App note from Vishay on variable resistors primer. Link here (PDF)
A potentiometer is a mechanically actuated variable resistor with three terminals. Two of the terminals are linked to the ends of the resistive element and the third is connected to a mobile contact moving over the resistive track. The output voltage becomes a function of the position of this contact. Potentiometer is advised to be used as a voltage divider.
We have to admit that when we first saw [Ajoy Raman]’s Instructables post, we figured that he used a universal motor to generate a voltage from the anemometer. But [Ajoy]’s solution to the coaxial shafts problem is far more interesting than that. A discarded universal motor donated its rotor and bearings. The windings were stripped off the assembly leaving nothing but the commutator. 1kΩ SMD resistors were soldered across adjacent commutator sections to form a series resistance of 22kΩ with taps every 1k, allowing 0 to 2.2V to be read to the ADC of a microcontroller depending on the angle of the vane.
As clever as that is, [Ajoy] still had to pull off the coaxial part, which he did by drilling out the old motor shaft from one end to the other using just a drill press. The anemometer shaft passes through the hole in the shaft and turns a small DC motor to sense wind speed.
There might have been other ways to accomplish this, but given the constraints and the low cost of this solution, our hats are off to [Ajoy]. We’re a little concerned with that motor used for the anemometer, though. It could result in drag when used as a generator. Maybe a better solution would be a Hall-effect sensor to count rotations of a hard drive rotor.
If you have a traditional regulated power supply that you want to make adjustable, you’ll have somewhere in the circuit a feedback line driven by a potential divider across the output. That divider will probably incorporate a variable resistor, which you’ll adjust to select your desired voltage.
The problem with using a standard pot to adjust something like a power supply is that a large voltage range is spread across a relatively small angle. The tiniest movement of the shaft results in too large a voltage change for real fine-tuning, so clearly a better means of adjustment is called for. And in many cases that need is satisfied with a ten-turn potentiometer, simply a pot with a 10 to 1 reduction drive built-in.
His 10-turn pot is a fantastic piece of workbench improvisation to deliver the goods from what is at hand. A capstan and shaft from an old radio tuner drives a small rubber belt and pulley from a CD-ROM drive. This in turn has a gear which drives a much larger gear coupled to the shaft of a standard potentiometer. All the drive components are mounted on a frame made from an old aluminium heatsink.
To be strictly accurate it’s a 13-turn pot rather than a 10-turn one, but the principle is the same. And it’s an extremely effective component, especially for the price. We like it!
The Wheatstone bridge is a way of measuring resistance with great accuracy and despite having been invented over 150 years ago, it still finds plenty of use today. Even searching for it on Hackaday brings up its use in a number of hacks. It’s a fundamental experimental device, and you should know about it.
How It Works
Here’s an easy way to understand how the Wheatstone bridge works. In the schematic are two voltage dividers (pairs of resistors in series): R1 with R2 and R3 with R4. If you do the math, you’ll notice that the voltages across R1 and R3 are the same, as are the voltage for R2 and R4. That’s because the ratio of the resistances R2/R1 is the same as the ratio R4/R3. Both form divider networks and both divide the voltage the same way.
But the key is that if we were to connect point A to point C, there would be 0 volts between them, no current would flow. At this point we say that the Wheatstone bridge is balanced.
Now let’s say we don’t know the resistance of R4 (let’s instead call it Rx). And let’s connect A and C through a galvanometer as shown above. If the current is 0 amps then we can find the value of Rx. That’s because, as we said above, when no current is flowing between A and C, the ratio R2/R1 must be equal to the ratio Rx/R3. Rearranging, we get Rx = (R2*R3)/R1, and solving, we get (8*6)/4=12 ohms.
But what if there is current flowing between A and C, or we could ask what if there is a voltage across A and C? In that case we’d change R2 from a fixed value resistor to a potentiometer, as in the second diagram above. Note that points B and D were really electrically the same so we can make them the same point and redraw the schematic in the more common diamond shape for a Wheatstone bridge. When we turn on the circuit, the galvanometer would likely show current between A and C. We’d then adjust the value R2 until the galvanometer shows no current. We could then measure the resistance of R2, getting 8 ohms, and then do the above math again and see that Rx is 12 ohms.
In fact, that’s just what I’ve done as shown in the photos to try it out. Of course there was no doubt that it would work, but I couldn’t resist (get it?) seeing it for myself.
But the question arises, if we can measure R2 then why couldn’t we have just measured Rx instead? Well, Rx could be a strain gauge integrated into a building and so we couldn’t get at it to measure it. We’ll see strain gauges more below. Alternatively, the change in resistance could be so small that our meter couldn’t be relied upon, in which case we’d actually be amplifying and measuring voltage instead, but we’ll talk about that below too.