Pot impedance..

[jeth]

Apologies in advance for posting another basic question, I'm sure you good people don't mind (too much)...
Something I just can't get my head round... When inserting a level pot into a circuit between two op-amp sections, in this case a buffer and an inverting mix stage, how is the impedance affected? Does the source see the pot resistance (which I gather is actually the parallel R of the pot setting and the remaining R shunted to ground..e.g. 100K pot at 25%=18.75K) as part of the input resistance of the following stage..i.e. This value+series resistance of the mix resistor. OR, is the pot value seen by the mix input in series with thge source output R?
I'd be much happier understanding this rather than making arbitrary choices based on what I see in other circuits (which, by the way, seems to be totaslly inconsistent!?!).. I want to know how the pot affects my mix resistor value, and how to balance the impedance from source to pot to input correctly. Simple stuff to most of you I know, so simple that most reference material goes right over my head..Sorry!

 

[bcarso]

Most of the time the source output R is small if it is the output of a buffer. If it is some other sort of source then the below will be modified.

And when you say an inverting mix stage I assume it is an opamp, where the inverting input is at virtual ground. So, with those assumptions:

The source drives the paralleled resistance of input summing R and pot end-to-end R at maximum gain. At other gain settings, it drives the top-to-wiper pot R, in series with the {input sum R paralleled with the wiper-to-ground R}. So, the loading on the source is minimum at minimum (~0) gain.

The sum amp summing node sees an impedance of just the summing R at max gain, assuming a ~zero Z source. It sees the summing R plus the paralleled R of each portion of the potentiometer (top-to-wiper, wiper-to-bottom/ground) at other gains.

For example: I have a 100k pot and a 100k summing R. Let's say the pot is set with the wiper at the resistive midpoint. The resistance seen by the sum node is 25k + 100k = 125k (50k parallel 50k plus 100k). The source drives 50k + (100k parallel 50k), or 50k + 33.3k = 83.3k. Note that this example is also the maximum R seen looking out from the sum node, which can be important to consider for thermal noise and sum amp input current noise.

 

[jeth]

Many thanks for another decent reply, the calculations are making a lot more sense now.
The only thing that I am now a little confused about is(using your example) if the source is driving 83.3k, then this is obviously a reasonable bridging load for the op-amp output. BUT, the Summing amp is seeing 125k..which by my (ill-informed) assumptions, means that the summer is seeing too high a source input R, as it is possible the input R will not be more than 10x 125k. Though I guess that the summing node sees the parallel of all it's inputs, so If i were to have six channels feeding a summing amp with the values in your example, I still have around 20k (sorry, no calculator to hand) source R seen at the summing node. If this is the case then i can probably play with the values of the mix resistors to get a more suitable figure..But maybe I'm still not getting it, if so, what am I missing?
My problem may be with the term "The resistance seen by the sum node is"...I am confused as to if it "sees" its total source impedance, or it's own input impedance. Or am I missing something by just looking at it as a simple "source driving load" situation, when you only know the basic rules it's hard to know when they don't apply...

 

[PRR]

Open this image full-size and follow along.

At "A" we have a source V10, a pot R1, a mix-resistor R2, a high-gain inverter, and feedback R3.

I've shown the pot set at an odd value, 13%. We can replace it with two fixed resistors. The bottom one is 10K*0.13= 1.3K, and the top one must be 10K-1.3K= 8.7K. We now have "B".

Pretend the high-gain inverter will hold a Virtual Ground/Earth at its input, as dotted in "B". We will have to check that assumption along the way.

Then the input side can be re-drawn as in "C".

"D" replaces the 1.3K||10K with 1.1504K.

"E" sums the two resistors into one 8.7K+1.1504K= 9.85K resistor. We see at a glance that the total current out of the source due to a 1V drive is 0.1015mA.

In "F" we put back some resistors. If 0.1015mA is flowing in 8.7K, then it drops 0.8832V. The voltage at the pot wiper must be 1V-0.8832V= 0.1168V. Keep hoping the Virtual Ground is a good assumption. The current in the mix-resistor must be 0.1168V/10K= 0.01168mA.

In "G" we shove that 0.01168mA into the inverter. Pretend the triangle has zero input current. That 0.01168mA must be negated by 0.01168mA flowing through the 100K feedback resistor. For that to happen, the output must swing to -1.168V.

