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Dear all,

Given that the above is the instantaneous source current drawn by a buck converter, may I know how I should go about determining the corresponding average value of the source voltage?

Here's what I have attempted or understand could lead to a solution.
1) The gradient of the positive slope in the current waveform equals to the source voltage over inductance value. However, the inductance value is not given.
2) I could calculate the duty cycle of the converter using the on and off period. And possibly find source voltage if I have the output voltage but the output voltage is not given either.

  • \$\begingroup\$ Homework question without any attempt at solving? \$\endgroup\$ – winny Apr 29 '18 at 17:34
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    \$\begingroup\$ Definitely sounds like homework without an attempt at solution. What have you tried, and where are you stuck? Your question will be closed unless you can show some effort in answering the question. Start with V=L*di/dt. \$\endgroup\$ – John D Apr 29 '18 at 17:38
  • \$\begingroup\$ You can even calculate the output voltage too. So what have you tried to formulate so far? \$\endgroup\$ – Andy aka Apr 29 '18 at 18:13
  • \$\begingroup\$ Knowing the current is really cool, but it says absolutely nothing about the voltage, so there is no answer... \$\endgroup\$ – peufeu Apr 29 '18 at 18:32
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    \$\begingroup\$ @JohnD Have added in my solution attempts kind sir. \$\endgroup\$ – stephchia Apr 30 '18 at 0:55

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