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Here's a switching circuit. The input Vs is a 24V rectangular pulse at 20kHz frequency with 75% duty cycle. Since the time constant L/R is very much greater than the pulse width, we expect the inductor current to be a triangular wave (i.e. integral of a rectangular wave) at a certain DC offset which is the average current. My problem is, how do you find the average inductor current assuming Vbatt is 13.8V? (Please ignore questions b and c.)

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  • \$\begingroup\$ What is the average voltage of Vs? What is the Vbat initial voltage? The difference across 0.1 ohm controls the average current. It cannot be solved unless Vbat is known or you assume from b) Vbatt=13.8V or dead at 11.5V if an auto battery. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 8 '17 at 5:02
  • \$\begingroup\$ Okay, let's say Vbatt is 13.8V. \$\endgroup\$ – John Smith Dec 8 '17 at 5:02
  • \$\begingroup\$ Now use Vavg of input and Ohm's law \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 8 '17 at 5:03
  • \$\begingroup\$ Yes, I've thought about that but what about the voltage across the inductor? Shouldn't it be a series of discontinuous sawtooth pulses? \$\endgroup\$ – John Smith Dec 8 '17 at 5:04
  • \$\begingroup\$ No , They define Vs as 24 or 0 so it is continuous so all sources are 0 ohm as well as L \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 8 '17 at 5:05
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Only guidance, not something to copy as is:

The time constant L/R = 1 second. That's enormous when compared to the pulse interval. The current in the inductor does not change remarkably, probably not even measurably during one pulse period. You can well think your pulse source as a constant voltage DC source. The voltage of it is the average pulse voltage.

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