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I'm starting a circuit-bending project using the rhythm PCB from an old Viscount electronic organ. The idea is to make a portable rhythm box by adding some extra controls and processing. What this question is about is my plan to add a half-tempo switch, in addition to the existing exponential fader that controls tempo.

My initial thought is to add a switch that will double the resistance when activated.

I assume this isn't possible except by placing another variable resistor on a circuit path, as so:

schematic

simulate this circuit – Schematic created using CircuitLab

This would require a design that forces keeping the two potentiometers physically aligned, though. If there's some trick to do this where one potentiometer will automatically 'track' the other one, I'd like to use that.

Edit: Per questions, here's a photo of the PCB: PCB showing connections

And relevant part of the circuit diagram: Photo of the circuit diagram

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    \$\begingroup\$ There exist such things as potentiometers that share the same axis. Usually these are used for stereo control (where they have the same resistance) or in systems where one might want multiple controls on a same axis (such as the fine/course adjustments) on some powersupplies. Not an answer, but it might give you a place to start looking \$\endgroup\$ – Joren Vaes Aug 21 '17 at 9:46
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    \$\begingroup\$ Please explain why you might want to double the resistance because, at the moment this question smells of an XY problem. \$\endgroup\$ – Andy aka Aug 21 '17 at 9:49
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    \$\begingroup\$ @Andyaka I am controlling a circuit that outputs audio beats, and I want to be able to half whatever tempo is selected by the "tempo slider" i.e. existing linear potentiometer. (As suggested in the answers, I might have to replace this with a stereo rotary encoder to get my desired behavior.) \$\endgroup\$ – buildsucceeded Aug 21 '17 at 9:56
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    \$\begingroup\$ Whatever circuit produces the timing pulses for the beats, use a divide by 2 logic circuit on the output to halve the BPM. \$\endgroup\$ – Andy aka Aug 21 '17 at 10:03
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    \$\begingroup\$ It's usually best to first describe the problem then your thoughts on how to solve it with this type of question. Some times the best solution is a different approach. I'd be with @Andyaka here and generate the higher rate pulses first then if required divide this by two digitally. Less mechanical parts to go wrong and only one set of contacts to keep clean. \$\endgroup\$ – Warren Hill Aug 21 '17 at 10:12
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With the newly added original schematic, we can see that this is a RC oscillator circuit. To achieve your real goal of adding a half tempo switch, we can divide the tempo in half by adding a second capacitor with the same value as the original one. With the switch the resulting circuit segment would look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

When the switch is closed the capacitance is doubled, which means the frequency is halved. That sounds by far like the easiest option here - much easier than replacing the potentiometer with anything.

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  • \$\begingroup\$ And much more elegant too! \$\endgroup\$ – leftaroundabout Aug 21 '17 at 19:30
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    \$\begingroup\$ This is the solution which is straight to the point and should be bubbled up in favor of the other ones. \$\endgroup\$ – Janka Aug 21 '17 at 19:47
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    \$\begingroup\$ This is the correct solution. Mine only shows why! \$\endgroup\$ – Transistor Aug 21 '17 at 19:58
  • \$\begingroup\$ Very elegant solution. +1. \$\endgroup\$ – rayryeng Aug 21 '17 at 20:26
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    \$\begingroup\$ @buildsucceeded: No problem. A word of caution: Many capacitors are notorious for varying significantly from their nominal value, due to having high tolerance bands. If having the switch be truly half tempo is important (as opposed to being within say 20-30%) you will need want the capacitors to match. You could measure the existing one, and then measure a bunch of replacements of the same value to find one that matches. Or you could find two that match of the same nominal value as the existing one, and replace the existing one. That may work better, but is probably not strictly necessary. \$\endgroup\$ – Kevin Cathcart Aug 22 '17 at 14:45
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It is very unlikely that your system is configured as shown in your schematic. With the leftmost active pot wiper fully left you short out the voltage reference.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) A more likely arrangement in your rhythm box. (b) Dividing the output by two. (c) Dividing the input by two.

  • (a) shows the likely arrangement. Note that the load on Vref remains constant no matter where the wiper is set. (This assumes that the load on OUT is fairly high resistance so that it doesn't load Vref significantly.)
  • (b) divides the output voltage by two. Since your downstream circuit will have some loading effect you may find that this does not quite give you half the frequency. R3 and 4 need to be about ten times the pot resistance value to avoid loading it too much or the voltage will droop.
  • (c) is very simple. Use the potentiometer resistance track as half of a potential divider. This should be the simplest to implement too. It may just require cutting the track to the "top" of the pot and adding the series resistor and single-pole, single-throw switch.

