Abstract
If you want to get your friend's attention at a crowded sporting event with lots of people cheering, you need to shout. If you're trying to do the same thing in a quiet library, a whisper works. The detection limit for each of our senses depends on the amount of "background" stimulation that is already present. This project uses an LED control circuit to investigate detection of changes in light levels.Objective
The goal of this project is to compare linear vs. logarithmic potentiometers for controlling signals that people sense (e.g., brightness of a light, or loudness of a sound).
Introduction
Our five senses (sight, hearing, touch, taste, and smell) give us information about the environment around us. Our ability to detect small changes in these sensations (the "just noticeable difference"), is proportional to the original intensity of the sensation. For example, if you are in a very quiet room, you can hear a whisper. A second person whispering could also be heard: the added sound would be significant in relation to the existing sound level. On the other hand, if you're at a basketball game with a lot of people cheering, you would not be able to hear someone whispering two rows down, because now the added sound is insignificant in relation to the existing sound level. In other words, as sounds get louder, there needs to be a bigger change in intensity in order to detect it.
The same holds true for other sensations. On a dark night, with your eyes adapted to low light levels, you can see a single candle a mile away. In broad daylight, the candle has to be much closer before you can distinguish it, because the existing light level in daylight is so much greater.
The mathematical description of this phenomenon is called Stevens' Power Law (Staff, 2004). In words, Steven's Power Law says that the perceived sensation, R (loudness of a sound, brightness of a light), is an exponential function of the actual level of the stimulus, S, (measured sound level or brightness). In equation form, it looks like this:
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S0 is the threshold, or lowest detectable level, of the sensation, and k is a proportionality constant. The value of the exponent, α, depends on the particular sensation. Taking the logarithm of both sides of the equation, we get:
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So on a double-log plot, the perceived response, R, to a stimulus, S, will be a straight line, with slope equal to α (the exponent in the original equation) and y-intercept equal to log k. In this project, you will use an LED with a simple control circuit to investigate how changes in light intensity are sensed. Before getting started, you'll need some background information on how LEDs work.
An LED (light-emitting diode) is a special kind of diode that produces light (see Figure 1).
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| Figure 1. A red LED (top). The longer lead is the anode (+) and the shorter lead is the cathode (−). In the schematic symbol for an LED (bottom), the anode is on the left and the cathode is on the right (Hewes, 2006). |
As the current through the LED increases, the brightness also increases. Typically, the recommended current for an LED is 20 mA or less. Above this value, the lifetime of the LED will be decreased significantly. Far above this value, the LED will fail catastrophically, like a flashbulb.
To keep the LED current at a reasonable level, LEDs are typically connected in series with a current-limiting resistor, as shown in Figure 2.
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| Figure 2. Schematic diagram of an LED in series with a 1kΩ resistor (Hewes, 2006). |
The voltage drop across an LED is about 2 V (except for blue or white LEDs, where the voltage drop is about 4 V). In the circuit in Figure 2, the voltage drop across the resistor will be 9 − 2 = 7 V. Using Ohm's law, the current, I, through the resistor will be V/R = 7 V/1kΩ = 7 mA.
Figure 3 (below) shows you how to use Ohm's Law to calculate what size resistor you need to limit the current through the LED to the desired value. The voltage drop across the resistor will equal the supply voltage minus the voltage drop across the LED (or, VS − VL). You can then use Ohm's Law to calculate the resistance, R, needed to produce a desired current, I:
So, if the supply voltage is +7 V, what resistor would you need for a 15 mA current? R = (7 − 2)/0.015 A = 333Ω. The closest standard resistor value would be 330 Ω. For more details, and a set of online calculators, see the LED references in the Bibliography section (Hewes, 2006; Ngineering, 2003).
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| Figure 3. Schematic diagram showing how to use Ohm's Law to calculate the correct value for the current-limiting resistor (Hewes, 2006). |
For this project, we'd like to vary the output of an LED over a continuous range. Obviously, it would be a major inconvenience if we had to keep changing the current-limiting resistor for each LED! The circuit below (Figure 4) is a different approach to solving our problem.
