• GTO being a monolithic p-n-p-n structure just like a thyristor. In particular, the p-n-p-n structure of a GTO can be thought of consisting of one p-n-p and one n-p-n transistor connected in the regenerative configuration.
• Due to low internal generation in GTO, the GTO has both holding and latching current of a GTO are considerably higher compared to a similarly rated thyristor.
• Since the holding current of a GTO is considerably higher than that of a thyristor anode current variation can generate serious problem because the GTO might unlatch at an inappropriate moment.
• To avoid this problem the gate drive unit of a GTO must feed the gate terminal with a continuous “back porch” current during the entire on period of the GTO.
• This back-porch current must be larger than the gate trigger current.

Important formulas:

1) $Q_{RR}=\frac{1}{2}\times I_{RR}\times t_{rr}$

2) $I_{RR}={\left[2Q_{RR}\frac{di}{dt}\right]}^{1/2}$

3) $t_{rr}={\left[\frac{2Q_{RR}}{\frac{di}{dt}}\right]}^{\frac{1}{2}}$

Calculation:

$Q_{RR}=\frac{1}{2}{t}_{rr}^2\left(\frac{di}{dt}\right)$

$=\frac{1}{2}\times {\left(3\times {10}^{-6}\right)}^2\times 30\frac{A}{\mu s}$

$Q_{RR}=135\mu C$

Concept:

The average output voltage of a single-phase semi-converter is given by

$V_0=\frac{V_m}{\pi }\left(1+\cos \alpha \right)$

V_m is the maximum value of supply voltage

α is the firing angle

Calculation:

The motor voltage constant is K_b = 0.1 V/rpm

Speed of the motor = 1200 rpm

Back emf (E_b) = 0.1 × 1200 = 120 V

Armature current (I_a) = 16 A

V₀ = E_b + I_a R_a

$\Rightarrow \frac{230\times \sqrt{2}}{\pi }\left(1+\cos \alpha \right)=120+16\left(0.5\right)=128$

⇒ α = 76.33°

Concept:

The average output voltage is given by, $V_1=\frac{2V_m}{\pi }\cos \alpha$

The average output current, $I_1=\frac{V_0}{R}$

If SCR T₃ is damaged and gets open circuited, the circuit acts as half wave rectifier.

The average output voltage is given by, $V_2=\frac{V_m}{\pi }\cos \alpha$

The average output current, $I_2=\frac{V_0}{R}$

$\frac{I_2}{I_1}=\frac{1}{2}$

Calculation:

Given that, I₁ = 4 A, α = 60°

$I_2=\frac{I_1}{2}=\frac{4}{2}=2A$

During α to π, both v_s and i_s are positive, power therefore flows from ac source to load. During the interval π to (π + α), v_s is negative but i_s is positive, the load, therefore, returns some of its energy to the supply system. But the net power flow is from ac source to dc load because (π – α) > α.

The average value of the output voltage is given by

$V_0=\frac{2V_m}{\pi }\cos \alpha$

Note:

• If α > 90°, V₀ is negative. If the load circuit emf E is reversed, this source E will feed power back to ac supply. This operation of full converter is known as the inverter operation of the converter.
• The full converter with firing angle delay greater than 90° is called a line-commutated inverter. Such an operation is used in the regenerative braking mode of a dc motor in which case then E is counter emf of the dc motor.
• During 0 to α, ac source voltage v_s is positive, but ac source current is negative, power, therefore, flows from dc source to ac source.
• From α to π, both v_s and i_s are positive, power, therefore, flows from ac source to dc source. But the net power flow is from dc source to ac source because of (π – α) < α. 

For firing angle, greater than 60^o the freewheeling diode starts to conduct.

