Stagg EUB stock preamp mod

Discussion in 'Electric Upright Basses (EUB's) [DB]' started by M1234, Oct 14, 2017.

  1. M1234


    Sep 1, 2016
    I’ve modded the stock onboard preamp of my Stagg EUB and have been asked to share details. I’ve decided to post it here, as I think this may help others also in the future.

    This is a cheap and easy mod which will kill almost all functions of the stock preamp. If you like hiss, a useless subbass pot, a noisy headphone out or a poor quality aux in then don’t read further. But if you do the mod you’ll have a clean and simple FET buffer onboard with very little power consumption (something like fifth or tenth of the original) and no unusable functions. Battery life will be amazing, most of the noise gone and you already control your tone on your pedals/amp anyway :).

    My original idea was to build a new preamp - basically a simple FET buffer. Then I checked the stock preamp and realized that its front stage is just that. So a short description of what I did: I left the two FET buffers as they are (there is one for each of the two piezos), and after their output signals are combined through two resistors I’ve cut a few traces on the PCB in order to prevent further stages from affecting the signal (and drawing power). Then I modded the connections of the existing volume pot, connecting it to control the output of the buffer stage. From there the signal goes to the output of the board and to the jack. That’s it.

    Please see the images, numbers indicate the following parts/areas:

    1 - new connection from the output of the buffer to the volume pot
    2 - the volume pot
    3 - old connection to the volume pot cut
    4 - old connection cut from the last stage to the output of the board (towards the jack)
    5 - power supply to the later stages cut
    6 - old connection to the volume pot cut
    7 - new connection from the middle of the pot to the output of the board (towards the jack)
    8 - the two resistors through which the output of the two buffers for the two piezos are connected into one signal. The PCB hole next to them is marked number 1 on the image of the other side of the PCB
    9 - the original FET buffers, the only part of the PCB that remains functional after the mod. FETs are K30A. R1 is 11MOhm. R3 is 3k6.

    IMG-4343 copy.jpg

    IMG-4339 copy.jpg

    IMG-4341 copy.jpg

    Of course it would be better to have something like a HPF-Pre onboard, but that’s a lot more work. And you’d still need two buffers because there are two piezos and connecting them to a single input -either in series or parallel- is not the best idea.
    With this mod you have your buffers onboard, piezo wires are kept short and once the active signal gets out of the instrument you can do whatever tone shaping you want externally.
  2. flaquito


    Jan 22, 2012
    Central Indiana
    I really like this mod! I'm guessing that by cutting the power to the later stages that it also cuts the LED? It'd be great to still be able to use that as a battery indicator.
  3. flaquito


    Jan 22, 2012
    Central Indiana
    So, I went ahead and did this over my lunch break, with one small change. I'm not a guitar tech, so I don't know if my change is a bad idea or not. I left the trace at cut 3 connected, since that's taking the output of the buffer directly to the wiper of the volume pot. Then I connected a single wire from point 7 (the output) to the new output (your input) of the pot. From what I know, this should be fine on the buffer output, since that's already the way it was. And it should maintain a more uniform output impedance. My understanding is that jazz bass volume knobs (and some others) are wired up this way.

    I thought it sounded rather good. Is there some reason I should change it to have the output on the wiper instead of the input?

    Also, I now have .2mA of current draw from my battery, vs the 20mA I had before. Should help just a touch!
  4. M1234


    Sep 1, 2016
    Yes, as you've seen since, the LED won't work with the mod. I don't mind, my power draw dropped so much since the mod that I didn't need to replace the battery once. I just keep a spare battery with me at all times and don't worry about it.
    Regarding the wiring of the volume pot you may be right. To tell you the truth I don't really know, I think I got the idea from the FDeck HPF Pre schematics from the pdf manual available online. He connects the buffer output to the top of the pot and the wiper goes to the output. You're saying something with the output impedance though...
  5. flaquito


    Jan 22, 2012
    Central Indiana
    Yeah, I was pretty astonished that it's using only 1% of the power that it was before.

    You mean I'm supposed to understand what I'm saying when I mumble about impedance? Ok, so I know output impedance is supposed to match the input impedance it's connected to. And impedance is resistance, but... not, because it's AC, not DC. And it varies with frequency, so it's not even consistent. Despite having taken electronics courses in college, it still feels somewhat like magic to me. But apparently "all" you need to measure it is a signal generator, oscilloscope, and calibrated variable resistor.

    At any rate, my bass sounds great now, and the battery should last forever. Thanks for posting this mod!
  6. M1234


    Sep 1, 2016
    No problem. I am glad it works for you, now this makes two of us. I've been using mine like this for about a year, no problems and only one battery. Okay, most of the time when I'm practicing alone I don't connect it so the battery is used only for one or two band practice sessions per week.
    Regarding the volume pot wiring I think the jazz bass is connected like that because it has separate volume pots for each PU. Somebody with more knowledge will probably explain :)
  7. flaquito


    Jan 22, 2012
    Central Indiana
    I hope somebody will explain. I asked why over in the EB electronics forum, and all I've gotten so far are "because this is how you do it" answers. Which I'm not satisfied with. I want to understand the reasons behind why we do it that way, but not many people seem to know that.
  8. Not a good idea to put the buffer output to the wiper.
    That way you can effectively shorten the buffer output and that might damage the previous amplifier stage.

    Maybe not if there was a resistor in series, but the correct way is buffer output to one side of the pot, ground on the other side and the wiper goes to the output jack. That way you have more or less constant resistive load to the previous amplifier stage to prevent overloading it and a voltage divider for the pot.

