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Building a passive splitter + resistance mixer- tell me, will this work?

Discussion in 'Effects [BG]' started by Nic., Apr 29, 2010.


  1. Nic.

    Nic.

    Aug 28, 2009
    Singapore
    Okay, this is what I want to do. My current signal chain is:

    Bass -> Tuner (pitchblack) -> Foxrox Octron -> Rocktron Silver Dragon -> Zvex Distortron -> Electronix Submarine

    The Dragon and Distortron give very light overdrive, I'm mainly using the dragon's tube stage to warm the sound a little, and the distortron to add a bit of drive. When I need distortion, I hit the other switch on the Dragon.

    The purpose of the Octron is to fill out the sound, like doubling the guitar line.

    However, putting the Octron in front of the Dragon and Distortron drives it into distortion, while putting it after, I feel that it affects the tracking when I kick in distortion.

    What I'm thinking is that I'll put the Octron parallel, i.e. splitting my original signal into two parts, one for overdriving, and one to transpose one octave up with the Octron.

    While I'm at it, I want to have the option to split it into 3 parts, so that I can have the tuner continuously on while playing. I'm picking up an unlined fretless soon, and I think it would be helpful to have the tuner on so I can see once I'm off tune to help train my ears.

    The general idea I get from the forum is that a Y splitter works so long as the input signal is buffered. I use active basses. So this should not be an issue.

    From http://www.tkk.fi/Misc/Electronics/circuits/linemixer.html I found a diagram for a passive resistance mixer. I built one to feed two basses into one amp and it seemed to work fine.

    So I'm thinking of building a pedal like this:

    3-chan-looper.

    Splits the signal into 3 outputs, one for pedal, one for the overdrive circuit, one for the octave up circuit; later, combines them all together. Should I feel the need, I can make the octave up circuit into a "clean" circuit and run it parallel with the overdrive circuit in full distortion.


    Is there anything that I'm missing out? Are there any fatal flaws in this design?
     
  2. Silent Fly

    Silent Fly Supporting Member Commercial User

    May 8, 2006
    London - UK
    Owner/designer [sfx]
    Interesting approach to signal mixing...

    There are two problems:

    1. Split: the signal source sees 3 pedals in parallel. This lowers the impedance. If the pedals connected to the input have relatively low input impedance, the pedal that delivers the signal may start to cut some frequencies.

    2. Mix. This is trickier to explain but far more dangerous for the pedals.

    The vast majority of pedals have low output impedance. In other words, apart from the signal they produce, their output is more or less like a short circuit.

    Let’s imagine that all the pedals are connected and all the pots of your circuit are turned up.

    To simplify, let’s also assume that only the first pedal is delivering a signal.

    The signal goes out from the first pedal and it finds not resistance from the pots, it continues to travel along the wire and if finds that the other two pedals offer almost a short circuit. All the current delivered from the pedal is short circuited in the output of the other two.

    The first pedal is in overload, the other two pedals have the output stages that absorb energy instead of delivering it.

    I may find this reading interesting: http://www.rane.com/note109.html

    An except: “Here is the rule: Outputs are low impedance and must only be connected to high impedance inputs -- never, never tie two outputs directly together -- never. If you do, then each output tries to drive the very low impedance of the other, forcing both outputs into current-limit and possible damage. As a minimum, severe signal loss results.”
     
  3. Nic.

    Nic.

    Aug 28, 2009
    Singapore
    Thanks!

    1) Ah didn't think of that... If I were to string a resistor in series before the split, would this help? That should increase the impedance, regardless of whatever pedals are connected to the effects out, right?

    2) Mmm I'm aware of the short-circuiting outputs together bit, but I figure as long as the pots are put to at least some decent resistance, it should be fine. Maybe, to be safe, I'll add a 10k to each effect in? When I built my mini mixer previously, at first I just used what was pretty much a Y cable... which gave a lot of problems, but connecting the inputs to a pot each turned out fine unless I turned up the pots, like you said.
     
  4. Silent Fly

    Silent Fly Supporting Member Commercial User

    May 8, 2006
    London - UK
    Owner/designer [sfx]
    It would inevitably reduce the volume because of the voltage drop across the resistor in series. :( More pedals, lower impedance, higher load, more current, higher voltage drop.

    If the pedal that delivers the signal has low out impedance and the pedals have a high input impedance it should work without any resistor.

    In other words: “Outputs are low impedance and must only be connected to high impedance” [devices]

    “to be safe” you need an active mixer. Solid signal, no voltage loss, clear tone and more importantly no risk of damage.

    A resistor in series should help (but I am not taking any responsibility if something gets damaged).

    To reiterate: “Outputs are low impedance and must only be connected to high impedance” [devices]

    In the pedals world, I wouldn’t say that 10k ohm is high impedance. ;)

    ----

    I suspect this is quickly becoming a one to one conversation that may not interest other people. If you need more advice it might be a good idea drop me an email (the address is in the web site in my signature).
     
  5. bongomania

    bongomania Commercial User

    Oct 17, 2005
    PDX, OR
    owner, OVNIFX and OVNILabs
    Take Max's words seriously--passive mixers are a terrible idea in most cases. The only time they work well is when combining keyboards or other active, low-output-impedance devices--especially when those devices have fairly high output. For bass guitar, and typical effects, passive mixers suck bigtime.

    Adding resistors is the wrong idea.

    Passive splitters are almost as bad. There are more circumstances when they will work fine, but still having plenty of instances where they will do more harm than good.
     
  6. Jazz Ad

    Jazz Ad Mi la ré sol Supporting Member

    Looks like we have another customer for a Wounded Paw Super Blender.
    [​IMG]
     
  7. Nic.

    Nic.

