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Potentiometers are renowned for wearing out (at least in my experience); the little wiper eventually just wears its contact and nolonger has a firm electrical connection. For an audio device, this can manifest as a crackle when changing volume. The wear isn't necessarily even and there may be positions that have worse contact than others. I've noticed it be worse usually near the upper limit (full volume; full brightness; etc.), but the wear distribution can probably mostly be attributed to how the device has been used.
    Having a component with such friction seems like a very bad idea to me (and evidently it is) and I often wonder whether there are commercially available designs that do not have a sliding contact (excluding digital potentiometers[1]), and whether they're economical. I envision that one such wiperless design would be based on ball-bearings or epicyclic gears, with at least one of the balls or planet gears being conductive, the rest being insulative, and the tracks in which they roll, or the annulus or star/sun gear, having the resistive gradient element(s). But is anything like this currently available?


Note 1: It should behave similarly to an ordinary passive potentiometer. Digital potentiometers require a power supply and draw power so, as I understand, are not necessarily drop-in replacements (a 3-pin digital potentiometer would require that the end pins double-up as the power supply, which isn't always the case). I'm specifically interested in knowing whether such components as wiperless passive potentiometers exist which in their simplest form have 3 pins where the sum of the resistances between pins 1 and 2 and between pins 2 and 3 is intended to be constant (i.e. a 2-pin variable resistor is not by itself a potentiometer).

George
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James Haigh
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    James, I've never come across or heard of epicyclic (planetary) potentiometers. You could have come up with something novel. – Nick Alexeev Mar 08 '15 at 20:10
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    You mean like a strain gauge? – Ignacio Vazquez-Abrams Mar 08 '15 at 20:15
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    Why would a pot 'wear out' usually at the full volume setting where it is rarely set? Crackles in pots in audio are usually due to dirt or DC. I repait vintage audio, 40-60 years old, and I've never seen the failure mechanism you describe. I've only seen the wiper lift off the track completely, or mechanical failure of an associated power switch. – user207421 Mar 08 '15 at 21:37
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    @EJP Pots crackle the most at the ends of travel, because that's where you shove all the dirt to. Ideally, there should be no current flowing through a pot's wiper - whether DC or AC. This makes it extremely hard to apply a potentiometer in a low-noise audio circuit: high impedance buffers are plenty noisy. – Kuba hasn't forgotten Monica Mar 08 '15 at 21:39
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    Just a question - are digital pots out of the question for your design? They are controlled by a microcontroller so you can get around the "scratchiness" issue with the tradeoff of a microcontroller input/output and perhaps a way to simulate linear/log control of it. – cowboydan Mar 08 '15 at 22:13
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    How about adding a clutch? Lift the contact while moving it. If you must maintain constant contact, make it a wheel with bearings. – JDługosz Mar 09 '15 at 07:50
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    @Adam Haun: “Are there potentiometers that don't wear out?” is a ridiculous title rename, because the answer can only be ‘no’. You could indeed say ‘Are there potentiometers that don't wear out faster than similar frictionless components?’ (like variable capacitors), but then why not just stick with the original title which expresses this more concisely? – James Haigh Mar 09 '15 at 11:56
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    @JamesHaigh: On the other hand, my initial reaction to the current title ("*Do there exist frictionless potentiometers?*") was basically "*Of course not, and if there were, how would you make one hold its position, anyway?*" Maybe something like "*Are there potentiometers that don't have sliding contacts?*" would be the most accurate and descriptive title. – Ilmari Karonen Mar 09 '15 at 12:32
  • @Ilmari Karonen: Sure, I considered this too, but I decided that the friction required to hold position can be considered part of the interface that the potentiometer is being used for. Potentiometers can also be used to measure angles between mechanical components and so there's no reason that the potentiometer itself is required to have friction. However, I like your suggestion. I'll clarify the terminology, removing the vague term ‘frictionless’ and speaking in terms of ‘without sliding contacts’ which is specifically what I mean. Hey, how about ‘wiperless’? – James Haigh Mar 09 '15 at 13:19
  • @cowboydan: Not really for my own designs, but quite often I'm repairing, modifying, or otherwise tinkering with old electronics, and as I understand, digital potentiometers are not simple drop-in replacements for ordinary 3-pin potentiometers. My own design choices likely involve open hardware boards with a microcontroller anyway, such as a BeagleBone Black or Arduino, so I'd use a better input method and, if used for signal amplitude, then: analogue output; → parallel diode (for flyback); → series inductor → parallel capacitor (low-pass filter); → voltage-controlled attenuator or amplifier. – James Haigh Mar 09 '15 at 15:12
  • @JamesHaigh Apologies for the confusion. My thinking was that no mechanical system is without friction, so the answer is an obvious "no", as Majenko said. Your first paragraph referred to "wear" several times, so I changed the title to reflect that. (Kuba Ober's existing answer about non-contact sensors seemed to fit that meaning, but perhaps that's not the sort of answer you're looking for.) I agree that "wiperless" is a much better choice. – Adam Haun Mar 09 '15 at 15:59
  • I don't understand the obsession with a 3-terminal device that requires no extra power. In many situations where such devices are applied with the control presented to the user, their use is bad practice. For example, you don't want to use a potentiometer to directly control audio volume - it won't ever work right. About the only use where a potentiometer can be applied directly in a circuit is for trimming, there the user doesn't constantly tweak the knob. Otherwise, it should only generate a control voltage, and you can certainly replace it with something fancier that requires a power supply – Kuba hasn't forgotten Monica Mar 09 '15 at 20:08
  • To put it differently: you propose an extremely complicated, and expensive to make mechanical contraption to replace a reliable, powered, solid state circuit. That makes no sense to me. You must first ask yourself whether you even *should be using a potentiometer at all* in the signal path. I see no vindication for a potentiometer obsession in your case :( I've stopped designing in potentiometers 15+ years ago, at least in professional products. They are way too much trouble. Even back in 2000 I'd much rather use a Xicor digital pot for trimming, with an external trim dongle with buttons. – Kuba hasn't forgotten Monica Mar 09 '15 at 20:09
  • Do you know Vactrols? It is just a LED with a LDR in one package. You still need a power source for it and you need something to control the LED. Nice is the natural behaviour of Vactrols, it acts like a variable resistor. You can control these with a potentiometer without the noise of potentiometer. Vactrols are 'slow' devices, it takes some time to react on changes so it has a more natural feeling. You are talking about audio, these are pretty suitable for audio however audio pots are different, there are linear or logarithmic (aka "audio taper") pots. The last one is used for audio ... – Codebeat Aug 15 '19 at 06:38
  • ... not a simple device because it is logarithmic. But I don't unerstand, you want to control audio so there must be some kind of power source. If a power source is not possible, use some coin cells. Controlling something doesn't need much power. If you don't want to use a power source, use a LDR with a gradient window/knob or something, however you still need something, in this case light, to control it. – Codebeat Aug 15 '19 at 06:38

