28

Best way would be to panel mount it, and then solder on three wires which you bring to the board, either directly solder them or use a 3 pin connector. This pot is not designed to be PCB mounted. Or choose another pot, of course.


14

Mount the pot to the board (using its intended mounting nut) and wire it from the reverse side. Edit: I suggest mounting it to the board because it is a "customer requirement" that this board and pot be used. Of course board-mounting a panel-mount pot is not ideal, but I have seen a few instances of this. Usually these are "trimpots" with ...


12

A potentiometer gives output (resistance) in response to a physical input (position). So that puts it pretty squarely in the ‘sensor’ category. And they all go to 11... At the risk of dissecting the live frog, from here: https://en.m.wikipedia.org/wiki/Up_to_eleven


6

It's not normally that much of a consideration. Many pots are used as potentiometers (as opposed to rheostats) with negligible wiper current and/or used way, way below their rated dissipation as trimmers. The pot dissipation rating is for the entire element, and any situation that exposes part of the element to a large proportion of the dissipation the ...


6

Justme gave the answer, you have to clock it. continuous conversion without having to initialise and command the conversion? That would mean the ADC would change the level of its output pins every time it does a new conversion. Counting uneven propagation delays, skew, etc, if the device that reads these pins reads them at the wrong time, it will get some ...


6

You cannot put a 22uF capacitor directly on the output of the op-amp, it will oscillate. Very low capacitances (like up to ~50 pF) are okay in your (worst-case) application as a voltage-follower. As it says in the datasheet: Capacitive loads which are applied directly to the output of the amplifier reduce the loop stability margin. Values of 50 pF can be ...


6

An electrical sensor is a device that converts a specific physical parameter to an electrical signal. A potentiometer as a level sensor, for example, would fit into this category. An electrical actuator is an electromechanical device that converts an electrical signal to a mechanical movement. A solenoid actuator would be one such example.


6

I would place 3 individual header pins snapped off from from these: https://www.digikey.com/catalog/en/partgroup/break-away-0-1-36-pin-strip-male-header/88887 positioned on your PCB such that the hole in each of the solder tabs on the the pot slides over a pin, and the back of the pot lies flat against the PCB. I'd probably also glue the pot onto the PCB - ...


5

You don't provide the circuit but in general the tolerances on pots are very loose, like 10-20% so 47K is close enough to 50K in most every practical case. Nothing will "fry" as a result.


5

No it won't work like that, page 15 of the datasheet describes how the chip bus works. At least some of the control signals need to have transitions, this chip is intended to be sitting on a memory bus so it looks like RAM or ROM chip.


5

I designed a product some years ago that used a similar potentiometer. I designed a snap-off board that had slots for the terminals. A flat cable connected the sub-board to the main board. The assembly could be tested before the sub-board was snapped off. You can connectorize it. Or just use it the way it is intended, soldered wires to the board (with heat ...


5

There may be "bouncing" if it's very fast. There will be CRV (contact resistance variation) and some voltage noise as well. CRV affects the reading if the wiper output is loaded, and has little effect if the pot is being used as a voltage divider into a high-impedance circuit. You can minimize those effects by using a pot designed for motion such ...


5

The 6-pin pot is dual channel. This means you have two pots in one enclosure. The top level 3-pins are the first pot and the bottom level 3-pins are the second pot. You can choose to replace them with 4 regular 25K resistors. Replacing the 50K variable value with the center value of 25K. This gives you the 'center' value of the original setup.


4

Holding CS and RD low permanently will force the ADC to make one single conversion right after power is applied. That value will be presented to the output bus until you remove power. No further conversions will be performed: -


4

Since it's for a musical MIDI application a smooth response may help you justify an analogue potentiometer. The simplest, which you have already considered, is a 10k rotary pot with belt drive. Typical 1-turn pot track swept angle is 330°. You're looking for 25 cm slide. From \$ s = \pi d \frac {330}{360} \$ we can calculate the required belt pulley ...


