Simple Y cable passive mixer, help needed

[AmpliFuzz]

I want to experiment with blending clean signal with the distortion and fuzz boxes on my bass pedalboard, while keeping it as simple as possible. The idea would be to have a passive Y splitter cable coming out of the compressor pedal (Behringer CL9). One side (clean signal to retain lows) would go straight to the last pedal in the chain (Behringer Bass EQ), the other would go to the distortion boxes, and ending in the Behringer Sansamp clone that's on all the time, and from it to the Bass EQ pedal.

All the pedals are of the buffered bypass variety or always turned on. No need for switches, and the 'level' control on the BDI21 should be enough to control the relative blend of clean and dirt, with the level of the graphic EQ would become the master.

I just need to know if there's something I overlooked, and the best value of the two blocking resistors at the tail end of the Y mixer cable that goes into the last pedal to make it one signal again.

Thanks!

[ashcat_lt]

As long as the signals are a) the same polarity or b) sufficiently different from one another that phase cancellation is unlikely, it should work fine...unless you end up having to attenuate the clean side...

Assuming youre driving a good high Z input, the two "mixing" resistors in series will make up the bulk of the impedance seen by each of the sources, so they want to each be at least 5x the larger of the two out-Zs and/or at least half the lowest impedance that the weaker of the two can drive.  You're probably safe with 4.7K each, but a little bigger shouldn't hurt much. 

Of course, matched resistors means something like 6db attenuation for each, so you might want a bit of makeup gain at the end.  This EQ you're running into should be able to handle that.

Edit - a trim pot in the 10-20K range might be ideal, and give you a chance to fine tune the mix a little more than the sansamp can do on its own.  Like if the sansamp can't get loud enough or something...

[AmpliFuzz]

The signals are in phase, I experimented with this setup with a Boss LS-2, but I'm shooting for the most compact board possible. The BDI21 has the higher out-Z at 1k, so 4.7k should be good. What would change from 4.7k to 10k, further volume attenuation?

[PRR]

Since you don't actually know the interface impedances, and resistors are really cheap (you should own a 100-pack all common values), I'd just tack it up loose with 2K 5K 10K 22k and see what difference it makes.

For some "reasonable assumptions", the difference will be small, and any loss easily overcome by bumping some knob 1 or 2 notches. But it could turn out a Acme Whoa pedal doesn't like 2K load, or a Kay 998 amp loads-down 22K too much. I'd spend the 10 minutes of mock-up before sweating it all together too small to change.

[fatecasino]

I would like to ask the oposite, if you want to mix two signals in the same way, should I insert a resistor 2K-22K after each input? Is this called "buffering resistors?

[ashcat_lt]

I would like to ask the oposite, if you want to mix two signals in the same way, should I insert a resistor 2K-22K after each input? Is this called "buffering resistors?

I think that's the same question.

Look out from one of the outputs call the other one ground.  You will see a voltage divider.  This output's own impedance (usually some little resistance and a cap) plus this one's mixing R is the "top resistor", and the other mixing R plus the other's out-Z in parallel with the load's input-Z (usually some large R and a cap) is the "bottom resistor".

Ignoring the caps for a minute, the mixing Rs want to be big enough that the contribution of the out-Zs is insignificant, but not so large that the parallel load is significant.  1:10 or 10:1 is a good ballpark for both.  If you can get it where the mixing Rs add up to at least 10x each output, and are no more that 1/10 the load, you'll get the most possible voltage transfer.  With most pedals meant for guitars (1M input, very low output), there's plenty of room.

If you start looking at the caps, things get a bit more complicated, but in the case that the two out-Zs (both R and C) are equal, the caps' impedance rises at the same rate per frequency, so the filters shouldn't actually move.  That's a big assumption, but those caps are usually plenty big to not really have to worry much.

Knowing what you're working with makes it pretty easy to sim out, but if things could change across some extreme range, it's really just easier to buffer before mixing so that you do know what you're working with.

[AmpliFuzz]

So I built the cable. It works, except it starts squealing uncontrollably when I add distortion. The same pedal chain behaves as it should without the extra routing. I tried both 5.6k and 10k resistors, where's the problem?

Do I need resistors at the beginning of the cable, not on the 'mixer' part but on the 'Y splitter part' as well?

[Phoenix]

So I built the cable. It works, except it starts squealing uncontrollably when I add distortion. The same pedal chain behaves as it should without the extra routing. I tried both 5.6k and 10k resistors, where's the problem?

Do I need resistors at the beginning of the cable, not on the 'mixer' part but on the 'Y splitter part' as well?

Sounds like you're getting some positive feedback. Likely what is happening is that the mixing resistors are still allowing enough of the output of the distortion pedal back into its input to begin oscillating. A simple unity gain buffer in the clean path of the Y-cable would probably fix this.

[ashcat_lt]

A buffer on the clean side will almost certainly fix it.  Bigger mixing Rs will help, but the buffer is the right way to go.  Actually, ideally you'd have a buffer for each so that you can plug in anydamnthing on either side and it'll still work predictably.  I know you were hoping for simple and passive, and I was rooting for you, but...

[AmpliFuzz]

I'll try the buffer on the clean side and report. Do I still need mixing resistors if I use the buffer?

[ashcat_lt]

Need and want are two different things.   

Put 'em in there.