When thyristor is ON, the circuit at steady state is shown below.

At steady state, inductor acts as short circuit.

Initial current, i_L(0) = 0 A

Steady state current, i_L(∞) = 10/10 = 1 A

Time constant $\tau =\frac{L}{R}=\frac{5\times {10}^{-3}}{10}=0.5\times {10}^{-3}$

Equation of inductor current,

$i_L\left(t\right)=i\left(\mathrm{\infty }\right)+[i\left(0\right)-i\left(\mathrm{\infty }\right)]e^{-\frac{t}{\tau }}$

$\Rightarrow i_L\left(t\right)=1-e^{\frac{t}{0.5\times {10}^{-3}}}$

Given latching current = 10 mA

$\Rightarrow 10\times {10}^{-3}=1-e^{\frac{t}{0.5\times {10}^{-3}}}$

⇒ t = 5 μs

The voltage and current waveforms are shown below.

$V_{dc}=\frac{1}{2\pi }{\int }_{{\alpha }_1}^{{\alpha }_2}{V}_m\sin wtdt$ 

$=\frac{V_m}{2\pi }{\int }_{{\alpha }_1}^{{\alpha }_2}\sin wtdt$ 

$=\frac{V_m}{2\pi }{\left[-\cos wt\right]}_{{\alpha }_1}^{{\alpha }_2}$ 

$V_{dc}=\frac{1}{2\pi }\left[{\int }_{\alpha }^{\pi }{V}_m\sin \omega td\omega t+{\int }_{\pi }^{2\pi }-V_m\sin \omega td\omega t\right]$ 

$=\frac{V_m}{2\pi }\left[{\left[-\cos \omega t\right]}_{\alpha }^{\pi }-{\left[-\cos \omega t\right]}_{\pi }^{2\pi }\right]$ 

$=\frac{V_m}{2\pi }\left[1+\cos \alpha +1+1\right]$ 

$=\frac{V_m}{2\pi }\left[3+\cos \alpha \right]$ 

For α = 60°, $V_{dc}=\frac{300}{2\pi }\left(3+\cos {60}^{\circ }\right)$ 

Concept:

Buck-Boost converter:

Mode I: 0 ≤ t ≤ T_ON

CH → ON

D → OFF

The inductor is charged from the source

Mode II:

T_ON ≤ t ≤ βT (in case of continuous land)

Inductor discharges through capacitor & load i.e. i_L = i_c + I_o

Once is reduces to zero; the inductor stops conducting until the chopper is again turned ON.

Application:

0 < t < αT, V_L = V_S

αT < t < βT, V_L = -V_o

Average voltage across inductor is zero; therefore

V_s α T + (-V_o) [β - α] T = 0

V_s α T - -V_o βT + V_o α T = 0

V_s (αT) = V_o (βT - αT)

$V_o=\left(\frac{\alpha }{\beta -\alpha }\right)V_s$

For given question α = 0.3, β = 0.6

$V_o=\left(\frac{0.3}{0.6-0.3}\right)V_s$

⇒ V_o = V_s

Input voltage (V_s) = 36 V

Resistive load (R) = 18 Ω

Frequency (f) = 200 Hz

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

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:

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

$I_o=\frac{48}{18}=2.67A$ 

$V_o=\sum _{n=1,3,\dots .}^{\mathrm{\infty }}\left\{\frac{4V_s}{n\pi }\sin \frac{n\pi }{2}\sin nd\right\}\sin n\omega t$

RMS value of the fundamental component

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

$=\frac{2\sqrt{2}{V}_s}{\pi }\sin d$ 

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

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

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

V_EE = η V_BB + V_D + I_P . R_E

= 0.67 × 20 + 0.7 + 12 × 10^-6 × 1 × 10³

= 13.4 + .7 + 0.012

= 14.112

V_s = 150 V

The current flows through diode only from θ to (π – θ) because only during this interval diode is forward biased.

