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General License Course Chapter 4

General License Course Chapter 4. Lesson Plan Module 11 – AC Power. AC Power. Definition and Measurement Root Mean Square (V RMS ) The RMS value for an ac voltage is equivalent to the value of a dc voltage that results in the same power dissipation in a resistor. Definition & Measurement.

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General License Course Chapter 4

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  1. General License CourseChapter 4 Lesson Plan Module 11 – AC Power

  2. AC Power • Definition and Measurement • Root Mean Square (VRMS) • The RMS value for an ac voltage is equivalent to the value of a dc voltage that results in the same power dissipation in a resistor 2015 General License Course

  3. Definition & Measurement • RMS value is 0.707 times the ac waveform’s peak voltage (VPK) • VRMS = 0.707 x VPK = 0.707 x (VP-P / 2) • VPK = 1.414 x VRMS • VP-P = 2 x 1.414 x VRMS = 2.828 x VRMS 2015 General License Course

  4. Measurements • Relationships between RMS, average, peak-to-peak values 2015 General License Course

  5. Doing the RMS Math • A sine wave with a peak voltage of 17 V has an RMS value of? • VRMS = 0.707 x 17 V = 12 V • A sine wave with a peak-to-peak voltage of 100 V has an RMS value of? • VRMS = 0.707 x (100 / 2) = 35.4 V • A sine wave with an RMS voltage of 120 V has a peak-to-peak voltage of? • VP-P = 2 x 1.414 x 120 = 2.828 x 120 = 339.4 V 2015 General License Course

  6. PEP Definition & Measurement • 1500 W PEP (peak envelope power) is full amateur power limit • PEP is the average power of one complete RF cycle at the peak of the signal’s envelope • PEP is an easy way to measure or specify the maximum power of an amplitude modulated signal • PEV (peak envelope voltage) is the voltage at the peak of the signal envelope 2015 General License Course

  7. PEP Doing the Math • Example 10: • Keystrokes: • 50 x 0.707 = 35.35 [x2] = 1249.6225 / 50 = 24.99245 W • Round the answer to 25 W 2015 General License Course

  8. PEP Doing the Math • Example 11: • Keystrokes: • 1200 x 50 = 60,000 [square root symbol] = 244.9489 • Round answer to 245 V 2015 General License Course

  9. PEP Doing the Math • Example 13: • Keystrokes: • 0.707 x 200 = 141.4 / 2 = 70.7 [x2] / 50 = 99.968 • Round answer to 100 W 2015 General License Course

  10. PEP Doing the Math • What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50 Ω resistor connected to the transmitter output? • Keystrokes: • 0.707 x 500 = 353.5 / 2 = 176.75 [x2] / 50 = 624.81125 • Round answer to 625 W 2015 General License Course

  11. PEP & Average Readings • PEP is equal to average power if an amplitude modulated signal is not modulated • An unmodulated AM signal, a key down CW signal, and a FM signal are all constant power modes for which PEP = Average Power. • Single sideband PEP is not equal to average power • If your CW transmitter generates 1060 W average output power, the PEP is also 1060 W 2015 General License Course

  12. Practice Questions 2015 General License Course

  13. What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output? A. 1.4 watts B. 100 watts C. 353.5 watts D. 400 watts G5B06 2015 General License Course

  14. What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output? A. 1.4 watts B. 100 watts C. 353.5 watts D. 400 watts G5B06 2015 General License Course

  15. What value of an AC signal produces the same power dissipation in a resistor as a DC voltage of the same value? A. The peak-to-peak value B. The peak value C. The RMS value D. The reciprocal of the RMS value G5B07 2015 General License Course

  16. What value of an AC signal produces the same power dissipation in a resistor as a DC voltage of the same value? A. The peak-to-peak value B. The peak value C. The RMS value D. The reciprocal of the RMS value G5B07 2015 General License Course

  17. What is the RMS voltage of a sine wave with a value of 17 volts peak? A. 8.5 volts B. 12 volts C. 24 volts D. 34 volts G5B09 2015 General License Course

  18. What is the RMS voltage of a sine wave with a value of 17 volts peak? A. 8.5 volts B. 12 volts C. 24 volts D. 34 volts G5B09 2015 General License Course

  19. What would be the RMS voltage across a 50 ohm dummy load dissipating 1200 watts? A. 173 volts B. 245 volts C. 346 volts D. 692 volts G5B12 2015 General License Course

  20. What would be the RMS voltage across a 50 ohm dummy load dissipating 1200 watts? A. 173 volts B. 245 volts C. 346 volts D. 692 volts G5B12 2015 General License Course

  21. What is the output PEP of an unmodulated carrier if an average reading wattmeter connected to the transmitter output indicates 1060 watts? A. 530 watts B. 1060 watts C. 1500 watts D. 2120 watts G5B13 2015 General License Course

  22. What is the output PEP of an unmodulated carrier if an average reading wattmeter connected to the transmitter output indicates 1060 watts? A. 530 watts B. 1060 watts C. 1500 watts D. 2120 watts G5B13 2015 General License Course

  23. What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50 ohm resistive load connected to the transmitter output? A. 8.75 watts B. 625 watts C. 2500 watts D. 5000 watts G5B14 2015 General License Course

  24. What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50 ohm resistive load connected to the transmitter output? A. 8.75 watts B. 625 watts C. 2500 watts D. 5000 watts G5B14 2015 General License Course

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