Homework 11 Sketch The Waveforms For Load Current And Voltag

Homework 11sketch The Waveforms For The Load Current And Voltage Forf

Sketch the waveforms for the load current and voltage for Figure 2–73. Show the peak values. Consider the circuit in Figure 2–75. (a) What type of circuit is this? (b) What is the total peak secondary voltage? (c) Find the peak voltage across each half of the secondary. (d) Sketch the voltage waveform across RL. (e) What is the peak current through each diode? (f) What is the PIV for each diode? To what value must R be adjusted in Figure 2–79 to make I Z = 40 mA? Assume that V Z = 12 V at 30 mA and Z Z = 30Ω. Assume the output of a zener regulator circuit drops from 8.0 V with no load to 7.8 V with a 500Ω load. What is the percent load regulation? Each part of Figure 2–84 shows oscilloscope displays of rectifier output voltages. In each case, determine whether or not the rectifier is functioning properly and, if it is not, what is (are) the most likely failure(s). Assume all displays are set for the same time per division.

Paper For Above instruction

The electrical engineering concepts surrounding waveforms of load current and voltage, as well as the characteristics of various rectifier and regulator circuits, are fundamental to understanding power supply design and analysis. This paper discusses the behavior of load waveforms pertinent to given circuits, analyzing the type of circuits, voltages, currents, and component ratings involved. Through a comprehensive examination of the specified figures, the study elucidates the waveform characteristics, peak values, diode parameters, and regulation efficiency, integrating core principles with practical circuit analysis.

Waveform Sketching for Load Current and Voltage

In assessing the waveforms of load current and voltage for the circuit depicted in Figure 2–73, it is essential to understand the type of rectification involved. If Figure 2–73 illustrates a single-phase half-wave rectifier, the voltage waveform across the load will resemble a unidirectional pulsating DC signal, with the current wave following a similar pattern but dependent on the load and circuit components. The voltage waveform across the load peaks at the maximum secondary voltage (V_s), which is influenced by diode drops and transformer turns ratio. The load current sharply rises during the conduction period and drops to zero during off periods, creating a pulse of peak value superimposed on the constant secondary voltage. Conversely, for a full-wave rectifier, the waveform would be symmetrical and pulsating at twice the frequency of the AC source.

The peak load voltage can be approximated by considering the secondary peak voltage minus diode drops in the conduction path. For example, in a half-wave rectifier, the peak load voltage equals the secondary peak voltage less the forward drop of the diode, typically around 0.7 V for silicon diodes. The waveforms are characterized by sharp peaks at regular intervals, and proper sketching involves plotting the voltage and current to reach their respective maximum values while noting the waveform shape’s smoothness and ripple.

Analysis of the Circuit in Figure 2–75

(a) Circuit Type Identification

Without the actual diagram, typical assumptions classify such circuits as either transformer-based rectifier circuits, which could be half-wave, full-wave, or bridge rectifiers. The configuration in Figure 2–75 likely represents a center-tapped secondary rectifier due to the inquiry about total peak secondary voltage and individual half voltages. This architecture involves a transformer secondary split into two halves, each feeding a diode, culminating in combined output across the load.

(b) Total Peak Secondary Voltage

The total peak secondary voltage (V_peak_total) is derived from the transformer's rated secondary voltage, multiplied by the turns ratio, and adjusted for diode drops. Typically, V_peak_total equals twice the peak voltage of each half secondary if center-tapped, which is determined by the secondary RMS voltage multiplied by √2. For example, if the secondary RMS voltage is 15 V, then the peak voltage per half would be 15 V × √2 ≈ 21.2 V. The total peak secondary voltage would be approximately 42.4 V before diode drops.

(c) Peak Voltage Across Each Half of Secondary

The peak voltage across each half of the secondary in a center-tapped configuration is equivalent to its respective half's peak secondary voltage, calculated as V_secondary_RMS × √2. For instance, with a 15 V RMS secondary, each half would peak at approximately 21.2 V. This voltage is crucial for designing the rectifier's peak current capacities and for ensuring safe diode operation.