V_m sin θ = 150

⇒ 300 sin θ = 150

⇒ θ = 30°

The average value of the current can be calculated by the waveform as:

$I_0=\frac{1}{2\pi R}\underset{\theta }{\overset{\left(\pi -\theta \right)}{\int }}\left(V_m\sin \omega t-E\right)d\omega t$

$I_0=\frac{1}{2\pi R}{\left[-V_m\cos \omega t-E\left(\omega t\right)\right]}_{\theta }^{n-\theta }$

$I_0=\frac{1}{2\pi R}\left[2V_m\cos \theta -E\left(\pi -2\theta \right)\right]$

Replacing the value of θ = 30°, R = 10 Ω we get

I₀ = 3.269 A

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

${\alpha }^{\prime }=\frac{T_{on}}{T^1}=\frac{70}{80}$

New Average output voltage

V₀ = α‘V_S

$=\frac{70}{80}\times 200$

= 175 V

Concept:

The Fourier analysis of phase voltage

${\mathrm{V}}_{a0}=\sum _{n=6K\pm 1}^{\mathrm{\infty }}\frac{2V_s}{n\pi }\sin n\omega t$ 

Where K = 0, 1, 2…

∴ Fundamental component of phase voltage

${\mathrm{V}}_{a1}=\frac{\sqrt{2}{V}_s}{\pi }$ 

Fundamentals component of line voltage

${\mathrm{V}}_{L1}=\sqrt{3}{V}_{a1}$

Calculation:

$\begin{array}{l}{\mathrm{V}}_{a1}=\frac{\sqrt{2}\times 220}{\pi }=99.035V\\ {\mathrm{V}}_{L1}=\sqrt{3}\times 99.035=171.534V\end{array}$

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 $=I_o-\frac{\mathrm{\Delta }{I}_L}{2}$

Maximum value of load current $=I_o+\frac{\mathrm{\Delta }{I}_L}{2}$

Average load current $I_o=\frac{V_o}{R}$

$\mathrm{\Delta }{I}_L=\frac{V_{dc}}{L}D(1-D)T$

Calculation:

Given that, Source voltage (V_dc) = 100 V

R = 0.5 Ω

L = 1 mH

T_ON = 1 ms

T = 3 ms

$D=\frac{T_{On}}{T}=\frac{1}{3}$

$V_o=100\times \frac{1}{3}=33.33V$

$I_o=\frac{V_o}{R}=\frac{33.33}{0.5}=66.66A$

$\mathrm{\Delta }{I}_L=\frac{100}{1\times {10}^{-3}}\times \frac{1}{3}\times (1-\frac{1}{3})\times 3\times {10}^{-3}$

= 66.66 A

Minimum value of load current $=66.66-\frac{66.66}{2}=33.33A$

Concept:

Boundary condition

${\left(i_L\right)}_{avg}=\frac{\mathrm{\Delta }{I}_L}{2}$

As the value of α is increased the conduction enters discontinuous mode.

Calculation:

$I_{omax}=\frac{P}{V_o}=\frac{120W}{48V}=2.5A$

V_s = 12 V; V_o 48 V

${\alpha }_1=1-\frac{V_s}{V_o}=1-\frac{12}{48}=0.75$

V_s = 36 V; V_o = 48 V

${\alpha }_2=1-\frac{V_s}{V_o}=1-\frac{36}{48}=0.25$

The Maximum permissible value of α = α₂ = 0.25

At boundary condition $I_L=\frac{\mathrm{\Delta }{I}_L}{2}=\frac{V_s\alpha }{2L_f}=\frac{V_o\left(1-\alpha \right)\alpha }{2L_f}$

By law of conservation of power V_oI_o = V_sI_s (I_s = I_L)

$\therefore I_o=\frac{V_s{I}_L}{V_o}$

$I_o=\frac{V_s}{V_o}\cdot \frac{V_o\left(1-\alpha \right)\alpha }{2L_f}$

$I_o=\frac{V_s\left(1-\alpha \right)\alpha }{2L_f}$      …(V_s = (1 - α) V_o)

$I_o=\frac{V_o{\left(1-\alpha \right)}^2\alpha }{2L_f}$

$2.5=\frac{48{\left(1-0.75\right)}^{2}0.75}{2L\times 50\times {10}^3}$

$\Rightarrow L=\frac{0.0225}{2.5}\times {10}^{-3}=9\mu H$

The circuit shown is the circuit of the Boost converter.

