Inverters Questions and Answers

Answer: D

A voltage source inverter needs switches that can be turned on and off under control: SCRs with forced commutation circuits, GTOs, power transistors, MOSFETs and IGBTs can all serve as the switching device.

A DIAC is a bidirectional trigger diode with no gate terminal; it breaks over at a fixed voltage and is used to trigger a TRIAC, so it cannot be controlled as a switching device in an inverter. The answer is DIAC, option D.

Answer: D

In sinusoidal pulse width modulation the gating signals are produced by comparing a sinusoidal reference (modulating) wave, whose frequency sets the output frequency of the inverter, with a high frequency triangular carrier wave. When the sine is above the triangle the switch is on, and the widths of the pulses vary sinusoidally, which puts the lower harmonics at higher frequencies where they are easy to filter.

So the carrier is triangular and the reference is sinusoidal, the order given in option D. A square wave is the reference in a different scheme, and the options that make the carrier sinusoidal or square reverse the roles. The answer is option D.

Answer: B

In single pulse width modulation of an inverter, one pulse of width \(2d\) is placed symmetrically in each half cycle. The Fourier series of the output has the harmonic amplitudes proportional to \(\sin(n d)\), so the \(n\)th harmonic is eliminated when \(\sin(nd) = 0\), that is when the pulse width equals \(360^\circ / n\) (or a multiple of it).

\[ 2d = 72^\circ = \frac{360^\circ}{5} \]

So the fifth harmonic is eliminated. For the third harmonic the width would be \(120^\circ\), for the seventh about \(51.4^\circ\) and for the eleventh about \(32.7^\circ\). The answer is the 5th, option B.

Answer: C

The output of a single phase full bridge (full wave) voltage source inverter is a square wave of amplitude \(\pm V_{dc}\). Its Fourier series has the fundamental component with peak value \(4V_{dc}/\pi\), so the RMS value of the fundamental is

\[ \begin{aligned} V_{1} = \frac{4V_{dc}}{\pi\sqrt{2}} = 0.9\,V_{dc} \\ &= 0.9 \times 100 = 90\ \text{V} \end{aligned} \]

The value 100 V is the RMS value of the whole square wave (it includes the harmonics), not the fundamental. The load resistance is not needed for the voltage. The answer is 90 V, option C.

Answer: A

In a three phase bridge inverter the maximum value of the phase voltage depends on the conduction mode. In the \(180^\circ\) conduction mode the phase voltage is a six step wave with levels of \(\pm V_{dc}/3\) and \(\pm 2V_{dc}/3\), so its maximum is \(2V_{dc}/3\). In the \(120^\circ\) mode, the phase voltage takes the values of \(0\) and \(\pm V_{dc}/2\), so its maximum is \(V_{dc}/2\).

\[ \frac{(2/3)V_{dc}}{(1/2)V_{dc}} = \frac{4}{3} \]

The input voltage of 400 V cancels out in the ratio. The answer is 4/3, option A.

Answer: C

For uniform pulse width modulation with \(p\) pulses per half cycle, each of width \(d\) (in radians), the RMS output of a full bridge inverter is

\[ V_{rms} = V_{dc}\sqrt{\frac{p\,d}{\pi}} \]

Here \(p = 5\), \(d = 32^\circ = 0.5585\) rad and \(V_{dc} = 400\) V.

\[ \begin{aligned} V_{rms} = 400\sqrt{\frac{5 \times 0.5585}{\pi}} \\ &= 400\sqrt{0.8889} = 400 \times 0.9428 \\ &= 377.1\ \text{V} \end{aligned} \]

The output must be smaller than the 400 V input but not far below it, since 5 pulses of 32 degrees cover 160 of the 180 degrees of the half cycle. That excludes the values of 37.71 V, 3.771 V and 0.3771 V. The answer is 377.1 V, option C.

Answer: A

The load power for a resistive load depends on the square of the RMS voltage, and for uniform pulse width modulation \(V_{rms}^2 = V_{dc}^2\,p\,d/\pi\). To keep the same power with the same number of pulses, \(V_{dc}^2 d\) must be constant.

\[ V_{dc}' = 0.8 \times 600 = 480\ \text{V} \]

\[ \begin{aligned} 600^2 \times 25^\circ \\ &= 480^2 \times d' \Rightarrow d' \\ &= 25^\circ \times \left(\frac{600}{480}\right)^2 \\ &= 25 \times 1.5625 = 39.06^\circ \end{aligned} \]

So the pulse width must be increased to about \(39^\circ\) to make up for the lower supply. The smaller angles of 35, 32 and 30 degrees would reduce the output further. The answer is option A.

Answer: A

The output of a single phase full bridge inverter with a DC input \(V_{dc}\) is a square wave that alternates between \(+V_{dc}\) and \(-V_{dc}\). The square of this wave is constant, \(V_{dc}^2\), so the RMS output voltage is

\[ \begin{aligned} V_{rms} = \sqrt{\frac{1}{T}\int_0^T v^2\,dt} \\ &= V_{dc} = 220\ \text{V} \end{aligned} \]

A half bridge inverter would give \(V_{dc}/2 = 110\) V, which is the other value in the list, and 100 V and \(200/\sqrt{2}\) V have no basis in the data. The answer is 220 V, option A.

Answer: B

In the 180 degree conduction mode of a three phase bridge inverter, the line voltage is a quasi square wave with the amplitude \(V_{dc}\) and its fundamental component has the RMS value

\[ V_{L1} = \frac{\sqrt{6}}{\pi}V_{dc} = 0.7797\,V_{dc} \]

\[ V_{L1} = 0.7797 \times 400 = 311.88\ \text{V} \]

The RMS value of the whole line voltage (with harmonics) would be \(0.8165 V_{dc} = 326.6\) V, and the rest of the options are factors of 10 away from 311.88 V. The answer is option B.

Answer: A

The input signal to the inverter circuit is a constant DC voltage. An inverter converts direct current into alternating current, so it is supplied from a DC source such as a battery or a rectified supply, and its switches then produce an AC output.

A pulsating DC voltage and AC voltages of low or high frequency are not what an inverter takes at its input, because the AC would be the output. So the first option is correct.