Your modified schematic still shows series connected variable resistors. You have forgotten to connect the wipers to one end of the pots. At the moment they do nothing as they are open circuit.

schematic

simulate this circuit

Figure 2. Stereo potentiometer arrangements. (a) What the OP intends. (b) This circuit will not work in the OP's application.

  • Figure 2a will only work if the original potentiometer is wired not as a potentiometer but as a variable resistor or "rheostat". This may be the case but is unlikely. Note that you can simplify this further by simply shorting out one of the pots using an SPST switch rather than an SPDT type.
  • Figure 2b will not work as the result would be a square-law. Fully clockwise the output would be 100%. At mid-position the output would be 50% x 50% = 25%.

To resolve your problem you need to reverse engineer the circuit to understand how the original potentiometer is being used.


Your latest edit shows you had the schematic all along. Oh, man!

schematic

simulate this circuit

Figure 3. The resistor solution based on the original design.

We can see from this that it is a heap of trouble. @KevinCathcart's solution is the way to go.

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  • \$\begingroup\$ Thanks for your help! I've updated the schematic again, please let me know if I'm still missing a trick. \$\endgroup\$ – buildsucceeded Aug 21 '17 at 13:37
  • \$\begingroup\$ Is the wiper of the original pot actually connected to one end of the pot as you have shown? You should be able to trace this on the PCB or measure the resistance from the wiper to each pot end with wiper set at 50%. If my hunch is right you will find one end of the pot permanently connected to the ground plane. You can confirm this by multimeter as well. \$\endgroup\$ – Transistor Aug 21 '17 at 13:40
  • \$\begingroup\$ Audio potentiometers tend to be intentionally nonlinear, for a reason, on top of that... \$\endgroup\$ – rackandboneman Aug 21 '17 at 19:22
  • \$\begingroup\$ @Transistor thought I'd lost the 'real' schematic— spent ages digging it out today! :) \$\endgroup\$ – buildsucceeded Aug 21 '17 at 21:38
  • \$\begingroup\$ Can you expand on why the solution you've posted is 'a heap of trouble'? I have no idea and it would be helpful for me to get a bit of advanced insight into this. \$\endgroup\$ – buildsucceeded Aug 21 '17 at 21:39
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If you need two potentiometers to be aligned, you should get a dual pot where two pots are mounted on the same rotary handle. No schematic will keep them aligned if you can turn them independently. Depending on what your actual schematic is, you may be able to divide the voltage V1 by two using a switch, and connect a single pot to it. This may have the same effect as doubling the pot resistance.

BTW, your schematic as it is presented is quite dangerous, since turning the pots to the leftmost position will result in a short circuit.

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  • \$\begingroup\$ Amended the schematic— I'm new to this! :) \$\endgroup\$ – buildsucceeded Aug 21 '17 at 12:30
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IF you are using potentiometers as variable resistors (not as real three-terminal voltage dividers as your schematics show), you can always use a potentiometer that is double the value, and make your switch control a fixed parallel resistor of the same value.

PS, not connecting the third terminal at all when using a potentiometer as a variable resistor is usually ill advised: potentiometers tend to develop intermittent contacts, especially when operated. An unconnected third terminal will make the arrangement look like a straight open circuit during such problems, whereas connecting that end to the wiper will make it look like the maximum resistance of the variable resistor - which usually will create less crass noises and keep the impedance in the circuit defined.

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You can use dual potentiometer (usually used in stereo audio applications) which consists of two potentiometers mounted on the same axis.

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  • \$\begingroup\$ How? If the original circuit is a variable resistance only then he could. If it's a potentiometer it is more complex. Add a schematic of how you would approach it. (There's a button on the editor toolbar.) \$\endgroup\$ – Transistor Aug 21 '17 at 11:40
  • \$\begingroup\$ OP's question already has the circuit shown (also, judging by the comments on the question, he probably needs entirely different solution). \$\endgroup\$ – Algimantas Aug 21 '17 at 15:10
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    \$\begingroup\$ Yes, it does now. His original one was bonkers. In any case he doesn't seem to know much about electronics and I reckon it's unlikely to be a series variable resistor setup as he indicates. \$\endgroup\$ – Transistor Aug 21 '17 at 15:14
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You want to use a stereo potentiometer. This is basically 2 potentiometers with just one shaft, so they always have the same value. To use it you also need a 2 position switch:

schematic

simulate this circuit – Schematic created using CircuitLab

The switch would either connect stereo pot 1 to the output directly, or to stereo pot 2.

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  • \$\begingroup\$ You have shown resistors instead of potentiometers. Given the OP's schemetic this isn't going to help much. There's a built-in schematic editor on the toolbar. It's much easier to use than Pbrush (or whatever you used) and will create an editable schematic. \$\endgroup\$ – Transistor Aug 21 '17 at 12:33

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