Electrical Engineering Technology EET 355 Web-Based Electric Circuits Laboratory
Experiment 6 – Diode Circuits
- 1) Explore various diode rectifier circuits.
- 2) Build and analyze a zener diode voltage regulator circuit.
Note: In this experiment, you will show all results in one Microsoft Word document named Exp06xxxx.doc (or .docx for Word2007) where xxxx is the last 4 digits of your ODU UIN. Each part of the experiment will be on a separate page of the document. You can force Word to start a new page by pressing CTRL+Enter.
In electrical circuits, a diode is analogous to a fluid check valve. That is, it allows electrical current to flow in one direction only. Any attempt to force current in the reverse direction through a diode will result in near-zero current flow.
Diodes are not perfect:
a) When they are reverse biased, the current is not zero (called reverse leakage current IR). However, this is usually small (microamperes) and can be ignored.
b) When they are conducting (called forward biased) the voltage drop is not zero (called the forward voltage drop Vf). For this reason, when a signal passes through a diode, the signal’s voltage is reduced by Vf.
c) Third, if we apply too much reverse voltage (the peak reverse breakdown voltage VRRM), the diode will break-down and conduct, similar to a pressure relief or “pop-off” valve. However, in a diode, reverse breakdown does not normally damage the diode unless the breakdown current is high. In fact, some diodes (called zener diodes) are designed to be operated in the breakdown region and are used as voltage regulators.
Since a diode acts as an electrical check valve, it can be used to convert ac voltages (bipolar signals) to dc voltages (unipolar signals). In this experiment we will be constructing and testing some of these rectifier circuits.
In Multisim, we have available both ideal diodes and real-world diodes. Ideal diodes have zero forward voltage drop and zero reverse leakage current. Real-world diodes have a forward voltage drop ranging from 0.5 V to 1.5 V, and a reverse leakage current that is generally lower
than 1 milliampere. In these experiments, we will be using a real-world diode, namely the industry standard 1N4001.
One of the most common uses of diodes is to convert ac signals to dc signals, a process called rectification. Rectifier circuits come in many varieties, more than can be explored in this experiment, so we will limit our investigation to two types, the half-wave rectifier and the full- wave rectifier. Then later in the experiment, we will investigate an application of a zener diode used as a voltage regulator.
A. HALF-WAVE RECTIFIER
1. In Multisim, construct the half-wave rectifier circuit in Figure 1. In Multisim, the 1N4001GP diode can be found in the “Diodes” group, “Diode”family.
2. In this circuit the input to the circuit is the ac voltage source (node 1 in Figure 1), and the output is the voltage drop on the resistor (node 2 in Figure 1). Perform a transient analysis, plotting the input and output voltages on the same graph. Set the time start and end time of the plot to start at zero and run for approximately two cycles of the waveform (i.e., you must calculate the end time using the input frequency).
4. In the grapher, switch on the cursors and set cursor 1 to the peak of the waveforms. In the cursor window, determine the peak values of the two waveforms, i.e., the y1 values for the two signals. Subtract these two values to determine the difference.
- Once you are satisfied the graph is set as specified, capture it using CTRL+PrtScreen.
- Open a new Microsoft Word document to be your report document. At the top put
“EET355 Experiment 5” and your name. Skip a few lines and paste the captured graph into it using CTRL+V.
7. Under the graph, answer the following questions:
7a) What is the difference in the peak value of the source voltage and the peak value of the resistor voltage?
7b) Explain why there is a difference in the two voltages.
8. In your report document, start a new page (CTRL+Enter) and then save your work to this point.
B. FULL-WAVE RECTIFIER
9. In Multisim, construct the circuit shown in Figure 2. In the figure, notice that the wires crossing inside the green circle do not connect to each other. For this circuit, the input voltage is the ac voltage source (in Figure 2, this is the node 2 voltage minus the node 1 voltage, but your node numbers may be different), and the output voltage is the voltage on the resistor, node 3.
10. Perform a transient analysis on this circuit plotting the input voltage and the output voltage on the same graph. Capture the graph and paste it into your report document.
11. Below the graph, answer the following question:
Comparing your plot of the half wave rectifier output and the full wave rectifier output, which circuit do you think is more efficient? Why?
C. ZENER DIODE VOLTAGE REGULATOR
12. In Multisim, construct the circuit in Figure 3. The zener diode can be found in the “Diodes” group, “Zener” family. Set the multimeter to read dc voltage. For this circuit, the input voltage is the dc power supply, and the output voltage is the voltage on the right-side resistor (R1 in the figure). To prevent the 150-Ω resistor from burning out, you must increase its maximum rated power to 1 watt. Do this by double clicking on the resistor and setting the wattage.
V(in) 12V 13V 14V 15V
In your report document, start a new page and type the following:
Vary the dc power supply, V(in), in 1 volt steps from 12V to 15V and measure and record the resistor voltage, V(out), for each case. Note that you cannot change the dc power supply voltage while Multisim is switched on. In your report document, enter the V(out) voltages in your table.
- Under your voltage table, answer the following questions:
14a) Based on the values in the table, what general statement can you make about the performance of the zener diode regulator circuit when the input voltage varies?
14b) In automotive electrical systems, depending on the battery condition and the engine RPM, the battery voltage can vary from 12V to 15V. However, onboard digital circuits (dashboard instruments, engine control system, CD player, GPS, etc.) require power that must be within the range of 4.75V to 5.25V. If we apply 12-15 volts, these circuits would be damaged. How can your circuit solve this problem?
- For part C, you do not need to include your circuit in the report document. Only include the voltage table and answers to questions 14a and 14b. Save your report document. Check to be sure it contains the following:
Page 1: Part A graph, answers to questions 7a and 7b.Page 2: Part B graph, answer to question 11. Page 3: Answers to questions 14a and 14b.
- Upload your report to Blackboard.