Using Voltage Dividers to Design a Photo-Sensitive LED Circuit ( 2009 - Doug Oliver & Jackie Kane. May be reproduced for non-profit classroom use.) Purpose: After completing the module students will: 1. Understand the linear nature of carbon resistors and the non-linear nature of LED s, 2. Be able to calculate and graph an I/V plot for the performance of an LED, 3. Understand that a mathematical model of a system is not a precise representation. 4. Understand that an LED may be modeled as on when V LED = V f, and off when V LED < V f,. 5. Be able to read and use log-log graphs, 6. Plot the resistance for a photoresistor as a function of the light intensity, and 7. Design, construct, and troubleshoot a light-switched LED circuit. Some lights are part of a circuit that is activated by a change in the ambient light intensity. For example, many night-lights switch on when they are not exposed to enough ambient light. What follows is an open-ended design exercise with a goal of designing and constructing a circuit where a light is switched on when a sensor is in an area with too little ambient light. Two new components are introduced in the following sections: the light-emitting diode (LED) and the photoresistor. 1. LED s Light-Emitting Diodes (LED s) have been used extensively recently. The object of this design problem is to build on your knowledge of voltage dividers to design a photosensitive circuit that will activate an LED in low-light conditions. Figure 1: Schematic and Illustration of an LED. 1
The Current/Voltage Relation for an LED An LED behaves quite differently than a resistor. The resistance value for most resistors is nearly independent of the current passing through them. Hence, a plot of the current, I, as a function of the voltage across the resistor, V, is a line passing through the origin. This is not the case for an LED. Figure 2: Current vs. Voltage for a resistor and an LED Notice that the slope of the curve for the current in a resistor is straight. That is a resistor is a linear device where the current is proportional to the voltage. This proportionality is illustrated by Ohm s law: V I = R The current across an LED is not linear. Rather, an LED has very little current flow until the voltage across the LED approaches a maximum value call the forward voltage or V f. The value for V f depends on the type of LED. Typical values for V f are in the range of 1.6 V to 5 V: 1.6V V f 5V In addition, an LED has a maximum current, I max. Currents above I max may cause the LED to fail. Hence, care should be taken not to exceed I max. Typical values for I max are in range of 20 to 40 ma. However, some LED s have maximum currents outside of this range. 2
Experimentally Obtaining an Estimate for V f for an LED. One means for estimating V f is to create an I/V plot similar to the right side of Fig. 2. Below is a schematic that may be used for this test. Items Needed: Figure 3: Schematic for Creating and I/V Curve for an LED. 1. Breadboard 2. DC power supply with voltage V s (5V < V s < 20V) 3. A fixed or constant resistor with a resistance value in Ohms of Rc 50Vs. That is, if V s = 10V, then use a constant resistor of about 500Ω, (50 x 10).. 4. Resistors with values of 0 (a wire), 100, 220, 470, 1,000, 2,000, 4,700, 10,000, 20,000, 50,000 and 100,000 Ω 5. an LED and a 6. multimeter. Procedure: 1) Wear safety goggles while testing LED s. 1 Always connect a resistor with an LED in a circuit. Never test for resistance with a battery or voltage source in the circuit. 2) Determine and record R c based on V s : Ω 3) Construct the circuit illustrated in Fig. 3. However, for the initial configuration, use a wire instead of a resistor for R v. (This is equivalent to R v = 0). 1 Care should be taken to protect your eyes. Some LED s can fly apart if I max is exceeded. 3
4) Record the following: a) V sa, (the voltage across the resistor R c ), b) V LED, (the voltage across the LED), c) The light intensity of the LED. (very bright, bright, dim, very dim) Vsa 5) Calculate the current in the circuit using I =. R 6) Repeat Steps 4 and 5, replacing the wire used for R v with the resistors indicated in the table below. 7) Plot the current as a function of V LED on the graph below. c Table 1: Experimenal Values for an LED Circuit x-axis RV (Ω) VSA (V) VLED (V) 0 220 1,000 4,700 10,000 47,000 100,000 y-axis LED intensity I color. (A) V = R c sa 4
