Single Phase Voltage Source Inverters MCQ Quiz - Objective Question with Answer for Single Phase Voltage Source Inverters - Download Free PDF

Explanation:

Single-Phase Full Bridge Inverter:

Definition: A single-phase full bridge inverter is an electronic device that converts direct current (DC) into alternating current (AC). It consists of four switches that control the direction of the current flow through the load, allowing the output voltage to alternate its polarity. The switches are typically implemented using transistors or thyristors.

Working Principle: The operation of a full bridge inverter involves turning on pairs of switches in a specific sequence to generate a square wave AC output. The switches are arranged in an H-bridge configuration, where each leg of the bridge has two switches. By controlling which switches are on and off, the inverter can produce positive, negative, or zero voltage across the load.

Switching States: The four switches in the inverter are denoted as S1, S2, S3, and S4. The output voltage across the load depends on the combination of switches that are turned on. The possible states are:

• S1 and S2 ON: Positive voltage (+V_dc) across the load.
• S3 and S4 ON: Negative voltage (-V_dc) across the load.
• S1 and S3 ON: Zero voltage (0 V) across the load (both ends connected to +V_dc).
• S2 and S4 ON: Zero voltage (0 V) across the load (both ends connected to ground).

Correct Option Analysis:

The correct option is:

Option 4: -V_dc

When switches S2 and S3 are turned on, the output voltage across the load is determined by the connections of these switches. Here's the detailed analysis:

• S2 ON: This switch connects the positive terminal of the DC supply to the left side of the load.
• S3 ON: This switch connects the right side of the load to the negative terminal of the DC supply.

With S2 and S3 turned on, the current flows from the positive terminal of the DC supply through S2, then through the load from left to right, and finally through S3 to the negative terminal of the DC supply. This results in a negative voltage (-V_dc) across the load, as the positive terminal of the load is connected to the negative terminal of the supply and vice versa.

Conclusion:

The correct output voltage when switches S2 and S3 are turned on is -V_dc, making option 4 the correct answer.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: -2V_dc

This option is incorrect because the output voltage of a single-phase full bridge inverter is either +V_dc, -V_dc, or 0 V. The inverter cannot produce an output voltage of -2V_dc as it is beyond the voltage range of the DC supply.

Option 2: V_dc

This option is incorrect in the context of switches S2 and S3 being turned on. The output voltage V_dc is produced when switches S1 and S2 are turned on, allowing current to flow from the positive terminal of the supply through the load to the negative terminal.

Option 3: 2V_dc

This option is incorrect because, similar to option 1, the inverter cannot produce an output voltage of 2V_dc. The maximum output voltage is limited to the magnitude of the DC supply voltage.

Option 4: -V_dc

This option is correct, as explained in the detailed analysis above. When switches S2 and S3 are turned on, the output voltage across the load is -V_dc.

Conclusion:

Understanding the switching states and the resulting output voltages of a single-phase full bridge inverter is essential for analyzing its operation. The correct output voltage when switches S2 and S3 are turned on is -V_dc, as this configuration connects the positive terminal of the load to the negative terminal of the DC supply and vice versa.

The output of a single-phase half bridge inverter on an R load is ideally a square wave. This is because the inverter switches the DC input voltage between two levels, resulting in a square wave output.

In a Voltage Source Inverter (VSI), the internal impedance of the DC source is negligible. This is because the VSI is designed to provide a stable and consistent voltage output, which requires a DC source with very low internal impedance to ensure that the voltage does not drop under load.

A voltage source inverter (VSI) is employed when source inductance is small and load inductance is large because higher value of source inductance will increase the overlap angle and cause commutation issues.

Important Points:

Voltage Source Inverter	Current Source Inverter
It is fed from a DC voltage source having small impedance	It is fed with adjustable current from a DC voltage source of high impedance
Input voltage is maintained constant	Input current is constant but adjustable
Output voltage does not dependent on the load	The amplitude of output current is independent of the load
The waveform of the load current as well as its magnitude depends upon the nature of load impedance	The magnitude of output voltage and its waveform depends upon the nature of the load impedance
It requires feedback diodes	It does not require any feedback diodes
The commutation circuit is complicated	Commutation circuit is simple as it contains only capacitors.
Power BJT, Power MOSFET, IGBT, GTO with self-commutation can be used in the circuit.	They cannot be used as these devices have to withstand reverse voltage.

1ϕ half-bridge inverter:

Case 1: 0 < t < T/2:

Thyristor T₁ conducts.

 \(V_o = {{V} \over 2}\)

Case 2:  T/2 < t < T:

Thyristor T₂ conducts.