4 Answers4

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How to get the most out of a potentiometer?

In many precision, low-noise designs, it's a bad idea to start with to even have the signal routed through the front panel. So, at the very least, the control element should merely produce a voltage signal that governs a voltage-controlled amplifier/attenuator. With a potentiometric source, you can buffer and low-pass-filter the control signal, so that wiper dropout effects are minimized.

schematic

simulate this circuit – Schematic created using CircuitLab

Here, a voltage reference feeds the potentiometer. The variable wiper resistance is modeled by Rw, which can vary by 9 orders of magnitude, but is mostly "low" and on the order of an Ohm. R2 keeps the time constant above 50ms. Since R2>>R1, R1's influence is small. C2 forms a low pass filter with R1+R2, but also acts as a hold capacitor. U2 is an op-amp set up in non-inverting mode, so that its input has a very high impedance. The output of U2 goes to a voltage-controlled amplifier.

C2 should be a low-leakage type with NP0 or plastic dielectric, and U2 should have a FET or CMOS input stage. So, don't use 741 for U2 with expectation that it would work all that great - although it will still work better than the naked potentiometer would.

If the wire from R1 to the circuit is long, you might need a bootstrapped shield. Some experimentation is necessary to ensure stability of the circuit then, though, as the shield-to-signal capacitance adds positive feedback to the system.

That already gives you a much better performing circuit than using a potentiometer directly on the signal. Even with a fairily short 50ms time constant, you can get rid of crackle even on the most ridiculously dirty potentiometers. You can always trade off response time for insensitivity to crackle.

Routing audio to front panels is usually an EMI nightmare and it's often not cheap at all to do it properly.