A boost converter is a DC to DC power converter that steps up voltage (while stepping down current) from its input to its output.

Duty cycle of Boost converter D = 0.7 supply voltage V_S = 40 V

Now, for the Boost converter

Output voltage, $V_o=\frac{V_S}{1-D}=\frac{40}{0.7}=57.14V$

Now the capacitor current,

I_C = I_D - I_O

I_O = (1 - D) I_L avg

= (1 – 0.7) 8 = 3 × 8 = 2.4 Amp

When switch is ON, the value of I_D = 0 (diode is open circuit)

And I_C = -I_O

⇒ (I_C)_On = -2.4 A

The waveforms are shown below.

$i_1=\frac{V_m}{R_1+R_2}\sin \omega t$ 

$=\frac{200}{400}\sin \omega t=0.5\sin \omega t$ 

$i_2=\frac{V_m}{R_1}\sin \omega t=\frac{200}{200}\sin \omega t=\sin \omega t$ 

$I_{dc}=\frac{1}{2\pi }{\int }_0^{\frac{\pi }{2}}0.5\sin \omega td\omega t+\frac{1}{2\pi }{\int }_{\frac{\pi }{2}}^{2\pi }\sin \omega td\omega t$ 

α = 45°

β = 235°

180 + α = 225° < β

Hence, the current is continuous.

The average load voltage V_dc is

$V_{dc}=\frac{2V_m}{\pi }\cos \alpha$ 

$=\frac{2\times 230\times \sqrt{2}}{\pi }\cos {45}^{\circ }$ 

= 146.5 V

With the freewheeling diode is used across the load, the waveform of the voltage will be as shown.

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

$=\frac{230\times \sqrt{2}}{\pi }\left(1+\cos {45}^{\circ }\right)$ 

= 176.67 V

Without a freewheeling diode, the load voltage is as shown below:

$V_{dc}=\frac{3}{2\pi }\underset{\frac{-\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\sqrt{2}{V}_s\cos \omega td\omega t$ 

$=\frac{3}{2\pi }{\left[\sqrt{2}{V}_s\sin \omega t\right]}_{\frac{-\pi }{3}+\alpha }^{\frac{\pi }{3}+\alpha }$ 

$=\frac{3\sqrt{2}{V}_s}{2\pi }\left[\sin \frac{\pi }{3}\cos \alpha +\cos \frac{\pi }{3}\sin \alpha +\sin \left(-\frac{\pi }{3}\right)\cos \alpha -\cos \left(-\frac{\pi }{3}\right)\sin \alpha \right]$ 

$=\frac{3\sqrt{2}{V}_s}{2\pi }\times 2\times \sin \frac{\pi }{3}\times \cos \alpha$ 

= 0.584 V_s

With the freewheeling diode, the negative voltage areas are eliminated.

$=\frac{3}{2\pi }\underset{\frac{-\pi }{3}+\alpha }{\overset{\frac{\pi }{2}}{\int }}\sqrt{2}{V}_s\cos \omega td\omega t$ 

$=\frac{3}{2\pi }{\left[\sqrt{2}{V}_s\sin \omega t\right]}_{\frac{-\pi }{3}+\alpha }^{\frac{\pi }{2}}$ 

$=\frac{3}{2\pi }\sqrt{2}{V}_s\left[\sin \frac{\pi }{2}-\sin \left(-\frac{\pi }{3}+\alpha \right)\right]$ 

$=\frac{3}{2\pi }\sqrt{2}{V}_s\left[\sin \frac{\pi }{3}\cos \alpha -\cos \frac{\pi }{3}\sin \alpha +\sin \frac{\pi }{2}\right]$ 

$=\frac{3}{2\pi }\sqrt{2}{V}_s\left[1-\sin \left(\alpha -\frac{\pi }{3}\right)\right]$ 

= 0.675 V_s

For mid-point rectifier,

$V_{dc}=\frac{2V_m}{\pi }\cos \alpha -\frac{\omega L{I}_{dc}}{\pi }$ 

And V_dc = I_dc R

$\Rightarrow R{I}_{dc}=\frac{2V_m}{\pi }\cos \alpha -\frac{\omega L}{\pi }{I}_{dc}$ 

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

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

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

= 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°

For the given full-bridge inverter, the output waveform for voltage is given below.