Are real-parts able to handle these voltages and current and work nearly like our assumptions?

Gain: the 100K:10K is about a 10:1 loss, the inverter's gain must be much more than 10. If it is not much-much-more than 10, we may have to check more closely. But many opamps have gain over 100 at the top of the audio band. (This assumes one input: in fact in multi-input mixers you soon run into troubles...)

Source loading: at 1V the 0.1015mA loading is trivial for most chips. We may put 10V in, the system is (we hope) linear, so we could need 1.015mA, still trivial. We have calculated for a 13% pot setting: a little foolin will show that 100% pot setting causes larger load, check that.

Output swing: at 1V input and 13% pot-set, the -1.168V and 0.01168mA are negligible. The virtual-ground assumption is probably safe. At 10V in and 13% pot-set, the output must drop to -11.68V at 0.1168mA: not all opamps will swing this, but that 13% pot-set was pulled from thin-air. In real life, if the output is clipping, you should pull the pot down some.

However, I'm not sure this is what you really wanted to know. Still, the same techniques work in reverse to find the impedance "seen" by the mix-amp.

 

[jeth]

Thanks PRR for a very thorough reply, I hope it didn't take up too much of your time.
I'm not sure it's what I wanted to know either, but I'm gonna print it off and attempt to drive the excessive loads of my neural pathways in an effort to see what i can glean. I tend to learn at least a little everytime i try to learn alot.
What I really want to understand (in my own, newbie language) is what are sensible values for such a circuit, and why. If I could see a working example of, say, a six input mix stage fed from buffer outs with level pots inbetween....
Sorry but thats just the way my brain (mal)functions.

As a beginner I'd always hoped I could achieve enough working knowledge to build some useful stuff but avoid the heavier maths..I'm discovering that was one of a long line of stupid assumptions, but I'm actually enjoying the (slow) learning curve.
Thanks again for your input(s)...!!

 

[bcarso]

[quote author="jeth"]Many thanks for another decent reply, the calculations are making a lot more sense now.
The only thing that I am now a little confused about is(using your example) if the source is driving 83.3k, then this is obviously a reasonable bridging load for the op-amp output. BUT, the Summing amp is seeing 125k..which by my (ill-informed) assumptions, means that the summer is seeing too high a source input R, as it is possible the input R will not be more than 10x 125k. Though I guess that the summing node sees the parallel of all it's inputs, so If i were to have six channels feeding a summing amp with the values in your example, I still have around 20k (sorry, no calculator to hand) source R seen at the summing node. If this is the case then i can probably play with the values of the mix resistors to get a more suitable figure..But maybe I'm still not getting it, if so, what am I missing?
My problem may be with the term "The resistance seen by the sum node is"...I am confused as to if it "sees" its total source impedance, or it's own input impedance. Or am I missing something by just looking at it as a simple "source driving load" situation, when you only know the basic rules it's hard to know when they don't apply...[/quote]

"Sum node," to me, means a virtual ground as shown in PRR's example. With sufficient loop gain in the op amp the inverting input with feedback is a very low impedance (how low---see the thread on half-voltage supplies a while back). If you ask for more than the maximum currents available then the opamp can no longer keep the loop closed. But the open-loop input impedance of the op amp is generally of little importance when feedback is applied to the inverting input. Feedback keeps the inverting input at roughly the same potential as the noninverting input, and if the latter is at ground so is the inverting input.

 

[jeth]

Thanks for your patience and continued help on this thread...
So now I'm gonna show you just how basic my knowledge is...

If the summing node is very low impedance, and it's source is high impedance, why doesn't that break the rules of bridging loads and cause the summing amp to load the source...???

Sorry, I really want to get it, I do, i do...

 

[bcarso]

The bridging impedance rule is a useful rule-of-thumb but not really applicable to the situation described. In a sense you are already in the midst of your amplifier when you hit the resistor tied to the arm of your potentiometer. And you have already posited a buffer amp for each input, so your original input sources' voltages are not loaded down. You have nice robust voltage sources that can be loaded within the constraints of the available output currents.