 \(V_o = -{{V} \over 2}\)

The RMS output voltage is given by:

\((V_o)_{RMS} = {V \over 2}\)

Given, V = 100

\((V_o)_{RMS} = {100 \over 2}\)

\((V_o)_{RMS} = 50\space V\)

Total harmonic distortion

The presence of AC harmonic components in DC output is known as total harmonic distortion.

The output current is the combination of fundamental (DC) + AC harmonics components.

\(I_{or}=I_{or1}+I_{or2}+I_{or3}........\)

\(I_{or}-I_{or1}=I_{or2}+I_{or3}........\)

where, I_or1 = RMS value of the fundamental component

\(I_{or2}+I_{or3}........\)= RMS value of net harmonic current

\(THD=\sqrt{({1\over CDF})^2-1}\)

\(THD=\sqrt{({I_{or}\over I_{or1}})^2-1}\)

\(THD=\sqrt{I_{or}^2-I_{or1}^2\over I_{or1}^2}\)

\(THD=\sqrt{({I_{or2}+I_{or3}........\over I_{or1}})^2}\)

\(THD={I_{or2}+I_{or3}........\over I_{or1}}\)

\({I_{or2}+I_{or3}........}=THD\times I_{or1}\)

\({I_{or2}+I_{or3}........}= 0.04 \times \frac{4}{{\sqrt 2 }} = 0.08\sqrt 2 \;A\)

1ϕ half-bridge inverter:

Case 1: 0 < t < T/2:

Thyristor T₁ conducts.

 \(V_o = {{V} \over 2}\)

Case 2:  T/2 < t < T:

Thyristor T₂ conducts.

 \(V_o = -{{V} \over 2}\)

The RMS output voltage is given by:

\((V_o)_{RMS} = {V \over 2}\)

Given, V = 100

\((V_o)_{RMS} = {100 \over 2}\)

\((V_o)_{RMS} = 50\space V\)

Power drawn, \({P_d} = {V_{01}}{I_{01}}\cos \phi\)

\({V_{01}} = \frac{{4{V_s}}}{{\pi \sqrt 2 }}\)

\({I_{01}} = \frac{{200}}{{\sqrt 2 }}mA\)

\({P_d} = \frac{{4 \times 220}}{{\pi \times \sqrt 2 }} \times \frac{{200 \times {{10}^{ - 3}}}}{{\sqrt 2 }}\cos 45^\circ = 19.8\;W\)

For the given Quasi square waveform,

\({V_n} = \frac{{4{V_s}}}{{n\pi }}{\rm{cos}}\left( {n\sigma } \right)\)

Given that, the magnitude of 3^rd harmonic component from the output voltage is zero i.e. V₃ = 0

\({V_3} = \frac{{4{V_s}}}{{3\pi }}\cos n\sigma = 0\)

⇒ cos (3σ) = 0

⇒ σ = π/6

The magnitude of the fifth harmonic component from the output voltage is,

\({V_5} = \frac{{4{V_s}}}{{5\pi }}\cos \frac{{5\pi }}{6}\)

The magnitude of the fundamental harmonic component from the output voltage is,

\({V_1} = \frac{{4{V_s}}}{\pi }\cos \frac{\pi }{6}\)

\(\frac{{{V_5}}}{{{V_1}}} = \frac{{\frac{{4{V_s}}}{{5\pi }}\cos \frac{{5\pi }}{6}}}{{\frac{{4{V_s}}}{\pi }\cos \frac{\pi }{6}}} = - 0.2\)

Now, the value of the magnitude of the 5th harmonic component as a percentage of the magnitude of the fundamental component is

\(\% \left| {\frac{{{V_5}}}{{{V_1}}}} \right| = 20\%\)

 

Concept:

From sinusoidal PWM control technique:

The peak value of fundamental voltage is given as,

Vm1 = MVdc    ---(1)

Explanation:

Given that,

RL load: R = 40 Ω, \(L = \frac{0.3}{π} \) H

f = 50 Hz

P = 1.44 kW

Modulation index (M) = 0.6

At 50 Hz,

XL = 2πfL = 100 × \(\frac{0.3}{\pi} \) = 30 Ω

Z = R + jXL = 40 + j30

|Z| = 50 Ω

∵ P = 1.44 × 103

⇒ \(\rm I_{rms}^2 R = 1440\)

\(⇒ \rm I_{rms}^2 R = \frac{1440}{40}\)

⇒ Irms = 6 A

we know that,

\(\rm Z = \frac{V_{rms}}{I_{rms}}\)

⇒ Vrms = Irms × Z

⇒ Vrms = 6 × 50

⇒ Vrms = 300 V

so, peak value Vm = 300√2

From equation (1)

300√2 = 0.6 VDC

⇒ VDC = \(\rm \frac{300√ 2}{0.6}\)

