# Help with understanding potentiometers?

In summary, the problem involves determining the resistance and voltage output of a potentiometer circuit in relation to the angle of wire A relative to wire B. The resistance can be described by the equation R(θ) = 100 kΩ * (θ/120), where θ is the angle and 120 is the maximum angle. To design the circuit, an input voltage of 5V and resistance values that result in a maximum output of either +5V or -5V must be chosen. The circuit should be designed so that the output voltage increases with the angle. A potentiometer can be used to measure resistance and varies with the position of a sliding contact.

## Homework Statement

Consider 2 wires, A and B, tied together at both ends, and call the point they are tied together point X. Wire B does not bend (it stays fixed) and only wire A bends. You need to measure the angle of wire A relative to wire B. When the wires are fully bent around point X, and are side by side, the angle is 0. When wire A is extended outward in the same plane as the top of wire B, wire A would be at 90 degrees. Assume that you have a large single–turn potentiometer (figure 1, maximum resistance = 100 kilo-ohms) where the wiper can rotate freely from a center on point X . Assume the maximum angle to be 120 degrees.

1) Derive an equation to describe the resistance measured across the output voltage (see figure 1) as a function of the angle.
2) Design an op-amp based circuit to output a voltage corresponding to the angle of wire A relative to wire B. Choose an input voltage and resistance values such that the maximum output of the circuit is either +5V or -5V.
3) Derive an equation that relates the voltage output to the angle of wire A relative to wire B.

## Homework Equations

This is part of my problem...I'm not really sure what the relevant equations are. I know that for a basic potentiometer, it can be thought of as a voltage divider. One end of the resistor where the output is placed is R1 and the other end is R2, so RL = R1 + R2. And then you have a fraction setting x (Im not really sure what this is, but its in my notes..). So R1 = x*RL and R2 = (1-x)*RL. So the voltage divider equation comes out as Vo = Vi(R2/RL) = (1-x)*Vi. This is all I know.

## The Attempt at a Solution

Firstly, I can draw and attach a picture of the setup of the two wires if anyone feels I need to clarify it. My professor said I am thinking of it correctly.

For 1, I have a feeling that if I can use the above I can relate R and theta. The problem is that I'm not fully understanding how the potentiometer circuit works, or what x represents and if x is analogous to theta. Surprisingly, I'm having a hard time finding some resources online which further explain the concepts I need.

2 is the scary one. I don't even know where to begin with this one. I am not sure what concept my professor is looking for here but I have no idea how I can design such a circuit. The thing is, I know that 3 shouldn't be bad with 2, I think I OK with deriving circuit equations.

I would really appreciate any help or hints or even resources someone could give me for this problem. Its very weird for me to so completely lost on a problem.

Edit: I am not sure if this should go in engineering or advanced physics. Could a mod help? Sorry I am still new to this.

#### Attachments

• Figure 1.jpg
7.2 KB · Views: 398
The idea of a potentiometer is like this - a length L of (Ohmic) resistor material is laid along the x-axis from x=0 to x=L. Electrical contacts are placed at the origin and at x=L and a variable contact can be moved between these limits.

If the resistance is measured between the origin and point x: 0<x<L then R(x)=Rx/L
If L=1 unit, then x will have a fractional value.

In your case, the "x axis" is laid out on the circumference of a circle.

Have you never seen a potentiometer before - never touched one?

Do this: get a ruler and fix a length of nichrome wire to it.
Put an alligator clip at each end, and another that can be slid clipped to the middle of the wire.
That is a potentiometer. Check that the resistance between either end and the slider varies with the position of the slider.

1 person
Simon Bridge said:
The idea of a potentiometer is like this - a length L of (Ohmic) resistor material is laid along the x-axis from x=0 to x=L. Electrical contacts are placed at the origin and at x=L and a variable contact can be moved between these limits.

If the resistance is measured between the origin and point x: 0<x<L then R(x)=Rx/L
If L=1 unit, then x will have a fractional value.

In your case, the "x axis" is laid out on the circumference of a circle.

Have you never seen a potentiometer before - never touched one?

Do this: get a ruler and fix a length of nichrome wire to it.
Put an alligator clip at each end, and another that can be slid clipped to the middle of the wire.
That is a potentiometer. Check that the resistance between either end and the slider varies with the position of the slider.

Hey Simon,

OK, I think I'm starting to understand how they work a little more. And you're correct, I've never seen a potentiometer before, much less use one. I'll absolutely do what you suggested once I have access to some wires and a multimeter (I'm snowed in at the moment).

So regarding what you first said, for #1, in this case since x has a fractional value then x = theta/120 (since max angle is 120). And since the max resistance is 100 k-ohms, I am thinking the equation should be:

R(θ) = 100 kΩ * (θ/120).