Voltage Controlled Gain

A good bang-for-the-buck voltage-controlled gain element can be made by using a photoresistor illuminated by a LED. Photoresistors, if you select them, can have very low voltage coefficient of resistance and thus very low distortion, certainly beating most simple multiplier circuits by an order of magnitude or more. They are available as self-contained units, known as Vactrols, from Excelitas. They need to be applied with some care, as you don't want to exceed about 100mV across the photoresistor, but otherwise they are wonderfully powerful devices for about $5 each.

There are decent integrated voltage-controlled amplifiers, such as the last-time-buy (sadly) SSM2018, or newer AD8338, THAT2181, etc.

How about rolling contact?

If you still have a mechanical mouse, open it up. Take the ball out and look at the rollers. Invariably they'll be covered with a hardened track of grime. Rolling contact isn't all it's cracked up to be if you can't control the environment quite well. Sliding contacts have a self-cleaning property. Rolling contacts, in a potentiometer, would have the exactly opposite behavior - they'd be self dirtying. That'd be a very bad idea.

Mechanically there's another aspect you seem to forget: rolling contact is wonderful at concentrating the stresses, and requires sufficiently hard surfaces to prevent wear. It's kinda hard to make a low-power resistive sensor where the surface needs to interface with a metal ball/roller while having any sort of useful life expectancy.

If you really don't care about the power of the circuit, you're fee to make the resistive track, C-shaped, out of hardened steel. Feed it a couple of amperes, in pulses, use a sample-and-hold circuit to get the pulse amplitude, and you're set. It'll work as long as you house it in a dust proof enclosure. Note that dust-proof is usually harder than water-proof (!).

TL;DR: The rolling contact would be possibly the worst thing you'd ever wish in a potentiometer wiper.

So, what other options are there?

You can obtain the signal from other sources. They all work by converting shaft angle into a voltage, using a variety of techniques. I present them in no particular order.

Noncontact Potentiometers

Suppose that you start with a basic, C-shaped resistive track of a potentiometer. Choose a large one, so that it's easy to work on. Open it up. Bend the wiper so that it's lifted up from the track, but just ever so slightly. Feed the track with an AC signal, say a 1MHz square wave, with the other end of the track at 0V. The wiper is capacitively coupled to the track, and will pick up a signal whose amplitude is proportional to the position on the track. You will need to tweak it to get rid of the worst parasitic capacitances, but work it will. You can use a FET follower or an op-amp to lower the impedance of the wiper's signal, then use a synchronous demodulator to convert the amplitude back to baseband. It might sound fancy, but for such a simple sensor you can do it on a couple dollars worth of parts, nothing fancy needed at all. You don't need any better than 10% linearity anyway.

Variable Transformers

A very precise, and perhaps an over-the-top source would be a RVDT (a rotary cousin of an LVDT). For a one-off "vanity" project, it'd be a nice choice - these things are virtually indestructible, and with luck you can get them cheaply from surplus. For a volume control, you could make a very simple RVDT conditioner (the circuit is same as for an LVDT).

Variable Capacitors

Another vanity option would be an old, heavy, rotary capacitor. The better ones have a pair of ball bearings. Similar to an RVDT, they have no other contacting parts to wear out. Put the capacitor in a multivibrator circuit, hook up to a voltage-to-frequency converter circuit (LT app notes have plenty of those), and you're set.

Magnetic Sensors

A much lower cost option would be a Hall sensor. Suppose you have a magnet oriented radially on a shaft, and a Hall transducer next to it. As you rotate the shaft, the magnetic flux passing through a properly placed sensor will vary. This is a good source of a control voltage - cheap to implement, too.

Optical Sensors

You can also have an optical sensor: print a V-gap, with X-Y mapped to polar coordinates, on a sheet of transparency foil. Install on the shaft. Put a LED-photodectector pair so that it "sees" through the gap. Condition the photodetector (either a transistor or a diode) with an op-amp.

Another optical option that doesn't need a V-gap would be to have a tilted disc mounted on the end of a shaft, so that it's not quite perpendicular to the shaft's axis. Then use a reflective sensor (LED + photodetector) to obtain a continuous signal proportional to the angle.

Another optical option is to have a multiphase pattern printed on a cylinder on the shaft, and use multiple optical sensors, with their outputs summed, provide the output. The pattern might look as follows:

axial distance
^
|   █████████
|      ██████
|         ███
|0---------360--> angle

As the cylinder turns above the sensors, their outputs get progressively lower. By judiciously tweaking off the number of detectors/stripes, and the detection distance, you can get by with a simple black-and-white pattern. Sometimes that's easier to manufacture than something fancier.

