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Simple passive crossfader/ balance circuit?
MUFF WIGGLER Forum Index -> Music Tech DIY Goto page 1, 2  Next [all]
Author Simple passive crossfader/ balance circuit?
Navs
I'd like to blend two sources passively, using a single pot as control.

Is it as simple as attaching the tips of both sources to the outer lugs of a pot and then tapping the centre lug?

I'm not concerned with the accuracy of the balance or that there might be slight signal loss - I'd just like something one step more sophisticated than a switch that isn't going to cause interference at source or destination.

Thanks!
sicpaul
I am no electronic expert, but i think it is not as simple as wiring a single pot beween two signals.

If you want to have it passive and simple i think you'll have to take a stereo pot (9mm 2x 100k lin, 1,70€ at reichelt). Solder one pots left lug to ground and the right lug of the other pot. To the other left / right lug you connect the outputs you want to crossfade between. The wipers can be connected to get the crossfade result.
gde
Navs wrote:

Is it as simple as attaching the tips of both sources to the outer lugs of a pot and then tapping the centre lug?


That will work. You'll see that used for the skew function on Thomas Henry's xr vco and also in some filters to switch between lp to notch to hp. I also assume the synthwerks slider modules do the same.
roglok
The external inputs on the Oberheim SEM VCF are passively mixed via a pot too:
http://www.synthfool.com/docs/Oberheim/Oberheim_SEM1A/Oberheim_SEM_1A_ Schematics.pdf

I'm not sure what the real life downside is compared to an op-amp mixer...
daverj
If you're just looking for a simple mix/balance, then that will work. If the modules feeding it have the typical resistor on their outputs then you'll never completely silence one side or the other. There will be some signal coming through even when the pot is turned all the way.

And as with all passive mixers, if the modules used have series resistors on their outputs, and you mult them to other places, you'll probably get some bleed from the second signal mixed in to the places you mult it to.

If the pot value is too large you'll have a bit of a volume dip in the middle. With a 10K or maybe 25K it shouldn't be too bad. With a 100K there will certainly be a dip. Of course, the smaller the pot the more likely to get bleed to mult'd signals, and the more you'll hear both signals with the pot at either end.

So there's no perfect solution when doing it passive, but it will work.
roglok
daverj wrote:
If you're just looking for a simple mix/balance, then that will work. If the modules feeding it have the typical resistor on their outputs then you'll never completely silence one side or the other. There will be some signal coming through even when the pot is turned all the way.

And as with all passive mixers, if the modules used have series resistors on their outputs, and you mult them to other places, you'll probably get some bleed from the second signal mixed in to the places you mult it to.

If the pot value is too large you'll have a bit of a volume dip in the middle. With a 10K or maybe 25K it shouldn't be too bad. With a 100K there will certainly be a dip. Of course, the smaller the pot the more likely to get bleed to mult'd signals, and the more you'll hear both signals with the pot at either end.

So there's no perfect solution when doing it passive, but it will work.


Thanks, that's some good info. So on the SEM VCF, for example, you would swap the 50K mix/balance knob for a 10K? Or will this improve the dip but cause more bleed?
mcop
Interesting info indeed.

I'm wondering if using say a 10K dual gang pot with both gangs wired in opposite directions might work here. So signal one end, output on the wiper and the other end to ground for each gang in different directions.

roglok wrote:
Thanks, that's some good info. So on the SEM VCF, for example, you would swap the 50K mix/balance knob for a 10K? Or will this improve the dip but cause more bleed?
Navs
Thanks guys. I ran out of time before my gig yesterday and made the space for an extra mixer. This project can wait for a day when I can do it 'properly' thumbs up
daverj
roglok wrote:

Thanks, that's some good info. So on the SEM VCF, for example, you would swap the 50K mix/balance knob for a 10K? Or will this improve the dip but cause more bleed?


Which pot? In the original SEM VCF there are three balance pots for the signal inputs (ext 1/2, VCO1 saw/pulse, and VCO2 saw/pulse). Those are each feeding into 220K resistors, which are several times higher than a typical module input so probably won't have much of a dip. The other 50k balance pot is on the modulation input, but it's a positive/negative signal with zero signal in the center. So it not only dips in the center, on purpose, but cuts out completely in the center (on purpose).

On the SEM variation in thread by moogah, his schematic has a 50K balance pot marked "response" that blends the two outputs into a third output. That one won't have a dip because the pot goes to a non-inverting amp with extremely high input impedance.

mcop wrote:
Interesting info indeed.

