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[ECEN 1400] Introduction to Digital and Analog Electronics R. McLeod HW #3: LEDs and Transistors (100 pts) 1 Light Emitting Diode IV Response (20 Points) Create a multisim circuit and include a schematic including a voltage source (Source → POWER SOURCES → DC POWER), an ammeter (Indicators → AMMETER → AMMETER H) and a generic red light emitting diode (Diodes → LED → LED red) and ground (Sources → POWER SOURCES → GROUND). Your circuit should look like this: You will now create a I/V response graph of a LED. To do this, increment the voltage of the DC POWER appropriately and record the current through LED .Take enough data points to give you a smooth curve of the I/V response, but dont waste your time by taking more data than required. Concentrate points where something interesting is happening. Current limits on LEDs vary, but a typical upper limit is 20 mA, so use this to find an appropriate voltage range. To gather each data point you will: 1. Start the Simulation 2. Record I and V 3. Stop the simulation 4. Increment the voltage appropriately 5. Repeat Note what voltage causes the LED to turn on to an accuracy of at least 0.05 V. (Turn on is indicated by the arrows to the right becoming red.) Make a plot of this IV curve in a graphing program such as excel including both negative and positive voltage. Show the turn-on voltage where the LED lit up on the graph. What do you notice about the I/V characteristics at this particular voltage? Version 3.2, September 24, 2014 Page 1 [ECEN 1400] 2 Introduction to Digital and Analog Electronics R. McLeod Limit the Current Through an LED (15 Points) The resistance of the LED is sufficiently low that it would draw too much current and be permanently damaged when hooked across a typical voltage source (e.g. 5 V). A solution is to include a current limiting resistor in series with the LED. Since the LED response is nonlinear, its not completely trivial to solve for the required resistance. A simple method is to use the LED I/V response that you found in problem one. Assume you want to limit the current through the LED to be 10 mA when the supply voltage is 5V. Use your graph to find the voltage drop across the LED at this current. From that, solve for the required resistance using Ohms law for the resistor. Check your work with multisim. 3 Regular Diode IV Response (20 Points) Delete the LED and replace it with the diode in your circuit kit which is a 1N4148 (Diodes → DIODE →1N4148). Repeat the experiment, but now that youve done it the tedious way, use multisim to do the hard work for you. In the top menu, select Simulate → Analyses → DC Sweep. You should see a dialog box like this: Set the Source in the first tab (shown above) to be V1 (the default) and let it sweep from -0.5 V to 2 V in increments of 0.01 V. This will sweep the voltage of your source over the specified range. Then select the Output tab. On the left is a list of variables you can calculate as the voltage is swept. Select I(D1[ID]) which is the current through the diode. Click Add to move it to the right-hand list. Then click Simulate at the bottom of the dialog box. This should plot a graph for you like you did by hand in the previous problem. No light is emitted from this device, but it will have a similar IV shape including a turn-on voltage. Which device has a larger turn-on voltage? Version 3.2, September 24, 2014 Page 2 [ECEN 1400] 4 Introduction to Digital and Analog Electronics R. McLeod Current Limit Your Diode Circuit (15 Points) Add a 1.69K resistor in series with your 1N4148 diode to create this circuit (the actual value of the voltage source wont matter since well use a DC sweep analysis). Use the same DC sweep analysis as you did on the previous problem, but now sweep from 0 to 5 V. You can plot either the current through R1 or through the diode since they are the same. What is the current through the diode when the supply is at 2V? 5 LED Controlled by a Transistor (30 Points) Create the following circuit in multisim. The LED is the same red LED used in problem 1. The transistor (labeled Q1, part number 2N2222A) is the one in your kit, found under (Transistors → BJT NPN). It is the top entry on the component list. For R2, use the value you found before that limits the current through the LED to 10 mA when using a 5V supply. Part a: Use the DC Sweep Analysis to sweep V1 from 0 to 5V and plot the current through R1. Compare this to your result with the 1N4148. What do you conclude about the base/emitter terminals of this transistor? Part b: Change your DC sweep analysis to plot the current through R2 as a function of sweeping V1 (not V2 leave this at 5V). This is a very interesting result. Imagine you had a voltage source, V1, which could deliver only 1 mA of current, was also noisy, and the value of the voltage varied randomly with an amplitude of 100 mV. What advantages would this circuit have as an LED driver over that of problem 3? These advantages are the foundations of digital logic. Version 3.2, September 24, 2014 Page 3