Bucknell ELEC 350 - Lab #6: Simple DC Power Supply (6 pages)

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Lab #6: Simple DC Power Supply



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Lab #6: Simple DC Power Supply

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Lecture Notes


Pages:
6
School:
Bucknell University
Course:
Elec 350 - Electronics I

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ELEC 350L Electronics I Laboratory Fall 2011 Lab 6 Simple DC Power Supply Introduction In almost all types of DC power supplies a rectifier circuit and a filter capacitor are used to convert an AC voltage to an almost constant DC voltage Because most electronic circuits are designed to be powered by DC sources rectifier circuits are almost always found in equipment that operates from 60 Hz AC outlets In practice sophisticated voltage regulation circuits are usually employed to maintain DC supply voltages at specified levels however the simple circuit that you will study in this lab exercise forms the foundation of most commercial DC power supplies In fact this basic circuit with a few minor modifications is sufficient for many applications in which a small amount of power supply ripple noise can be tolerated Theoretical Background A very basic power supply circuit is shown in Figure 1 A source of AC such as the 120 V rms available at a wall socket and represented by vin drives a transformer which might step up or step down the voltage Consequently the transformer s secondary voltage represented by vsec in Figure 1 is AC as well If no load resistor RL is present i e the load is an open circuit the capacitor charges during the first half cycle that the diode is forward biased a positive half cycle of vsec The capacitor voltage vo in Figure 1 essentially follows vsec minus the approximately 0 7 to 1 0 V drop across the diode because the internal resistance of the secondary winding Rsec and of the diode RD are negligibly small In other words the time constant Rsec RD C is usually much less than the period of the sinusoidal voltage waveform If the load resistance RL is finite then the time constant is RL Rsec RD C Do you know why During the next half cycle when vsec is negative the capacitor remains charged because the diode is reverse biased and there is no path available through which charge can leave the capacitor if RL is an open circuit The diode is also reverse



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