Therefore, the conduction angle for the freewheeling diode can be represented by the formula:

θ = α - $\frac{\pi }{3}$

For firing angle α = $\frac{\pi }{3}$; conduction angle for the freewheeling diode is 0^o

Taking Laplace at t = 0

$I\left(s\right)=\frac{\frac{20}{s}}{Ls+\frac{1}{cs}}$

$=\frac{\frac{20}{L}}{s^2+\frac{L}{C}}$

Taking inverse Laplace transform,

$i\left(t\right)=V_s\sqrt{\frac{C}{L}}\sin \frac{t}{\sqrt{LC}}$

$V_c\left(s\right)=I\left(s\right)\times \frac{1}{Cs}$

$V_c\left(s\right)=\frac{\frac{20}{L}}{s^2+\frac{L}{C}}\times \frac{1}{Cs}$

$=\frac{20}{LCs[s^2+\frac{1}{LC}]}$

Taking inverse Laplace transform,

$V_c\left(t\right)=V_s[1-\cos \omega t]$

$\Rightarrow V_c\left(t\right)=V_s[1-\cos \frac{t}{\sqrt{LC}}]$

The thyristor will allow current in positive direction only, i.e. till π.

$V_c(\omega t=\pi )=V_s[1-\cos \pi ]=2V_s$

At steady state, V­_c = 2 × 20 = 40 V

Concept:

For a step-down chopper, average voltage and current is given by

V_0(avg) = δV_s

$I_{0\left(avg\right)}=\frac{\delta V_s}{R}$

rms voltage and current is given by

$V_{0\left(rms\right)}=V_s\sqrt{\delta }$

$I_{0\left(rms\right)}=\frac{\sqrt{\delta }{V}_s}{R}$

Where δ is duty cycle

Calculation:

We have given

δ = 0.49

Average load current $I_{0\left(avg\right)}=\frac{0.49\times 200}{10}=9.8A$

Rms value of load current $I_{0\left(rms\right)}=\frac{\sqrt{0.49}\times 200}{10}=14A$

Concept:

In a single phase fully controlled bridge rectifier without freewheeling diode,

The average output voltage, $V_0=\frac{2V_m}{\pi }\cos \alpha$

The average output current, $I_0=\frac{V_0}{R}$

In a single phase fully controlled bridge rectifier with freewheeling diode,

The average output voltage, $V_{0,fd}=\frac{V_m}{\pi }\left(1+\cos \alpha \right)$

The average output current, $I_{0,fd}=\frac{V_{0,fd}}{R}$

$\frac{V_{0,fd}}{V_0}=\frac{1}{2}\left(1+\sec \alpha \right)$

$\frac{I_{0,fd}}{I_0}=\frac{V_{0,fd}}{V_0}=\frac{1}{2}\left(1+\sec \alpha \right)$

Calculation:

Given that, firing angle (α) = 60°

$\frac{V_{0,fd}}{V_0}=\frac{1}{2}\left(1+\sec 60\right)=1.5$

Input voltage = V_dc

Output voltage = V₀

Inductance = L

Capacitance = C

When switch ‘sw’ closed (0 < t < DT)

$\frac{d{i}_L}{dt}=\frac{V_{dc}-V_0}{L}$ 

$d{i}_L=\frac{V_{dc}-V_0}{L}dt\Rightarrow i_L\left(t\right)=\int \frac{V_{dc}-V_0}{L}dt+I_{Lmin}$ 

$i_L\left(t\right)=\frac{V_{dc}-V_0}{L}t+I_{Lmin}$

When switch is open (DT < t < T)

$\frac{d_L}{dt}=\frac{-V_0}{L}$ 

$d{i}_L=\frac{-V_0}{L}dt$ 

$i_L=\frac{-V_0}{L}\int dt+I_{Lmax}$ 

$i_L\left(t\right)=I_{Lmax}-\frac{V_0}{L}\left(t-DT\right)$

Now ripple current in inductor

Δi_L = I_Lmax – I_Lmin

at t = DT ⇒ i_L(t) = I_max

So $I_{Lmax}=\frac{V_{dc}-V_0}{L}DT+I_{Lmin}$

Now $\mathrm{\Delta }{i}_L=I_{Lmax}-I_{Lmin}=\frac{V_{dc}-V_0}{L}DT{V}_0=D{V}_{dc}$

$\mathrm{\Delta }{i}_L=\frac{V_{dc}-D{V}_{dc}}{L}DT=\frac{V_{dc}\left(1-D\right)}{L}DT$ 

Δi_L α D (1 - D)

To get maximum, $\frac{d\mathrm{\Delta }{I}_L}{dD}=0$

$\frac{d\mathrm{\Delta }{I}_L}{dD}=0\Rightarrow \frac{V_{dc}}{L}T{\left(1-2D\right)}_{max}=0$ 