I'm wondering if using say a 10K dual gang pot with both gangs wired in opposite directions might work here. So signal one end, output on the wiper and the other end to ground for each gang in different directions.


The problem with using a dual gang pot is that when it is in the center then both signals are 1/2 strength. Then when you passively mix them they are 1/2 strength again. So the dip in strength is much worse that way. But there is no bleed when the pot is all the way at either end.

The dual gang idea would be good for feeding the two gangs into an active mixer. That way you don't get the second strength cut from the passive mixer, and maintain a constant signal level after the active mixer.
daverj
Here's some graphs to show the differences of pot value in a typical situation where modules have the common 1K output resistor and a typical 100K input resistor. The amount of bleed is much lower if they have a smaller output resistor or no output resistor. And the effects of the "dip" at the center are also worse if the input is less than 100K, which it sometimes is.

You can see that the dip gets better by using a 10K pot instead of a 50K pot. But the amount of bleed into the input signals and the minimum levels obtainable at the ends of the pot are worse with the 10K. The bleed into the input signals only matters if you mult those signals to other places. But the minimum signal levels stay the same (meaning if there is a resistor on the output of the modules feeding in, you never get completely rid of one signal or the other at the ends of the pot).





Navs
Thanks for taking the time to explain this and the graphs, Dave!

I shall just have to try it with the module in question to see if the levels are acceptable and, failing that, build an active version.

It would be for a RES-4, to feed it either pulse trains or white noise - the 'missing' mixer. I've done this before with an A-112 delay/ sampler: 4HP solutions when you don't want to 'waste' a four channel mixer.

http://navsmodularlab.blogspot.de/2011/11/feedback-looper.html
roglok
daverj wrote:
roglok wrote:

So on the SEM VCF, for example, you would swap the 50K mix/balance knob for a 10K? Or will this improve the dip but cause more bleed?


Which pot? In the original SEM VCF there are three balance pots for the signal inputs (ext 1/2, VCO1 saw/pulse, and VCO2 saw/pulse). Those are each feeding into 220K resistors, which are several times higher than a typical module input so probably won't have much of a dip.


Yes that's the one I was referring to. I'm actually building the moogah version, but planned on having a balance pot for a couple of three total inputs.

Thanks for doing those graphics - very helpful. I agree with navs - it's probably makes most sense to experiment with different values and then decide what you can get away with (or if it needs an active solution). After all it only requires a few alligator clips smile
Navs
FWIW and for my application, this solution sounded worse than it looks on the graphs lol

Active it is, then ... cry
roglok
Navs wrote:
FWIW and for my application, this solution sounded worse than it looks on the graphs lol

Active it is, then ... cry


How would an active solution look like? Do you already have a design for this?
Navs
Bump!

Using the Geofex article linked here: http://music-electronics-forum.com/t32035/

I tried this:



The equal power aspect is for volume, I guess, so the result with DC voltages/ CVs isn't what I'd first thought. With the pot centred and two 5V inputs, the output is 6.8V, at one extreme or the other it's 4.9V in the simulation.

Much simpler would be this:



but I'm not sure if that's ok. It's too simple hihi

The last option is to do the obvious and build a two input mixer: two pots, two mixing resistors, and inverting amp followed by another.

I really want something with just one pot to control the balance but I can see why these sorts of circuits average rather than sum. Also, if I try to compensate with gain, I run the risk of hitting the rails.
daverj
Your second drawing above solves the "dip" issue in the center that I described on the previous page because it is now an active output with a high impedance load on the center of the pot. As opposed to the fully passive circuit in the first post of the thread, which would have seen a load from the module you patched into.

There remains the issue that the output resistor that exists on a lot of modules might make it so the circuit above will never completely kill the input signal when turned fully one way or the other. That's the bleed issue I described on the previous page.

That can be improved by adding input buffers like those shown in the upper drawing of your post above. By running each input into a buffer before the pot, the inputs are isolated from the output resistor of the module being patched in. Then connecting the pot directly to the outputs of those buffer amps you should be able to fade out the opposite input much more. There might still be a tiny amount of bleed since the op amps do have output resistors internally. But it would be 20 times smaller than connecting modules directly to the pot.

The only remaining issue would be the slight signal loss on the output because of the 1K resistor on the output. You can compensate for that by giving the output buffer circuit a small amount of gain. Of course since you don't know the actual load of whatever module you are plugging into, you would have to guess on how much gain to add. Assuming a 100K input load on the next module, having a gain of 1.011 would compensate for the loss. (or you just live with the 1.1% loss and leave the amps at unity gain).

