Re: stereo amp cleaning

Dave wrote:
"Mark D. Zacharias" <spammenot@xxxxxxxxxxxx> wrote in message
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Dave wrote:
"Mark D. Zacharias" <spammenot@xxxxxxxxxxxx> wrote in message
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Dave wrote:
"Mark D. Zacharias" <spammenot@xxxxxxxxxxxx> wrote in message
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Man.

This might be really easy. Q 405 conrols the turn-on delay. HK's
don't use relays, so they mute the signal til the amp stabilizes.
On this model it is Q405 and Q406 respectively. The transistor
could be bad, but I would be especially concerned with D401,
R405, C405, C407, and D403, which is a 15 volt zener.

Could just be solder connections relating to the above, but in
any case you need to see that Q405 turns off hard a few seconds
after turn-on. The -12.5 or so volts at the base is critical.

Mark Z.
Something is still failing in my amp. As I noted, I put the
original caps (C401, 403, 405, 407) back in and the problem went
away. For awhile. It's back... I measured voltages after the
right channel had cut out.
The collector of Q407 was 0V. So was the collector and base of
Q409, and the base of Q413. Makes sense as they're all tied
together. The collector of Q405 was also 0V. the collectors of
Q411 and Q413 were ALSO 0V. I noted that although R437 tested
fine, it looks like it's been run hot. As in it's dark brown and
you can't really read the color bands any more.
I don't understand how Q405 works as far as the turn-on delay, but
if it were bad could it be causing no voltage where I've
indicated? I tested the other, working channel and all voltages were
correct.

Thanks

Dave

I need the base and emitter voltages on Q405 versus Q406. I wasn't
thinking clearly before, but the basic principle is the same. As
the power supply comes up, a positive voltage is passed through
D401 and D402 to turn ON Q405 and Q406. In the case of Q405, you
can see that there should be a negative voltage at the emitter,
and a somewhat LESS negative voltage at the base. This turns ON
the transistor and should pass a negative voltage through the
transistor, so that the negative 7 volts or so appears at the
collector. If there's no voltage at the collector of Q405, it's
likely there is no negative voltage at the emitter either.
Possibly R429 or R429 are opening up, or there could be bad solder
connections on one or both of them. Let me know what you find.

Oops I guess I omitted information... had typed it then deleted it
as insignificant. I did test the negative power rail at R417 where
the schemo shows -14.2V. It was, I think around -14.9V. In the
zone anyways which would rule out R427 or R429 as culprits. If one
of these had opened up would I not likely see a positive voltage at
the base of Q405?
Base Collector Emitter
Q405 -14.2 -0.4 -14.7
Q406 -12.7 -6.9 -13.5

As you can see the base vs. emitter of the two transistors has a
comparable delta... say a half-volt lower at the base which is what
you'd expect I think.

So, if I have the correct voltages at the base and emitter of Q405,
but the wrong collector voltage, I should be looking hard at Q405? I
would think that the -0.4V seen at the collector would be derived
from the 2 x -0.6V coming out of Q401 and Q403 attenuated by
R413/R415/R483 with no contribution from Q405.

Dave


I do think I'd replace Q405 at this point. You say the B-E voltage is
comparable but I don't think so. There's a big difference between 0.5
volts and 0.8 volts when it comes to turning on a transistor. If the
transistor isn't the problem I'd still say it's not turning on
sufficiently. Could be a resistor is not open but the value has
changed, for example.


Well, I'll lift the legs of R427 and R429... and, I suppose, R407,
R417 and R405 just to rule out the resistors as I likely have
replacements kicking around for them whereas the transistor will need
to be procured. Something has got to be causing the ~1.5V
differential between the R and L amp channels in the first place.

I could add a small ~40-ohm resistor to the base, too and see if Q405
turns on with a 0.8V B-E drop... FYI the spec drop is only 0.6V and I
doubt the designer would cut it so close that a 0.1V difference
causes failure... but them I'm often surprised.

Will post results, might be awhile before I get my hands on the
transistor.
Out of curiousity, what does the Q5/Q6 circuit which biases Q405 do? As
near as I can tell, it takes 30VAC prior to rectification, runs it
through a diode to separate out the negative half of the wave, then
past a 12V zener to drop the voltage. Not sure what the function of
D9 is, it is oriented the reverse of D8. you end up with -1.1V at
the base of Q5. The net result being -0.56V out of the collector of
Q6 to bias Q405/6. Seems like a lot of work to derive a half-volt of
regulated power but I guess they wanted it fully independent of the
amp channel power rails.
Thanks again.

Dave

The function of Q5 and 6 is just to provide a more positive bias to Q405
after a delay (mainly determined by R24 and C22), enabling the amp channel
to "un-mute". It neen't be actually "positive" just more positive than the
negative rail at the emitter 0f Q401. Diode D9 is a question mark to me
also, but I suspect that since the capacitor C21 is only 33uF then some
ripple would still be present, and the diode might then act similar to a
zener and clamp the voltage drop across R24. I have to admit that since I'm
no engineer I can lose it when trying to figure the theory aspects. I just
fix them when they break.

You can swap Q405 and Q406 and see if the problem moves to the other
channel. Resistor R417 (560 ohm) can be checked in circuit. R405, a 68K,
would probably need to be checked out-of-circuit. Experience tells me that
high-value resistors can change value or go open-circuit for no good reason.
Worth keeping in the back of your mind sometimes as you're troubleshooting
problems such as this.

Not sure how you would use a 40 ohm added to the base of Q405.... in series
with R405 (68K) would do nothing, and in parallel with the BE junction of
Q405 would just keep the transistor from ever turning on in this
application.

At this point I think R427 and 429 are probably OK, or you wouldn't see the
voltage at the emitter of Q405. Our problem seems to involve getting 405 to
turn on, which is to say, getting the base of Q405 a bit more positve (less
negative) than it is right now.


Mark Z.


.

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