Graph 1: Experimental Values for an LED Circuit 5
2. Photoresistors In order to design a circuit that changes with light level, at least one component of that circuit must be a light-sensitive device. One such device is a photoresistor. A photoresistor is a light-sensitive resistor. Below is a plot of a resistance vs. light intensity for a photoresistor. Notice that the scale for the resistance (vertical) and the light intensity (horizontal) is logarithmic. A small change in a logarithmic scale can indicate a large change in the value of the parameter. Figure 4: Resistance vs. Light Intensity for a Photoresistor Unfortunately, due to variations in manufacturing of photoresistor, apparently, identical photoresistors may be different resistance values. Hence, it is useful to experimentally determine the curve by plotting the resistance as a function of the light intensity for the specific photoresistor to be used. Experimental Determination of the Performance of a Photoresistor Equipment Needed: 1. Photoresistor, 2. Multimeter, 3. Access to varying light levels, and 4. Optional: light meter. Procedure: Measure the resistance of the photoresistor, R PR, as a function of the light level. 6
Table 2: Experimental Values for Photoresistor Low Light Medium Low Light Medium Light Medium High Light High Light Resistance (Ω) Figure 5: Log-scale graph for Photoresistor Performance 7
3. Schematic of a Light-Sensitive LED Circuit The electrical schematic for a light-sensitive LED circuit is illustrated below. When designing a device, it is often useful to develop a conceptual model of the system of components that makeup the device. In the following paragraphs a mathematical model for this circuit. This mathematical model may be used to establish the ranges for appropriate values for the voltage source and the fixed-value resistor. Figure 6: Schematic for an LED that switches on at decreased ambient light levels. Most conceptual models include some approximations to simplify the mathematics associated with the model. One reasonable assumption that can be made for the illustrated circuit is that the LED is either off or on. That is, when the voltage across the LED is less than V f, it is assumed that the current through the LED is zero. Additionally, the voltage across the LED never exceeds V f. Mathematically this may be stated as: V b < V f, Assumptions: If V a < V f then I LED = 0, and the LED is in off mode, If V a = V f then the LED is in on mode. 8
Figure 7: Conceptual Circuit Model: off mode and on mode. Discussion Questions: Compare Figures 6 and 7. 1. Why is the LED missing in the off mode model? 2. What are the similarities and the differences between the two modes? 3. Why does no mode in Figure 7 assume that V a > V f? 4. What assumptions are made in the conceptual model? Design Problem: 4. Designing a Circuit Apply what you have learned about voltage dividers, LED s, and photoresistors to design and implement the circuit in Fig. 6. In your circuit, the LED should switch on at a pre-determined level of ambient lighting. Your design variables include: 1. The ambient light level desired 2. The voltage of the DC power source, V s 3. The value of the constant resistor, R c. Design Discussion Questions: 1. Which component of Fig. 6 is affected by the level of light? 9
2. What information can be used to determine the value of R PR to use for your design? 3. How can you determine what values of R c to use? 4. How can you determine what voltage to use for the DC power source? 5. Which design factor should be selected first the voltage of the DC power source, or the value of R c? Why? Instructor s Initials Hint: Model the circuit as a voltage divider where V a = V f and R PR is the resistance at the light level at which the LED is to switch at. V f = V s R R PR PR + R Example: Suppose V r = 1.9 V and V s = 9.1 V with R PR = 1,100Ω. The above equation would become: 1.9V 1,100Ω = 9.1V 1,100Ω + The solution to this is R c = 4,168Ω. (As a practical matter, a 3.9kΩ or a 4.7kΩ would be used.) Acknowledgement: This was developed while Douglas Oliver was an AAAS Science & Technology Fellow working at the National Science Foundation. All opinions are those of the authors. Contact: Doug Oliver Associate Professor Mechanical Engineering University of Toledo doliver@eng.utoledo.edu c R c. 10