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: Use the standard result for a linear ramp.
For a current that ramps linearly and symmetrically between $-I_{pk}$ and $+I_{pk}$ over a time span $T_{on}$, and is exactly the ramp shape throughout that span, its mean-square value averaged over just that span is a known result:
\[ \frac{1}{T_{on}}\int_0^{T_{on}} i^2\,dt = \frac{I_{pk}^2}{3} \]
This comes from the fact that a linear ramp centered at zero has the same mean-square behavior as a uniformly distributed variable between $-I_{pk}$ and $I_{pk}$.

Step 2: Apply this to the $S_1$ ON-interval.
Here $I_{pk}=5$ A and the ramp lasts $T_{on}=10\ \mu$s (from $t=0$ to $t=10\ \mu$s, while $S_1$ conducts). So the mean square of $i_{S1}$ during just this interval is
\[ \frac{25}{3}\ \text{A}^2 \]

Step 3: Bring in the duty ratio.
Since $S_1$ is OFF (current exactly zero) for the remaining $10\ \mu$s of the $20\ \mu$s period, the true rms must be computed over the WHOLE period, not just the ON time. The overall mean square is the ON-interval mean square scaled by the fraction of the period during which $S_1$ conducts, that is the duty ratio $D=T_{on}/T=10/20=0.5$:
\[ I_{rms}^2 = D\times\frac{I_{pk}^2}{3} = 0.5\times\frac{25}{3} = \frac{25}{6} \]

Step 4: Take the square root.
\[ I_{rms} = \sqrt{\frac{25}{6}} = \frac{5}{\sqrt6} = \frac{5\sqrt6}{6}\approx2.04\text{ A} \]

Step 5: Sanity check against the extremes.
If $S_1$ conducted for the entire period without ever switching off, the rms would be the full $I_{pk}/\sqrt3\approx2.89$ A; since $S_1$ only conducts for half the period, the actual rms must be smaller than this, and indeed $2.04$ A is smaller than $2.89$ A, which confirms the answer is reasonable.
\[ \boxed{I_{rms}\approx2.04\text{ A}} \]