Strain-to-angle Converters

Yet another option, quite sensible if you know how to deal with strain gages, would be to have the shaft interface with a long spiral spring. Slap a 4-gage strain gage bridge somewhere on the spring, with the sensitive axis along the spring's length, and you get a very nice signal proportional to the shaft angle. You'll need to add a bit of friction into the mechanical circuit so that the shaft stays put when you release the knob.

Odds-and-ends

Yet another option, if you want to get funky, would be to have a variable acoustic capacitor. Have the shaft go through a flat toroidal box. It can have a rectangular cross section, of course. Make a radial slot through the inside of the box, and extend a radial pin from the shaft through the radial slot. Attach a paddle that almost fills up the box's cross section to the end of the pin. At the zero point in the box, add a partition and an acoustic transducer. Attach it to an oscillator, and you've got an electro-acoustic angle-to-period converter.


The above are only the things I have tried, with some degree of success, at some point in life. There's almost an infinite supply of other ideas, if you want to have some transduction fun.

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    If you can handle the additional complexity, a rotary optical encoder is probably your best bet. They're used extensively in the space industry because they're low power, reliable, and don't wear out. (I'm working on a payload that's using a potentiometer to save cost, but let me tell you it's not worth it.) – 2012rcampion Mar 08 '15 at 21:49
  • @2012rcampion The major problem with encoders is that they offer a discrete output. If one cares, as some people do, about a truly step-less output, that's only time-discrete if at all, all those other methods can work quite well. Everything depends on the degree of overboardiness you wish the project to be. For a vanity project, the funkier the better :) – Kuba hasn't forgotten Monica Mar 08 '15 at 21:54
  • You're right, I wouldn't use an encoder to replace a pot in any analog circuit (say, the volume pot on a guitar for example). I'd use it if you're digitizing the output anyway (e.g. to read into a uc or to control the output gain on a dac). – 2012rcampion Mar 08 '15 at 22:00
  • @2012rcampion Fankly said, I would not apply a pot directly to the audio signal even on a guitar. In fact, I would never ever apply it directly to an audio signal period. It never is a durable solution, and replacing crackly pots is not my favorite pastime. Probably the simplest low-distortion controlled resistance is a photoresistor - it would be a way better element for volume control than a pot. – Kuba hasn't forgotten Monica Mar 08 '15 at 22:03
  • Unfortunately a pot is really your only option on the guitar itself since it's unpowered. Even effects pedals (at least the cheap ones I've dismantled) use the pots directly in an opamp buffer. I'll have to look into the photoresistor design. – 2012rcampion Mar 08 '15 at 22:10
  • @2012rcampion You can certainly supply phantom power to the guitar, if you can live with <10mA of current available from most supplies. The less you use, the better. If you want to use a stereo cable and a custom supply breakout box, it's easy enough to feed as much power as you wish. – Kuba hasn't forgotten Monica Mar 08 '15 at 22:37
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    You can build a fully continuous optical encoder by gradually occluding a light or changing its incident angle. It's just that then you're hostage to the linearity of your detector. – pjc50 Mar 09 '15 at 12:40
  • @pjc50 I mentioned just such a device in my bucket list. I didn't think such a device should be called an encoder. When you have coding, you - by definition - transform discrete information. In most cases, the linearity of the detector is vastly better than that of the process modulating the light intensity. Even a 741 hooked up to a phototransistor or a photodiode is much better than the cheap V-slit feeding it would ever be. – Kuba hasn't forgotten Monica Mar 09 '15 at 12:47
  • Sorry, I'd missed that. It's a very impressive list, the sort of thing that should be issued as further reading to EE undergraduates! – pjc50 Mar 09 '15 at 13:16
  • @pjc50 I'd say that any EE undergrad should be building such things (not necessarily from any particular list, of course), like crazy - it's a lot of fun, and it's amazing how many ways any particular problem can be cracked. It's also amazing how useful the modern semiconductors are. You can get a lot done for ten bucks :) – Kuba hasn't forgotten Monica Mar 09 '15 at 13:31
  • Wow, thank you for this incredibly comprehensive answer! I couldn't keep-up reading your large expansion of edits or everyone's comments yesterday. Your answer has a lot of useful information. Particularly the noncontact potentiometers are interesting, because they can be modelled by an ideal potentiometer with a capacitor connected to the middle pin, right? I'd never route audio through a potentiometer in my own designs (I'd use a circuit fragment similar to the one that you describe), but I'm asking about the individual component rather than how you'd implement a volume control knob or such. – James Haigh Mar 09 '15 at 17:58
  • Terrific answer, though I'm not sure rolling contacts are as bad an idea as you say. A mechanical mouse's rollers have to work under _very_ adverse conditions, namely, they're directly fed the grime from the surface you're moving the mouse on. In a potentiometer, this sticky kind of dirt shouldn't ever enter in the first place, only much lighter dust. And a rolling design might be easier to make dust-proof than a sliding one: if the rolling action is actually just a couple of degrees of large-radius circle segments, you could vacuum-seal this and actuate it with a screw moving only a few mm. – leftaroundabout Mar 09 '15 at 19:34
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No, they don't exist. Simply because they can't.