1 – 2D_max = 0

⇒ D_max = 0.5

Concept:

Three-phase power $=3{\left(I_{ph}\right)}_{rms}^2\times R$

${\left(I_{ph}\right)}_{rm}=\frac{{\left(V_p\right)}_{rms}}{R_{load}}$ 

${\left(V_L\right)}_{rms}=V_{dc}\sqrt{\frac{2}{3}},{\left(V_{ph}\right)}_{rms}=\frac{V_{dc}}{\sqrt{3}}$ 

${\left(V_{ph}\right)}_{rms}=\frac{V_{dc}\sqrt{2}}{3}$ 

Calculation:

Load resistance, R_L = 20 Ω

Source voltage, V_s = 450 V

Mode of operation = 180°  

${\left(V_{ph}\right)}_{rms}=\frac{450\times \sqrt{2}}{3}$ 

= 212.132 V

${\left(I_{ph}\right)}_{rms}=\frac{212.132}{20}$ 

${\left(I_{ph}\right)}_{rms}=10.6A$ 

Three-phase power is

$=3{\left(I_{ph}\right)}_{rms}^2\times R=3\times {10.6}^2\times 20$ 

= 6741.6 W = 6.74 kW

Concept:

Calculation:

The inverter is operating in 120° conduction mode.

Supply voltage (V_s) = 420 V

Load resistance (R) = 15 Ω

The RMS value of thyristor current, $I_{Tr}=\frac{420}{2\sqrt{3}\times 15}=8.08A$

Given that, V_s = 200 V

V_DS(sat) = 1.9 V

Load resistance (R) = 20 Ω

Duty cycle (m) = 0.8

Steady on-state current, $I_l=\frac{V_s-V_{DS\left(sat\right)}}{R}=\frac{200-1.9}{20}=9.9A$

Conduction losses, $P_c=\frac{1}{T}\underset{0}{\overset{t_{ON}}{\int }}{V}_{DS\left(sat\right)}{I}_{D}dt$

Concept:

The load current, $I_l=\frac{V_s-V_{GTO\left(ON\right)}}{R}$

The load power, $P=\frac{V_l^2}{R}=\frac{{\left(V_s-V_{GTO\left(ON\right)}\right)}^2}{R}$

The total power gain $=\frac{P}{P_G}$

Calculation:

Given that, DC supply voltage (V_s) = 600 V

Load resistance (R) = 30 Ω

the on-state voltage drop, V_GTO(ON) = 2.2 V

The average gate power, P_G = 10 W

The load current, $I_l=\frac{600-2.2}{30}=19.93A$

The load power, $P=\frac{{\left(600-2.2\right)}^2}{30}=11912W$

The gate power, P_G = 10 W

Given that,

I_g(min) = 100 mA

V_g(min) = 2 V

Simplifying the above circuit diagram, we get below the circuit diagram where we assume SCR as a voltage device.

Note: The diode in the given circuit is just for current reversal protection.

Now applying KCL, we get

$\frac{V_g-V_{trig}}{30}+\frac{V_g}{30}+100mA=0$

⇒ 2V_g – V_trig + 3 = 0

⇒ 2V_g + 3 = V_trig

Concept:

In a full wave rectifier though a transformer,

Average output voltage, $V_0=\frac{2V_m}{\pi }$

Average output current, $I_0=\frac{V_0}{R}$

RMS value of output voltage V_or = V_s

RMS value of load current, $I_{or}=\frac{V_s}{R}$

Average value of diode current, $I_d=\frac{I_m}{2}$

RMS value of diode current, I_dr = I_or

Peak value of diode current, I_dm = √2 I_or

Power delivered to the load = V_or I_or

Input voltamperes = V_s I_or

Calculation:

Given that, DC output voltage (V_DC) = 400 V

Average output voltage of rectifier (V₀) = 400 V

$\Rightarrow \frac{2V_m}{\pi }=400$

$\Rightarrow \frac{2\times \sqrt{2}\times V_s}{\pi }=400$

⇒ V_s = 444.28 V

Load resistance (R) = 10 Ω

RMS value of load current, $I_{or}=\frac{444.28}{10}=44.428A$

The output voltage waveform of the 3ϕ half-wave rectifier is given below.

