Here’s an attenuverter:

You find this design in lots of places. When the wiper is fully counterclockwise, it’s equivalent to this:

`R3`

doesn’t matter, it doesn’t change the input voltage, so we can ignore it (aside from the fact it draws a little current, but it doesn’t change the output). So it’s just an inverting amplifier with gain equal to -1 times the ratio of `R2`

to `R1`

, which is -1: The output voltage is minus the input voltage. Which is one thing you (normally) want from an attenuverter.

When the wiper is fully clockwise it’s equivalent to this:

Here again `R3`

doesn’t matter. The voltage on the `+`

input is `Vi`

, so the voltage on the `-`

input is `Vi`

too. That means there is no current flowing through `R1`

since the voltage is the same on both sides. Then there’s no voltage flowing through `R2`

either, which means it also has the same voltage on both sides. So the output voltage is equal to the input voltage. Which is another thing you want from an attenuverter.

Finally, if the wiper is in the middle, it’s equivalent to this:

`R4`

and `R3`

form a voltage divider. The voltage on the `+`

input, and hence the voltage on the `-`

input, is half the input voltage. For the current through `R1`

to equal the current through `R2`

the voltage drops must be the same, so the output voltage is zero. Which is the third thing you want from an attenuverter.

Can you generalize this? What if the resistors (including the pot) have different values? With the wiper counterclockwise you have $l V_o = -(R2/R1)V_i$. The gain is $l g = -R2/R1$. But with the wiper clockwise you still have $l V_o = V_i$; the gain is still 1. So we need to have $l R2=R1$. As for the pot, its value doesn’t change anything — it just acts as a voltage divider for the voltage on the `+`

input.

Now here’s another attenuverter. This is used in the Befaco A*B+C module:

This is rather more complicated. With the wiper fully counterclockwise (notice the pot is the other way around this time) it’s equivalent to:

`R1`

doesn’t matter here. The `+`

input connects to ground, so the `-`

input is at 0 V. There is no current through `R3`

, so it doesn’t matter either, and it all comes down again to an inverting amplifier with gain equal to minus the ratio of `R4`

to `R2`

, or -1. Output equals minus input. That’s good.

With the wiper clockwise:

`R1`

and `R5`

are a voltage divider; the voltage on the `+`

input is half the input. Same on the `-`

input. Now it gets a little messy, we have to balance the currents through `R2`

and `R3`

with that through `R4`

:

$l \displaystyle (V_i-V_i/2)/R2 + (0-V_i/2)/R3 = (V_i/2-V_o)/R4$

or

$l \displaystyle (V_i/200\mathrm{k}) - (V_i/100\mathrm{k}) = (V_i/2-V_o)/100\mathrm{k}$

Which works out to:

$l \displaystyle V_o = 100\mathrm{k}\times V_i(1/100\mathrm{k}-1/200\mathrm{k}+1/200\mathrm{k}) = V_i$

Right! And in the center,

Same topology, different values. The voltage on the `+`

input is $l 1/3$ the input voltage. Now

$l \displaystyle (V_i-V_i/3)/R2 + (0-V_i/3)/R3 = (V_i/3-V_o)/R4$

or

$l \displaystyle (V_i/150\mathrm{k}) - (V_i/300\mathrm{k}) = (V_i/3-V_o)/100\mathrm{k}$

Which works out to:

$l \displaystyle V_o = 100\mathrm{k}\times V_i(1/300\mathrm{k}-1/150\mathrm{k}+1/300\mathrm{k}) = 0$

Right again.

Since there are more components, there are more possibilities to generalize this. For instance, the gain, $l g = R4/R2$, can be non unity. To make it work, you want the output voltage with the wiper clockwise equal to $l gV_i$, and if I’ve done the algebra right (who knows?) this leads to the constraint

$l \displaystyle R4/R3 = 2g\frac{RV1+R1}{RV1}-g-1$

Similarly you want the output to be 0 when the wiper is halfway, which leads to another constraint. It all gets very messy, though. You sure you want a non unity gain attenuverter that badly?

There is one other reason to consider the second of these attenuverters. In the first one, when the wiper is fully clockwise, the input voltage is connected directly to the `+`

input. With a TL07x op amp, you can get unwelcome behavior known as phase reversal when the `+`

input voltage is below -8 V (with ±12 V power). Most of the time synth signals are in the -5 to +10 V range, in which case you’re okay, but -8 V is possible.

In the second attenuverter, though, the largest the voltage on the `+`

input can be is $l RV1/(RV1+R1)$; if R1 and RV1 are both 100k, and if the input voltage is strictly limited to ±12 V, then the voltage on the `+`

input can go no lower than -6 V. No risk of phase reversal there.

But again, -8 V rarely occurs, and if it does nothing gets fried; you just get some unexpected and undesired output values. And the first attenuverter has an advantage in that it can be modified to give it a usefully nonlinear response — see this Kassutronics blog post: https://kassu2000.blogspot.com/2018/04/precision-attenuverter-mixer.html . The mod consists of adding two resistors:

Its effect is shown here: Green is the unmodified attenuverter response, blue is the modified one. (You can get more or less pronounced nonlinearity using different resistor values — larger is more linear.)

As for the second attenuverter, the response actually is slightly S shaped already:

Maybe there’s a way to change the amount of nonlinearity, but probably not without a whole lot of treacherous algebra if so.

And finally, as noted above, the value of the pot in the first attenuator doesn’t affect the output voltage. That’s not the case for the second attenuator. Pots have very loose tolerances; nominally 20% is common, though in reality they tend to be better than that. But here in green are the second attenuverter outputs with 80k, 100k, and 120k for the pot — fully counterclockwise is on the left, fully clockwise on the right. In blue is the same for the first attenuator — you see only one line because all three coincide.

A 20% variation in the output at the fully clockwise position, versus no variation at all. That’s a strong argument in favor of the first attenuverter.

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