Fender reverse taper volume pots

[Xris]

Hi all.
It's been known that fender installed reverse taper pots on their volume controls on certain amps.
The code is 50k-30C.
Everyone complains that the volume goes from
0-"jesus, that's loud!!!" with little movement, and not much after.
On the amp there is a tone control pot labeled 50K-15A,
Which is a forward taper pot.
So, the question is,
Can I swap these pots over so I can have a usable volume control?

Cheers.

[Leo_Gnardo]

It's been known that fender installed reverse taper pots on their volume controls on certain amps. The code is 50k-30C. Everyone complains that the volume goes from 0-"jesus, that's loud!!!" with little movement, and not much after.

Must be solid state amps with volume control in an op amp feedback loop? Swapping as you suggest, will be making matters worse. You can "bend the curve" with just one resistor. That may limit the maximum volume you can dial but . . . max would be way too much for any conceivable use, so what.

Here's R.G.'s excellent article on what to do. You can skip the first half & jump down the the spot you see the graph with pot curves on it, start reading there, get a couple resistors ready to hand & start heating up your iron. Here you go, I've copied the text below but not charts. You'll know what you are doing and why, and you can adjust the curve to your own needs.

http://geofex.com/

Go to 8/31/99 "the secret life of pots", skip down to the spot you see pot curve charts and read starting about here:

Which leads us to tapering.

What is taper? It's just the ratio of the resistance already passed as the pot turns to the total resistance of the pot, described as a curve. For instance: we want to make a variable power supply with an adjustment pot that smoothly varies the voltage from one to ten volts, so we want a control that lets us do that. We have no idea whether we'll want mostly low voltages or high voltages, so we want to adjust it equally well anywhere in the range. In this case, it's most natural for the control pot to have an equal change in resistance or voltage divided per unit of rotation - we want the control to feel linear. This much of a turn is one volt, no matter whether it's near 0V or near 10V.

Volume controls are different. The human ear does not respond linearly to loudness. It responds to the logarithm of loudness. That means that for a sound to seem twice as loud, it has to be almost ten times the actual change in air pressure. For us to have a control pot that seems to make a linear change in loudness per unit of rotation, the control must compensate for the human ear's oddity and supply ever-increasing amounts of signal per unit rotation. This compensating resistance taper is accurately called a "left hand logarithmic taper" but for historical reasons has been called an audio or log pot. In these pots, the wiper traverses resistance very slowly at first, then faster as the rotation increases. The actual curve looks exponential if you plot resistance or voltage division ratios per unit of rotation.

If you used an audio/log taper pot for the control of the power supply we mentioned, the output voltage would increase very slowly at first, creeping up to maybe 10% of the final output at 50% of the pot rotation. It would then blast the other 90% in the last half of the rotation - very hard to control. Likewise, if we used a linear pot for volume control, the volume would come up dramatically in the first half of pot rotation, and then do very little change in the last half.

The dark horse taper is reverse audio, or more strictly "right hand logarithmic" taper. This taper traverses resistance very quickly at first, then more slowly as it is turned further. It's the inverse of the audio taper. This is used in some bias circuits and in controlling the speed of certain RC oscillators, which is where the audio tinkerer runs into it most.

The following diagram shows the three main kinds of pot tapers, along with one common approximation to an audio taper. Curve 1 is linear taper. If we clip one lead of our Ohmmeter (Hey! There he is again!) onto the leftmost lug, and the other lead on the center lug, then the resistance we read as we rotate the pot clockwise will fall on the curve that goes diagonally upwards. The proportion of the total pot resistance we traverse as we turn the pot is linearly proportional to the amount of rotational travel we turn.

Curve 2 shows what happens with an audio or logarithmic taper. As we turn the shaft, the proportion of resistance we traverse increases slowly at first, more slowly than the percentage of rotation. As we get past half the available rotation, the rate of resistance traversed speeds up as we get closer to the furthest rotation. This compensates for the human ear by increasing sound levels very slowly at first, then faster as the ear's sensitivity falls off at higher sound levels.

When we buy "audio taper" pots, we usually get something like Curve 3. For less expensive pots, manufacturers use a two or three-segment approximation to Curve 2. It's not perfect, but it usually works OK. Curve 4 is the typical resistance versus rotation curve for reverse log pots. In real life - that is, if you ever found one of these in real life - it is usually a two or three segment approximation, too.

