Does your setup permit you to measure and annotate the observed voltages in the schematic at bottom of #19, with PA boards disconnected? You might tack ~2k2 resistor loads on the collectors of Q7 and Q15 to present them with modest loading.
My preference is to trouble live circuits when circumstances permit.
My preference is to trouble live circuits when circumstances permit.
I’m sorry, I guess I don’t know what you were asking. The two boards are disconnected. Those last measurements I took are with the boards disconnected. I can tack resistors there, but I do have the -15 V at the collector of Q7, just missing the voltage at the collector of Q15.
I actually replaced ZD2 and ZD3 with 1N4744A since they were getting so hot and I had no idea what kind of abuse they’ve been through. But I know that they are not shorted because I’m getting the plus and -15 V at pins 7 and 8 which come directly from those two Zener diodes.
Dan
I should let saabracer23 respond directly, but some of the voltages he notes in #17, #18 suggest that large currents are being drawn from negative supplies and are pulling the +15 rail to negative voltages where they are clamped to ~ -1V when ZD2 conducts in its forward direction.
As soon as I check C 27 and get those two transistors back into circuit, I will power it up and measure. This will be with the two amplifier boards still disconnected.
I’m actually going to see if I can get my daughter to sleep and sneak down there in a bit too do all these tests.
Dan
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What I was trying to say is that my preference is to troubleshoot symptoms without removing components for out of circuit testing. I find it's faster, less laborious, and less likely to cause PCB damage.
I see what you were saying. The only reason I moved the two amplifier boards is because I wanted to see if the power supply had the proper voltages with nothing loaded to it.
Dan
Check the ground connection at the filter capacitors. A bad ground can cause weird voltages.
So C27 is fine, it’s actually brand new. I just put it in there. Because of the voltages I was seeing I took out the 16 V rated caps and put in 25v.
Collector of Q8 is 18.17v
Q15
E: 18.2v
C: 12 mV
B: 18.2v
Dan
Collector of Q8 is 18.17v
Q15
E: 18.2v
C: 12 mV
B: 18.2v
Dan
The filter caps are new as well and have fresh solder on them. I do have continuity between the filter, ground all the way to the ground pins on both of the pin headers.
Dan
So Q8 is not conducting, so Q15 isn't turning on. Next question is why isn't Q8 active. I believe you said earlier that Q7 was presenting -15 at its collector. That would suggest that base current in Q7 would also bias Q8 into conduction--- but maybe something's wrong.
Would you report voltages at "B" and base of Q8, and emitter, collector for sanity check.
P.S. What is status of +12 PS output?
Would you report voltages at "B" and base of Q8, and emitter, collector for sanity check.
P.S. What is status of +12 PS output?
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It is
I do have a -17.8ish or so at the collector of Q7.
At position “B” it’s fluctuating between -2.45 and -2.49
Q8
E: 0
C: 18.192
B: 0
The 12 PS? I have a +12 V present at both pin headers if that’s what you mean. Well 11.49v
Dan
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Good morning, Dan.
I must have been half asleep when I made that comment--- D2 points in the wrong direction to get its base current from Q7's bias current. 😳
Q7 is probably getting its bias current from R58; the emitter of Q2 should be about 9.3V, base about 8.7V. Assuming this is the defacto path, -2.5V at "B" seems about right. So for the B node to provide operating bias to D2 and Q8, there has to be an additional path that I don't yet understand. The B node appears to pass through one of the PA boards from J7-A to J4-B. From there I lose track of where it goes, but I imagine having the PA board absent prevents bias being applied to Q8. Can you help me sort out this mystery?
With B lacking proper drive, naturally Q8 and Q17 aren't enabled. They may be OK, but we don't know yet, and have to get them enabled.
Would you confirm my estimates of Q2 voltages? Further, Q2 should be conducting and powering a multivibrator, Q3 and Q5. The output "A" should be generating a low frequency square wave, but I don't know its purpose.
I must have been half asleep when I made that comment--- D2 points in the wrong direction to get its base current from Q7's bias current. 😳
Q7 is probably getting its bias current from R58; the emitter of Q2 should be about 9.3V, base about 8.7V. Assuming this is the defacto path, -2.5V at "B" seems about right. So for the B node to provide operating bias to D2 and Q8, there has to be an additional path that I don't yet understand. The B node appears to pass through one of the PA boards from J7-A to J4-B. From there I lose track of where it goes, but I imagine having the PA board absent prevents bias being applied to Q8. Can you help me sort out this mystery?
With B lacking proper drive, naturally Q8 and Q17 aren't enabled. They may be OK, but we don't know yet, and have to get them enabled.
Would you confirm my estimates of Q2 voltages? Further, Q2 should be conducting and powering a multivibrator, Q3 and Q5. The output "A" should be generating a low frequency square wave, but I don't know its purpose.
I think you may be on the right track with "B". I traced it a bit and looked like it ended up being driven by a 339 comparator. Maybe some kind of enable.
Good morning!!!
Q2
E: 8.81
B: 8.08
Here is what I see on “A”. I don’t know if it was the signal or what, but my scope was lagging. It was taking a good amount of time for the signal on the screen to catch up with what I was doing as far as time base and such.
Should I hook up one of the boards?
Dan
Q2
E: 8.81
B: 8.08
Here is what I see on “A”. I don’t know if it was the signal or what, but my scope was lagging. It was taking a good amount of time for the signal on the screen to catch up with what I was doing as far as time base and such.
Should I hook up one of the boards?
Dan
Is your scope AC coupled?
I guess I'd install the board that passes the B signal for quick insight.
Good luck!
I guess I'd install the board that passes the B signal for quick insight.
Good luck!