4

The picture shows two potentiometers, each of value 100k. The wiper of each potentiometer is connected to a fixed resistor, one of 100k and one of 1M. Each fixed resistor is connected to what looks like a thermistor (the circled things marked 10k.)


4

solder a steel bracket to the PCB and mount the potentiometer in the bracket. you could have slots in the the PCB and solder the wiring tabs in there but that would probably be unreliable.


4

You could do something like this: simulate this circuit – Schematic created using CircuitLab Adjust trimpot R9 to get the output voltage just above 0V with the minimum usable setting on the input R8. The gain is 100K/R2 = 33.3, so the output will be from (say) 200mV to 3.5V for the input from R8 from 2-3V. You can change R2 if you like. If you want to ...


4

A "loudness" control was intended to be "bass boost" circuit - increasing the bass component of the input audio at low volume levels. Typically it was just a switch. Treble and bass were intended to be set for mid-volume (normal) listening levels and loudness set off. Then loudness turned on if listening at low volume levels. The more ...


3

simulate this circuit – Schematic created using CircuitLab Figure 1. Equivalent circuits. R1 allows selection of a bias voltage between 0 and -12 V. R2 limits the current from the bias to the next stage of the circuit. Note that your coarse pot wiper resistor is 100 kΩ and the fine is 1 MΩ. If they both feed into the same circuit the fine pot ...


3

Yes, a bad pot could very well be the issue here. I have a similar unit and I find that I got what I paid for which, unfortunately, is not very much. My voltage display drifts over time so that if I set the output for 3.3V the display will drift up by 2-3 volts over time even though the output itself it pretty stable. I've purchased a higher quality ...


3

In this diagram it seems left end of the red wire (The blue dot) has highest potential and the PD should drop as we move to the right. That would be correct if you took battery negative as your reference or "ground" point. In the illustration the author isn't measuring voltage from a ground point, s/he's just measuring difference between two ...


3

As we don't have a schematic, I'll take a guess. Looks like a simple triac-diac "dimmer" type circuit. In such a case, the potentiometer is connected as a rheostat. In such a case, the center and the left pin (from the front, pins down) of the pot are either connected to each other or the left pin is unconnected. So you can try the 1M pot with 1M 1/...


3

ATTINY13A with only 512 program words is a bit small for arduino and it has only 2 PWM outputs (so you'll be doing the third in software) if you want a challenge it's probably possible to accomplish your task.


3

In case you are working with MeanWell LED drivers (like their HLG/XLG/ELG family) or any other driver that provides a constant current at the DIM input, you just need to select a potentiometer with suitable resistance. Assuming the driver sources a constant current of 100uA (like MeanWell drivers do), 100kOhm are required to produce a dimming voltage of 10V: ...


3

I see you've asked a few related questions about LED drivers, so I'll try and give a big picture answer. You put current through LEDs and photons are generated at some rate, roughly 1 photon per 1.5-3 electrons. To move that current you apply some voltage, with small changes in voltage leading to very large changes in current. Unfortunately, the specific ...


3

Maybe the older pot just has a different electrical angle, nothing to do with wear. Normally there's a region in the mechanical range of motion at either end where the resistance does not change much. The electrical angle, the more-or-less linear portion between those end bits, is less than the mechanical angle, but how much less can vary. For example, this ...


3

The hard way: make an analog path that does the following: normalize each pot for span and offset. Let's assume that's done mechanically, when the sensor is assembled. Add the normalized signals to each other. Their voltage sum should equal the reference voltage. The sum should not vary. use a pair of comparators that trip at -10% and +10% of reference. &...


3

simulate this circuit – Schematic created using CircuitLab If you've connected the potentiometer in line with one terminal of the speaker, then the "ground" would be the other speaker terminal. If it's connected as below, then the signal can never go to zero. simulate this circuit


2

I have not thought I would need to adjust the current while welding. But it sounds like a good idea. Honestly, I am only a beginner, stick welder. Compliment to your schematics drawing skill as a welder. Soon, welders will take over EEs' job. Anyway, this is the electrical symbol for your last try. simulate this circuit – Schematic created using ...


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