Help Wirring Bass Guitar on board preamp

[ilcaccillo]

Hi,
I'm repairing at the moment an Oakland & Co Bass. I'm having a lot of difficulties in trying to wire the active preamp circuit.
There's not much info around on these bass guitars, and no schematics.
The Active Bass preamp circuit uses 2 IC's, the TL064 and a TL061.
It seems a similar layout to a MEC bass preamp, although the MEC uses TL064 and TL062, couldn't find a schematic on the MEC unfortunately.

Someone in the past messed around with the wiring of the active circuit and some wires were cut also.

I did a reverse engineering on the circuit and drawn a schematic of it (attached) , with the color of all the wires that come out of the board.
I would like to ask if someone could help me out of point me in the right direction of where to connected the wires.
The active part has a DPTP 3 way switch and a Dual gand 56K Pot.
The passive part has an 250K volume pot and an 500K Tone pot. The passive circuit is wired and working (picture of the current layout attached).







I just don't really know where to connect the wires coming out of the preamp board and how to integrate this circuit into the passive circuitry already in place.

I know how to read schematics, and I'm able to build fx pedals and other audio electronic projects although I'm still learning and I don't know electronically how an amplifier, buffer or active tone circuit works. That's why I'm asking for your help.

The wires coming out of the preamp board have this colors:

Yellow
Green
White
Blue/Black
Grey
Black (ground connection to VOL Pot case)
Brown
Yellow Stripped (ground connection to Tone Pot case)
Orange stripped
Orange
Violet
Blue Stripped
Blue
Red (+Voltage from the Batteries)
PAD??? (pad on the pcb , it seems a wire was cut from here)

Some of them should connect to the 3 way DPDT switch (the switch that turns the active circuit on or of)
Others to the Dual gang pot
and the rest to the others part of the circuit like the pickup selector switch, volume pot, tone pot, output jack

The 3 Way DPDT Switch has already a Jumper soldered between 3 pins, picture attached.



It would be great if someone could have a look into the schematic and give some tips and direction

Thank you so much

[ilcaccillo]

I redrawn the schematic with Eagle and replaced the hand drawn one in the first post.

Here are also some pics of the PCB top and bottom:

[ilcaccillo]

[dwmorrin]

You'll need to connect the Brown wire to a V- power source before you can do anything with this circuit.
The orientation of polarized components C10 and D2 are a clear indicator that this should be a voltage more negative than your ground voltage (0V).
I don't think this is supposed to wired 18V, 2x 9V in series. The batteries should be "grounded" in the middle so you have a +9V output and a -9V output. This is called a "bipolar supply."
Your "+18V" pad should really be a "+9V" pad, and Brown should be -9V.
You'll need to connect the battery snaps that way. Where the two snaps are joined in the middle should be connected to the 0V/Ground node.
When powering a circuit with two batteries this way, the "ring-terminal-as-switch" trick for the input jack doesn't work. The circuit will still draw power when the ring terminal is open circuit. Typically, a DPDT switch is used to open circuit both battery connections. I would wire up the DPDT to do that. Remove the wire that is already on the jumper and give each battery its own pole of the switch. Maybe there is a missing DPDT from where ever this was originally installed?

Once you've got this circuit powered correctly, it should be able to amplify and pass audio so you can mess around with it.

PAD?? is the input for IC2, and I would suspect this is the passive signal input due to the high impedance. None of the other inputs have a high impedance, so PAD?? is the only spot I would try connecting the input to.

ORANGE_STRIPED looks like the main output.

IC1 needs to get connected together to form a EQ circuit. All those colored wires probably go to pots.
Yellow, violet, blue-striped, blue, green, blue-black, and grey are all wires that connect through pots or switches to make IC1 into an EQ. OK, that's really just a good guess, but I can't think it could be anything else.
As shown, IC1A and IC1C will just be a fuzz tones, but with global negative feedback they will be low pass filters.

Get the power, input and outputs sorted first and start messing around from there. Maybe someone else can figure out the EQ circuit.

[MarkusBass]

The batteries should be "grounded" in the middle so you have a +9V output and a -9V output. This is called a "bipolar supply."

