FCC Exam Question: 40E5
What are the input/output voltages of a “BUCK” switching power supply with a 50% duty cycle?
Explanation: A "BUCK" switching power supply is a step-down converter, meaning its output voltage will always be lower than or equal to its input voltage. The output voltage ($V_{out}$) of an ideal buck converter is directly proportional to its input voltage ($V_{in}$) and the duty cycle (D) of its switching element. The relationship is given by the formula: $V_{out} \approx D \times V_{in}$. With a 50% duty cycle (D = 0.5), the output voltage will be approximately half of the input voltage. * **D) 10 volts in, 5 volts out:** Here, 5V is 0.5 times 10V. This perfectly matches the formula for a 50% duty cycle buck converter. * **A) 120 volts in, 240 volts out** and **B) 24 volts in, 36 volts out:** These options show an output voltage *higher* than the input, which describes a "BOOST" converter, not a BUCK converter. * **C) 12 volts in, 12 volts out:** This would imply a 100% duty cycle (D=1) for a buck converter, meaning the switch is always on, allowing the input voltage to pass directly, not a 50% duty cycle.
8A5
12B2
25C4
39E6
19B2
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Includes Elements 1, 3, 6, 7R, 8, and 9.