59

If you do this, then it will waste a lot of power: simulate this circuit – Schematic created using CircuitLab You'll also need a very large and heavy potentiometer because it will have to handle all the power that the speaker can handle. If you do this, then almost no power will be wasted: simulate this circuit The signal is very low power, and you ...


28

If you connect to the top and wiper, you can use the potentiometer as a simple variable resistor. If you connect to the top and wiper, AND short wiper to bottom, you are still using it as a variable resistor, but with one important difference : if the wiper goes open circuit, the resistance value is 10K rather than infinity. That can: reduce crackling ...


23

The "A" part stands for audio taper. "B" would be linear taper and "C" is a reverse audio taper: - Image from this website What taper means in this context: - Audio Taper Potentiometers are log taper potentiometers that are specifically designed for use in audio applications. Audio taper potentiometers are the potentiometers ...


8

If it's a "B" type, it should be linear. From your plots, it's not very linear, but it's definitely not logarithmic either. So it's a 5K linear potentiometer. Given the tolerances, a 4.7K one would probably also work if you can't find 5K. As per the comment by Hearth, if it's got 6 pins, it will be a dual-gang 5K pot.


8

"20K" is the resistance - 20 kilo-ohms. "A" means logarithmic taper. Probably.


7

That's a multi-section variable and semi-fixed (trimmer) capacitor out of a radio. Here is pretty good photo of a similar vintage part (from here): Typically one variable section would be used for the AM band and one section for the FM band. You can't practically use it directly to control an LED or speaker volume, but you might be able to use it as an ...


7

That would appear to be condensed specification of a 20 kilo-ohm potentiometer with a non-linear "audio taper". Because hearing is roughly logarithmic, a useful volume control must be as well.


7

\$\color{red}{\boxed{\text{Will it waste electrical energy?}}}\$ Potentiometers are variable resistors and resistors waste power: - $$P_{dissipation} = \dfrac{V_{applied}^2}{R}$$ Or $$P_{dissipation} = I_{applied}^2\cdot R$$ Power wasted is also energy wasted if you regard heat as a waste product. If we use potentiometers as volume controls, don't they ...


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 ...


5

The formulas do not match the picture. In the picture, the resistance between ground and 5V can be set between 0R where the wiper bypasses the resistance and 10k where the wiper selects full resistance. To whatever position you set the wiper, it will short out the unused portion of the resistance when it is connected like in the picture, so it will have no ...


5

If the wiper is at the bottom of the resistance element in your schematic, the whole 10K resistance element is in circuit. At any point in the wiper's travel, you will only have one part of the resistance element in the circuit. There is no way to get two parts of the resistance element in parallel, as your calculation would require.


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

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.


4

That six pin pot is really just two normal three pin pots controlled by the same shaft. Each PC board is one pot, so you just wire to one board as you would for a normal single pot - there should be no internal connections between the two pots.


4

Here's how, if I have understood it right.


4

It's not a potentiometer but a ganged 4 section variable capacitor (generally referred to as a 'Polyvaricon' or 'PVC Gang Condenser / Capacitor'), with each capacitor having a trimmer capacitor in parallel. Its intended for use in AM / FM superheterodyne receivers for tuning the antenna and oscillator circuits. The AM sections would be 365 pF each and the ...


4

It is literally the metal shield surrounding the case. While not electrically connected to any of the internals, it protects them from electromagnetic interference. In most cases you should connect it to ground.


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

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

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.)


3

The potentiometer is a resistor component. When you try to create a voltage reduction through a resistor the voltage drop is 100% dependent upon the amount of current flow through the resistor. A load like an Arduino is not a fixed current load so it is not possible to use a single resistor value to reduce the 9V to 5V suitable for proper operation of an ...


3

It's a dual-potentiometer (e.g. volume control for a stereo amplifier). The volume control wiring is shown for one channel and is to be duplicated for the other.


3

Format your code with Ctrl+T. it is than easier to read Write in after "fade out" analogWrite(TR, 0); The fadeout value has in 4 of 5 cases still a value in it. HTH


3

How much room do you have on the PCB? You could make a meander out of thin (1/4oz?) narrow copper traces, and have heavy pads for the slider/brush to contact. The narrow lines are covered in soldermask to protect them (not shown). Then you do a 4-wire measurement of the resistance of the meander. You drive current through the meander at the "I+", &...


3

The switch and the potentiometer are completely independant. the 2-pin connector is for the switch and the 3-pin for the potentiometer. VCC and ground are connected to the lateral pins of the pentiometer the middle one is the output variable voltage.


3

simulate this circuit – Schematic created using CircuitLab Figure 1. This is what you've got. It has 5 pins instead of 3 as a normal potentiometer: Look into it. It should be very clear that there are two independent layers or wafers. ... two of them are connected to VCC and GND, Make sure you don't connect the switch wires to VCC and GND. If you ...


3

To the right of the current knob is a knob to select voltage. It is set to 6V in your picture. Rotate that knob so that it points to "3-15" or whatever the right most marking is. That is the variable voltage setting. With the selector knob set to the variable range, the coarse and fine voltage knobs will set the output voltage.


3

As I suspected, your question is a XY problem. You don't need 100 k potmeter because there are other ways to get the same result using a 10 k potmeter. You think that you need a 100 k potmeter for a 555 timer circuit. The timing of a 555 circuit is determined by a resistor (100k ohm) and a capacitor (of for example 100 nF). That 100 kohm + 100 nF will give ...


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