Kirchhoff's Laws ELECTRONICS LABORATORY Experiment 2: Diode Applications: Half Wave and Full Wave (Bridge) Rectifier Circuits Objective: The purpose of this lab is to investigate the use of diodes in a half wave rectifier and bridge rectifier circuit. Equipment: Analog Discovery Design Kit (module) Analog Discovery Component Kit 4.7 KΩ Resistor 4 small signal diodes (1N914) Theory: A rectifier is an electrical device that converts alternating current (AC) to direct current (DC), a process known as rectification. Half-wave rectification In half wave rectification, either the positive or negative half of the AC wave is passed, while the other half is blocked. Because only one half of the input waveform reaches the output, it is only 50% efficient if used for power transfer. Half-wave rectification can be achieved with a single diode in a single phase supply as shown in Figure 2.1. Figure 2.1: Half wave rectifier using one diode Full-wave rectification A full-wave rectifier converts both the positive and negative halves of the input waveform to a single polarity (positive or negative) at its output. By using both halves of the AC waveform full-wave rectification is more efficient than half wave. When a simple transformer without a center tapped secondary is used, four diodes are required instead of the one needed for half-wave rectification. Four diodes arranged this way are called a diode bridge or bridge rectifier as shown in Figure 2.2. Figure 2.2: Full wave Bridge rectifier using four diodes If the transformer is center-tapped, then two diodes back-to-back (i.e. anode-to-anode or cathode-to-cathode) can form a full-wave rectifier (Figure 2.3). Twice as many windings are required on the transformer secondary to obtain the same output voltage compared to the bridge rectifier above. This is not as efficient from the transformer perspective because current flows in only one half of the secondary during each positive and negative half cycle of the AC input. Figure 2.3 Full-wave rectifier using a center tapped transformer and 2 diodes. Pre-Lab Simulation: In Multisim, construct the half-wave rectifier circuit shown in Figure 2.4. Use a 1N914 diode along with a 4.7k resistor. At the input use an input ac voltage source (sinusoidal) having 4 Volt amplitude and 0 Volt offset. The frequency should be adjusted to 100 Hz. Obtain both the input and the output waveform (remove dark background, use white background). Include circuit schematics and the waveforms. Comment on your results. Is the maximum value of the rectified output less than the max value of the AC input and by how much? Construct the full-wave bridge circuitry shown below in Multisim (you will later use the same circuit in Analog Discovery). Use a 4.7k resistor as the load. Set input ac source to 4V maximum as in step 1 with 0V offset. Obtain the plots of the input waveform and output waveform (the voltage across 4.7k resistor). In displaying the resistor waveform, you need to add an expression on the output tab of the transient analysis in Multisim. Include your own version of the circuit schematics along with input and output waveforms. Again, remove dark background, use white background in the plots. Please comments on your results. Is the maximum value of the rectified output less than the max value of the AC input and by how much? Procedure (Analog Discovery (AD)): 1) Half-wave Rectifier a) Set up the breadboard with waveform generator output W1 attached to anode terminal of the diode (1N914). The other end of the diode (cathode) is connected to the load resistor as shown in Figure 2.4. The other end of the resistor is connected to ground. Connect waveform generator output W1 to 1+ terminal of the scope so that the input waveform is observed. 1- terminal should also be connected to ground (GND) to avoid any noise effects. Single ended input of scope channel 2 (2+) is also connected to the upper end of the resistor (2- input need to be grounded). In other words, we are observing both input and the output waveforms with the scope. Figure 2.5 shows the breadboard picture of the half-wave rectifier. Note that the yellow wire is waveform generator output. As you know, orange and blue wires are the 1+ and 2+ scope inputs. In this picture, Channel 2 is used for input and Channel 1 is used for the output waveform in contrary to the one shown in Figure 2.4. However, this would not make any difference. You could swap the Blue and Orange wires, if you wish to make it correspond to Figure 2.4 where 1+ (orange) used at the input, 2+ (blue) used on the output node. Figure 2.4. Connection diagram for a half wave diode rectifier Figure 2.5. Breadboard picture of Figure 2.4. (Remember to include breadboard picture of your setup in lab report.) b) The waveform generator should be configured for a 100 Hz Sine wave with 4 Volt amplitude and 0 Volt offset. See Figure 2.6. Needless to say include screen capture of your waveform generator window in your report. Figure 2.6: The waveform generator setup. c) Set both scope channels’ range properly so it displays both waveforms nicely. Below is an example of what I obtained using 1V per division for input and output waveform (this one uses 2V for amplitude). Yours should use something like 600 mV or 700 mV per division to display waveforms even larger. Include the screen capture of your scope screen in your lab report. Figure 2.7: Input and output waveforms for a half-wave rectifier for a random 2V peak input. 2) Full-wave Rectifier using a Bridge configuration Four diodes can be arranged in a bridge configuration to provide a full-wave rectification from a single AC phase as shown below in Figure 2.8. However, note that output voltage is across the resistor RL, hence 2+ and 2- should be connected across RL as shown. When connecting diodes, make sure to connect properly by noting where the anode and cathode terminal is. Figure 2.8: Bridge Rectifier Circuit Note that breadboard picture is NOT provided. But, as a guide, the figure is redrawn with actual diodes. Repeat procedure 1- b) and 1-c), include breadboard picture and screen-capture of input and the output waveforms Questions for Lab Report: Referring to both circuits you implemented in AD: Why is the peak value of the rectified output less than the peak value of the AC input and by how much? At what point in the input waveform does the rectified waveform become positive i.e. something other than zero? Compare experiment results (AD) to Multisim simulation results. How much the output voltage value is different compared to the maximum value of AC voltage in each? Is there any discrepancy between AD and Multisim results? State possible reasons for discrepancy (if exists). What happens if the direction of the diode is reversed in half wave rectifier? 6