# creating 'realistic' log pot taper with VCA

#### pgo1

Joined Nov 7, 2012
67
I'm making an amplifier with a VCA and I want to create a pot control that responds like a normal log pot. Obviously when you control the volume with a linear pot into the control port of a VCA, it never goes to 0 without altering how steeply the volume cuts off as you turn the pot.

I read elsewhere that a log pot is actually two tapers, a log taper and a linear taper which goes to 0 at the end of its range. My question is at what point of the taper does the linear taper start? Its a little difficult to measure this reliably with a physical pot

additionally, does anyone know a schematic for VCA control which does this? I can't have been the first..

#### Papabravo

Joined Feb 24, 2006
21,322
If I understand you correctly you want to create a voltage source that is a linear to log converter. In other words the output voltage is the logarithm of the input voltage. Did I get that correct?

#### MrChips

Joined Oct 2, 2009
31,094
A physical log taper pot is still a resistor.
Hence it should go to 0Ω at the bottom of its travel.

#### crutschow

Joined Mar 14, 2008
34,847
A common way to get a quasi-log response from a linear pot is to add a resistor, with a value of 10-20% of the total pot resistance, from the wiper to ground.

Below is the LTs simulation of that configuration:
The output is plotted with a dB (log) scale, so a perfect log output would have a straight line.
The output is effectively logarithmic (for audio purposes) from about 10% to 90% of pot rotation.
Below 10% it drops off rapidly as the output approaches an attenuation of infinite dB at 0% rotation.

The value of R2 determines the output curve shape.
Here a 10% value for R2 gives a 50% pot rotation output attenuation of ≈-17dB.
A linear pot has an output of -6dB at 50% rotation, of course.

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#### pgo1

Joined Nov 7, 2012
67
ah hmm i think some of you are misunderstanding what I mean.

I have a VCA which is scaled at -33mV / dB. If I use a linear pot to control the voltage to the CV control then at it's minimum the signal will not be 0V/-infdB, it will only be low and in some cases audible.

If I use a normal log pot as a volume control the signal will actually be 0V /-infdB (ok I know it's not infinity but it's small enough that it's buried in noise)

I want to make the VCA with a lin pot control behave like a log pot i.e. go to 0 at the end of the taper. I could make the linear control slope steeper which would result in a lower minimum but it would make the pot "feel" different.

I'm going to rephrase the question a bit - i heard that pot tapers are composites of more than 1 taper and I was planning on making a circuit which used a linear control voltage but "swaps" tapers (both linear voltages but one is "steeper") at low volume. I wondered at what point in the taper (at what gain in dB) i should swap between tapers to achieve my goal and retain a normal feel, and if anyone is familiar with any circuits which do this

Most log pots are just an approximation using three linear sections.
what are the three sections?

#### crutschow

Joined Mar 14, 2008
34,847
I have a VCA which is scaled at -33mV / dB. If I use a linear pot to control the voltage to the CV control then at it's minimum the signal will not be 0V/-infdB, it will only be low and in some cases audible.

If I use a normal log pot as a volume control the signal will actually be 0V /-infdB (ok I know it's not infinity but it's small enough that it's buried in noise)
Not true.
Both types of pots go to 0V (wiper shorted to the end terminal) when they are at the zero position.
There is no difference between a log pot and an linear pot at that position.
Why do you think there is?

#### pgo1

Joined Nov 7, 2012
67
Not true.
Both types of pots go to 0V (wiper shorted to the end terminal) when they are at the zero position.
There is no difference between a log pot and an linear pot at that position.
Why do you think there is?
you're still not understanding me. im not talking about a lin vs a log pot, im talking about *controlling a VCA* with a lin pot

#### Ian0

Joined Aug 7, 2020
10,277
what are the three sections?
For a 270° rotation potentiometer, they are about 90° per section, as close a fit to the log curve as they can manage.

