Fig. 3. Definition of percent rippleIII. Pre-Laboratory ExerciseIV. Laboratory ExerciseEE 462: Laboratory # 4DC Power Supply Circuits Using DiodesbyDrs. A.V. Radun and K.D. Donohue (2/14/07)Department of Electrical and Computer EngineeringUniversity of KentuckyLexington, KY 40506(Lab 3 report due at beginning of the period) (Pre-lab4 and Lab-4 Datasheet due at the end of theperiod)I. Instructional Objectives- Design and construct circuits that transform sinusoidal (AC) voltages into constant (DC) voltages.- Design and construct a voltage regulator based on the characteristics of the Zener diode.- Evaluate the performance of simple rectifier and regulator circuits.See Horenstein 4.3 and 4.4II. BackgroundElectric power transmits best over long distances at high voltages. Since P = I V, a larger voltageimplies a smaller current for the same transmitted power. And smaller currents allow for the use of smaller wires with less loss. The high voltages used for power transmission must be reduced to be compatible with the needs of most consumer and industrial equipment. This is done with transformers that only operate with AC (DC does not pass through a transformer). However, most electronic devices powered by a home outlet require DC (constant) voltages. Therefore, thedevice must have a power supply that converts AC voltages into a DC (constant) voltage.The terminology "DC" is somewhat ambiguous. DC can mean the voltage or current always has the same polarity but changes with time (pulsating DC), or it can mean a constant value. In this lab assignment DC will refer to a constant voltage or current. Voltages or currents that maintain the same polarity, but change with time, have both a DC and AC component. The process of changing an AC signal to a signal with only positive values is called rectification, and circuits that perform this operation are called rectifier circuits. The rectifier circuit operates similar to theclipping circuits used in a previous lab. Figure 1 a) shows a half wave rectifier and Fig. 1 b) shows a full wave rectifier.RL Vs Vout (a) + - RL Vs Vout (b) + - Fig. 1 a) Half wave rectifier. b) Full wave rectifier.Although the output of a rectifier is always positive, it is generally not constant, often going fromzero to a peak value. Thus, the output of the rectifier must be filtered to remove the AC component so as to pass only the DC (constant) component to the output. Since DC has a frequency of 0 Hz, a low-pass filter can be applied to attenuate or block the higher frequency signal components. The simplest low-pass filter is a capacitor. Figure 2 shows examples of passing rectified signals through a low-pass filter. Low-pass filtering a waveform is sometimes called smoothing because it smoothes-out fast or sharp voltage jumps. Real-time filter can not have a cut-off sharp enough to totally eliminate unwanted frequencies, so the actual output of thefilter will always have some AC content, often called ripple (ripple voltage or ripple current). A rectifier combined with a filter forms a simple DC power supply. RL Vs Vout (a) + - RL Vs Vout (b) + - Fig. 2 a) Half wave rectifier with capacitor filter. B) Full wave rectifier with capacitor filter.One performance measure of a DC power supply is the percent output ripple computed from the ratio of the (peak-to-peak) output voltage to the average (DC) output voltage. Output ripple can be expressed as r in the equation below: opopVVr(1)where popV is the peak-to-peak output voltage and oV is the mean of the output voltage, which is equivalent to the DC component. Multiply r by 100 for percent output ripple. A typicaloutput signal is illustrated in Fig. 3. The best performance occurs when the percent ripple is zero(a battery produces ideal DC). This lab examines and compares the two rectifier schemes in Fig. 2, and demonstrates the contributions of the different stages to the final output.Peak to peak ripple Average DC oV t popV oV Fig. 3. Definition of percent rippleA DC power supply provides constant DC voltage to a load, which can be modeled as a resistor. Ideally the constant DC output should be independent of the load and input voltage fluctuations. In an actual power supply, however, when a load is applied to the output (as in Figs. 1 and 2), theoutput voltage decreases due to increased current drawn and the increased internal voltage drops.A voltage regulator circuit is used to prevent/limit these output voltage changes. A Zener diode can be used to make a voltage regulator circuit (as shown in Fig. 4) by taking advantage of the Zener diode’s reverse breakdown characteristic. Recall that once a Zener diode breaks down, its voltage remains essentially constant independent of its reverse current. The regulator's resistor, Rreg, limits the current through the Zener to reduce the power dissipated in the Zener. This is done, however, at the price of limiting the maximum load current that can be supplied with a regulated output voltage. Vin Vout + - + - Rreg Fig. 4. Zener voltage regulatorAn important characteristic of a voltage regulator is its percent regulation defined as the difference between the average no-load voltage (implies zero load current and thus infinite load resistance) and the average full-load voltage (the load draws its maximum (or rated) current and thus has its minimum (or rated) resistance) divided by the average full-load voltage. Regulation can be expressed in the equation below:oFLoFLoNLVVV Regulation(2)where oNLVis the average output no-load voltage and oFLV is the average output full-load voltage. Percent regulation is obtained by multiplying Regulation in Eq. 2 by 100. The best regulation performance is achieved with a 0 % regulation.A typical DC power supply consists of 3 stages, which are a rectifier, a filter, and a voltage regulator. A power supply using this combination is shown in Fig. 5. Vin Vout + - + - Rreg RL Vs + - Rectifier Filter Regulator Load AC source DC Power Supply Fig. 5. Basic power supply consisting of a rectifier, filter, and regulator.III. Pre-Laboratory ExerciseRectification Waveforms:1. For the half-wave and full-wave rectifiers in Fig. 1, determine the output voltage and the current through a 2.2k- load with a sinusoidal, 6.4V rms, 60Hz input voltage. Use suitable approximations. Write a Matlab program to plot the voltage and current on the same graph and then write a script to numerically compute the average power in