I don't think that this is the case. There are some exceptions but usually guitar preamps supplied from 2x9V batteries have them in series and the ground rail is obtain with so called "virtual ground" circuit (usually two resistors and a capacitor). Like in this Carvin premp: http://carvinmuseum.com/pdf/guitarba...itarmodule.pdf
Of course you may discuss whether the rail should be marked +9V, GND and -9V, or +18V, V/2 and GND. This is actually the same. But there is important difference between these two solutions. With symmetric supply you would need to disconnect two batteries when the preamp is not used. You cannot do it with a standard stereo Jack. With +18V (and virtual ground circuit) you can do it. That's why "bipolar supply" is never/or very rarely used in guitar electronics. Additionally, in this case +18V is connected to the board. With "bipolar supply" one battery would work all the time so I don't think that this is the case. The problem is that I don't see the "virtual ground" circuit. Is it possible that to made a mistake while drawing the schematic? I think so because on the schematic there are 16 resistors but on the PC board photo I see 17 resistors.
For starters, I would connect BROWN to RING_OUT_JACK, assume that PAD??? is input, plug it in and check if you get output signal on ORANGE_STRIPPED. The rest of the circuit is most probably related to the equalizer part. I would leave it for now until you draw a better schematic.

EDIT: I just thought that if D1 and D2 were Zener diodes, they would form kind of virtual ground circuit. But much better results could be obtained with just 2 resistors so maybe I'm wrong.

EDIT2: I'm surprised with the large values of the capacitors (C1, C2, C3). Are they correct? Maybe they used "state variable filter" circuit: http://en.wikipedia.org/wiki/State_variable_filter ?

Mark

[J M Fahey]

Ok, thanks for the reverse engineering effort, I thyink we still have a few details to clear.

1) your bass is already wired passive, and obviously it works, so for now it does not *need* the preamp.

Just thinking aloud, if you wanted to use it you would have to pull current wiring, probably direct each pickup to each preamp section (1/2 TL0x4 each?), somehow connect the selector switch to choose between individual preamp outputs (1/2 TL0x4 each?) and feed it into the active mixer (TL061?) .
Somehow some kind of tone control, either passive or active will work too.

2) if so, yellow and green would be the preamp inputs, each receiving 1 pickup, each preamp would be IC1a+b and IC1c+d .

That said, the schematic as shown raises a few concerns, you should recheck schematic against actual PCB.

3) Op amps NEED a DC path between output and - input so either C1 and C2 are actually resistors, or a resistor (a pot?) is wired in parallel with them.
A pot makes sense because it would become an active gain control.

4) Op Amps NEED biasing at the + input (IC1 pins 3 and 10, IC2 pin 3), yetv they are shown connected to "ground".

Either said "ground" is a virtual ground, usually created by 2 resistors and a cap end to end of the battery, in which case battery b- terminal is ground, or it's actual ground but then battery must supply +/-9V .

I think the first (single supply) case is most probable, doubly so because output jack switching is easy.

In a nutshell, I think the schematic must still be perfected.

One useful way is what the pedal cloning guys do:

they somehow kludge a fixed mount (a tripod? a couple books? a piece of wood?) so the camera is fixed (that or they have incredibly firm hands) and take pictures of both sides of the PCB (what you did) but making certain that the board itself is exactly perpendicular to the camera, and at exact same distance , so one picture can be flipped , set to 50% transparency (so you can see through it) and superimposed exactly over the other.

Imagine you look at your PCB from the component side but the PCB itself is made out of transparent acrylic or at least semi transparent tracing paper. so you can see parts and tracks at the same time.

That way it's much easier to trace what goes where, and many can work at home, at the same time, tracing the circuit and comparing notes.

As a side note, the values I see suggest that the original pickups (not sure they are still there or somebody replaced or rewound them) were very low impedance.

Let's see what the pickup guys think.

David Schwab ... Kojak ... Salvarsan .... Big Tee ... are you reading this?

[MarkusBass]

Is it this bass: Oakland bass | TalkBass.com ?

The capacitors are way to big. IC1A and IC1C work as integrators and they most probably won't work in standard arrangement. This is either a mistake, or something more exotic like "state variable filter". The advise to make transparent photos from JM is very good.

EDIT: on Talkbass one of the guys has similar bass. What if you ask him to make a photo of the preamp in his bass?

