Consider a buck converter connected to a PV module. The average PV voltage is 40[V] and the average load voltage is 4[V] . If the time the transistor is closed during a switching period is 1[μs] , what is the switching frequency of the converter? Give the answer in [kHz] and as an integer.
To find the switching frequency of the converter, we can use the formula:
Switching frequency = 1 / (Time transistor is closed during a switching period)
Given that the time the transistor is closed during a switching period is 1 μs, we can substitute this value into the formula:
Switching frequency = 1 / (1 μs)
To convert microseconds (μs) to seconds (s), we need to divide by 1,000,000 (since 1 μs = 1/1,000,000 s):
Switching frequency = 1 / (1 μs) = 1 / (1/1,000,000 s) = 1,000,000 s
Finally, we need to convert the switching frequency from Hz (s^-1) to kHz (s^-1 * 1000). Therefore:
Switching frequency = 1,000,000 s = 1,000 kHz
Therefore, the switching frequency of the converter is 1,000 kHz or 1 MHz.