kiev wrote:The voltage on pin 6 seems to be from the internal 30uA current source, which i think gets turned on when the short circuit protection gets triggered;
Yes. The long time constant of the Rs and C on pin 6 (≈ 0.5 sec) seems to be the short circuit protection. The logic seems to be: if there is no output for 500 ms, something is wrong, likely a short circuit (either at the IC702 output or in the transistor it drives), so best to shut it off and prevent either IC failure or meltdown of other components like the "transformer" (really a multi-winding inductor). So whenever there is a failure in this circuit, you would find it in this state: 0 V in pin 1, ≈2.5 V on pin 2, ≈2.5 V on pin 6, and no drive (0 V) on pin 4. So see the action, you need to trigger the DSO at power-up, or as Kenny suggests below, force a "reset" with a resistor or similar and capture voltages on the DSO.
The 2.5 volts on pin 2 seems to be the output of the error op amp due to the 0.52 internal reference on the (+) input and nothing on the inverting input from pin 1. i have an idea for a test, briefly touch a resistor (eg 100k - 1M) between pins 1 and 2 to see if it causes the chip to start switching.
I don't believe that will work; the short circuit protection is acting because something is wrong elsewhere. But the idea of a "reset" is good. My suggestion is put one channel of the DSO on pin 6, put the DSO in one-shot mode and trigger on that channel at about 1 V, low-going edge. Put the other channel on whatever you want to test, perhaps start with the output, pin 4. Use a clip lead or whatever to short the 4.7 μF capacitor on pin 6; this should trigger the DSO, and you should capture some 300 μs of activity. That should allow you to figure out what is happening in the initial half second before the short circuit protection kicks in. It would be best to have all the parts in place, so that you'll be able to see why no feedback gets to pin 1 (if that's the case). It might turn out for example that there is too much load on one of the "transformer" outputs (so check those as well). Pin 1 would be the next place to capture from, and work back from there.
The voltage on pin 2 is causing the PWM to stay locked high and also sets the short circuit protection hi which latches the chip and stops the output.
I believe it's the continued
high voltage at pin 2 (higher than the "battery" connected to the PWM op-amp) that allows pin 6 to ramp up to higher than the rightmost "battery" that is the problem. When pin 2 occasionally goes low, it should stop the 30 μA current source from ramping up the voltage at pin 6, which should be enough to keep pin 6 low enough to stop the short circuit protection.