Woogie SS power amp troubles

[Steve Conner]

Sorry, I don't remember the band name. Just looked like a bunch of noisy art students. I remember the Woogie was used with a Big Muff on bass and the noise was incredible. Maybe it's the same one and that's how it got blown

You could always just replace the whole mess with a power amp kit from Rod Elliot. He's Australian, so it would still be authentic. The single supply might be tricky mind you.

[teemuk]

You could always just replace the whole mess with a power amp kit from Rod Elliot. He's Australian, so it would still be authentic.

Especially if you replace with this kind of design... ;-)
Simple 60 Watt Power Amplifier

[J M Fahey]

Agree, that's why rather than "repairing" it, my suggestions in practice amount to "rebuilding" it.
Not my usual approach, but in this particular case many hands seem to have been uncomfortably close to it.
Working on "remote control" I much prefer to know what's actually happening.

[teemuk]

For a -complete- rebuild I'd rather just throw a LM3886-based amp in it. The thing is likely going to need a new circuit board if built from the scratch anyway so what the heck. That would offer far superior amp with protections and all...

Otherwise, the rebuild circuits is with quite good probability very much identical to what's there already.

So, if the amp honestly just could be repaired with "old school" method of tracking down the problem and fixing it, then I don't see why rebuilding is even neccessary. The reason, I see, we're suggesting the original poster should do this and that to the amp is that so far a lot of the traced sketches haven't seemed to make much sense or provide a working circuit.

Now, I very much doubt the amp was like that when it left the manufacturer.

I'm interested to see the schematic, before that I'm afraid the suggestions we can offer are dubious at best.

[Swampy]

Ok, here's my latest tracing.

I don't have a camera but you guys would shit the bed if you could see inside of this thing...it's had a rough life to say the least.




I've left out the 330 ohm and alleged cap but I'll reinstall them as per your thoughts teemuk.

And to reiterate, I'd really apperciate if someone could explain the functions of Q8 and Q9 in this circuit, I know they most likely drawn in incorrectly but that's how they sit in the board at present.

Thanks,
Swamp

[J M Fahey]

Well, to fit there, the new amp should like the single supply available.
Most (99.9%) of modern amps need split supply.
That's why I suggest rebuilding it, using a known good schematic, to make best use of what's available.
Can you please post a couple pictures, specially showing the power board (both sides) plus the heatsinking?
Thanks.
EDIT: OOOOPS !!! Simulposting !!

[Swampy]

Well, to fit there, the new amp should like the single supply available.
Most (99.9%) of modern amps need split supply.
That's why I suggest rebuilding it, using a known good schematic, to make best use of what's available.
Can you please post a couple pictures, specially showing the power board (both sides) plus the heatsinking?
Thanks.
EDIT: OOOOPS !!! Simulposting !!

The power transistors are bolted to a 200 x 100 mm heatsink which is mounted to the chassis, none of the drivers are heatsunk.

[Steve Conner]

Q9 is the input stage. This kind of input stage is called a singleton. It takes the difference between the input signal and the feedback, and feeds it to Q8, which amplifies it enough to drive the remaining stuff, which has a gain of 1. It is often called the voltage amp or VAS.

It kind of worries me that the PNP driver doesn't have an emitter resistor. The negative half of the output stage should be connected the way it is in that link Teemu posted to Rod Elliot's site: a Sziklai pair with the ballast resistor connected to give feedback around the whole thing. Indeed the circuit in that link shows more or less how the whole amp ought to be.

[teemuk]

Q8 and Q9 are both voltage amplifiers. In the latest schematic they are drawn correctly.

Q9 amplifies the input voltage and also receives global negative feedback from the output to its emitter. This basically combines two transistor amplifier modes: first being common emitter (inverting) and second being common base (non-inverting), due to amp mode for negative feedback signal being a common base, “negative feedback” signal is actually in phase with the input signal. The "opposing forces" of common emitter amp and common base amp cancel each other, allowing the feedback to work like negative feedback. The feedback signal controls the overall closed loop gain of the amplifier and also servo controls the DC offset of the output.

Think of the whole thing as a giant opamp: You have non-inverting input (at base of Q9), inverting input (at emitter of Q9) and the output. The -open loop gain- of the circuit can be, and preferably is, very high but the overall gain is decreased to much lower -closed loop gain- once the global negative feedback loop is introduced.

The magnitude of global negative feedback signal, and therefore the whole closed loop voltage gain, is defined by the ratio of the voltage divider that is the feedback circuit. Roughly it’s 4K7 / 150R, but there’s also capacitor bypass for higher frequencies so the gain will be lower at such. The loop is also totally AC coupled by the capacitor in series with R13 so gain for DC is just one. However, it’s also a RC circuit that decreases gain when the signal frequency decreases. To ensure proper operation you want that to filter operate below audible bandwidth, or at least at the very limits of it.

