This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “PWM Inverters-1”.

1. In the single-pulse width modulation method, the output voltage waveform is symmetrical about __________

a) π

b) 2π

c) π/2

d) π/4

View Answer

Explanation: The waveform is a positive in the first half cycle and symmetrical about π/2 in the first half.

2. In the single-pulse width modulation method, the output voltage waveform is symmetrical about ____________ in the negative half cycle.

a) 2π

b) 3π/2

c) π/2

d) 3π/4

View Answer

Explanation: In the negative half the wave is symmetrical about 3π/2.

3. The shape of the output voltage waveform in a single PWM is

a) square wave

b) triangular wave

c) quasi-square wave

d) sine wave

View Answer

Explanation: Positive and the negative half cycles of the output voltage are symmetrical about π/2 and 3π/2 respectively. The shape of the waveform obtained is called as quasi-square wave.

4. In the single-pulse width modulation method, the Fourier coefficient b_{n} is given by

a) (Vs/π) [ sin(nπ/2) sin(nd) ].

b) 0

c) (4Vs/nπ) [sin(nπ/2) sin(nd)].

d) (2Vs/nπ) [sin(nπ/2) sin(nd)].

View Answer

Explanation: The Fourier analysis is as under:

b

_{n}= (2/π) ∫ Vs sin nωt .d(ωt) , Where the integration would run from (π/2 + d) to (π/2 – d)

2d is the width of the pulse.

5. In the single-pulse width modulation method, the Fourier coefficient a_{n} is given by

a) (Vs/π) [ cos(nπ/2) cos(nd) ].

b) 0

c) (4Vs/nπ) [sin(nπ/2) sin(nd)].

d) (2Vs/nπ) [sin(nπ/2) sin(nd)].

View Answer

Explanation: As the positive and the negative half cycles are identical the coefficient a

_{n}= 0.

6. In the single-pulse width modulation method, when the pulse width of 2d is equal to its maximum value of π radians, then the fundamental component of output voltage is given by

a) Vs

b) 4Vs/π

c) 0

d) 2Vs/π

View Answer

Explanation: The Fourier representation of the output voltage is given by

Put 2d = π & n = 1.

7. In case of a single-pulse width modulation with the pulse width = 2d, the peak value of the fundamental component of voltage is given by the expression

a) 4Vs/π

b) Vs

c) (4Vs/π) sin 2d

d) (4Vs/π) sin d

View Answer

Explanation: For the fundamental component put n = 1.

Vo = (4Vs/π) sin (d) sin (ωt)

Hence the peak value is (4Vs/π) sin d.

8. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the nth harmonic from the output voltage

a) d = π

b) 2d = π

c) nd = π

d) nd = 2π

View Answer

Explanation: To eliminate, the nth harmonic, nd is made equal to π radians, or d = π/n.

From the below expression,

when nd = π. sin nd = 0 hence, that output voltage harmonic is eliminated.

9. Find the peak value of the fundamental component of voltage with a pulse width of 2d = 90 and Vs = 240 V for single-pulse modulation in a full wave bridge inverter.

a) 305 V

b) 216 V

c) 0 V

d) 610 V

View Answer

Explanation: The peak value of the fundamental component of voltage is given by (4Vs/π) sin d.

10. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the 3rd harmonic from the output voltage waveform, the value of the pulse width (2d) must be

a) 0°

b) 60°

c) 120°

d) 180°

View Answer

Explanation: To eliminate the nth harmonic, nd = π.

Therefore, d = π/n = π/3 = 60°

Hence, 2d = 120°.

**Sanfoundry Global Education & Learning Series – Power Electronics.**

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