Power Electronics Test 2

Output is a square wave with an amplitude of 100V.

RMS value of a square wave is equal to its peak value.

So, rms value of output voltage is

Input voltage (V_s) = 36 V

Resistive load (R) = 18 Ω

Frequency (f) = 200 Hz

Time period (T) \( = \frac{1}{f} = \frac{1}{{200}} = 5\;ms\)

Duty ratio \( = \frac{{{T_{ON}}}}{T} = \frac{{1.25}}{5} = 0.25\) 

For buck converter:

V_o = δV_s = 0.25 × 36 = 9 V

\({I_o} = \frac{9}{{18}} = 0.5A\) 

Load power = 9 × 0.5 = 4.5 W

For boost converter:

\({V_o} = \frac{{{V_s}}}{{\left( {1 - \delta } \right)}} = \frac{{36}}{{0.75}} = 48V\) 

\({I_o} = \frac{{48}}{{18}} = 2.67A\) 

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

RMS value of the fundamental component

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

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

Fundamental rms component of output voltage

= 60% of input DC

= 0.6 × 100 = 60 V

\( \Rightarrow 70 = \frac{{2\sqrt 2 }}{\pi } \times 100 \times \sin d\) 

⇒ d = 51.03°

Pulse width = 2d = 102°

For a three-phase semi converter, the average dc output voltage is

\({V_{dc}} = \frac{{3\sqrt 3 {V_m}}}{{2\pi }}\left( {1 + \cos \alpha } \right)\) 

\({I_{dc}} = \frac{{{V_{dc}}}}{R} = 15\) 

⇒ V_dc = 15 × 10 = 150 V

\( \Rightarrow 150 = \frac{{3\sqrt 3 \times \frac{{400 \times \sqrt 2 }}{{\sqrt 3 }}}}{{2\pi }}\left( {1 + \cos \alpha } \right)\) 

The power factor at the ac source is

\(pf = \sqrt {\frac{1}{{2\pi }}\left( {\pi \alpha + \frac{1}{2}\sin 2\alpha } \right)} \) 

\( = \sqrt {\frac{1}{{2\pi }}\left( {\pi - \frac{\pi }{3} + \frac{1}{2}\sin 120^\circ } \right)} \) 

The fundamental output voltage for a modulation index of M is

A CSl can be either load commutated or force commutated.

Concept:

For step-down chopper

Average output V₀ = α V_S

Where α = duty cycle

\(\alpha = \frac{{{T_{On}}}}{{{T_{on}} + {T_{off}}}}\)

Time period \(T = {T_{on}} + {T_{off}} = \frac{1}{f}\)

Calculation:

Given

V₀ = 140 V

V_S = 200 V

T_on = 70 μsec

\(\alpha = \frac{{{V_0}}}{{{V_S}}} = \frac{{140}}{{200}} = 0.7\)

\(\frac{{{T_{on}}}}{{{T_{on}} + {T_{off}}}} = 0.7\)

0.3 T_on = 0.7 T_off

⇒ 0.3 × 70 = 0.7 T_off

T_off = 30 μ sec

Time period T = T_on + T_off = 70 + 30

= 100 μ sec

Frequency \(f = \frac{1}{T} = \frac{1}{{100}} \times {10^6}\)

= 10 kHz

Now frequency is increased by 25%

Hence f¹ = 1.25 × 10

= 12.5 kHz

New time period \({T^1} = \frac{1}{{{f^1}}} = \frac{1}{{12.5}} \times {10^3}\)

= 80 μ sec

Conduction Time is constant

T_on = 70 μ sec

New duty cycle

\(\alpha ' = \frac{{{T_{on}}}}{{{T^1}}} = \frac{{70}}{{80}}\)

New Average output voltage

V₀ = α‘V_S

\( = \frac{{70}}{{80}} \times 200\)

= 175 V

Concept:

The Fourier analysis of phase voltage

\({{\rm{V}}_{a0}} = \mathop \sum \limits_{n = 6K \pm 1}^\infty \frac{{2{V_s}}}{{n\pi }}\sin n\omega t\) 

Where K = 0, 1, 2…

∴ Fundamental component of phase voltage

\({{\rm{V}}_{a1}} = \frac{{\sqrt 2 {V_s}}}{\pi }\) 

Fundamentals component of line voltage

\({{\rm{V}}_{L1}} = \sqrt 3 {V_{a1}}\)

Calculation:

\(\begin{array}{l}{{\rm{V}}_{a1}} = \frac{{\sqrt {2\;} \times 220}}{\pi } = 99.035\;V\\{{\rm{V}}_{L1}} = \sqrt 3 \times 99.035 = 171.534\;V\end{array}\)

Resistance firing circuit:

Diagram

The voltage drop across R is V_gk and

I₁R = V_gk = I_gmax R

⇒ I_gmax = I₁

I = I₁ + I_gmax = 2 I_gmax

By applying KVL,

V = IR₁ + V_D + V_gk

\( \Rightarrow {R_1} = \frac{{V - {V_D} - {V_{gk}}}}{I}\)

