IntroductionExperimental ProcedureReferenceELEC 350L Electronics I Laboratory Fall 2011Lab #6: Simple DC Power SupplyIntroductionIn 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 circuitthat 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 BackgroundA 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-cycleof 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 whenvsec is negative, the capacitor remains charged because the diode is reverse biased, and there is nopath available through which charge can leave the capacitor if RL is an open circuit. The diode is also reverse biased (that is, vD ≤ 0.7 V) during the subsequent positive half-cycles, because the capacitor remains charged at a value near the peak positive voltage. The capacitor retains its charge and the voltage across it stays constant thereafter because the diode remains reverse-biased at all times. No additional current can flow into the capacitor, but no current can flow out either (again, assuming the load is an open circuit). The voltage across the capacitor remains at the peak positive value of the voltage across the secondary winding, minus the value of the diode’s turn-on voltage.Figure 1. Basic power supply circuit using half-wave rectifier.1 of 6vin+vsec−RLC+vo−iL+ vD −Of course, a load must be connected to this circuit if it is to serve any useful purpose. In this case, the load resistance RL is finite, and vo is no longer constant because the capacitor dischargesthough the load whenever vsec falls below the level necessary to keep the diode forward-biased. This circuit is called a rectifier because it allows current to flow in only one direction though the load. Although the load current iL is unidirectional, its level nevertheless varies as the capacitor voltage rises and falls.The major disadvantage of the circuit shown in Figure 1 is that the diode conducts during only one of the AC voltage’s half-cycles, and for this reason the circuit is more specifically called a half-wave rectifier. During the half-cycle when the diode does not conduct, the capacitor continues to discharge through the load. The rate of discharge depends on the value of RL. If RL isrelatively small, the current flowing out of the capacitor when the diode is off can be significant. As a result, the voltage across the capacitor can drop to an intolerably low level. Another way to view this is that the time constant that governs the discharge rate is given by RLC. The smaller the value of RL, the smaller the time constant, and therefore the more quickly the capacitor discharges. The alternating charge and discharge cycles cause the output voltage of the power supply to fluctuate, creating what is called a ripple voltage that is superimposed on the average DC output voltage. This effect is shown graphically in Figure 2. Since most loads require a nearly steady supply of DC, the presence of a significant ripple voltage Vr is undesirable. The ripple voltage is sometimes expressed as a percentage and is calculated using the formula [1] voltage DCaverage voltageripplepeak -to-peak100ripple % .Figure 2. Ripple voltage in the output of a DC power supply.The presence of ripple in the output of a power supply can be mitigated partly by using a full-wave rectifier circuit like the one shown in Figure 3. In this circuit, diodes D2 and D4 conduct during positive half-cycles (when vsec is positive), and diodes D1 and D3 conduct during negative half-cycles. During both half-cycles current flows downward through the load resistor. The full-wave rectifier is an improvement over the half-wave circuit because the capacitor is charged during both half-cycles. Consequently, the capacitor does not have as much time to discharge, and the output voltage does not drop as far as with the half-wave rectifier. For a given target 2 of 6votVP = VmaxVmin0.5TVrunfiltered,rectifiedAC waveformripple voltage only half the capacitance is needed in the full-wave rectifier circuit relative to that needed in the half-wave rectifier.Figure 3. Full-wave bridge rectifier circuit with filter capacitor. The “+” symbol next to the upper plate of the capacitor indicates that it is an electrolytic type. Electrolytic capacitors are commonly used in power supply circuits because they are available in large values. They are also polarized; that is, the voltage across them must have the indicated polarity for proper operation.The capacitors shown in Figures 1 and 3 are often called filter capacitors because their purpose is to reduce the ripple voltage to a tolerable level. As the sinusoidal input voltage rises to its peak, the capacitor voltage rises with it, as explained earlier. When the input voltage begins to fall again, the capacitor starts to discharge. However, if the capacitor is large enough and if the load resistance is not too small, the rate of discharge is much less than the rate at which the input voltage falls. Thus, there will be a period of time during each