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| Figure 4. This circuit uses an operational amplifier (op-amp) to control the current through an LED (Calvert, 2002). |
The large triangle marked "LF411" in the center of the schematic is an operational amplifier, or op-amp, for short. An op-amp has two inputs and one output. (It also has positive and negative supply inputs, typically +/− 15 V, which are usually not shown in schematics.) In the schematic, the first input (marked +, and called the non-inverting input) is connected to a potentiometer. The second input (marked −, and called the inverting input) is connected to two circuit elements. It connects to the cathode of an LED, and to a 330 Ω resistor that connects to ground. The output of the op-amp is connected to the anode of that same LED. So the output of the op-amp is fed back to the inverting input. Op-amps are almost always used with some type of feedback.
What does the op-amp do? For a good basic understanding of how op-amps work in circuits, you just need to understand two simple rules (Horowitz and Hill, 1989, 177):
These rules are simplifications—for example, the inputs do draw some current. For the LF411 op-amp used in this project, 0.2 nA. You won't be able to measure that with your DMM! So these op-amp rules, as the authors state, are "good enough for almost everything you'll ever do," (Horowitz and Hill, 1989, 177).
- "The output attempts to do whatever is necessary to make the voltage difference between the two inputs zero.
- "The inputs draw no current."
What do the op-amp rules mean for our LED control circuit? Let's examine the inputs to the op-amp to figure it out. The potentiometer connected to the non-inverting input is a variable resistor. When wired as shown in the schematic (Figure 4), the output from the center lug of the potentiometer provides a voltage, ranging between 0 and +5 V, depending on how far the knob is turned. Simple enough. OK, what about the non-inverting input? This is the feedback connection from the output. So, by rule 1, the op-amp output should do whatever is necessary to make the voltage at the inverting input match the voltage we set with the potentiometer.
Let's consider three examples:
Example 1 is the easiest. With 0 V at the non-inverting input, there should be 0 V at the inverting input (rule 1), so the op-amp does nothing. No current flows through the LED.
With the potentiometer half-way up, the non-inverting input of the op-amp sees +2.5 V. How does the op-amp match that at the inverting input? It does it by passing current through the LED. How much current? Enough current so that the voltage at the inverting input equals +2.5 V. Where does the current go? The current flows to ground through the 330 Ω resistor (by rule 2, we know that the inputs draw no current, so the resistor is the only path to ground). By Ohm's Law, we have I = V/R = 2.5 V/330 Ω = 7.6 mA.
With the potentiometer all the way up, there is now +5 V at the non-inverting input. By the same analysis, we conclude that the current through the LED is now 15.2 mA. Perfect! The circuit has just the right range for a typical LED. The LF411 can supply 15 mA without a problem, so the behavior we expect from our op-amp rules is exactly the behavior that we get.
You can see from Ohm's Law that the LED current increases in direct proportion to the voltage from the potentiometer. Do some more calculations for more potentiometer settings to convince yourself that this is true. Then build the circuit and see how it actually performs.
The potentiometer in the op-amp circuit gives you a way to control the current, and thus the brightness, of the LED. It turns out that there are two types of potentiometers that you could use, linear or logarithmic (also called "linear taper" and "audio taper"). A linear potentiometer changes resistance in equal steps all around the dial. A logarithmic potentiometer changes resistance in successively larger steps as you advance the dial. Which one is a better match for controls related to human senses, like brightness or volume?
