Essay Project: Rectifier and Acme Manufacturing Company

Submitted By bigred892
Words: 1061
Pages: 5

Introduction / Assignment As per the teacher's instruction we have been assigned to design and build a 5 Volt DC power supply for ACME Manufacturing Company. The power supply has several specifications that must be met in order to ensure a satisfied customer. The manufacture required that the power supply is small and of course cost effective. For simplistic design we used a transformer to step down the voltage. We then used a full wave rectifier and a voltage regulator in order to filter the voltage. Lastly, we used capacitors in order to fine tune the voltage. Of course there are many details involved in the process that will be explained later in the report.
Design

The first thing you must do before you build the project is to design the circuit that would provide the desired results. We used the material we have learned in past experience in order to successfully design our circuit. We had to begin realizing that we were working with a standard U.S. outlet which would be a 120 V rms AC power source. We put a fuse after the outlet in order to protect our device. We decided to go with a 1 A fuse. We chose the 1 A because it was the smallest fuse we could acquire and it would work for our application. Ideally the fuse should be much smaller. We then had to decide what type of transformer to use. After looking at the data sheet for the LM7805 regulator we found that the regulator can handle an input voltage of 5-18 V. This allowed us to decide to go with a basic 10:1 transformer. The secondary side of the transformer is connected to a full wave rectifier in order to eliminate negative voltage, thus changing from alternating current to direct current. After the rectifier we put 2 capacitors and a resistor in parallel. The capacitors are used to eliminate the voltage ripple. We calculated the size of the capacitor using the voltage ripple equation VR=ILT2C. The two capacitors we used were 330µF and .1µF. The large capacitor is used to reduce the voltage ripple, while the smaller capacitor is used while the large capacitor is discharging. The resistor is used to limit the current going to the voltage regulator just slightly. The voltage regulator then cleans up the voltage and reduces it to a constant 5 volts. The specifications state that it will produce anywhere from 4.8 – 5.2 V. We included a diode between the output and input of the voltage regulator. This is to provide protection in case the input voltage collapses faster than the output voltage. After the voltage regulator we placed another 100µF and 0.1µF capacitor to remove any ripple that may be remaining. Finally before the load we added 2 1KΩ resistors in parallel before the LED. The resistors limit the current to the LED as it is only rated for 50mA. The 500 Ω resistance allows the LED to light up while allowing the most current to the load.

Simulation We used a program called Multisim in order to create and test the circuit we designed. Multisim was an extremely useful tool because it had digital versions of the testing instruments that we had available to us in the lab. We found that our circuit performed without flaw during the simulation. However, a simulation is always ideal so we could not trust the values to be the same as what our actual results would be. The result from our simulation was an output voltage of 5 volts. The current across the load was 100.06 mA. The ripple voltage across the load was 229.03 µV. The results from the simulation are also shown below:

Production Finally, we were able to use our circuit design in order to produce the product. We began by first bread-boarding the circuit in order to test the circuit in real life. Also, we wanted to verify that it was being assembled correctly before we began soldering onto a board.…