With respect to the above waveform:

$V_{0rms}=\frac{V_{mph}}{\sqrt{2\cdot \frac{2\pi }{3}}}{\left[\left[\pi -\left(\frac{\pi }{6}+\alpha \right)\right]+\frac{1}{2}\left[\sin \left(2\alpha +\frac{\pi }{3}\right)-\sin \left(2\pi \right)\right]\right]}^{\frac{1}{2}}$

$V_{0rms}=\frac{\sqrt{3}{V}_{mph}}{2\sqrt{\pi }}{\left[1.832+\frac{1}{2}\left[\sin \left(\frac{\pi }{2}+\frac{\pi }{3}\right)\right]\right]}^{\frac{1}{2}}$

V_{0 rms} = 126.67 V

Power command by the load; $P=\frac{{\left(V_{0rms}\right)}^2}{R}=\frac{{\left(126.67\right)}^2}{20}=0.802kW$

For a single-phase semi-converter, the peak value of the n^th component of the current

$I_{s_n}=\frac{4I_0}{n\pi }\cos n\frac{\alpha }{2}$

${\left(I_{s_1}\right)}_{rms}=\frac{2\sqrt{2}}{\pi }{I}_0\cos \frac{\alpha }{2}$

(I_s)_rms = RMS value of source current

$I_{s_n}=I_0{\left(\frac{\pi -\alpha }{\pi }\right)}^{1/2}$

Current distortion factor, $g=\frac{I_{s_1}}{I_{s_n}}=\frac{\frac{2\sqrt{2}}{\pi }{I}_0\cos \frac{\alpha }{2}}{I_0{\left(\frac{\pi -\alpha }{\pi }\right)}^{\frac{1}{2}}}$

$g=\frac{2\sqrt{2}\cos \frac{\alpha }{2}}{\sqrt{\pi \left(\pi -\alpha \right)}}$

Replacing the value of $\alpha =\frac{\pi }{6}$ we get,

$g=\frac{2\sqrt{2}\cos \left(\frac{\pi }{12}\right)}{\sqrt{\pi \left(\pi -\frac{\pi }{6}\right)}}=0.9526$

Concept:

Effect of source inductance: The presence of source inductance affects the rectifier output voltage

V₀ = V_d0 cos α – ΔV_d0

Where $\mathrm{\Delta }{V}_{d0}=\frac{V_{d0}}{2}\left[\cos \alpha -\cos \left(\alpha +\mu \right)\right]$

μ is the overlap angle

$V_{d0}=\frac{2V_m}{\pi }$ for 2 pulse

$=\frac{3V_{mL}}{2\pi }$ for 3 pulse

$=\frac{3V_m}{\pi }$ for 6 pulse

Also, ΔV_d0 = f L_s I₀ (1 pulse)                                                      

= 4 f L_s I₀ (2 pulse)

= 3 f L_s I₀ (3 pulse)                                                                    

= 6 f L_s I₀ (6 pulse)

Calculation:

V₀ = -E₀ + I₀ R

V_d0 cos α – 3 f L_s I₀ = -200 + (20)(2)

$V_{d0}=\frac{3V_m}{2\pi }$

$\frac{3V_m}{2\pi }\cos \alpha -3f{L}_s{I}_0=-160$

$\frac{3V_m}{2\pi }\cos \alpha =-154$

On solving we get cos α = - 0.57

⇒ α = 124.76°

For three pulse converters:

$\frac{V_{d0}}{2}\left[\cos \alpha -\cos \left(\alpha +\mu \right)\right]=3f{L}_s{I}_0$

$\frac{3V_m}{2\pi \cdot 2}\left[\cos \left(127.76\right)-\cos \left(124.76+\mu \right)\right]=3f{L}_s{I}_0$

On solving the above equation for overlap angle (μ) we get

cos (α + μ) = -0.6145

μ = 3.15°   

Concept:

For a buck-boost converter

$V_0=\left(\frac{\alpha }{1-\alpha }\right)V_s$

Capacitor voltage ripple:

$\mathrm{\Delta }{V}_C=\frac{\mathrm{\Delta }Q}{C}=\frac{I_0{T}_{on}}{C}$

$\mathrm{\Delta }{V}_C=\frac{I_0\alpha }{fC}$

V₀: output voltage

V_s: source voltage

α: duty ratio

Calculation:

