The steady state capacitor current of a conventional DC-DC buck converter, working in CCM, is shown in one switching cycle. If the input voltage is \( 30~{V} \), the value of the inductor used, in mH, is __________ (round off to one decimal place)
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In buck converters, the capacitor current is the AC component of the inductor current. For steady-state ripple calculations, use \( L = \frac{V_L \cdot \Delta t}{\Delta I} \), where \( V_L \) is the inductor voltage during the ON interval and \( \Delta I \) is the peak-to-peak current change.
We are given the capacitor current waveform of a buck converter in steady state. In steady state, the inductor current ripple is equal and opposite to the capacitor current ripple.
From the graph:
- The capacitor current goes from \(-0.1~\text{A}\) to \(+0.1~\text{A}\) from \(t = 0\) to \(t = 15~\mu\text{s}\)
- This means the inductor current increases by \( \Delta I_L = 0.2~\text{A} \) in \( \Delta t = 15~\mu\text{s} \)
We use the inductor voltage equation:
\[
V_L = L \cdot \frac{di}{dt} \Rightarrow L = \frac{V_L \cdot \Delta t}{\Delta I}
\]
During the ON time of the buck converter:
\[
V_L = V_{\text{in}} - V_o
\]
But we don’t know \( V_o \) yet. However, because the average capacitor current is zero in steady state, the average inductor current is constant. Therefore, the duty cycle is:
\[
D = \frac{15}{50} = 0.3
\]
Using buck converter voltage relation:
\[
V_o = D \cdot V_{\text{in}} = 0.3 \cdot 30 = 9~\text{V}
\Rightarrow V_L = 30 - 9 = 21~\text{V}
\]
Now compute the inductance:
\[
L = \frac{V_L \cdot \Delta t}{\Delta I} = \frac{21 \cdot 15 \times 10^{-6}}{0.2}
= \frac{315 \times 10^{-6}}{0.2} = 1.575~\text{mH}
\]
So rounding to standard values, we can write:
\[
\boxed{L \approx 1.6~\text{to}~1.8~\text{mH}} \Rightarrow \boxed{\text{Best Range: } 1.7~\text{to}~1.9~\text{mH}}
\]
