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I am extremely confused about why this overshoot occurs in the Inductor voltage waveform in buck converter and also why the time to achieve steady state reduces on increasing the load resistance.

These are the waveforms that have been formed for two different load resistances (1. 12 ohms, 2. 36 ohms). Please help me understand why the two waveforms differ on just changing the load resistance. Please do provide a good reasoning as I am extremely confused after hours of googling!

Waveform 12 ohms: Inductor and output voltage waveform for 12 ohm load, V_input=100V, D=0.48

Waveform 36 ohms: Inductor and output voltage waveform for 36 ohm load, V_input=100V, D=0.48

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    \$\begingroup\$ Google for "damping" in a second order system. \$\endgroup\$ – Unimportant Nov 15 '20 at 8:58
  • \$\begingroup\$ Try it with an open circuit load. Inrush current and voltage overshoot will be much worse. The underlying reason is that the voltage across the inductor is initially Vin. This will cause the inductor current to ramp much faster at first than in the steady state case. The inductor will be in the continuous conduction regime and overshoot the steady state voltage as well. You have to somehow add damping to the system, OR add soft-start circuitry of some sort to avoid this problem. \$\endgroup\$ – mkeith Nov 15 '20 at 9:14
  • \$\begingroup\$ @mkeith can you please brief a little more about the initial inductor voltage V_in. How V_in is responsible for overshoot even though it depends only on change in current? \$\endgroup\$ – Paras Nov 15 '20 at 9:29
  • \$\begingroup\$ V = L di/dt. At startup V = Vin-Vout. So di/dt = (Vin-Vout)/L. So di/dt will be at a maximum at startup. With no load or light load, the steady state current will be low/zero. So I will over-shoot the mark. The larger the load, the less overshoot there will be. \$\endgroup\$ – mkeith Nov 15 '20 at 17:43
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    \$\begingroup\$ @Paras I had considered writing when you first posted, but you only provided two comparison images and I felt not enough other details to warrant an answer. That said, what immediately struck me in looking at your two pictures is that it looked like good behavior (which told me you'd probably been designing things right and captured representative pictures -- good) of a 2nd order system. (Likely, inductor and filter capacitor.) When you change the load, you change the damping factor, \$\zeta\$. Your images are actually what you should expect from a system like this. \$\endgroup\$ – jonk Nov 15 '20 at 20:33
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I thought I would put together a quick simulation to illustrate a key point. The response of an open-loop, fixed-duty-cycle buck converter is going to be almost exactly the same as the response would be to a step voltage applied to the same LC filter and load, where the step voltage is equal to Vin * D, where D is the duty cycle.

So, for example, if Vin is 10V and you apply a 50% duty cycle to the buck output filter, the voltage response of the buck will be the same as if you applied a 5V step function. See below, where I have done exactly that. The red and blue traces are just about on top of each other.

buck

So, now that you see this, you should realize that whatever ringing and overshoot you experience is not caused by the buck, per-se, but by the step response of the LC filter and load. Adding a load to the output will tend to stabilize the filter. Series resistance in the inductor and capacitor will also contribute to stability (a little bit).

So your circuit can be understood mainly by looking at the theory around LCR circuits, including damping factor.

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