If you have an unknown pot, you can figure out what taper it is. You measure the resistance from end to end, then turn the pot exactly to half its rotation and measure the resistance from the counterclockwise lug. The crosses on curves 1, 2 , and 4 show the most probable values. If the resistance is 50% of the total resistance, then the pot is linear. If you measure only 10% to 20% of the total resistance, the pot is an audio taper. If you measure 80%-90% of the total resistance, the pot is a reverse log taper.

At this point I should probably explain what a counterclockwise lug is. Of the three contacts on the pot, the wiper is easiest to pick out. If you turn the shaft fully counterclockwise, the wiper lug will show very small resistance to one of the other contact lugs. This is the counterclockwise or "cold" terminal. Turning the shaft fully clockwise, the wiper will show very small resistance to the most clockwise lug, also called the "hot" lug.

There are other tapers, but they have very specialized uses.

Our problem is this: If we need a specialized taper to recreate some effect, how do we get it if we can't go buy one? That leads us to tapering resistors.

If we set up the easiest, simplest pot to get, a linear taper pot of resistance R, and then connect a "tapering resistor" across the wiper and CCW lug, we get the situation shown in the following diagram. For clarity, I've separated the resistance above the wiper and the resistor below the wiper into two separate resistors. It makes the calculations much easier.

We're assuming the total pot resistance R is split into an R1 at the CW side and R2 on the CCW side, with R3 paralleled with R2. We'll let "a" represent the fraction of the total resistance R that the wiper has turned, and "b" be the fraction of R that R3 is. When we get out the algebra books and do the math, we find out that we can show that the ratio of output voltage to input voltage is that odd looking fraction in the picture. When we calculate out the results, we find that the divider ratio of Vout to Vin is shaped something like a true logarithmic tapered pot if we pick the right value for b. If b happens to be 1/4 to 1/5, the resulting voltage division is remarkably close to a true logarithmic pot, probably closer than a two segment approximation that we could buy! Wow! No more waiting for volume control pots!

Here's what we see when we do the math:

Unfortunately there's a gotcha in there. It's true that the voltage division ratio of this rig is arbitrarily close to that of a log taper pot. However, neither the load seen by whatever drives Vin or the source resistance as seen by the input of whatever is connected to Vout is close to what would exist for a real log pot of value R. In fact, the load on Vin varies from 1/(1+1/b)*R up to R. That means that if we're trying to do a log taper with b = 1/4, the load on Vin will be as much as 0.2* R. This may be OK, but you have to keep it in mind.

In general, if you have a voltage source that can drive a load of 1/4 to 1/5 of R and a load on Vout that has an input impedance much higher than that same 1/4 to 1/5 of R, this is a good replacement for an audio or log taper pot.

In a bit of good fortune, if you hook up the tapering resistor from the CW or hot side of the pot to the wiper, the pot emulates a reverse log pot, just as well as it did a log pot when hooked to the CCW side.

There are two ways to hook up a pot. You can hook it up as a three terminal voltage divider as we've seen above, or as a variable resistance, the two-terminal connection (sometimes called a rheostat connection for historical reasons).

In the two terminal connection, the resistance through the pot is what we're interested in, not the voltage divider ratio.

It turns out that the tapering trick works here, but only partially. If we want to make a reverse log tapering pot, we're in! However, there is no way to get a simulation of a log taper pot in the two terminal connection. For that we have to buy real audio taper pots.

Here's what you get if do the series resistor connection:

Note that the CW terminal is unused. This is usually tied to the wiper, although that is not seen here. You can't simply put the tapering resistor from the CW terminal to the wiper and get a log taper pot emulation, like you could with the voltage divider connection. That just gives you the reverse of this graph, with the resistance starting to decrease slowly and then faster. The two terminal connection is non-polar; it looks the same however you hook it up. The only thing that changes is which end of the graph you start from.

In the math examples I left b, which is the fraction of the pot resistance that the tapering resistor is, as a parameter rather than making it a fixed ratio. Usually, people pick a value of b of about 4 or 5. Those curves are close to the classical mathematical description of a log or reverse log pot. I left b a parameter to show you that you can make your own taper by selecting a different value of b. For a semi-log taper, use a b of about 2.