0
\$\begingroup\$

To my understanding the reactive components in a converter should not store energy over a switching period, otherwise they will blow up. An in inductor for example abides by volt-second balance, which translates to zero average voltage over a switching period- no flux linkage/ no average energy stored. That being said, the inductor current waveform will be switching at rate Fs ( switching frequency) and will introduce a ripple on the current waveform. Now if we consider the transient time of buck converter. There will be no energy stored in the capacitor by the inductor at t=0. My question is how do we calculate the transient time of the buck converter? And what is the circuit analysis reasoning were the inductor current increases and settles at a DC offset? (Why doesn’t the waveform ripple go back to zero at period [DTs - Ts] - how does the capacitor hold the charge in period [DTs-Ts] and not discharge through the load connected on the output terminals

inductor current waveform

\$\endgroup\$

2 Answers 2

0
\$\begingroup\$

This is an inductor current graph, not an output capacitor voltage graph, so it's difficult to tell from it whether the output capacitor discharges through the load (it does).

The current between \$DT_s\$ and \$T_s\$ is not constant, but the derivative of the current is proportional to the voltage difference between the inductor terminals, which during "off" time is the voltage of the output capacitor at this point (not much) plus the forward voltage of the diode (also not much). Compared to the voltage during "on" time, the change in current isn't as pronounced, so it appears flat in this graph (but the \$i_L(T_s)\$ marker is applies only to the time \$T_s\$, not to the interval).

\$\endgroup\$
0
\$\begingroup\$

The output LC filter has a cutoff frequency, fc, much lower than the switching frequency, fs, and the transient response is the same as a regular RLC lowpass. Therefore what you see there is the transient of the RLC lowpass superimposed with the switching ripple due to fs. Here is a quick test:

buck vs RLC, transient

Below is an idealized buck stage, with fs @100 kHz. The LC filter has an fp of ~2.25 kHz. It's 40 times lower, so fc << fp. Above is the same RLC output, but with a 5 V source. As you can see, the two outputs, V(o) and V(o2), are almost overlapping, their only differences being the extra parasitics of the switches and the fs ripple, itself. The current through the inductor maintains the same transient, with the additional switching ripple -- which is exactly what you see in the OP. Note: this relation only holds as long as there is CCM involved or, at the very least, BCM (borderline, or critical conduction mode).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.