Hello folks,
I have a problem understanding the attached circuit, a DC-DC voltage converter.
It says in the book where its depicted that "the average voltage across an inductor has to be zero, otherwise the magnitude of its overall current is rising without limits. From this it follows that the average output voltage is half the input voltage." (The switching ratio is 50-50)
What I dont understand now is the role of the inductor. Why do we need it in the circuit? Cant we just switch 50-50 without the inductor and also get a "time-average" voltage of half the input voltage?
And what would happen if we, say, attach a bulb between ground and the output voltage? Then this would mean that because through the inductor flows no current on average the bulb discharges the capacitor C2 and will never be recharged since the average current through the inductor is 0 and the circuit would only be useful for digital applications.
I have a problem understanding the attached circuit, a DC-DC voltage converter.
It says in the book where its depicted that "the average voltage across an inductor has to be zero, otherwise the magnitude of its overall current is rising without limits. From this it follows that the average output voltage is half the input voltage." (The switching ratio is 50-50)
What I dont understand now is the role of the inductor. Why do we need it in the circuit? Cant we just switch 50-50 without the inductor and also get a "time-average" voltage of half the input voltage?
And what would happen if we, say, attach a bulb between ground and the output voltage? Then this would mean that because through the inductor flows no current on average the bulb discharges the capacitor C2 and will never be recharged since the average current through the inductor is 0 and the circuit would only be useful for digital applications.
Attachments
-
11 KB Views: 16