GMDSS Trainer
Subelement C: Receiving Systems – 10 Key Topics – 10 Exam Questions – 4 Drawings— Topic 26: Power Supplies

Question 8-26C2

Element 8 (RADAR)

A constant frequency switching power supply regulator with an input voltage of 165 volts DC, and a switching frequency of 20 kHz, has an “ON” time of 27 microseconds when supplying 1 ampere to its load. What is the output voltage across the load?

Explanation
A switching power supply (SMPS) regulates its output voltage by rapidly turning an input voltage ON and OFF, controlling the ratio of the "ON" time to the total switching period. This ratio is known as the **duty cycle (D)**. For a step-down (buck) converter, which is typical when the output voltage is less than the input, the output voltage (Vout) is the input voltage (Vin) multiplied by the duty cycle. 1. **Calculate the total switching period (T):** T = 1 / frequency = 1 / 20 kHz = 1 / 20,000 Hz = 0.00005 seconds = 50 microseconds (µs). 2. **Determine the duty cycle (D):** D = ON time / Period = 27 µs / 50 µs = 0.54. 3. **Calculate the output voltage (Vout):** Vout = Vin * D = 165 V * 0.54 = 89.1 V DC. This efficient voltage regulation method, minimizing resistive losses, is fundamental in modern radio equipment. The other options are incorrect as they either assume a different type of converter (e.g., boost) or an incorrect understanding of the duty cycle's effect on output voltage.

Ready to test your knowledge?