A potentiometer consists of a carbon track with a wiper moving up and down it. You can't have that wiper moving over the carbon track without friction. Yes, you could reduce the friction with bearings and such, but there will always be that friction.

So people use a rotary encoder instead - most often an optical one if you want low friction - a disc with slots in it that breaks a number of infra-red beams.

Majenko
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    Typical potentiometer has sliding friction. If I understand the the O.P correctly, he is proposing to use different mechanics that would replace sliding friction with rolling friction. This may or may not be feasible or economical. But the idea seems neat, at least from a blue-sky standpoint. – Nick Alexeev Mar 08 '15 at 20:15
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    “You can't have that wiper moving over the carbon track without friction.” – Sure, a car's tires have a tiny bit of rolling resistance, due to deformation of the rubber, cornering, imperfect shape and angle of the wheel and road, minor misalignment with the other wheels, etc., but this is somewhat different to an exhaust pipe scraping behind on the road. :-] – James Haigh Mar 08 '15 at 20:32
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    Roll a ball bearing over a piece of carbon for a while. Watch the groove appear. Then imagine how you are going to prevent that ball bearing ending up loose and breaking contact. Pressure. Increased pressure. That equals increased friction, and increased groove depth. – Majenko Mar 08 '15 at 20:36
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    Only the cheapest of potentiometers use a carbon track. There are conductive polymer designs, as well as cermet ones. All are way too fragile to roll anything on them. Sliding contact actually *is* the most gentle way to interface with the resistive track. Most potentiometers, improperly applied, don't wear out but merely get crackly. That's due to dust trapped on the track, between the slider and the track. It doesn't indicate wear, but merely the reality of how hard it is to keep the dust out - and is **is hard**. The OP somewhat chases a solution without defining the problem. – Kuba hasn't forgotten Monica Mar 08 '15 at 20:54
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    If you use a rolling cylinder instead of a ball bearing, then you would have a lot more surface area spreading out the applied force and reducing friction. To see what the professionals do to make a precision decade resistance box, you'll see that they don't avoid friction: EEVblog #461 - Genrad Decade Resistance Box Teardown - at https://www.youtube.com/watch?v=fKrvtYS_6fI&t=10m18s – MicroservicesOnDDD Feb 25 '19 at 00:10
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It is very difficult to avoid having wiper resistance vary arbitrarily with wiper position. In a good design, however, wiper resistance will have minimal effect on circuit behavior. Each tenfold reduction in the amount of current carried by the wiper will cause a tenfold reduction in the amount of voltage superimposed by its resistance. Likewise each tenfold increase in the voltage carried by the pot will cause a tenfold reduction in the significance of any voltage superimposed by the resistance.

If a device tries to drive a 1/8 watt 8-ohm speaker (1VRMS) using a 10-ohm pot as a volume control, a one-ohm variation in wiper resistance will manifest itself as a 1/8-volt variation in the signal. Nasty. If one were to use a 50:1 step-up transformer to scale the voltage from 1V 1/8A to 50V 1/400A before passing it through a 500-ohm pot, then a one-ohm variation in wiper resistance would manifest itself as a 1/400-volt variation in the signal at the pot; passing it through a 1:50 step-down transformer to drive a speaker would make it appear there as a 1/20,000-volt signal (a 2,500-fold reduction versus controlling the speaker directly). A major improvement.

supercat
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On an more engineering aspect, to achieve the effect of a "friction-less pot", you can control a digital pot (or something similar) with a contactless measurement tool.

For example, you can get one of those sonar modules and control a d-pot by translating the distance between the sensor and the moving target as measured contactlessly using the sonar into a resistance (or wiper position) on the d-pot.

Maxthon Chan
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