    A logarithmic pot is needed and to make the pot control "linear" to the ears. It matters which side goes to ground and which to the buffer output. If you are not sure try both kinds of connection, play and turn the knob a bit several times from one end to the other and you understand what I mean.
  9. flaquito


    Jan 22, 2012
    Central Indiana
    Thanks! That makes sense. That is the way it was wired up to begin with (I didn't change that part of the circuit, although M1234 did). I think because the two buffer outputs are combined through a resistor before going through the pot, so it can't completely short.

    And I did notice that it rolls off very quickly at the bottom end of volume, more like a linear pot. That difference in volume curve is the most sensible answer to "why" yet. So one of these days I'll open it back up and swap them around. Thanks!
  10. You are probably right about the possible short. Since both pickup outputs can have inverse polarity that would be even worse than a short, so the resistors in front of the pot should be high enough to avoid any damage.

    If you swap the outer connection at the pot keep in mind that it also swaps the silent and full volume sides at the pot.
    If nothing changes beyond the end swapping you have a linear pot which you can connect as you like since the mechanical control linearity does not change.
  11. M1234


    Sep 1, 2016
    With this mod we are using the front part of the stock preamp as-is. I doubt that the two piezos/buffers would be connected out of phase as they come out of the factory. (If they were I think we'd have a very weak signal).
    I can't find my quick sketch of the schematics at the moment but I think the two resistors through which the two buffer outputs are connected are 10kOhms.
    The volume pot in mine (pics above) works in the correct direction and with a linear "feeling" - I guess it's a tapered pot since this was the volume pot of the stock electronics as well.
    flaquito likes this.
  12. This is not a question of electronic design but can result as a time delay from the source of vibration to the two pickups. This is dependent on the distance, transfer medium and frequency components in the signal. And since we want a lot of different frequencies pass through and cannot put every vibrational source at the same distance to the two piezos, different polarities are possible.

    10 KOhms make sense as the resistors from buffer output to connection point of the two signals. If the pot has 10 KOhms too, half of the amplitude is lost that way, if the pot is 100 KOhms the loss is about 9 percent because of the two 10 KOhms coupling resistors.
  13. M1234


    Sep 1, 2016
    For a moment I was considering this :) - makes a lot of sense.
    I'd disassemble mine now to measure the pot and R12 - R13 but I can't - I'm practicing :)
  14. flaquito


    Jan 22, 2012
    Central Indiana
    I don't think propagation distance matters that much with audio. It definitely does with antennas, where we're dealing with the MHz-GHz range, but we're only worried about Hz-KHz. Rod Elliot, of Elliot Sound Products, an expert in audio electronics, says (talking about speaker cables):

    "The equal length cable theory is a bit of a myth really. You can prove this to yourself by running a 10 metre and a 5 metre cable in parallel (or other numbers that remain passably sensible grin.gif )

    If the 'rule' were true, then you would hear the difference, but the propagation delay is only nanoseconds, so you won't hear any change. Electric current flows along cables at around 2.25E8 metres/second worst case, so if one had a 1 metre and a 100 metre cable in parallel, this represents a 440ns (nanosecond) delay between the two. This would create a 1 degree phase shift (delay actually, but let's not split hairs grin.gif ) at a little over 6 kHz. Even at over 20 kHz, the shift is only about 4 degrees - you get more than that acoustically by moving your head a few millimetres.

    The cancellation caused by a 440 ns delay between parallel conductors is 0.08 dB at 100 kHz (which is negligible), so at normal audio frequencies it may be completely ignored."
  15. M1234


    Sep 1, 2016
    I think DoubleMIDI means natural phase relationships within the instrument. Anything is possible between the two piezos, there can even be moments when they're completely out of phase.
  16. Propagation speed very (!) much depends on the medium. Mechanical speed is a lot slower than electrical speed.

    BTW, one of the piezo preamps might be phase inverted because (!) the mechanical pressure is inverted on the two piezo sensors because the bridge wiggles from G-side to E-side and back. But that would result to have both signals in phase (but not necessarily with identical frequency content and amplitudes).

    Anyway, it is wise to design electronics with the worst theoretically possible input in mind.
    M1234 likes this.
  17. M1234


    Sep 1, 2016
    Here's the data guys. R12 and R13 are 16k, and the volume pot is 100k.
    I've just disconnected the volume pot from mine. Don't really use it anyway. Now my buffer coupling resistors look out through the output capacitor to the jack. During two minutes of testing didn't hear a difference. I don't know how my output impedance changed with this now.
  18. If you would connect the bass to a line input (around 10 KOhms to 47 KOhms) you would notice a differences in control linearity. Maybe not too much but at least some loss of volume towards the middle and lower regions of the pot.
    A lower output impedance means that this output can take a higher input load (= lower impedance) without any change of signal and amplitude. That doesn't hurt.
    Connected directly to a passive piezo crystal the impedance load may influence the vibrational behavior of the crystal and there might be a range that does not overload the output capabilities of the crystal and also helps to avoid too much self oscillation on resonant frequencies of the crystal. But since you are working behind the buffer stage that cannot influence the crystal and therefor a lower output impedance is better, since you can connect to many kinds of different inputs.
    (Well, there is still the amplitude you have to keep in mind, so the sensitivity of the input needs to match the output voltage of the bass.)
  19. M1234


    Sep 1, 2016
    Luckily I still have a clean signal without the volume pot. (I mean my amplitude is not clipping the input of my compressor). What I can't figure is if I am better off now without the volume pot impedance-wise. I certainly don't miss the function.
  20. It doesn't matter.
    The output impedance is the value of the resistors between buffer opamp or transistor output and pot input or output jack (without the volume pot).
    The resistance of the volume pot is there to reduce volume. That's not really part of the output impedance (unless you always set the volume pot to a rather low volume).
    A low input impedance may change the response of the volume control (linearity) and in worst case reduce the output volume up to almost silence, but the sound should not be affected since this is all after the buffer stage.