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I understand that the voltage across the inductor is constant so the current will be linear, but what about the capacitor in the circuit(Sw off state)? Will it not affect the current waveform? I think in DC transient response of an LC circuit we get sinusoidal waveforms, so why not here as well?
in DC transient response of an LC circuit we get sinusoidal waveforms, so why not here as well?
Short answer is "we do get a sinusoidal waveform" but it's only for a short period of time.
Quite literally, the boost converter switching speed is so much higher than the natural resonant frequency of the inductor (and output capacitor), that you never get an opportunity to actually see any sinusoidal shape emerging in the charging waveform.
So, if we look at the waveforms of an LC circuit that is based around the topology of a boost converter, we would see a sinusoidal shape emerging: -
In the above example, a 10 μH inductor is initially charged to 1 amp and an output capacitor (1 μF) is initially charged to 20 volts. The input DC voltage is 10 volts i.e. we have the relevant parts of a boost circuit but with an important feature missing; the output diode is shorted out.
The above scenario represents the 2nd phase of the switching cycle i.e. the MOSFET (also not shown) has charged the inductor with 1 amp and has then gone "open-circuit" hence, I've not bothered to show it because it plays no role in the 2nd phase.
The upper waveform (red) is the output voltage and, initially it starts at 20 volts and peaks a little bit higher due to the energy transfer from the inductor (1 amp). It's no coincidence that it peaks in time exactly when the inductor current passes through zero amps.
The lower waveform is the inductor current and, as you can see, it starts at 1 amp and falls as it transfers energy into the capacitor. In a boost converter, we would expect the inductor current to fall with the same trajectory as above but, refrain from further changes when it hits 0 amps. This of course is because the output diode prevents the current reversing.
So, if we put in the diode (S1 wired as an ideal diode) we would see this: -
The inductor current can be mistaken for a linear discharge to zero amps but, given the previous circuit, it is clearly part of a sinusoidal response.
This circuit behave as an Underdamped second order system if we consider voltage source as an input and inductor current as an output.
And we know that for an Underdamped second order circuit, initially output is proportional to input (you can verify by looking a graph of step response of 2nd order Underdamped system)
but after some time output start oscillating(see image) but due to very high switching frequency (less transient time) we don't reach at oscillating state and hence a linear approximation of input vs output is good.
If the output ripple is 'small', then the voltage across the inductor is approximately constant, and its current has a linear ramp. This linear-ramping current will generate a parabolic ripple on the capacitor (the net capacitor current is the inductors ramp minus the (constant) load current).
If the ripple is very large (equivalent to C being small, and switching frequency being similar to the L.C resonant frequency), the above approximations don't hold, and there will be segments of sinusoids on the capacitor.
Note that in real applications, the ESR and ESL of the capacitor also have an effect, adding a triangular (ESR) and square wave (ESL) component to the capacitor voltage.
