Question

Consider the single-phase voltage source inverter circuit feeding an inductive load (\(L\)). Assume that the power MOSFET switches are ideal. \(S_1\) and \(S_2\) are switched on during the first \(10\ \mu\)s, and \(S_3\) and \(S_4\) are switched on during the next \(10\ \mu\)s in a switching cycle. The switches in the same leg are thus switched in a complementary fashion. Neglect the dead time. The waveform of the inductor current (\(i_L\)) in the steady state is triangular with a peak value of \(5\) A as shown.

The rms value of the current through the switch \(S_1\) is:

• A. 2.88 A
• B. 2.04 A
• C. 3.54 A
• D. 2.50 A

Hint:

\(S_1\) carries the inductor current only during the first 10 microseconds of each 20 microsecond cycle (and zero for the rest); find the rms of this waveform over the full period.

Answer 1

The Correct Option is B

Step 1: Read the inductor current waveform.
From the graph, \(i_L\) is a symmetric triangular wave of period \(20\ \mu\)s: it rises linearly from \(-5\) A at \(t=0\) to \(+5\) A at \(t=10\ \mu\)s (while \(S_1,S_2\) are ON), then falls linearly from \(+5\) A back to \(-5\) A over the next \(10\ \mu\)s, from \(t=10\ \mu\)s to \(t=20\ \mu\)s (while \(S_3,S_4\) are ON), and this pattern repeats.

Step 2: Identify when \(S_1\) actually carries current.
\(S_1\) is turned ON only during the first \(10\ \mu\)s of every \(20\ \mu\)s cycle. When \(S_1\) and \(S_2\) are ON (and \(S_3,S_4\) are OFF), the entire load current path runs through \(S_1\) and \(S_2\), so the current through switch \(S_1\) equals the inductor current itself during this interval, whatever its sign. During the next \(10\ \mu\)s, \(S_1\) is OFF and carries no current at all.

Step 3: Write \(i_{S1}(t)\) over one period.
\[ i_{S1}(t) = \begin{cases} i_L(t) = -5+t, & 0\le t<10\ \mu\text{s (}t\text{ in }\mu\text{s)} \\ 0, & 10\ \mu\text{s}\le t<20\ \mu\text{s} \end{cases} \]
(Here \(i_L(t)=-5+t\) is just the equation of the straight line joining \((0,-5)\) to \((10,5)\).)

Step 4: Set up the rms integral over the full period.
\[ I_{rms}^2 = \frac{1}{T}\int_0^T i_{S1}^2(t)\,dt = \frac{1}{20}\left[\int_0^{10}(t-5)^2\,dt + \int_{10}^{20}0\,dt\right] \]

Step 5: Evaluate the integral.
Substitute \(u=t-5\), so as \(t\) runs from \(0\) to \(10\), \(u\) runs from \(-5\) to \(5\):
\[ \int_0^{10}(t-5)^2\,dt = \int_{-5}^{5}u^2\,du = \left[\frac{u^3}{3}\right]_{-5}^{5} = \frac{125}{3}-\left(\frac{-125}{3}\right) = \frac{250}{3} \]

Step 6: Substitute back.
\[ I_{rms}^2 = \frac{1}{20}\times\frac{250}{3} = \frac{250}{60} = \frac{25}{6} \]
\[ I_{rms} = \sqrt{\frac{25}{6}} = \frac{5}{\sqrt{6}} \approx 2.041\text{ A} \]

Step 7: Rule out the other options.
\(3.54\) A does not correspond to any consistent calculation for this waveform. \(2.88\) A is close to the rms of the ramp computed only over its own \(10\ \mu\)s ON-time (which would be \(5/\sqrt3\approx2.89\)), without accounting for the fact that \(S_1\) is off (contributing zero) for the other half of the period. \(2.50\) A also does not follow from this waveform. Only accounting for both the shape of the ramp and the \(50\%\) duty cycle over the full period gives the correct value.

Final Answer:
\[ \boxed{I_{rms} = \frac{5}{\sqrt{6}}\approx2.04\text{ A}} \]