∴ VDC = 500√2 V

Concept:

Parameter	180° conduction mode	120° conduction mode
Phase voltage (Vph)	\(\frac{{\sqrt 2 }}{3}{V_s}\)	\(\frac{{{V_s}}}{{\sqrt 6 }}\)
Line voltage (VL)	\(\sqrt {\frac{2}{3}} {V_s}\)	\(\frac{{{V_s}}}{{\sqrt 2 }}\)
RMS load current (Ior)	\(\frac{{\sqrt 2 }}{{3R}}{V_s}\)	\(\frac{{{V_s}}}{{\sqrt 6 R}}\)
RMS thyristor current (ITr)	\(\frac{{{V_s}}}{{3R}}\)	\(\frac{{{V_s}}}{{2\sqrt 3 R}}\)

 

Calculation:

Given that, V_oc = 600 V

R_L = 20 Ω/ph

RMS value of phase voltage (V_ph) =\(\frac{{{V_s}}}{{\sqrt 6 }}\) = 0.4082 V_dc

 Vph = 0.4082 × 600

Vph = 244.92 V

Load power \(= \frac{{3{\rm{V}}_{{\rm{ph}}}^2}}{{\rm{R}}} = \frac{{3 \times {{\left( {244.92} \right)}^2}}}{{20}} = 8.99{\rm{\;kW}}\)

Concept:

In single pulse modulation, the Fourier series expansion for the output voltage is given by,

\({V_0} = \mathop \sum \limits_{n = 1,3,5}^\infty \frac{{4{V_s}}}{{nπ }}\sin \frac{{nπ }}{2}\sin nd\sin n\omega t\)

The RMS value of the fundamental component \({V_{01}} = \frac{{2\sqrt 2 }}{π }{V_s}\sin d\)

The RMS value of the output voltage is, \({V_{or}} = {V_s}\sqrt {\frac{{2d}}{π }} \)

Where 2d is the pulse width

Vs is the dc input voltage.

Solution:

As in the question, it is not talked about the pulse width, then which means full-wave rectification is taking place

That means in positive half +Vs (0 to π) and in negative half -Vs (π to 2π)

So 2d = π 

The RMS value of the output voltage is, \({V_{or}} = {V_s}\sqrt {\frac{{2d}}{π }} = {V_s}\sqrt {\frac{{\pi}}{π }} = V_s=V_{DC}\)

Since, DC supply of 48 V, hence the rms value of output is 48 V

Concept:

The output voltage of a single-phase inverter is given by,

\({V_0}\left( {rms} \right) = \frac{{4{V_s}}}{{n\pi }} \times \frac{1}{{\sqrt 2 }}\)

Fundamental component of output voltage is \({V_{01}} = \frac{{4{V_s}}}{\pi } \times \frac{1}{{\sqrt 2 }}\)

Calculation:

Given that, input dc voltage (V_S) = 48 V

Fundamental component of output voltage is \({V_{01}} = \frac{{4 \times 48}}{\pi } \times \frac{1}{{\sqrt 2 }} = 43.2\;V\)

Concept:

The output voltage of a single-phase full-bridge voltage source inverter is controlled by unipolar PWM with one pulse per half cycle.

\({V_o} = \mathop \sum \limits_{n = 1,3 \ldots }^\infty \left\{ {\frac{{4{V_s}}}{{n\pi }}\sin \frac{{n\pi }}{2}\sin nd} \right\}\sin n\omega t\;\)

RMS value of the fundamental component

\({V_{01}} = \left\{ {\frac{{4{V_s}}}{\pi }\sin \frac{\pi }{2}\sin d} \right\}\frac{1}{{\sqrt 2 }}\)

\(\Rightarrow {V_{01}} = \frac{{2\sqrt 2 {V_s}}}{\pi }\sin d\)

Calculation:

Given that, V₀₁ = 75% of V_s

\(\Rightarrow 0.75\;{V_s} = \frac{{2\sqrt 2 {V_s}}}{\pi }\sin d\)

⇒ d = 56.41°

pulse width = 2d = 112.82°

DC power, P_dc = 5 kW

AC power, P_ac = V_s1 I_s1 cos θ

= 220 × I_s1 × 1 = 220 I_s1

DC power is equal to AC power.

220 I_s1 = 5000

⇒ I_s1 = 22.72 A

X_S = ωL = 2πfL = 2π × 50 × 5 × 10^-3

From the given phasor diagram,

\(\tan \delta = \frac{{{I_s}{X_s}}}{{{V_s}}}\)

\( = \frac{{22.72 \times 2\pi \times 50 \times 5 \times {{10}^{ - 3}}}}{{220}} = 0.162\)

⇒ δ = 9.2°