What you are doing at a summing node is dumping current in---not into a black hole, or (for the most part) into the interstices of the amplifier, but into a node that the amp is valiantly working to keep at ~ground. How it does this is by counteracting any net current, that by itself would make the node voltage move, with current from the feedback network---in this case a resistor. It may make it easier to think of your input resistors as transforming voltages into currents, and your sum node as summing those without loss.

So what the voltage on the output end of that feedback resistor tells you is what the algebraic sum of all the currents from all of your summing resistors is. And that is a voltage again, that you can use within the limits of the amp's available output current, for anything you like.

Another way to mix a bunch of signals together is just by tying a bunch of resistors together on the mix end and feeding the free ends of each with each signal. No buffers, no summing node, nada. But if you are using the signals for other things and they don't each have a very low output impedance, you will get crosstalk at a given output based on the combination of the other signals interacting with the given output. Also, you will have a mix out impedance of some nonzero value---for all identical resistors of value R, the mix out will be at least as high as R/n, where n is the number of mixed channels---and that may not be convenient to use downstream. And you may not like the attenuation on each signal due to one resistor loaded by all the others. So, the common practice is to use a feedback amp as described.

 

[PRR]

> what are sensible values for such a circuit, and why.

> if the source is driving 83.3k

Then, in modern solid-state design, it is NOT sensible.

Mixing with resistors is a lossy process. If you are mixing junk, loss won't kill you. But if you are mixing for quality, loss is a problem and the solutions are not perfect.

Since we have loss, we want to suck the maximum possible signal from the source. Say it is chip-amps, rated 2K load. We make the mix-resistors 2K, to suck all that the amps will give.

Oh, you want to control the mix? With the basic pot, for 2dB conformance to marked taper, you can load the pot with about its own value. That rule is not unbreakable, but often a good starting place when you need pot-control and you need a lot of output.

So 2K pot with 2K mix resistor. Wait, that could be 2K||2K= 1K when the pot is all the way up, and we postulated 2K-min amps, so we use 4K pot and 4K mix resistor. That's an odd pot; 5K and 4K7 would be fine.

Ah, but the one source could be driving Left, Right, Reverb, and Monitor buses. With 5K pot and 5K mix resistor, source loading could be as low as 0.6K (without a calculator, and you should NOT need a calculator to rough-in a design). Up-size again so the 2K amp is not strained: 16K pots and mix-resistors. (20K and 25K pots are very useful mixer parts.)

Or: admit that you have good busses and crap busses. The Left and Right are precious. The Reverb will get contaminated by hiss and brought back probably at lower level: it need not be as good. And who cares if the headphone monitor mix is not infinitely clean? So some boards use 10K pots and mix-Rs for the good buses, 100K for the other buses. Noise will be higher on the "other" busses, NBD. The total worst-case load can be held in the area of 2K (5K||5K||50K||50K), allowing use of cheap chips.

I have implicitly assumed the mix resistors go to a low impedance, the Active Mixer. If that opamp is any good, the Virtual Ground impedance will be much-much-less than 10K or so, and the low impedance assumption is good enough.

If there is no active summer, then the mix-node impedance is (neglecting the pots for a moment) the parallel array of all the mix resistors. If you have 2 10K resistors, it is 5K, and low-Z is not a good assumption. If you have 10 mix resistors of 10K each, the node is 1K, and low-Z is not a bad assumption.

If you are not using a Virtual Earth summing amp, then you feed this sorta-low impedance into a plain bridging amp. If you design your own, no big problem. If you have loaded the heck out of the sources, and have more than a few inputs, the bus impedance will be reasonably low and amp design is not too fussy.

However another plan is to use an existing mike-amp. This is very different. Mike amps need to see a specific low source, 100-300 ohms, or they get noisy and un-flat. And mike-amps normally have a LOT of gain. If the sources are putting out line-level, and we want line-level out of the mike/mix-amp, and the amp has a typical 40dB (1:100) of gain, we better have 40dB of loss in the mix network or the mike-amp output will be much too hot (probably overloaded). Fortuitously, these two problems are synergistic. We use a reasonably high mix resistor value, 10K or 20K, which makes pot and source loading small. The output impedance and level will be too high: for 8 inputs the loss for one input will be about 8:1 or 18dB, if the resistors are 16K the mix node impedance will be around 2K. Now stick a 200 ohm resistor on that. Level drops about 10:1 or 20dB, we have total 38dB loss, in the ballpark for a 40dB amp and unity gain from one source to mix-amp output. The mix bus impedance will be around (16K/8)||200= 180 ohms, in the ballpark for a "200 ohm" input. This general scheme works out to about 100 inputs: 100 input mixer has a loss near 40dB, and if the resistors are 15K then the output impedance is 150 ohms. And conversely, you can start with a 100-input design (or whatever your mike-amp gain is), pick the mix resistors 100 times higher than your desired node impedance, cross-out the 92 mix-resistors you don't need for an 8-in design, and replace them with one R/92 resistor to ground.

