Blend, mix and pan

Started by marcelomd, November 01, 2017, 08:23:41 AM

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marcelomd

Hi,

I'm playing with blending two signals and I found a few ways of achieving that, but I'm having a hard time understanding the practical implications of each circuit.



Can you guys share your experiences with it? Comment on the pros and cons of each circuit?

Why, or in which situation, would you use one over the other?

Thanks a lot!

diffeq

My 2¢:
A) Simple panning circuit with output buffer, allows to pan between input a & b. Does not amplify inputs, also seems that another signal always bleeds through.
B) Amplifying mixer circuit with individual volume control (you can mute both inputs) Gain is set by R2,R1 to R3 ratio. Also inverts the signal phase.
C) Same as B but for panning between two inputs, dual pot solution. No bleedthrough.
D) Same as B, except that single knob pans between inputs, no bleedthrough too.

Solution B is for mixing, others are panning circuits (that is, they mix proportionately, 50%/50%, 30%/70% and so forth)
D is a bit confusing though.

GGBB

#2
As drawn, C is not a blend control. The added dotted line joining the two pots indicates a single dual-gang pot instead of two separate pots. Remove the dotted line and it is identical to B, just drawn differently. But with a dual-gang pot, and assuming pin 1 is counter clockwise for both pot sections, the pin numbering makes this a stereo input, mono output volume control. If you reverse the pin order on one of the pot sections - so that pin 1 (counter-clockwise) of one section and pin 3 (clockwise) of the other section are connected together to ground - then it becomes a blend control - same circuit as B but a single knob operates both pots simultaneously in opposite level directions. I think that is what C was intended to be, but the creator messed up the pin numbers (or they are properly described somewhere else).
  • SUPPORTER

anotherjim

With A, if the inputs are from low impedance sources (such as directly from op-amp outputs), there is negligible bleed at the control extremes. The low impedance of the selected source will actively zero out any other signal coming via the full resistance of the pot from the other source.

Although the pot in D looks like a pan control, the way it's connected is just a variation of a 2 input virtual earth mixer, which is what B & C are. For D to be a panner, it needs 2 separate outputs from R8 & R10.

None of those are panning controls, they are all mixers/blenders.


marcelomd

Hi,

Circuit D came from RG's Panning for Fun article. I think the advantage over circuit A is having constant power for the entire sweep range, with no bleedthrough. I saw it in VFE's Dragonhound.

The pots in circuit C are supposed to be a dual gang pot and the idea is raising volume from one side while lowering the other. Sorry about the confusion =)  I saw this implementation in a few circuits (Way Huge Pork Loin is one, IIRC). I know about MN taper pots. Is this where they are used?

So, for blending two signals, why would one use circuits A, B, C or D?

Thanks for your responses!

ElectricDruid

Quote from: marcelomd on November 01, 2017, 03:33:00 PM
So, for blending two signals, why would one use circuits A, B, C or D?

For me, circuit B is the odd one out - it's the full mixer. You can mute either or both channels, and you can mix both channels at any output level (you could turn both channels up to 25% and have a quiet output, for example). The others don't allow those options, but the others have the advantage of only needing a single control. Circuit B is the only one with two knobs.

Circuit C doesn't seem like a great option. It's simple to understand since (ignoring the dodgy pin numbering) it's just the simple inverting mixer but with a dual gang pot. But there's the problem; who wants to use a dual gang pot if they don't have to?

Circuit D is the classic panner circuit found in many mixers since the Year X. It's a good circuit and does what it claims. A good option. I've considered it for various uses in various circuits, but (as yet) I never have found a perfect need for it! But I'm sure that it'll happen that I have a problem to which this circuit is the solution sooner or later. It's one of those.

Circuit A is the "cheap and simple" option and depends on the sources feeding it having nice low impedances - like op-amp outputs. If that's the case, it's definitely the easiest way to go. But if I wanted something with external inputs and couldn't guarantee what was being fed in, I wouldn't use this, or I'd add buffers ahead of each side of the pot.

HTH,
Tom


Kipper4

I've used this (single gang pot crossfader) before to good effect. With LM358.
IIRC all the ground connections in Toms diagram where referenced to V/2 in my usage.

http://electricdruid.net/single-vca-crossfader/
Ma throats as dry as an overcooked kipper.


Smoke me a Kipper. I'll be back for breakfast.

Grey Paper.
http://www.aronnelson.com/DIYFiles/up/

PRR

Pan is one signal to two (L&R) outputs. Placing 16 tracks across a stereo mix.

Cross-fade (also "blend") is two signals to one output. DJs running the end of one song into the start of another song.

All the plans in the first post are crossfades.

