The picture above is an example of a buck converter. It converts a high voltage DC signal to a low voltage high current DC signal with great efficiency.
During the ON state, the inductor is charged until the supply is cut off. When that happens, the OFF state is achieved, and the inductor supplies the load in turn.
To maximize the usage of the inductor (that is to make it fully charged (at approximately 99%) in the ON state and fully discharged in the OFF state), the frequency of the triggering signal sent must be considered. If looking back resistance of the circuit as seen from the inductor is with the inductor having an inductance of , determine the frequency (in ) of the PWM triggering signal required to put this circuit in operation.
Details and Assumptions :
Neglect also the capacitor effects.
Ignore the diode voltage drops and the internal resistances of the components,
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For the inductor to assume a "Full-Charge" state, that is, at least 99% charged (bc theoretically it will take forever for it to be fully charged), it takes five time constants. That is roughly
5 × 1 5 0 0 0 3 0 × 1 0 − 3 = 1 0 μ S
So, that is the time it will take for the inductor to fully charge. Similarly, it will take the same amount of time for it to fully discharge.
So the frequency is
f = t c h a r g e + t d i s c h a r g e 1
f = 1 0 μ S + 1 0 μ S 1
f = 5 0 0 0 0 H z