Terms, Concepts and Questions to Start Background Research
To do this project, you should do research that enables you to understand the following terms and concepts:
Questions
Bibliography
The following references are recommended for more advanced students:
Materials and Equipment
To do this experiment you will need the following materials and equipment (we've supplied part numbers from Jameco Electronics; similar parts are available from other suppliers):
| description | part number | quantity | total price (US $, March, 2007) |
| solderless breadboard with regulated power supply (+/-5 V, +/-15 V) | 1537264 | 1 | 84.00 |
| wire jumper kit for solderless breadboard | 19290 | 1 | 16.00 |
| basic digital multimeter (DMM) | 318781 | 2 | 20.00 |
| LF411CN integrated circuit (IC) | 23018 | 2 | 2.70 |
| high-output LED, red 642 nm, 3.5 cd | 334481 | 2 | 0.60 |
| potentiometer, 10 kohm, 1/2-watt (1/4-watt would be fine instead), linear taper | 29081 | 1 | 1.00 |
| potentiometer, 10 kohm, 1/4-watt, audio taper | 255426 | 1 | 1.20 |
| resistors: 330 ohm, 1/4-watt (1 needed) | 690742 | 100 (min. order) | 1.00 |
| resistors: 10 kohm, 1/4-watt (1 needed) | 691104 | 100 (min. order) | 1.00 |
| red knob | 265050 | 1 | 0.65 |
| green knob | 265084 | 1 | 0.79 |
| blue knob | 265077 | 1 | 0.92 |
| solid hook-up wire, #22 AWG, red | 36856 | 1 | 5.50 |
| solid hook-up wire, #22 AWG, black | 36792 | 1 | 5.50 |
| alligator-clip test leads | 10444 | 1 pkg. (10 leads) | 5.00 |
| heat-shrink tubing | 71871 | 1 | 0.55 |
| TOTAL (to nearest $5, does not include tax & shipping) | 150.00 |
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Experimental Procedure
| Note Before Beginning: This science fair project requires you to hook up one or more devices in an electrical circuit. Basic help can be found in the Electronics Primer. However, if you don't have experience in putting together electrical circuits you may find it helpful to have someone who can answer questions and help you troubleshoot if your project isn't working. A science teacher or parent may be a good resource. If you need to find another mentor, try asking a local electrician, electrical engineer, or person whose hobbies involve building things like model airplanes, trains, or cars. You may also need to work your way up to this project by starting with an electronics project that has a lower level of difficulty. |
Wiring the Potentiometers
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| Figure 5. Wiring the potentiometers. You can daisy-chain the +5 V wire (red) between the pots (right hand lug, when viewed from beneath, as in the photo). You can also daisy-chain the ground wire (black; left-hand lug, when viewed from beneath, as in the photo). The center lug is what goes to the non-inverting input of the op-amp (pin 3). To keep your connections straight, use a wire color that resembles the color of the knob (or label the wires). |
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| Figure 6. Completed potentiometer controls mounted in a tunafish can. |
Soldering Leads to the LED and Light-to-Voltage Converter
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| Figure 7. Completed LED with wires soldered to leads. The yellow bar marks where the needle-nose pliers should be clamped to protect the LED during soldering. Use heat shrink tubing after soldering is completed to fully insulate the component leads. |
Building the LED Control Circuit
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| Figure 8. Pin numbering scheme for the LF411 DIP IC. The small semi-circular depression marks the top end of the chip. Pin 1 is at the top left. Often, the chip will also have a small circular depression next to the pin to mark pin 1. You count pin numbers down the pin 1 side, and then back up the other side. |
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| Figure 9. Schematic of a single LED control circuit (Calvert, 2002). |
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| Figure 10. Detail view of completed circuit on breadboard. Notice how the LF411 ICs span the gap between the holes in the breadboard. Power connections to the chip are made to pin 7 (+15 V) and pin 4 (−15 V). The 330 Ω resistor is connected between pin 2 and the ground bus. The anode (+ lead, red) of the LED connects to pin 6, and the cathode (− lead, black) connects to pin 2. The wire from the center lug of the potentiometer connects to pin 3. Don't forget to connect the +5 V and ground leads of the potentiometer to the corresponding bus lines on the breadboard. |
Building the Light Detection Circuit
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Figure 11. Light-to-voltage converter functional block diagram (TAOS, Inc., 2006). |
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| Figure 12. Drawing of light-to-voltage converter package (TAOS, Inc., 2006). |
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| Figure 13. Light-to-voltage converter circuit schematic (TAOS, Inc., 2006). |
Making Light Measurements
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| Figure 14. One possible setup for making the light measurements. This photo shows both the circuit assembly and the LED and light-to-voltage converter. A better way to do it would be to use the long lead wires so that the circuit and potentiometers are supported separately. That way you won't disturb the alignment of the LED and light detector when you make adjustments to the potentiometers. |
| pot. type:___________________ | ||||||
| pot. setting |
Subjective brightness (compared to previous setting) |
Vr (V) |
Vp (V) |
ILED (mA) |
Vp − V0 (V) |
ΔV (V) |
| 0 | ||||||
| 1 | ||||||
| 2 | ||||||
| etc. | ||||||
Analyzing Your Results
Variations
Credits
Andrew Olson, Ph.D., Science Buddies
Last edit date: 2009-03-15 14:28:00
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