#### crutschow

Joined Mar 14, 2008
34,847
you're still not understanding me. im not talking about a lin vs a log pot, im talking about *controlling a VCA* with a lin pot
And you don't seem to be understanding me.

I showed you in post #4 how to use a lin pot to generate a quasi-log output for your "VCA".
And the lin pot goes to 0V output at 0% rotation [which can't be shown on a dB (log) scale].

So what more do you want?

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#### Ian0

Joined Aug 7, 2020
10,277
The output of the VCA is already log, why would you want to control it with a log pot?
A good VCA will go down to -90dB (provided you don't get crosstalk due to poor pcb layout, or coupling through the power supply etc.) How much attenuation do you need?

#### LowQCab

Joined Nov 6, 2012
4,311
Here's some interesting info on Volume Pots ..........
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#### pgo1

Joined Nov 7, 2012
67
I showed you in post #4 how to use a lin pot to generate a quasi-log output for your "VCA".
There's no need for the quotation marks - it's not euphemistically a VCA, it's an actual VCA

ill try and explain it another way. Attached is a schematic with a pot which goes from 0 to -5V and derives a CV from that which controls a VCA. Below are the dB responses of

green) plot of how a normal log pot behaves (more or less, I used your loaded lin pot solution for this)

red) the VCA output with the gain scaled to approximately how a normal log pot behaves - you can see that at minimum it's about -33dB, this isn't quiet enough

pink) what happens if i feed the green trace into the VCA, with a few component tweaks. it's fine, workable, but not exactly what I was asking. In my particular circuit (of which this is a smaller part) I don't want to alter the response of that pot because it needs to feed other things which need the linear response

With those plots in mind what I was asking with my original question is at what point in the end of the taper will the green curve start to go negative compared to the red i.e. when the gain quickly reduces to 0. I've obviously answered my own question here by plotting it so don't worry about answering that

Additionally, I wanted to know a circuit where I could sort of swap between responses, have the red plot all the way up to the end of the taper of the pot then swap to the green plot for the end - some kind of op-amp processing *after* the pot - and I was wondering if there are schematics for anything like that

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#### pgo1

Joined Nov 7, 2012
67
Here's some interesting info on Volume Pots ..........
thanks! I know that site well. Still not exactly what I was asking though

#### Ian0

Joined Aug 7, 2020
10,277
Just use a linear pot and increase R1 to 68k.
You have only 1.1V control range, which will give you a maximum of 33dB attenuation. If you increase to 3.3V control range, you will get 100dB attenuation, and it will be perfectly logarithmic.
SSM2164 is a very good VCA.

#### pgo1

Joined Nov 7, 2012
67
Just use a linear pot and increase R1 to 68k.
but WITHOUT doing that
I could make the linear control slope steeper which would result in a lower minimum but it would make the pot "feel" different.

#### Papabravo

Joined Feb 24, 2006
21,322

#### pgo1

Joined Nov 7, 2012
67
Is there a reason for this obstinacy?
I could make the linear control slope steeper which would result in a lower minimum but it would make the pot "feel" different.
Having -50dB in the middle of the pot is not usable. Plus, i use this in different contexts so it's an interesting general question for me. nothing to do with obstinacy, that's was the original question and it's still the question.

#### Ian0

Joined Aug 7, 2020
10,277
Having -50dB in the middle of the pot is not usable. Plus, i use this in different contexts so it's an interesting general question for me. nothing to do with obstinacy, that's was the original question and it's still the question.
First decide on your minimum and maximum points, and set R1 accordingly. The pot will always be at 0V and 5V at the ends of its range. Then decide on what attenuation you need at the centre, and calculate the value of resistor to put between wiper and one end of the pot.

#### crutschow

Joined Mar 14, 2008
34,847
I don't want to alter the response of that pot because it needs to feed other things which need the linear response
That's a new requirement I wasn't aware of (would have been nice to know at the beginning).
So I'll let others guide you on your quixotic quest.