To add 1.1% gain to the output buffer put a 110 ohm resistor as feedback between the output and the minus input, and a 10K from the minus input to ground. That adds a tiny amount of gain to that amp.
EATyourGUITAR
what about something like this?
daverj
EATyourGUITAR wrote:
what about something like this?


It loses a huge amount of signal (about 70%), plus has the same bump up in the center of the pot he describes a couple of posts back for the active version of this same circuit.
Navs
Thanks, guys.

I went active because passive wasn't doing it. The equal power configuration is interesting, but as Dave suggests, is more trouble than it's worth for my application.

I have some of Matthias' cute Attenuverter chicklets. I wondered about his decision to use only a single opamp on the TL072, but it proved a blessing of sorts:



I lifted the legs to free one opamp and soldered everything in situ - screwed jacks and pots into place then wired, soldered resistors directly to IC pins etc. lol If you have switching jacks, you could normal the inputs to ground for instances where you're only using one input.

The good thing was it provided a very compact fix. The bad was it gave me the excuse to ignore Dave's advice about isolating the inputs. Some CV sources are happy with this arrangement, others not.

e.g. I passively mult a stepped CV and send it to a VCO and this mixer. I cross-fade the CV and a mult of an envelope and send this mix back to control the envelope. The VCO picks up the envelope's signal too. Does this have something to do with 'innie' and 'outie' protection/ feedback resistors on the stepped CV source?

I'm going to redo this project, probably as a simple two input mixer. What I don't understand is how an inverting mixer configuration would respond any differently if I don't buffer each input. hmmm.....
daverj
Navs wrote:
I'm going to redo this project, probably as a simple two input mixer. What I don't understand is how an inverting mixer configuration would respond any differently if I don't buffer each input. hmmm.....


A two input inverting amp mixer doesn't require input buffers because the two inputs are isolated from each other by the nature of the amplifier. So no bleed back out to the inputs. The place where the two inputs join is a "virtual ground" point with an extremely tiny signal at it. So no signal goes back out the other input resistor. (well, possibly a few micro-volts)
Blinkar blå
I was planning on building something similar, & I just stumbled upon RichardC64's pan/morph/vca: http://www.sdiy.org/richardc64/mmvca/index.html
(Technical info in this this thread @ e-m.)

Looks fairly simple & maybe one can leave out the input attenuators.
tomatoKetchup
Hi

What do you guys think of the following design for a crossfader with an out and its inverted equivalent?



I created a topic for this specific circuit design. I was already told about the 100K resistor at the inputs. I was wondering about the caps too. Should they be here if I want the ability to mix audio AND cv?
Dr. Sketch-n-Etch
Signals into taps of pot, wiper to + of opamp, - of opamp wired to output of opamp -- output is the buffered wiper. That'll do what you want, for less than $2.

I could lay out, etch and build you a board for a dual unit with both pots and all jacks on the board, a socket for a dual opamp, and a eurorack power header, in the next 20 minutes if you wanted.
tomatoKetchup
Dr. Sketch-n-Etch wrote:
Signals into taps of pot, wiper to + of opamp, - of opamp wired to output of opamp -- output is the buffered wiper. That'll do what you want, for less than $2.

I could lay out, etch and build you a board for a dual unit with both pots and all jacks on the board, a socket for a dual opamp, and a eurorack power header, in the next 20 minutes if you wanted.


What about reproducing the behavior of the M32's crossfader, where when nothing's plugged into the INs, the OUT is a sweepable DC voltage (that's the regulator's part of the circuitry above)?

EDIT, here's an updated circuit, without the capacitors and with +/- legs of the op-amps on the left swapped (seems to work better on Multisim's simulation):

euromorcego
[sorry to hijack the thread ... but I'll stay on topic]

Dr. Sketch-n-Etch wrote:
Signals into taps of pot, wiper to + of opamp, - of opamp wired to output of opamp -- output is the buffered wiper. That'll do what you want, for less than $2.

this is essentially the schematics of Navs a few posts up.

Would it be recommended to buffer both inputs? As reported above, it would avoid bleed.

And does the pot value matter? I previously looked at differential OpAmps and there the value of the pots will determine the shape of the response (linear, log, etc ...). But this configuartion here seems different.

No need to "lay out, etch and build" for me, though, I just need to understand how it works to build a small crossfade tile.

Here is the current schematics:

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