 

[jeth]

Thanks guys..seeing the math applied to a real world example has helped a lot. In fact, some of my original assumptions weren't far off the mark..it's just that with limited knowledge and experience it's difficult to have faith in your understanding. Unfortunately I'm not presently in the position to spend cash on plenty of parts to experiment with, otherwise I'd prefer to confirm things for myself that way. I just wish I hadn't had to leave my copy of Horowitz and Hill back home in the UK.

PRR..I worked through your design example last night and it tought me a lot...so thanks for that. But like I mentioned before, I'm new to the math, so I still do need a calculator at times I'm afraid.

 

[PRR]

> I'm new to the math, so I still do need a calculator at times

Round everything to one significant digit. Track the decimal-point on your toes. I said 2.5K/4 is 0.6K, because I remember that 24/4 is 6; calc says 0.625K. Do I care? In general, 0.625/0.6 is 4% "error". In audio, 1dB or 10% errors are small. In this case I don't care a whit: I was checking for loading on a 2K-min chip, 0.6K-something is far too low, pick a bigger number and try again.

Pick "nice" numbers, easy to multiply/divide on your fingers. This can even mean assuming a 10-input mixer, when you really want 8-input: the answers won't be very different, and "10" is so much easier. (Anyway, you always end up with a few odd inputs.)

Yes, I own a dozen calculators and have two at arm's reach, but it is a real help to be able to rough-out a design with sloppy-but-quick values. When you are totally lost, you may resort to picking random values and taking a quick whack: usually a very few iterations will get the right number of zeros on the resistors and some barely-significant first digits; then you can fire-up the calculator and check/refine.

 

[jeth]

I worked through some possibilities last night and you'll be pleased to know I'm already starting to need the calculator less... Still not quite up to working out the parallel value of say, 47k, 22k and 7.5k on my fingers and toes yet though.

As usual every revelation brings another question...whilst looking at the possibilities I noticed that while you had stated a worst case of (e.g.) 5k for a 10k pot into 10k mix R(pot all the way up), this doesn't actually seem to be the worst case.
According to my calculat(or)ions..the parallel combo of pot and resistor goes from 950R at 10% and even lower at settings closer to zero, at 50% we have 3.75K. So I guess I should really account for lower pot settings increasing the load to below 2k, as my design does indeed require the sources to feed a number of parallel loads.

One other thing (before you let go of my hand!) ..You state a general rule of loading the pot with mix R of equal value...what happens when this rule is broken..Is it just a deviation from the taper, or worse?

Thanks again for your input..you've pushed me a way up that slope these past few posts!

 

[PRR]

> you had stated a worst case of (e.g.) 5k for a 10k pot into 10k mix R(pot all the way up), this doesn't actually seem to be the worst case.
According to my calculat(or)ions.. the parallel combo of pot and resistor goes from 950R at 10% and even lower at settings closer to zero, at 50% we have 3.75K.

??? No, something wrong there. Have you mis-wired the potentiometer as a variable resistor shunting the source? Look up at my sketches.

> general rule of loading the pot with mix R of equal value...what happens when this rule is broken..Is it just a deviation from the taper, or worse?

If the load is infinite: taper is as-marked, pot input impedance does not vary.

If the load is very low: Warped taper, low input impedance when pot is full-up.

> working out the parallel value of say, 47k, 22k and 7.5k on my fingers and toes

7.5K is similar to six or seven 47K resistors in parallel. 22K is about two 47K resistors in parallel. So we have roughly six+two+one= nine 47K in parallel. 47K/9 is about 5K. Close enough for rock and jazz.

Oddly enough, the calculator says 4.9984K... somehow I got a 0.03% result without taking off my shoes. That's gotta be blind luck.

