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Hartke B300 bass amp repair

Started by DrGonz78, January 17, 2014, 05:29:47 AM

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DrGonz78

Okay I started some of this thread on MEF but that site is currently down or something...? Sorry to post in two places but I need a little advice at this point of the repair. I have a schematic that is close to the B300 but not exact. I am posting the output section of the amp here and deleting out components that are not in this circuit. Just a few components were not in this output section of the B300 and this edited version is very accurate. Not too concerned there but just thought showing what I am working on better with edited schematic. One components still on this page that is not on my board is C7. That cap is not on my board in the power output, but I did not delete it from this edited version.

I had a shorted 2SA1266 transistor, shorted collector to base, in the spot of Q4 on the schematic. Was getting 4.5vDC on the output. I matched a similar transistor(exact except for tiny bit higher in HFE) in it's place A1015 GR98 and it is doing it's job. Not really concerned about the HFE situation as I am only using it in place for some temporary testing. More concerned about my voltage readings on Q8 K D2059-y, one of the output chips. Here are my voltage readings of the output section.

(all measurements in DC voltage) Note: I am measuring about 350mv DC on the output.
Q2 E= -.63mv C= 35.75vDC B= -31.5mv
Q3 E= -.63mv C= 36.4vDC   B=17.8mv
Q4 E= 36.2vDC C= 1.48vDC B= 35.66vDC
Q5 E= .83mv C= 36.2vDC B= 1.48vDC
Q6 E= 350mv C= -36VDC B= -303mv
Q7 B= .83mv C= 36.4vDC E= 350mv
Q8 B= -35.82vDC C= 347mv E= -36.2vDC
Q9 E= -301mv C= 1.47vDC B= 319mv

So it looks to me that either Q6 is bad or it is Q8 that is bad. I tested Q6 (another 2SA1266) and it seems to be fine. I had another A1015 so I swapped it out just to make sure and Q8 still has HV on emitter and base. Here are measurements after the swap...

Q6 E= 350mv C= -36vDC B= -276.9mv
Q8 B= -35.7vDC C= 350mv E= -36vDC
Q9 E= -280mv C= 1.49vDC B= 338mv

Just wondering if someone can confirm my thoughts that Q8 is bad? I am pretty sure it is pointing towards that direction. Unless perhaps Q9 is the culprit? I had originally incorrectly thought that Q4 was part of the bias circuit but an MEF member informed me it is a predriver for Q5 and Q6. So is Q9 the main part of the bias circuit? Any help is greatly appreciated, thanks!
"A person who never made a mistake never tried anything new." -Albert Einstein

Roly

Quote from: DrGonz78Just wondering if someone can confirm my thoughts that Q8 is bad? I am pretty sure it is pointing towards that direction. Unless perhaps Q9 is the culprit? I had originally incorrectly thought that Q4 was part of the bias circuit but an MEF member informed me it is a predriver for Q5 and Q6. So is Q9 the main part of the bias circuit?

Q4 is also known as the VAS - Voltage Amplifier Stage.

Q9 is the temperature compensated bias circuit.

I wouldn't be too worried about exactly matching transistors as long as the substitute has sufficient voltage withstand (the sum of both rails or 70-something volts) and adequate current and power rating.  I've used BD139/BD140's in every position in circuits just like this (except the output pair, obviously) with no problems at all.  The very high negative feedback in these amps tends to sort out any minor difference like hfe.


Forgive me Doc, I may be missing something obvious but I can't see what the problem is.  You don't give the actual half rail voltage, but the DC feedback loop now seems to be re-balancing the half rail around +350mv.  This is a mite higher than ideal, but not all that unusual - the main point being that it seems to be re-balancing the output/half rail to something reasonable which strongly suggests that all the devices are okay.

Quote from: DrGonz78Q8 still has HV on emitter and base.

As it should have.

You seem to be bothered by the voltages around Q6 and Q8 but I don't understand why.  The emitter of Q8 is connected to the -36V rail, so its base will be about half a volt above that, and therefore so will the collector of Q6.

The emitter of Q6 is connected to the collector of Q8 so it will be close to the half rail (less the idle current drop across R51 which will be minimal, only a few tens of mV, this we expect the base of the driver Q6 to be about half a volt lower.  Assuming a half rail of around +300mV the base of Q6 at -300mV is about 600mV lower - what we expect.

In a quasi-complimentry the lower driver/output pair from a synthetic PNP transistor - the PNP driver makes it act like a PNP and the NPN following simply gives it grunt.

 
Basic NPN cell (left); basic PNP cell (right)

More detail here.

Sorry if I'm missing the point (we are in the middle of a 40ÂșC heatwave and bushfire emergency so I may not be at my sharpest at the moment); can you be specific about what you think is wrong, or how it is misbehaving?
If you say theory and practice don't agree you haven't applied enough theory.

Enzo

Gonz, buddy, WHAT'S WRONG WITH THE AMP?  What are we trying to fix?  You found a bad part and replaced it.  OK, so now what is left to fix? 

Q2,3 are the differential input pair. Q4 is your voltage amp, Q5,6 are drivers, and Q7,8 are outputs, and yes, Q9 is the bias circuit.

This is a common quasi-complementary circuit.  Note both outputs are NPN.  So I don;t know what you expect at Q8.  Emitter at V-, base at one junction drop more positive.  Collector at output bus voltage.  Q6 seems to be putting Q8 where it belongs, so what is the issue?   

DrGonz78

I guess it is fixed and ready to go!! Sorry to be so obsessed lol. I just hate seeing more than 250mv on the output and have not really dealt with quasi-complementary circuits before. I mean I probably have but did not understand them very well, obviously. Okay yeah I see now, just was confused by the output using the two NPN's. Thanks for setting me straight there as I sometimes get lost on my own road.  :duh
"A person who never made a mistake never tried anything new." -Albert Einstein

Enzo

The popular Peavey 400BH is a quasi.

Roly

Ah; the circuit has a small drafting error that might be the source of your confusion.  The line heading north from R46 beside J11-B and J12-B goes nowhere, not to the half rail (no dot).

The source of the output offset in this arrangement is the lack of match between the Base currents into the input diff. pair, Q2 and Q3.  If they were identical then the drops due to Base currents across R40 and R44 would be identical, and would cancel out, leaving an output offset of zero, but they aren't identical, giving a small offset between them.

Ideally Q2 and Q3 should be a matched pair, or they can be balanced by injecting a tiny extra current at the Base of Q3 from the +ve rail (filtered), or an identical tiny current sink at the base of Q2 to the -ve rail (also filtered) to bring it back to exactly zero, however while +300mV on the output isn't ideal it also isn't something I'd stress over.

You could try transposing Q2 and Q3, but I expect this will only trade a +300mV offset for a -300mV offset.  YMMV.
If you say theory and practice don't agree you haven't applied enough theory.

DrGonz78

Hey Roly thanks for that added bit of info too. Yeah that drafting error was my mistake. I will fix the post later on with updated schematic. Here is the original schematic from a A35 type Hartke. The A35 happened to be very very close to the B300 and I was trying to smoothly adjust the schematic. Looks like I got some drafting errors lol.
"A person who never made a mistake never tried anything new." -Albert Einstein

Roly

Thought that might be the case - just as long as you are aware.   :tu:
If you say theory and practice don't agree you haven't applied enough theory.