Mark

[dwmorrin]

Whoa fellas... let's not confuse the OP. He's not trying to build something from scratch.
For those in favor of a single 18V supply, you are asking him to remove C10 and D2, add 2 resistors and a cap to make a low z +9V rail, and then carefully check for traps like how that puts dc on the PAD?? (input) node.
Seems a bit more complicated than just wiring the battery snaps as they should be, no?
Using +/-9 means the OP can get straight to work on this thing.
Converting to +18 means putting 2 extra resistors and a cap on an already crowded board. Are you sure that's the advice you want to give?

This guy just wants to get it working. Let's try to keep it to that... and then we'll redesign it.

And yes, the 18V switched via the input jack is what any of us would design if we were building this from scratch... but I've worked in guitar repair a long time and I've seen many a +/-9V bass preamp, mostly older circuits like this, some use double switching normally open jacks instead of open stereo jacks.

The OP could help clear up some of this by explaining if the active circuit was always in the same bass or if this is some sort of transplant job or what... more details the better.

[MarkusBass]

I still think that this preamp is a little bit different from typical bass preamp. In the 70-ties Alembic used preamps based on "state variable filter" principle. In general it was kind of active "Varitone" circuit. Several companies copied their preamp. Maybe this is one of such preamps. Here is the original circuit:



If you look carefully, you will find large capacitors with opamps working as integrators, dual 50k pot. What is missing on the OP schematic is the virtual ground circuit. But, have I already mention that on the schematic one resistor is missing? I would make sense to find out what is the role of the resistor (and add it to the schematic).

Here: http://youtu.be/rFDSe6p6qK8 is a demo of a guitar with such a preamp.

Mark

[eschertron]

What is missing on the OP schematic is the virtual ground circuit.

Mark

Did the schem posted change (been known to happen)? I see D1 and C9 between GND and +18v, and D2 and C10 between 0v and GND; almost verbatim what you posted as an example. So Brown is connected to 0v (-9v from GND)?

[J M Fahey]

I still think that this preamp is a little bit different from typical bass preamp. In the 70-ties Alembic used preamps based on "state variable filter" principle. In general it was kind of active "Varitone" circuit. Several companies copied their preamp. Maybe this is one of such preamps. Here is the original circuit:

[ATTACH=CONFIG]32689[/ATTACH]

If you look carefully, you will find large capacitors with opamps working as integrators, dual 50k pot. What is missing on the OP schematic is the virtual ground circuit. But, have I already mention that on the schematic one resistor is missing? I would make sense to find out what is the role of the resistor (and add it to the schematic).

Here: http://youtu.be/rFDSe6p6qK8 is a demo of a guitar with such a preamp.

Mark

I think that you hit the jackpot or are very close to it.

Nobody's trying to redesign anything here but work with what's found there, as close as possible.

And the batteries in series making single +18V were there from the beginning.

The "Alembic" explains what instantly worried me: the lack of a DC path between output and - in , path which in this particular circuit is closed from end Op Amp to 2nd one through R6 .
Also the too low input resistors are explained because they are actually used with series pots to set frequency ... and are fed from buffers, not straight from the pickups.
Alsio the dozen mystery wires: now justified because besides pickups they must also go to a lot of frequency and gain setting pot connections.

All it needs is some virtual ground source ... which might be explained by the mystery resistor and rechecking the circuit, we might have missed something else.

EDIT: that said, and not trying to sound defeatist, not sure how much we can advance with this one because we don't have the original pots and external wiring.
As in: not sure about how many hours can be dedicated to solve this riddle, because even having a perfect PCB schematic we still are left with many unknown variables.

What's the actual target?

Did the owner want it reconnected?
Without extra data (such as an original schematic, including original wiring) it's more than he will gladly pay.

Is it for personal use?
Go ahead, dig original or similar schematics or get a user post a gut picture ... although I guess the original pots aren't there any more.

Personally, I would straight build the "Alembic style" preamp posted above, even on perfboard , or spend a rainy weekend afternoon with both schematics side by side, and see how to use the Oakland board in an "Alembic" preamp.

