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enter image description here

The above problem was asked in GATE 2018 Electrical paper. I got the answer as 50. But I am confused if my answer is correct. The funny part is all top 3 coaching institutes has given 3 different answers for this problem. Please help me with this problem.

Solution by Kreatryx enter image description here

Solution by Made Easy enter image description here

Solution by ACE Academy enter image description here

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    \$\begingroup\$ Since Kreatryx can't even spell 'Solutions' I'd toss that answer straight away. I believe Made Easy forgot to average the switch current (i.e. 50A for half the time, 0A the rest = 25A) and ACE has it figured out. \$\endgroup\$ – calcium3000 Feb 15 '18 at 19:56
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    \$\begingroup\$ Agreed on the bad spelling, but they are the only ones that got it right. The answer is 12.5A \$\endgroup\$ – John D Feb 15 '18 at 20:32
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schematic

simulate this circuit – Schematic created using CircuitLab enter image description here

Note the average inductor current is shown here. The average switch current is half that. That is, 12.5A per switch.

So Kreatryx got the right answer, despite dyslexia.


Intuitively you would think that Made Easy's point that the load is always connected to the source means the output voltage must be the input voltage, but that is not correct.

If you consider just one half, what is happening is you are creating a square wave voltage and passing it through a filter to retain the average DC voltage component. Since the wave is 50-50 mark-space, the average output voltage from both sides is 50V.

When you put them in parallel that voltage does not add. It is just the same as putting two 50V batteries in parallel, you still have 50V.

What does change, just like putting two batteries in parallel, is the current taken from each is divided by two.

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    \$\begingroup\$ Your simulation shows the average INDUCTOR current. The problem asks for the average switch current which is half that. So 12.5A. (The duty cycle of each switch is 50%) \$\endgroup\$ – John D Feb 15 '18 at 20:33
  • \$\begingroup\$ @JohnD ya I just realized that too. \$\endgroup\$ – Trevor_G Feb 15 '18 at 20:34
  • \$\begingroup\$ @Trevor_G Sir, can you please explain the working of the circuit. I'm still confused. \$\endgroup\$ – Nikhil Kashyap Feb 15 '18 at 20:42
  • \$\begingroup\$ @NikhilKashyap see addition to answer plus John D's answer. \$\endgroup\$ – Trevor_G Feb 15 '18 at 20:55
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    \$\begingroup\$ Now I got it. That two batteries in parallel logic is great !! \$\endgroup\$ – Nikhil Kashyap Feb 15 '18 at 20:59
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In a multi-phase continuous buck converter (which this is) the output voltage (ideally) is the input voltage times the duty cycle.

In this case the duty cycle is 50%. You can think of the LC filter as averaging the input voltage which is a 50% duty cycle 100V square wave. So Vout = 50V.

Output power is then 50^2/1 = 2500W.

Input power = output power (ideally) so input current is 25A.

Each switch will carry the input current for 50% of the time, so the average switch current is 12.5A.

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  • \$\begingroup\$ I'm not getting the Vo=DVs logic. I took Vo=100V and ended up getting 50A like the made easy solution. \$\endgroup\$ – Nikhil Kashyap Feb 15 '18 at 20:46
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    \$\begingroup\$ The output LC averages the input waveform. If the input waveform is a 100V square wave the output is 50V DC, assuming the switching frequency is well above the resonant frequency of the LC filter. \$\endgroup\$ – John D Feb 15 '18 at 20:53

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