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attenuverter circuit
MUFF WIGGLER Forum Index -> Music Tech DIY  
Author attenuverter circuit
nff
Hi!
I know attenuverters are an old matter, but I'm kinda stuck, as I was wondering, is there a way to have a non-unity gain in-built in the attenuverter itself? I mean, without the need of a supplementary gain stage.

I attached the schematic I'm using, very basic stuff, but somehow I can't get it to have the same gain (>1) both in the positive and negative region.

In fact, I don't think I'm fully grasping both operation modes, the attenuverter is basically an amplifier that can act as inverting or non-inverting, depending on the position of the pot.
So I'll try to redraw the circuit with the pot fully panned on either side, assuming a CW and a CCW direction (marked on the first schematic).

With the pot fully CCW it's pretty straightforward, it becomes an unity-gain inverting amplifier.
The problem is with the CW position; no current will flow on the negative input resistor of course, so the topology will be similar to a non-inverting amplifier, but there is still a resistor in the negative feedback path!
In the real world it does behave as a unity-gain voltage follower, tho, so I'm definitely missing something here...

I know, it's no big deal, it works, even if I don't fully understand it, and I can always use a gain stage after the attenuverter....but...I want to understand!

Thanks!
CLee
That's the circuit I use for processing inputs and I think it needs to be unity gain.

The circuit Ken Stone uses in his DC mixer may be able to be built with gain other than +/- 1.0 (I haven't built this, so theoretically...)

https://www.cgs.synth.net/modules/cgs04_mix.html

set the feedback resistor of both op amps to adjust gain. The stuff in red is the attenuverting circuit, which he explains building in the link.

cygmu
nff wrote:

The problem is with the CW position; no current will flow on the negative input resistor of course, so the topology will be similar to a non-inverting amplifier, but there is still a resistor in the negative feedback path!


Right, but again no current flows. The output comes up to the input voltage level, both ends of the resistor reach the same voltage because no current is flowing, so the op amp sees the same voltage at both its input terminals and it stops ramping up the output at that point.
loki
Here is an analysis for you to look at. You can have gain.
nff
loki wrote:
Here is an analysis for you to look at. You can have gain.


oh YESSS. that's wonderful, exactly what I wanted. thanks a lot!

from 43 years ago, wow! how cool is it to be able to use one single opamp to perform attenuvertion AND gain? How elegant. Love it.
nff
CLee wrote:
That's the circuit I use for processing inputs and I think it needs to be unity gain.

The circuit Ken Stone uses in his DC mixer may be able to be built with gain other than +/- 1.0 (I haven't built this, so theoretically...)

https://www.cgs.synth.net/modules/cgs04_mix.html

set the feedback resistor of both op amps to adjust gain. The stuff in red is the attenuverting circuit, which he explains building in the link.



yeah I'm familiar with this one, very well thought, but still using 2 op-amps, it looks like you CAN actually attenuvert AND gain with a single one (see loki's post). Gona give it a breadboard soon.
thanks!
nff
cygmu wrote:
nff wrote:

The problem is with the CW position; no current will flow on the negative input resistor of course, so the topology will be similar to a non-inverting amplifier, but there is still a resistor in the negative feedback path!


Right, but again no current flows. The output comes up to the input voltage level, both ends of the resistor reach the same voltage because no current is flowing, so the op amp sees the same voltage at both its input terminals and it stops ramping up the output at that point.


makes total sense. now I see.
Just curious, did this behavior look obvious to you from the beginning?
It's SO obvious now that you told me.
Thanks!
notmiserlouagain
Hi nff, the circuit you posted and loki´s circuit is a variaton of is called "switch hitter", but I personally would use the 2 opamp circuit anyway, because you can mix/sum various pots/inputs that can be individually configured for uni-/bipolar (which is what I usually want), the switch hitter only processes one input, so in a mixer application you´d need an amp for every input plus one sum amp vs. two amps whatever...

If you want gain in the 2 opamp version don´t change both feedback resistors (like proposed above), change only the second one!
nff
notmiserlouagain
hey hi!
I'm having a look at the switch hitter, there's a nice article here
http://leachlegacy.ece.gatech.edu/ece3050/sp04/OpAmps01.pdf (section 1.5.5) with all the equations.

I agree, doing it with two inverting stages lets you have an extra summing node, this can come in handy when the circuit is near the user interface, but since I'm using it embedded with other parts of the circuit, I really appreciate being able to spare an op-amp! Also, I have no problem driving the circuit from a low impedance, since I'm controlling exactly where the input comes from!

Quote:
f you want gain in the 2 opamp version don´t change both feedback resistors (like proposed above), change only the second one!

yeah, I always do it like that, keeps calculations much simpler, not sure about noise, but at a glance I'd say performance is the same.

thanks!
fitzgreyve
loki wrote:
Here is an analysis for you to look at. You can have gain.


I've just been playing with this circuit - in essence it works, but I think there is an error in the article:

It states that (for gain "n") R1 = R/n - I agree [standard gain formula for inverting op-amp configuration]

it also states that R2=R1/(n-1) - I think this is wrong and it should be R2=R/(n-1) [i.e. standard gain formula for non-inverting op-amp configuration]

I've just tried this for gains of n=2 and n=3:
n=2:
R=100k, R1=100k/2=50k, R2=100k/(2-1)=100k
this gave a +ve gain of x2, but a -ve gain of a bit under x2 (about 1.95 - could just be tolerances)

n=3:
R=100k, R1=100k/3=33k, R2=100k/(3-1)=50k
this gave +ve and -ve gains of x3.

interestingly, if you go for n=1, you get:
R=100k, R1=100k/1=100k, R2=100k/(1-1)=100k/0=infinite- i.e. open circuit, giving the more common "x1" attenuverter circuit.

So, while I haven't checked the linearity, it does appear to work.
Graham Hinton
fitzgreyve wrote:

I've just been playing with this circuit - in essence it works, but I think there is an error in the article:

It states that (for gain "n") R1 = R/n - I agree [standard gain formula for inverting op-amp configuration]

it also states that R2=R1/(n-1) - I think this is wrong and it should be R2=R/(n-1) [i.e. standard gain formula for non-inverting op-amp configuration]


You are correct.

The non-inverting gain will be 1 + R/R/(n-1) = 1 + n - 1 = n.

Quote:

So, while I haven't checked the linearity, it does appear to work.


The linearity is dependent on the pot and there lies the rub. Attenuvertors are difficult to zero because the cancellation point is not necessarily the mechanical centre of the pot and you halve the resolution of the pot anyway. Having gain is going to reduce the resolution further. I would rather have a gain control and an inverting switch, preferably integrated with the pot.
euromorcego
fitzgreyve wrote:

interestingly, if you go for n=1, you get:
R=100k, R1=100k/1=100k, R2=100k/(1-1)=100k/0=infinite- i.e. open circuit, giving the more common "x1" attenuverter circuit.

for n=1 it is exactly the schematics of the fonik attenuverting chicklets. And there he uses a trimpot to set the zero value, so the small offset seen by you is probably normal.

I agree with Graham Hinton that a switch to invert has the advantage that you keep the full resolution of the pot. But still, the single OpAmp circuit with some gain might be useful.
nff
still didn't have the time to breadboard it, thanks a lot fitzgreyve for pointing out the error!
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