(d) Waveform Sketch Across RL

The voltage waveform across the load resistor RL in a rectifier circuit manifests as a pulsating DC signal, akin to the rectified voltage shape. In a peak rectifier, the waveform rises sharply to the peak value and then drops sharply, producing a waveform with ripples. The shape is heavily dependent on the filtering components; a simple resistor load results in a waveform that reaches the peak voltage repeatedly. If smoothing capacitors are used, the waveform becomes less ripple-like, approximating a nearly constant voltage. Accurate sketching involves plotting the waveform with sharp rise times, gradual or abrupt drops, and noting the peak value.

(e) Peak Current Through Each Diode

The peak current through each diode in a center-tapped rectifier circuit is the maximum load current delivered during the conduction period, determined by the load resistor and the peak secondary voltage. Using Ohm's law, I_peak = V_peak / R_L. For example, if V_peak is 21 V and R_L is 1 kΩ, then I_peak ≈ 21 mA. This current must be within the diode’s maximum ratings to prevent damage. Additionally, during the conduction, the diode experiences transient peak currents, which must be considered in the diode’s datasheet specifications.

(f) PIV for Each Diode

The Peak Inverse Voltage (PIV) for each diode in a center-tapped rectifier is the maximum reverse voltage it must withstand when not conducting. For the diode connected to the positive half-cycle, PIV equates roughly to the total secondary peak voltage, because during the other diodes’ conduction period, the reverse voltage appears across them. As previously noted, if the secondary peak voltage per half is approximately 21 V, then PIV should be at least this value, preferably with a margin—say 30–50% higher—to ensure reliable operation under transient conditions.

Adjusting R to Achieve Desired Zener Current

In the circuit depicted in Figure 2–79, adjusting the resistor R to produce a Zener diode current (I_Z) of 40 mA involves Ohm’s law and the Zener diode's characteristics. Given V_Z = 12 V at 30 mA and Z_Z = 30Ω, the circuit must be configured such that the resistor R drops an appropriate voltage to achieve the target current. By first calculating the voltage drop across R at the desired current, R = (V_in - V_Z) / I_Z. Given the V_in, which is related to the secondary voltage and circuit configuration, R must be reduced or increased accordingly to meet the target current while ensuring the Zener diode operates within its regulation range.

Assessing Load Regulation in a Zener Voltage Regulator

The percent load regulation of a Zener regulator is calculated by measuring the change in output voltage as the load varies, relative to the no-load voltage. Given a no-load output of 8.0 V and a load voltage of 7.8 V with a 500Ω load, the load regulation is ((8.0 V - 7.8 V) / 8.0 V) × 100%, which equals 2.5%. Load regulation indicates the circuit’s ability to maintain a constant output voltage despite varying load currents, with lower percentages signifying better regulation characteristics.

Analysis of Rectifier Output Waveforms Using Oscilloscope Displays

Figure 2–84 depicts oscilloscope outputs for various rectifiers. Proper functioning rectifiers should display expected waveforms: half-wave rectifiers show a single pulsating waveform, full-wave rectifiers show two pulses per cycle, and bridge rectifiers produce a similar pattern but with slight variations due to diode drops and circuit topology. Any anomalies, such as missing pulses, irregular ripple, or distorted waveforms, suggest potential malfunctions like diode failures, opens, shorts, or faulty filters. Correct analysis involves comparing observed waveforms with theoretical expectations, considering the circuit configuration and operating conditions.

Conclusion

Understanding and analyzing load current and voltage waveforms are crucial for designing reliable power supply systems. The circuit analysis, spanning rectifier types, voltages, currents, and regulation characteristics, provides foundational insights into circuit functionality and performance. Proper waveform sketching, diode rating calculations, and regulator assessments enable engineers to optimize power supplies for efficiency, durability, and stability. These principles underpin advanced electrical engineering applications, ensuring safe and effective circuit operation in diverse environments.

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