In buck circuit, If I mount 0E resistor instead of inductor then what will be the problem ?

And what is the use of inductor in buck regulator ? whether just for filtering ?

Regards, Azlum

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    \$\begingroup\$ There is a ton of stuff on google about energy storage and filtering in switching power supplies. Far more stuff than what could possibly constitute an answer on SE.EE. \$\endgroup\$ – Andy aka Jun 18 '15 at 9:27
  • \$\begingroup\$ Have you already wondered how a buck regulator is supposed to work? Im saying in electrical therms. Do the circuit analysis and will be self-explanatory. \$\endgroup\$ – Pedro Quadros Jun 18 '15 at 10:53
  • \$\begingroup\$ In some Google stuff saying, Inductor is used for only just for filtering. But some of stuffs saying its for controlling desired output by storing energy. Which is correct ? \$\endgroup\$ – Mohammed Azlum Jun 18 '15 at 11:47

You can think of a switching regulator (such as a buck converter) as a digital pulse generator followed by a low-pass filter which averages out the pulses to create a DC voltage. The normal inductor-capacitor network that you see in a buck converter is simply a second-order low pass filter and indeed you could replace it with a first-order resistor-capacitor filter if you wanted.

However the quite obvious problem with doing this is that it would be extremely inefficient because the resistor creates an impedance by dissipating energy whereas an inductor creates an impedance by storing energy. Indeed you will have essentially just made a chopper style linear regulator, which doesn't offer any advantages I can think of over using a regular linear regulator while being much more complex.

This is why you don't see such a design for power supplies in practice.

  • \$\begingroup\$ I have mounted the 0E resistor instead of inductor. But I didn't get any desired output from the regulator (TPS54319RTER) and finally regulator spoiled. I am wondering how regulator (TPS54319RTER) has been spoiled due to resistor instead of inductor ? \$\endgroup\$ – Mohammed Azlum Jun 18 '15 at 11:43
  • \$\begingroup\$ The answer was a general response to your non-specific question. If the question is will XXX regulator IC work without an inductor the answer is more than likely no, because the regulator IC was not designed for that. Also if you mean a 0ohm resistor, then that is absolutely not going to work. Have a read of the wikipedia article on buck converters. If you work through the details you should be able to see what the problem is. \$\endgroup\$ – Jon Jun 18 '15 at 13:36

The inductor is required. It is a key component.

Simple Answer

During ON time current is coming from source. enter image description here During OFF time current is coming from the action of the inductor.

When ON turns OFF, current in inductor stops. No current, so magnetic field will collapse. Faraday's Law states a voltage will be induced across inductor (- +), which forward biases the diode and current flows through the load.

Now before you say, we are getting something for nothing in the OFF state, you must realize the inductor was charged during the ON state.

What we are trying to achieve is current in the inductor never reaches 0. No inductor and the circuit will NOT work.


You could, theoretically, use an RC filter instead of an LC filter on the output of a buck converter. As Jon mentioned, this is very inefficient. But in order for it to work there has to be some resistance. A zero-ohm resistor won't work. If you try to use a zero-ohm resistor, the capacitor will short out the pulse voltage during transitions. That may be what causes your regulator to fail over time.

  • \$\begingroup\$ I used exact 0E resistor. And which will pass the switching pulses to capacitor and capacitor may pass the pulses to ground. This I understood. But how the regulator is gone bad ? \$\endgroup\$ – Mohammed Azlum Jun 19 '15 at 5:33
  • \$\begingroup\$ There's nothing to limit the current through the capacitor. The pulse voltage changes very fast, which makes a very large current in the capacitor. It acts like a short circuit. That can burn out the regulator. \$\endgroup\$ – Adam Haun Jun 19 '15 at 15:08
  • \$\begingroup\$ Also, the output voltage would probably drift between the desired voltage and the input voltage. \$\endgroup\$ – Adam Haun Jun 19 '15 at 15:10

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