Q8 is another common emitter voltage amplifier that cascades to input amplifier to increase open loop gain and to invert the signal yet again so that output will be in phase with the input signal, something that is mandatory for this circuit setup to work properly. Q8 is loaded by a so-called bootstrap (two 4K7 resistors and a capacitor from their interconnecting node to output). This circuit act sort of like a constant current source providing more linear loading for the stage, it can also, figuratively speaking, “lift itself up from the bootstrap”, meaning it works like a voltage pump and allows peak voltage swings (at the intersecting node) that exceed the rail voltage.

In between you have two feed points for push-pull emitter follower circuit of the output that amplifies the current, allowing the amp to drive a low-impedance load with high voltage signals. The voltage amps couldn’t do it alone. A bias control circuit, which sort of like a discrete zener diode, is placed in between the inputs to create little bit of voltage potential between each half of the buffer and therefore ensure a little bit of idle current keeps transistors off their crossover regions when the amp idles.

[J M Fahey]

Hi Steve..
When you have a single output transistor (in the lower half), the preferred connection is as you say (the Rodd Elliott's example). Agree.
When you have multiple paralleled devices, they are usually connected as I have shown, so ballast resistors main function becomes equalizing current sharing.
Agree not the best theoretical solution, but good enough to have been used by thousands of amplifiers.
Just check those old Randalls and many more.
Connected as shown in some examples posted above, simply in series with collectors, ballast resistors have no useful effect.
If you are worried about the apparent lack of internal feedback caused by the lack of the (upper) "emitter resistor" of the Sziklai pair, remember it is still driving the speaker load which is usually , say, 4 to 8 ohms, so the loss of .1 to .47 ohms is negligible and the pair still behaves as a big virtual PNP emitter follower.
Maybe in a Hi Fi amp the TIM or some other parameter is affected, but on Musical Instruments such effects stay unnoticed.
The theoretical best solution would be to provide individual ballast emitter resistors , *and* same amount and value of them paralleled in the collectors (Sziklai new "emitter") so you can connect the PNP driver emitter there, but most amp makers just don't want to pay for them.
At most in some cases they provide *one*, in series with *one* collector, obviously not for current sharing but as current sampling for the short protection. (Peaveys, Lab series, etc.).

[Swampy]

Cheers Teemuk, my understanding is greatly increased thanks to your explanations.

Now I've managed to identify Q10 as responsible for pulling the V+ to about 20 volts. Can you explain it's function in this circuit? It's possible again that it's placed incorrectly on the circuit board.

Cheers,
Swamp

[teemuk]

As is drawn, it seems to be a emitter follower current amp serving a function in a circuit that determines the overall DC bias of the amp.

The input transistor's base needs to be biased to a DC potential that is about half the rail-to-rail voltage, which in such unipolar rail circuit means half of that 60VDC. In many amps you see this biasing done simply with a resistive voltage divider but in this case, the circuit is more complex and somewhat different, actually consisting of two divider parts.

First resistive divider is at the base circuit of Q10. The divider determines the DC voltage potential at Q10's base, thus DC voltage potential at Q10's emitter is always pretty much exactly one diode voltage drop lower than that. The trick is that an emitter follower amplifier circuit can source very high current in order to keep emitter steadily at that level. The emitter then acts as a reference voltage source to another divivider, of which's intersecting node then biases the base of Q9.

So, all in all its a voltage reference source with very low source impedance and a convenient way to implement a DC offset trimmer.

[Swampy]

Yeah good, that resembles my intuitive idea of it's function.

So if I yank Q9 V+ goes to 60 volts and I get around about half V+ at the node which Q9s base connects to, but once I put it back in V+ drops to about 20 volts (With about 10 volts at it's base, as we would expect)

If I yank Q10 V+ goes back up to 60 volts. but when I try measure voltage at the base of Q9 V+ drops to 20 volts (I'm metering with two voltmeters).

With both in circuit, Q9 B - E junction is only about .2 volts? Q9 tests good out of circuit.

Any thoughts?

Swamp

[teemuk]

Yes. Replace Q9. Wouldn't be the first time when a transistor tests good off circuit but fails under load.

[J M Fahey]

So if I yank Q9 V+ goes to 60 volts and I get around about half V+ at the node which Q9s base connects to, but once I put it back in V+ drops to about 20 volts (With about 10 volts at it's base, as we would expect)

If I yank Q10 V+ goes back up to 60 volts. but when I try measure voltage at the base of Q9 V+ drops to 20 volts (I'm metering with two voltmeters).

Why are you "yanking" transistors? (I understand you mean pulling them from the circuit board)

EDIT: sorry, I was sort of low in patience today.
Anyway, my personal view still is that the shortest and less frustrating path *in this particular case* is to rebuild it following a simple, no frills (no surprises) schematic which I may suggest if you wish.
That's why I asked for a couple pictures, to be certain it can be built on the original PCB .
We don't need to reinvent the wheel here, nor win Nobel Prizes, we just want to play a chord and hear it through the speakers.
By the way, is it a combo or head/cabinet?
Speakervtype and impedance?
One of the sketches showed part of a relay, what is it for?
Good luck.