    Aug 28, 2009
    Singapore
    Silent Fly & Bongomania- Thanks for the advice! I figured maybe I could get away without using any active electronics, guess not then. I'll try working with some FETs I had lying around then, wanted to avoid these since my past work with FETs weren't too good.

    Jazz Ad- That isn't the full diagram, right?
     
  8. UncleFluffy

    UncleFluffy

    Mar 8, 2009
    California
    Head Tinkerer, The Flufflab
    use a jelly-bean opamp, less hassle than FETs.
     
  9. bongomania

    bongomania Commercial User

    Oct 17, 2005
    PDX, OR
    owner, OVNIFX and OVNILabs
    If you can't find a jelly bean, then a Valnott will do. ;)
     
  10. Jazz Ad

    Jazz Ad Mi la ré sol Supporting Member

    Oh yeah the mighty Valnott.Oh yeah the mighty Valnott.
    It's a signal path diagram. It basically does what you want to achieve but it's active and fully buffered.
     
  11. Nic.

    Nic.

    Aug 28, 2009
    Singapore
    Thanks for the advice guys!

    Was extremely busy over the weekend... so I looked up some stuff on buffering, and I guess doing a unity gain over an op amp should be sufficient. I've a bunch of TL074s lying around, so I'm thinking of using a pair of them for this.

    [​IMG]

    The last time I worked with op amps, my circuit got rather noisy and in the end I didn't have time to isolate the noise and perfect the circuit. I think theoretically this circuit should be workable, though. I added in a 2pdt so that I can run the octave or drive in parallel or series, realized I might still need the octave in front when I'm doing the octave down to avoid downtuning.

    Will I be able to run both TL074s off the same 9v supply? I'm worried about the current draw.

    The Tuner Out loop is not going to have a receive when the tuner is in operation. This should not be an issue, right?
     
  12. UncleFluffy

    UncleFluffy

    Mar 8, 2009
    California
    Head Tinkerer, The Flufflab
    You're probably going to want to capacitively couple your inputs and outputs, especially if you intend to run this off the same supply your pedals are using. Also, why the capacitor to ground on the input? I'm guessing that the '+' means it's an electrolytic, therefore fairly meaty. Might want to consider what that would do to your frequency response.

    Here's a page with some simple-but-workable opamp mixers you might want to look at for inspiration:

    http://www.all-electric.com/schematic/simp_mix.htm
     
  13. Nic.

    Nic.

    Aug 28, 2009
    Singapore
    Thanks for the link!

    Ah my bad, I was thinking I needed a capacitor to get rid of any DC, but it kinda slipped my mind that it should be in series and not to ground. That's what you mean by capacitively couple inputs and outputs, right?

    If so, I'm thinking I'll put one all the way at the start before the voltage splitter, and one all the way at the end after the last op-amp. Otherwise, I'll be cutting out the 4.5v required, right?

    I've to admit I don't really understand buffering much, so what I was doing is throwing in a unity gain at every point so it feeds off the amp-ed signal rather than the original signal.

    All the splitters I've found are based on transformers. I'll rather use an op-amp. Is it possible for me to simply use the original signal to run 3 op-amps? Then, I can cut out the TL074 1a in the original circuit... seems reasonable to me, especially since I use active basses.

    Based on the page you gave me, it seems to make sense that I can do away with the buffering at the loop receive.

    I revised the circuit a little, do you think this is sufficient? I'll probably do it like the one on top.

    Also, something I wanted to check- if I'm connecting straight to the inverting input, do I still need to center the input in the middle of the voltage range of the opamp?


    [​IMG]
     
  14. UncleFluffy

    UncleFluffy

    Mar 8, 2009
    California
    Head Tinkerer, The Flufflab
    Couple of questions:

    - I'm not sure what the switch in the middle is doing. Is it allowing you to switch the octave/drive combination between series and parallel?

    - Where is your ground coming from? (i.e. what's the relationship between the battery contacts and the ground signal in the schematic?)
     
  15. Nic.

    Nic.

    Aug 28, 2009
    Singapore
    Yes, the switch is for series/parallel...

    I would be linking the signal ground with the battery ground. That's how it should work, right?
     
  16. UncleFluffy

    UncleFluffy

    Mar 8, 2009
    California
    Head Tinkerer, The Flufflab
    Normally in op-amp circuits the signal ground is half-way between the +ve and -ve supply rails. Remember, the input is AC so 1/2 of the waveform is "below ground". If your circuit can't swing below ground potential then that part of the signal will get lost.

    Do you have access to a copy of "The Art of Electronics" ?
     
  17. Nic.

    Nic.

    Aug 28, 2009
    Singapore
    Ah I see... no, don't have any access to any reference books for now at least. Some things I would like to clarify,

    1) Should the signal I deliver to a pedal be centered around 0v? That's why although I set the input to 4.5v before running through the opamps, I threw the capacitor between the opamp and the send.

    2) The signal coming back from the pedal should be centered around 0v, right? So it's fine to put the pot to ground at the receive?

    3) I guess that means that I have to link the 4.5v to the 3 receives?

    4) Likewise, the ground connected to the non-inverting should also be 4.5v, right?
     
  18. UncleFluffy

    UncleFluffy

    Mar 8, 2009
    California
    Head Tinkerer, The Flufflab
    *corrected*

    The opamp inputs you have at ground should be at 4.5V. The pots can go to 0V. The opamp supply rails can go to +9 and 0.

    It's often easier to think in terms of +4.5 / 0 / -4.5 than +9 / +4.5 / 0. Looking at it that way may make things clearer.
     

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