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How to calculate the input current consumption of the buck converter. It's seem that the output current only depends on the output voltage, duty cycles, input voltage, and coil inductance. Does the output current have any relation to the input current.

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For a lossless buck converter (reasonable first pass assumption), the input power his equal to the output power. So if the output voltage is half the input, then the input current is half the output current.

For a real buck converter, the efficiency will be less than 100%, so the input power will exceed the output power a little.

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It's seem that the output current only depends on the output voltage, duty cycles, input voltage, and coil inductance.

True but we generally use feedback to make a buck converter work in a predictable way. That feedback sets (fixes) the output voltage. The load on the output then determines the current. The feedback loop then controls the Duty Cycle such that the output voltage is correct for that current.

Due to the feedback we do not have to worry about the Duty Cycle, input voltage and coil inductance. As long as all parameters are such that the buck converter can work properly, it should (work properly). For example, if the input voltage increases a bit, the feedback loop will adjust the Duty Cycle such that the output voltage remains the same.

As Neil_UK states in his answer, for a simple calculation we can assume that the converter is 100% efficient so that input power and output power are the same (no power is lost). An example of an input current calculation would then be:

input: \$V_{in}\$ = 12 V

output: \$V_{out}\$ = 5 V, \$I_{out}\$ = 1 A => \$P_{out}\$ = \$V_{out}\$ * \$I_{out}\$ = 5 V * 1 A = 5 Watt

then at the input also 5 Watts go in: \$P_{in}\$ = 5 W

so the current will be: \$I_{in}\$ = \$P_{in}\$ / \$V_{in}\$ = 5 W / 12 V = 0.417 A

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It's seem that the output current only depends on the output voltage, duty cycles, input voltage, and coil inductance.

The output current depends on the output voltage and the load connected to the output. The output voltage depends on duty cycle and input voltage etc. but output current is equated directly to voltage and load resistance.

So, the output voltage and current convey a power to the load and, for a decent buck converter, the load power is about 90% of the overall power taken by the converter from the incoming DC voltage supply.

How to calculate the input current consumption of the buck converter

Hence, if output power is 10 watts, the input power will be about 11 watts. If input voltage is 30 volts then the input current will be 10 watts / 30 volts = 333 mA.

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