the old books out again. At least you got me thingking again What might have happened is that I have selected a value close to R14 but a bit higher which will give you a positive voltage on the base of Q1. Thanks Patrick :-\

I really dont want to get in a argument about Ohm's law but my multimeter measure the 32v minus the 5,6V which gives you the 26.4. What I said and measured is that Q1 was on with the values in the circuit diagram which pulled the output of U2 down which caused it to get hot!!! So you have to switch Q1 off to get the supply working Worked for me Hopefully this is the end of the story :-X

I think your assumption about the voltage is wrong If you messure with a voltmeter the voltage is 26,4 volts and not the too voltages together If you make R13 a 46K then the voltage on Q1 is less than 0.6 volts to switch it on so it stays off Only swithes on as the AC is removes as you say the positive voltage takes longer to get to 0V

Had the same problem with U2 getting hot The voltage devider circuit (ResR13,R14)that is suppose to hold the Transistor Q1 off is keeping it on and thats why U2 is getting hot. Get the voltage to the negative so that Q1 is off and only swithes on when AC is removed. Problem solved