Glad I'm not the only one confused.
meBigGuy said:
Something in that process won't make sense and will provide a clue as to the problem.
Good Troubleshooting...
Aha ! at 460% magnification, R404 is 6800 not 680K.
There's how gate current gets into the four-transistor trigger circuit Q401-404, not through R401 like i drew it...
Planobilly - is R404 blue-bray-red not blue-gray-yellow?
so here's my first shot at the gate drive circuit - doubtless wrong but can be corrected and polished up.
C405 causes voltage at Q401& 403's emitters to lag that at their bases by a small fraction of a cycle..
Between zero and 90 degrees while voltage is rising,
...Q401 is held off , its eb is reverse biased.
...Q403 is turned ON but D403 blocks gate current.
After 90 degrees when voltage starts decreasing from peak the situation reverses.
...Q401 is now forward biased and Q403 reverse...
...Q401 and 402 deliver gate current to the triac
So it CAN bootstrap up
AND we have achieved phase control because
Current from the right, controlled by optocoupler, advances or retards the firing angle.
Closing power switch sinks all the current that can get in through R404 removing gate drive.
So - R404, C405 and D401/PC301 are the muscle of phase control. Brains are in that column of opamps above.
R404 sees full voltage when system is off. Likely candidate to be open.R401 and R402 drop voltage to something within the photocoupler's 55 volt rating?
Sound plausible guys? You were doubtless there already...
old jim
Edit back to drawing board
R404 is 680K per partslist
so i do not see how that circuit as drawn can deliver trigger current more than peak volts/R401,
170v/51k = about 3 milliamps.
Does it use the charge stored in C405 for a trigger pulse?
dv/dt = I/C , it could deliver 50 ma for 10 microseconds with a 3.3 volt drop...
?
How to check C405 in circuit ?
Check my arithmetic, not to mention the logic.