 

[jeth]

PRR..after looking at this again last night I suspected you might have just given up on me. I realised I'd been calculating the pot-wiper R + the mix R, in parallel with the wiper to ground R. Checked through again with the correct calculations and of course, it made more sense.. It certainly makes the sums a bit simpler that way.

Your arithmetricks make perfect sense...I'd failed to see it so simply so thanks for passing on the insight. Should save my calculator from wearing out so fast.[/code]

 

[jeth]

Been mulling over what I learnt from this thread and I'm afraid it got me thinking ....

Say one of my sources is feeding several "virtual earth" inputs, all of which need to have level control at the channel input. In order to keep the load on my source reasonable, I'll have to use fairly large pots and mix R's..at least 50K/47K as a rough idea (I'm getting there PRR..no calc. today :wink: ) this is going to present a load of around 6K to my source. A smaller available value for the pots/R's will be increasing the load too much.
The question is... From a noise point of view, at what point am I better served by keeping my pots/R's as low in value as I can, and buffering between pot and mix R. Just thinking there must be a point at which the extra buffer stage will introduce less noise than the increase in Resistance values. Another possible advantage is more consistent gain in the mix amp itself.

Any thoughts...am I way of track again?

 

[PRR]

> I'm afraid it got me thinking ....

Oh, oh...

> am I way of track again?

No. Your analysis is quite right: many-bus consoles need buffer amps.

Where you draw the line is very much a matter of opinion, taste, and budget.

The added noise of the buffer should not be an issue. Cost, distortion, cost, heat, cost, space, and cost are the usual objections. Most "real" mixers buffer. A non-buffered line mixer is pretty much a minimalist do-no-harm (or maybe cheap-junk) idea.

 

[jeth]

So I guess i was right to assume the buffer should go between pot and mix R?

Bearing in mind that we're talking about just 4 busses here, are the buffers really necessary? As it's a small one off project cost/heat etc aren't too much of an issue. Seems like you suggest that an extra buffer is going to contribute less noise than an increase in pot/R values?
From your comments on "cheap junk" unbuffered designs perhaps you were thinking I planned on connecting external line sources straight to the mix bus. I will be using line input buffers with their own level pots, these will then feed 4 seperate mix sections, for which I need a level control at the input for the signal from some sources ( but not others, so a post mix stage master fader for each bus is no use)..

Thanks for the continued efforts to educate...

 

[jeth]

Another (dangerous) thought...

I'm assuming that the impedance of the pot into a buffers input is substantially higher than the pots own R. Otherwise I'm not really gaining anything. Am i correct to assume the input R will be the pot R (calculated as advised by PRR previously) in series with the input R of the opamp buffer stage? If so this should be a considerably higher impedance than the pot/mix R combo alone, therefore decreasing the load on the source with several such bus controls in parallel...

 

[bcarso]

[quote author="jeth"]Another (dangerous) thought...

I'm assuming that the impedance of the pot into a buffers input is substantially higher than the pots own R. Otherwise I'm not really gaining anything. Am i correct to assume the input R will be the pot R (calculated as advised by PRR previously) in series with the input R of the opamp buffer stage? If so this should be a considerably higher impedance than the pot/mix R combo alone, therefore decreasing the load on the source with several such bus controls in parallel...[/quote]

I think you mean to say the buffer's input Z is substantially higher than the pot Z.

Unity gain or even more-than-unity gain noninverting buffers are very high Z, typically. Unless they are very low input current parts though, you will want to capacitively couple to the pot wiper to minimize wiper noise, and thus will need a d.c. reference resistor at the opamp's input. This value will typically set the loading on the pot wiper at audio frequencies.

 

[jeth]

Sorry but i think i'm in slowmode again..

This value will typically set the loading on the pot wiper at audio frequencies.

Ok, before reading your post my main concern was the benefits to the loading on the source, rather than pot loading. So while the pot is loaded by the reference R at buffer input, what is the source seeing?

I'm assuming that the impedance of the pot into a buffers input is substantially higher than the pots own R.

What I meant is that I assumed the load of the pot followed by the buffer would be decreased compared with the pot alone between source and summing input.

Above what sort of input current will the arrangement you described be necessary and how would I implement it. Sorry, but at this stage of learning every explanation presents more questions..