Calculation:

Given that, V = 100, V_D = 1 V, V_gk = 1 V

I = 2 I_gmax = 2 × 50 = 100 mA

Resistive load = (supply voltage – drive transistor drop – gate cathode drop)/100mA

R₁ = (10 – 1 – 1)/100mA

⇒ = 80 Ω

Concept:

In a class – A chopper,

Average load voltage V_o = D V_dc

Minimum value of load current \(={{I}_{o}}-\frac{\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}}{2}\)

Maximum value of load current \(={{I}_{o}}+\frac{\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}}{2}\)

Average load current \({{I}_{o}}=\frac{{{V}_{o}}}{R}\)

\(\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}=\frac{{{V}_{dc}}}{L}D~\left( 1-D \right)T\)

Calculation:

Given that, Source voltage (V_dc) = 100 V

R = 0.5 Ω

L = 1 mH

T_ON = 1 ms

T = 3 ms

\(D=\frac{{{T}_{On}}}{T}=\frac{1}{3}\)

\({{V}_{o}}=100\times \frac{1}{3}=33.33~V\)

\({{I}_{o}}=\frac{{{V}_{o}}}{R}=\frac{33.33}{0.5}=66.66~A\)

\(\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}=\frac{100}{1\times {{10}^{-3}}}\times \frac{1}{3}\times \left( 1-\frac{1}{3} \right)\times 3\times {{10}^{-3}}\)

= 66.66 A

Minimum value of load current \(=66.66-\frac{66.66}{2}=33.33~A\)

Voltage control of an inverter

The waveform of the output voltage obtained from a single-phase inverter is rectangular in nature with an amplitude approximately equal to the input dc voltage.

However, in many applications, the output voltage of the inverter needs to be controlled due to the following reasons:

• The voltage required by ac loads may be constant or adjustable. 
• In motor control applications, inverters handle the control of circuit voltage along with frequency so that the saturation of motor magnetic circuits is avoided.
• Voltage control of inverters is employed in order to compensate for changes in input dc voltage.

Basically, there are three techniques by which the voltage can be controlled in an inverter. They are:

1. External Control of AC Output Voltage

• This method is also known as the series-inverter control method.
• In this method of control, an ac voltage controller is connected to the output of the inverter to obtain the required (controlled) output ac voltage. 
• The voltage control is primarily achieved by varying the firing angle of the ac voltage controller that feeds the ac load.

2. External Control of DC Input Voltage

• The external control of dc input voltage is a technique that is adapted to control the dc voltage at the input side of the inverter itself to get the desired ac output voltage at the load side.

3. Internal Control of Inverter

• The output voltage of an inverter can be adjusted by employing the control technique within the inverter itself.
• This control technique can be accomplished by the following two control methods: Series Inverter Control, and Pulse Width Modulation Control.

In case of singte-putse width modulation (PWM), the width of the pulse is adjusted to reduce the harmonic. However, a single phase IGBT bridge inverter produces a square wave. This square wave contains  harmonic, 20% 5^th harmonic and  harmonic.

The rms value of amplitude of harmonic voltage of a single, pulse modulated wave is given by

The output voltage wave form

For mid-point rectifier,

\({V_{dc}} = \frac{{2{V_m}}}{\pi }\cos \alpha - \frac{{\omega L{I_{dc}}}}{\pi }\) 

And V_dc = I_dc R

\( \Rightarrow R{I_{dc}} = \frac{{2{V_m}}}{\pi }\cos \alpha - \frac{{\omega L}}{\pi }{I_{dc}}\) 

\({I_{dc}}\left[ {R + \frac{{\omega L}}{\pi }} \right] = \frac{{2{V_m}}}{\pi }\cos \alpha \) 

\( \Rightarrow {I_{dc}} = \frac{{\frac{{2{V_m}}}{\pi }\cos \alpha }}{{\left[ {R + \frac{{\omega L}}{\pi }} \right]}}\) 

\( = \frac{{\frac{{2 \times 120 \times \sqrt 2 }}{\pi }}}{{15 + \frac{{100\pi \times 20}}{{1000\pi }}}}\) 

= 3.17 A

V_dc = 15 × 3.17 = 47.6 V

\(\cos \alpha - \cos \left( {\alpha + \mu } \right) = \frac{{\omega L}}{{{V_m}}}{I_{dc}}\) 

\( \Rightarrow \cos \left( {\alpha + \mu } \right) = \cos 60^\circ - \frac{{2\pi \times 50 \times 20 \times {{10}^{ - 3}}}}{{169.7}} \times 3.17\) 

⇒ α + μ = 67.5°

⇒ μ = 7.5°

In a 1-φ full bridge inverter if RLC load is underdamped, then the two thyristors (namely T₁ and T₂) shown in figure will get commutated naturally and therefore no commutation circuitry will be needed. Thus, load commutation will be possible.

The average voltage across the inductor for a complete cycle is zero and also the power dissipated across inductor per cycle is zero So,