For the next part, how do you suggested I go about making the circuit? I know that the input voltage needs to be 5V, and that one of the resistors (if there needs to be more than one) has to be R(θ). But should it be that output voltage increases with the angle or the other way around. And I how would I know where to the resistor(s) and the positive and negative supply for the op amp?

Sorry for the further questions, but I already really appreciate the help you've given me. Definitely feel a little bit better about the question right now.

1. that equation would be correct - you seem to have the right idea.

I'd write: ##r(\theta) = R\theta /\Theta : R=100\text{k}\Ohms, \Theta=120^\circ## ... but it's the same difference - I just don't like to put the numbers in until I absolutely have to.

2. The circuit:
Shorthand for "potentiometer" is "pot".

A pot has three terminals - one at each end of the resister, and another that can slide on the resistor.
The symbol reflects this.

Basically pick one end to measure resistance from - the resistance between that end and the "arrow" is ##r(\theta)##, the resistance between the arrow and the other end is ##R-r(\theta)##.

You are going to be rigging a voltage divider circuit where those are the two resistances.

3. physical pot
You should also find or buy commercial pots - get an old radio or TV perhaps.
The knobs on an old car stereo tape-deck are pots. Take one apart.
Surplus electronics stores are usually lousey with old salvaged pots.

I would suggest breaking down the problem into smaller parts and understanding the underlying principles involved. Here are some suggestions:

1) To derive an equation for the resistance measured across the output voltage, you can use the concept of voltage division. The resistance of the potentiometer can be thought of as two resistors in series, with the wiper acting as a variable resistor. Use the equations for voltage division to relate the voltage across the output to the total resistance and the resistance of the wiper at a given angle.

2) For designing the op-amp based circuit, you can use the concept of a differential amplifier. This circuit amplifies the difference between two input voltages, which in this case would be the voltage across the output of the potentiometer and a reference voltage. By choosing appropriate resistors, you can set the maximum output voltage to +5V or -5V. You can also use a voltage divider circuit to set the reference voltage.

3) To relate the voltage output to the angle of wire A relative to wire B, you can use the equation derived in part 1 and the output voltage of the op-amp circuit. This will give you an equation that relates the angle to the output voltage.

Some resources that might be helpful in understanding these concepts are basic circuit analysis textbooks or online resources such as Khan Academy. It might also be helpful to discuss the problem with your professor or a classmate to get a different perspective. Good luck!

## 1. What is a potentiometer and how does it work?

A potentiometer, also known as a pot, is an electronic component that functions as a variable resistor. It has three terminals, with the first and third being connected to a fixed resistance and the second being a wiper that can slide along the resistance. This allows the potentiometer to vary the amount of resistance in a circuit, which in turn can control the voltage or current flowing through it.

## 2. How do I choose the right potentiometer for my project?

The key factors to consider when choosing a potentiometer are its resistance range, power rating, and physical size. You should also consider the type of potentiometer, such as linear or logarithmic, and whether it has a single or multiple turn design. It's important to match the potentiometer's specifications to the requirements of your circuit or device.

## 3. Can I use a potentiometer as a voltage divider?

Yes, a potentiometer can be used as a voltage divider by connecting the first and third terminals to the power supply and the second terminal to the load. The wiper can then be adjusted to vary the voltage output. However, keep in mind that potentiometers are not precise and may have a significant tolerance, so they may not be suitable for applications that require high precision.

## 4. How do I wire a potentiometer into a circuit?

To wire a potentiometer into a circuit, you will need to connect the first and third terminals to the power supply or ground, and the second terminal to the load or signal input. It's important to note the orientation of the potentiometer, as some have a clockwise or counterclockwise rotation for increasing resistance. You may also need to use additional resistors to create a voltage divider or to limit the current through the potentiometer.

## 5. Can a potentiometer be used as a sensor?

Yes, a potentiometer can be used as a sensor in certain applications. For example, it can be used as a position sensor by attaching a mechanical linkage to the wiper and measuring the resistance change as the position changes. It can also be used as a force or pressure sensor by incorporating it into a strain gauge circuit. However, it's important to note that potentiometers may not have the same level of accuracy and reliability as dedicated sensors.

• Engineering and Comp Sci Homework Help
Replies
6
Views
1K
• Engineering and Comp Sci Homework Help
Replies
5
Views
860
• Engineering and Comp Sci Homework Help
Replies
18
Views
2K
• Engineering and Comp Sci Homework Help
Replies
34
Views
2K
• Engineering and Comp Sci Homework Help
Replies
2
Views
842
• Engineering and Comp Sci Homework Help
Replies
3
Views
807
• Engineering and Comp Sci Homework Help
Replies
17
Views
3K
• Engineering and Comp Sci Homework Help
Replies
1
Views
2K
• Engineering and Comp Sci Homework Help
Replies
8
Views
1K
• Engineering and Comp Sci Homework Help
Replies
1
Views
1K