The many different crossfades have different costs and different shortcomings. Often not obvious at a glance. Sometimes not important in a specific application. Much depends on context. "A", as said, works fine in an opamp implementation (all low-Z drive) but leaks badly in simple transistor (hi-Z drive) design unless buffers are added.
  • SUPPORTER

Rob Strand

#8
B vs C are doing different jobs.  B can do anything C can do and more as it is independent.  If you want B to act like C then B is just using  two pots with added inconvenience.

IMHO you should look what happens to the gain of the signal, when one input is active.  Suppose you have gain 1 when the control is centered.  Look what happens to the signal level when the control is at each extreme.

This is where the specific part values can determine the behaviour.

For example A, the gain at the center is 1/2 and when at the extreme it is 1.   You have to ask yourself do you want that behaviour?   On the other hand if you have the same signal feeding into each input  it doesn't matter where the control is set the output is the same.  If you have uncorrelated noise in each path then they don't add arithmetically

C on the other hand will behave a bit like A if the loading resistors R4 and R5 are large but if they are small the gain difference when you have the pots centered and the pots on full will be enormous.

With D you can control what happens to some degree by choosing the correct resistors.   This gives some flexibility to the design.

A and D have the advantage of only requiring a single pot.

There's other choices as well.  For case C we can choose A + C taper pots or M + N taper pots.   These put a new spin on things.   

[edit 1: fixed typos, added more detail]
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

intripped

There are dual pots specifically designed for the blending function, they are used especially on bass guitars for blending between two PUs.
Perhaps it's possible to use these in circuit C, avoiding the gain issues; I just don't know if the available values could do the job

http://www.stewmac.com/Pickups_and_Electronics/Components_and_Parts/Potentiometers/Alpha_Blend_Pots.html

bartimaeus

Any tips on how to get a different responses from circuit D? I like blends where one signal is at about full volume for 75% of the sweep, and attenuated for the last 25%, while the other signal is at full volume for that last 25%, and is attenuated slowly over the other 75%. I've mainly seen this on my Strymon Timeline, which probably does it with digital control, but anything similar would be really cool!

roseblood11


Rob Strand

Quoteny tips on how to get a different responses from circuit D? I like blends where one signal is at about full volume for 75% of the sweep, and attenuated for the last 25%, while the other signal is at full volume for that last 25%, and is attenuated slowly over the other 75%
That would mean from 25% to 75% both are full on?
I think you will find for 50% to 100% rotation A is full from while the B decreases linearly from full to zero.   For 0 to 50% rotation, it's a mirror image,  B is full and A decreases linearly from full to zero.   Because the curve is linear you don't hear a big difference in the centre.
Checkout M+N taper pots (maybe sometime called X+Z taper) wired like a blend-pot on a guitar/bass.   An A+C taper pot wired the same way will give an approximation to this.


Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Rob Strand

#13
Shows different level with position.
http://peaveycommercialaudio.com/mmhelp/Devices/Level/MotoCross.htm
The things to note are:
- the rise of the dominant channel from the centre position to the maximum position
- constant power approximates constant level when two uncorrellated signals (like two different audio tracks or noise) are combined.
- constant voltage gives constant output when the same signal is fed to each input.   It also applies when the inputs are nearly the same input (say when B is a processed version of A).

For circuit D you can get a rough approximation to constant power when the 4x input resistors are 1.5 times the pot values (IIRC you need a feedback resistors 7.5 times the pot value for unit gain at the centre position).   This is a very common circuit.

Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

intripped

Quote from: roseblood11 on November 02, 2017, 02:19:51 PM
Quote from: intripped on November 02, 2017, 10:40:10 AM
http://www.stewmac.com/Pickups_and_Electronics/Components_and_Parts/Potentiometers/Alpha_Blend_Pots.html

What makes this pot a "unique blend pot"? Isn't it a standard linear taper dual potentiometer, except for the ridiculous price?

In this kind of pot, at 50% of rotation (middle position) you get max, for both parts A&B of the pot; and they stay at max when you increase rotation till 100%.
Since the tapers are linear, you can wire-up one part inversely to the other.
In this way you have:
At 0% rotation (fully ccw) -> A=0 / B=max
At 50% (mid position) -> A=max / B=max
At 100% (fully cw) -> A=max / B=0

marcelomd

The "Unique Blend Pot" probably is an MN taper pot:



So, in the ideal world, for blending two signals I would use circuit A or D, depending on the impedance of the sources. B, the full mixer, only if I need that level of control. D depends on that potentiometer to be useful.

Still thinking about A/D vs B, if I need a volume control, these 3 options end using 2 potentiometers. I prefer having less controls on the box, so either one is good.

Thansk a lot for your responses!