Harmon Kardon Citation Sixteen

[J M Fahey]

Just to discard it as culprit, open Q9's base (lift its leg from the hole) and see what happens.
If the high current through 120r stays, we have to look elsewhere.
If it stops, reconnect it because it looks like Q9 is right (it's showing normal behaviour), so now I'm suspecting a BE short in Q11; because Q9's emitter should be *at least* one diode drop (0.6 to 0.7V) above output rail, which is at ground potential if everything else is fine, because it must turn Q11's base on.
If Q11 were open, you can't have 8V across the 120r resistor and only 0.4V across an 180r one.
To re-check, cut the trace to isolate Q11 base from R30 and Q9's emitter, then we'll see whether Q9 continues pulling so much current.

[lowell]

now I'm suspecting a BE short in Q11; because Q9's emitter should be *at least* one diode drop (0.6 to 0.7V) above output rail, which is at ground potential if everything else is fine, because it must turn Q11's base on.

Even with the limiter?

[lowell]

To re-check, cut the trace to isolate Q11 base from R30 and Q9's emitter, then we'll see whether Q9 continues pulling so much current.

Ok did this and voltage across 120ohm is reading much lower at 1.6v. Still seems high though right? FYI Q9 Ve reads 550mv now.

Just to discard it as culprit, open Q9's base (lift its leg from the hole) and see what happens.
If the high current through 120r stays, we have to look elsewhere.
If it stops, reconnect it because it looks like Q9 is right (it's showing normal behaviour)

I'm confused by this. It seems that both of the statements lead to Q9 NOT being the issue. It seems you're saying that with Q9's base lifted: If the high current persists - it's not Q9. If the high current stops, it's not Q9. ??

[lowell]

I'm trying to figure out why everyone suspects Q9 first with this repair. Why not Q3 or Q5? Just curious.

[Enzo]

because if Q3 or Q5 were shorted, then we might expect to see R15,16 burning up. Also, Q9 is a better suspect because it is more closely tied to the heavier current paths through driver Q11, and the output transistors.

In a circuit where a resistor burns up, that resistor is conducting excess curent flow. There has to be a complete path for that failure current. Your 120 ohm comes into the picture essentially at the collector of Q9 or at least at J1/P1-15. SO start thinking about where the current comes from and goes to.

[lowell]

Thanks Enzo.

JM did I lose you here?

[J M Fahey]

Not at all.
It's just that this @#"·$%&~€ computer froze after a long answer and so I got "unmotivated" to rewrite it, specially since we are getting into Friday and Saturday, with mobs of anxious musicians phoning and knocking the door desperate for their "tools of the trade".
Oh well, I chose to do this on my own will, nobody forced me.
Probably you are in the same situation.
Go finish what's needed today and tomorrow, we'll continue with that beautiful Citation from Monday on.
Post a couple pictures meanwhile.

[lowell]

I feel ya JM. Thanks for gettin' back to me... have a good weekend.

[lowell]

Ok so since the voltage across the 120 was brought down to a safe area with Q11's base disconnected, I removed the bulb limiter and remeasured. All 4 rails are around 2v across their respective 120 resistors. So it seems to me Q11 is the issue. Do you think a 2n5875 will be a good replacement here? http://www.centralsemi.com/PDFs/prod...n5875-5878.pdf

[lowell]

I have a MJ15015 in stock here, thinking that might work. What do you guys think? What should one consider when subbing a transistor other than the following?

Maximum ratings:
Vce
Ice

[lowell]

Alright now I'm chasing my tail. I threw the MJ15015 in there and it's back to 8v across the 120 ohm. I then left that transistor in there and disconnected its base, as was suggested before, and still 9v across the 120.