The given circuit represents a buck-boost converter

$\therefore V_0=\left(\frac{\alpha }{1-\alpha }\right)V_s=\left(\frac{0.4}{0.6}\right)\times 150=100V$

$I_0=\frac{V_0}{R}=\frac{100}{20}=5A$

Ripple voltage of capacitor $\mathrm{\Delta }V=\frac{I_0\alpha }{fC}$

$\mathrm{\Delta }V=\frac{5\times 0.4}{50\times 200\times {10}^{-6}}$

⇒ ΔV = 200 V

Input voltage, V_dc = 50 V

Output voltage, V₀ = 150 V

Inductance, L = 15 mH

Capacitance, C = 1 μF

Duty cycle, D = 0.4

Frequency = 100 kHz

When switch closed during 0 < t < DT, inductor will charge

t = DT

$\because V_L=V_{dc}=L=\frac{di}{dt}$ 

$di=\frac{V_L\cdot D\cdot T}{L}$

We know di become I_max at t = DT

$I_{max}=\frac{50\times 0.4\times }{15\times {10}^{-3}}\frac{1}{{10}^5}=\frac{5\times 4}{15\times {10}^2}$ 

$I_{max}=\frac{4}{3}\times {10}^{-2}A$ 

From DT < t < T i.e. when switch open, the inductor will discharge.

$I_L\left(t\right)=I_{max}-\frac{V_0}{L}\left(1-D\right)T$ 

$=\frac{4}{3}\times {10}^{-}-\frac{150}{15\times {10}^{-3}}\times 0.6\times \frac{1}{{10}^5}$ 

$=\frac{4}{3}\times {10}^{-2}-\frac{0.6}{10}$ 

I_L(t) = -0.0466 A

Important Observation:

Current cannot become negative because of Diode D­₁.

So, at the end of each cycle inductor current is zero.

So, inductor current becomes zero between DT < t < T

So, energy stored after 10^th cycle in the inductor

$=\frac{1}{2}\times 15\times {10}^{-3}\times 0^2$ 

= 0 mJ

Calculation:

V_o = I_oR_L = 3 × 10 = 30 V

V_b = V_o (1 – D)

$1-D=\frac{12}{30}=0.4$

D = 0.6

Periodic time $T=\frac{1}{f}=\frac{1}{50\times {10}^3}=20\mu s$

On time = DT = 0.6 × 20 × 10^-6 = 12 μs 

over an on-time the change in battery current will be

$\mathrm{\Delta }i=\frac{V_L}{L}\times Ton$

$\mathrm{\Delta }i=\frac{12}{20\times {10}^{-6}}\times 20\times {10}^{-6}=7.2A$

Concept:

When the transistor is switched on, current ramps up in the inductor and energy is stored.

When the transistor is switched off, the inductor voltage reverses and acts together with the battery voltage to forward bias the diode, transferring energy to the capacitor.

When the transistor is switched on again, load current is maintained by the capacitor, energy is stored in the inductor and the cycle can start again.

The value of the load voltage is increased by increasing the duty cycle or the on-time of the transistor.

During the transistor on-period, assuming an ideal inductor and transistor

V_b = V_L = Ldi/dt

V_b/s = (sL) i(s)

i(s) = V_b/s²L

i(t) = (V_b/L)t = kt

Over an on-time of T_ON, the change of battery current will be

Δi = k T_ON

During the off-period

V_o = V_b + V_L = V_b + L(di/dt) = V_b + L(Δi/Δt)

= V_b + L(V_b/L)T_ON /T_off = V_b(1 + T_ON /T_off)

Let the duty cycle

D = T_ON T, and T_off = T – T_ON = T(1 - D). Now

V_o = V_b (1 – DT/T(1 – D))

which simplifies to

V_o = V_b/(1 - D)

In an ideal circuit, V_bI_b = V₀I₀. Therefore, from eq.

I_b = (V_o/V_b)I_b = I_o/(1 – D)

I_b = I_b + Δi/2

I_bo = I_b – Δi/2

When the steady variation of battery current has been reached, the variation will be between l_o and I₁, as shown in.

Drawing output current waveform for 180° mode.

