17 Apr
2012
17 Apr
'12
12:52 p.m.
No Tony, for sure this is not such an circuit.
I've meausred the pieces of the wirewound resistor, it must had a
resistance with approx 4-5 Kohms before. The Place for the FET is marked
Ok, lets think for a second.. This wirewound resisotr is, perhaps, 4700
ohms. And it burnt up spectacualrly, so perhaps it disipated at least 5W.
Now, P=V^2/R, so V= sqrt(P*R). That implies a voltage across that
resistor of about 153V by my calculations. And a voltage that won't drop
even when supplying such a load.
Are you sure?
GDS on the PCB and lat but not least the circuit
isn't connected to the
Capacitors on the mains side.
What is it conencted to? Which side of the isulation barrier is it on
(mains or output side?)
-tony