$THD=\sqrt{{\left(\frac{V_o}{V_{01}}\right)}^2-1}$

$V_o=V_s\sqrt{\frac{\beta }{\pi }}=\sqrt{\frac{2}{3}}{v}_s$

$V_{01}=\frac{2\sqrt{2}{V}_s}{\pi }\sin \frac{\beta }{2}$

$=\frac{\sqrt{V_s}}{\pi }$

$THD=\sqrt{{\left(\frac{\sqrt{\frac{2}{3}}}{\frac{\sqrt{6}}{\pi }}\right)}^2-1}$

= 0.3108 = 31.08 %

Concept:

• Need of series connection of SCR is required when we want to meet the increased voltage requirement by using various SCR’s.
• When the required voltage rating exceeds the SCR voltage rating, a number of SCR’s are required to be connected in series to share the forward and reverse voltage.
• When the load current exceeds the SCR current rating, SCR are connected in parallel to share the load current.

Application:

According to the question:

Given that

SCR 1 voltage = 350 V; current = 6 A

SCR 2 voltage = 300 V; current = 9 A

SCR 3 voltage = 250 V; current = 12 A

Let us take the total current to be ‘I’

Current through resistor R in shunt with SCR 1 is

I₁ = I – 6

Similarly, current through resistor R is shunt with SCR 2

I₂ = I – 9

And, current through resistor R in shunt with SCR 3

I₃ = I – 12

Now, the string voltage becomes

V_s = I₁R + I₂R + I₃R

Vs = (I - 6) R + (I - 9) R + (I - 12) R

Vs = (I - 6) R + (I - 6) R – 3 R + (I - 6) R – 6 R …..(1)

Note that

We should consider the extreme case from the calculation of resistance R for voltage equalization in string of SCR.

In extreme case, the voltage drop across SCR 1 (or the one having highest voltage drop) will be maximum forward blocking voltage.

∴ V­_max = 350 = (I - 6) R     ----(2)

Put (I - 6) R = 350 in equation (1), we get

V_s = 350 + 350 – 3 R + 350 – 6 R

(350 + 300 + 250) = 1050 – 9 R

900 = 1050 – 9 R

9 R = 1050 – 900

$R=\frac{150}{9}\mathrm{\Omega }$ 

R = 16.66 Ω

Concept:

For constant load current I_o

$I_o=\frac{V_o-E}{R}$

The average thyristor current I_T is given by

$I_T=I_o\frac{T_{on}}{T}=\left(\frac{V_o-E}{R}\right)\propto =\frac{{\propto }^2{V}_s-\propto E}{R}$

This will be a maximum value when

$\begin{array}{l}\frac{d{I}_T}{d\propto }=\frac{2\propto V_s-E}{R}=0\\ \propto =\frac{E}{2V_s}\end{array}$

$\therefore I_{Tmax}=\frac{E}{2V_{S}R}\left[\frac{E}{2V_s}\times V_s-E\right]=\frac{E^2}{4V_{s}R}$

Calculation

i_a = i_A - i_C

$=\frac{V_A-V_C}{R}$

V_A = V_a0 – V_b0

V_C = V_C0 – V_a0

$\Rightarrow i_a=\frac{2V_{a0}-V_{b0}-V_{c0}}{R}$

$I_1=\frac{V_s}{R}{I}_2=\frac{2V_s}{R}$

$I_{a,rms}=\sqrt{\frac{1}{\pi }[I_1^2\frac{2\pi }{3}+I_2^2\frac{\pi }{3}]}$

Power delivered to the load. P₀ = V₀₁ I_or cos ϕ

V₀₁ is rms value of fundamental output voltage

I_0r is rms value of load current

$V_{01}=\frac{4V_{dc}}{\sqrt{2}\pi }$

Given that, V_dc = 400 V

$V_{01}=\frac{4\times 400}{\sqrt{2}\times \pi }=360.12V$

Power delivered, P₀ = V₀₁ I_or cosϕ

$=360.12\times \frac{200}{\sqrt{2}}\times \cos {30}^{\circ }$

= 44.1 kW