${\left(i_T\right)}_{rmsfor{180}^{\circ }}^2=\frac{1}{2\pi }\left[\underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{V_{dc}}{3R}d\omega t+\underset{\frac{\pi }{3}}{\overset{\frac{2\pi }{3}}{\int }}\frac{2}{3}\frac{V_{dc}}{R}d\omega t+\underset{\frac{2\pi }{3}}{\overset{\pi }{\int }}\frac{V_{dc}}{3R}d\omega t\right]$ 

$=\frac{1}{2\pi }\left[\frac{2\pi }{3}\times {\left(\frac{V_{dc}}{3R}\right)}^2+\frac{\pi }{3}{\left(\frac{2V_{dc}}{3R}\right)}^2\right]$ 

$i_T^2={\left(\frac{V_{dc}}{3R}\right)}^2\left[\frac{1}{3}+\frac{4\pi }{3\times 2\pi }\right]$ 

$i_T^2={\left(\frac{V_{dc}}{3R}\right)}^2\Rightarrow {\left(i_T\right)}_{tms}=\frac{V_{dc}}{3R}$ 

$\Rightarrow \frac{V_{dc}}{3R}=15A$

Drawing output and switch current for 120°.

${\left(i_T\right)}_{rms}^2={\left(\frac{V_{dc}}{2R}\right)}^2\times \frac{2\pi }{3\times \pi }\$$

${\left(i_T\right)}_{rms}=\left(\frac{V_{dc}}{2R}\right)\times \frac{1}{\sqrt{3}}$

$=\frac{45}{2}\times \frac{1}{\sqrt{3}}=12.99$ 

Concept: This method of pulse modulation is termed as multiple pulse modulation. r is an extension of single pulse modulation.

The amplitude of the n^th harmonic of two pulse waveform is given by:

$V_n=\left[\frac{8V_s}{n\pi }\sin \left(n\gamma \right)\sin \left(\frac{nd}{2}\right)\right]$

In general, $\gamma =\frac{\pi -2d}{N+1}+\frac{d}{N}$ where N is the number of pulses per half cycle (Here N = 2)

Calculation:

The amplitude of n^th harmonic $V_m=\frac{8Vs}{n\pi }\sin \left(n\gamma \right)\sin \left(\frac{nd}{2}\right)$

For peak value of fundamental voltage component

$V_1=\frac{8V_s}{\pi }\sin \left(\frac{d}{2}\right)\sin \left(\gamma \right)$

$\gamma =\left(\frac{\pi -2d}{N+1}\right)+\frac{d}{N}$

γ = 56°, V_s = 200 V, N = 2

$\left(\frac{\pi }{{180}^{\circ }}\right)56=\left(\frac{\pi -2d}{3}\right)+\frac{d}{2}$

Solving we get d = 24°

$V_1=\frac{8\left(200\right)}{\pi }\sin \left(\frac{24}{2}\right)\sin \left({56}^{\circ }\right)$

V₁ = 87.78 V

Concept:

General equation of n^th harmonic voltage waveform for a single-phase full-bridge inverter is given by:

$V_{On}=\frac{4V_s}{n\pi }\sin \left(n\omega t\right)$

3^rd harmonic waveform:$V_{03}=\frac{4V_s}{3\pi }\sin \left(3\omega t\right)$

$I_{03}=\frac{{\left(V_{03}\right)}_{rms}}{\left|Z_3\right|}$

Calculation:

Supply voltage (V_s) = 230 V

${\left(V_{03}\right)}_{rms}=\frac{4V_s}{3\pi }\times \frac{1}{\sqrt{2}}=\frac{4\times 230}{3\sqrt{2}\pi }$

(V₀₃)_rns = 69.02 V

$Z_3=R+j\left(3\omega L+\frac{\left(-1\right)}{3\omega C}\right)$

$Z_3=4+j\left(3\omega \left(35\times {10}^{-3}\right)+\frac{\left(-1\right)}{3\omega \left(155\times -6\right)}\right)$

$\left|Z_3\right|=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

|Z₃| = 26.44 Ω

∴ 3^rd harmonic current component

$I_{03}=\frac{{\left(V_{03}\right)}_{rms}}{\left|Z_3\right|}=\frac{69.02}{26.44}=2.61A$