They already match 80% by the way

There must be some reason why an earlier owner said WTF and just wired it passive

[dwmorrin]

I think I caught an error in the schematic.
I'm seeing C5, the output of the pickup buffer, connecting to 2 resistors. And neither of those resistors are going to pin 13 of the TL064.
What I'm seeing is a resistor to pin 6, and a resistor to pin 9.
Please double check this... this would solve the riddle if I'm correct.

The reason I started looking there anyways is that IC1B would be the input of a state variable filter. The dual gang pots will surely connect to the yellow and green wires... details yet to be worked out.

[MarkusBass]

I think I caught an error in the schematic.
I'm seeing C5, the output of the pickup buffer, connecting to 2 resistors. And neither of those resistors are going to pin 13 of the TL064.
What I'm seeing is a resistor to pin 6, and a resistor to pin 9.

I caught the same mistake. The reason for this is that the OP made a mistake while drawing the schematic: C2 is not between pins 8 and 9 of IC1 but rather between 13 and 14. So IC1C is actually IC1D and most probably IC1D should be IC1C. Does this solve the problem?
It's a pity that the OP seems to be not interested in solving the problem. This would be interesting due to the unusual schematic and maybe historical value of such a preamp.
Anyway, we already know a lot:
- this seems to be a copy (slightly redesigned) of Alembic "state variable filter" preamp. I think that Wall also used similar preamps.
- BROWN is -9V rail (if we assume that they are +9V, GND and -9V).
- it seems that the batteries were from beginning in series and the preamp was supplied with +18V. But then the virtual ground circuit is missing. So the idea from dwmorrin about the symmetric power supply may be correct (the batteries were connected differently in the beginning). For this you would need a special Jack with double isolated switches. Most probably Switchcraft manufactures such a Jack. What is more interesting is that Alembic used similar idea: they switched +9V with a Jack with single isolated switch (which very often failed, BTW). This would explain why the virtual ground circuit is missing. The OP could simply tell us what Jack is currently installed in the bass and this would solve the problem. In this case, D1 and D2 are simple a protection diodes (against inverted polarity of batteries).
- one resistor on the schematic is missing. There are 16 resistors on the schematic and 17 resistors on the board. 17-16 = 1 is missing.
- the integrators circuits with 22nF capacitors are part of the VSF. Their inputs are connected to dual 2x50K pot (and not 2x56K) like on Alembic schematic, The integrators must be connected in series so either IC1A is first, or IC1C (actually it should be IC1D - this is drawing mistake). So there are just two options. I would draw them both as see which one works. For me it seems that IC1A should be first because there is kind of a boost circuit on IC1B (probably switchable with a switch).
- IC2 is an input and should be drawn differently (currently it is drawn as equalizer circuit). The input is slightly different than on Alembic schematic. It is just a simple voltage follower.
- ORANGE_STRIPED is most probably the output af the preamp.

I would redraw the schematic using current findings and then most probably we would have to clarify only 5-7 wires more.

Mark

[dwmorrin]

Here's a quick sketch after reviewing the photos.
I think this is the topology of the preamp.

I only count 16 resistors on the circuit board from the photos. Maybe you're counting 1 of the 2 diodes as resistors since they're all vertical mounted and look similar.

[MarkusBass]

Very nice. Additional resistor is in the lower left corner of the board, beneath the yellow wire. I think that this is R1 on the schematic so some other resistor is missing. If you count them, there are 17 resistors on the PC board. The missing resistor may be critical to the final version of the preamp.

I see few minor problems on your drawing:
- there is one more resistor in the input buffer (R15 - to the ground after 100n cap),
- there is only one resistor to pin #5 of IC1B but in reality there are two resistors (R5 and R6) - maybe connected to the switch,
- on your drawing there are two resistors in series to pin #10 of IC1C. In reality there is only one resistor there,
- I'm not sure about the direction that IC1A is connected. It should be rather in series with IC1D. The basic idea of the SVF is that there are 2 integrators in series with variable resistors (pots) on their inputs.
- If you draw the pot on IC1A and IC1D inputs, the drawing would be more readable. To make it even more readable, you could add wire colours listed by the OP.
- on your drawing there are only 14 resistors. After adding the one that is missing in the input buffer, at least one is still missing,
- have you added C8 (22uF) to the drawing? This may be a part of the power supply, or maybe not,

Anyway, you are very close to the final schematic. Great job.

Mark