Power Electronics Test 1

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

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

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:

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

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

\({\left( {{I_{ph}}} \right)_{rms}} = 10.6\;A\) 

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, \({I_{Tr}} = \frac{{420}}{{2\sqrt 3 \; \times \;15}} = 8.08\;A\)

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, \({I_l} = \frac{{{V_s} - {V_{DS\left( {sat} \right)}}}}{R} = \frac{{200 - 1.9}}{{20}} = 9.9\;A\)

Conduction losses, \({P_c} = \frac{1}{T}\mathop \smallint \limits_0^{{t_{ON}}} {V_{DS\left( {sat} \right)}}{I_D}dt\)

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, \({I_l} = \frac{{600 - 2.2}}{{30}} = 19.93\;A\)

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

The gate power, P_G = 10 W

Concept:

In a full wave rectifier though a transformer,

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

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

\( \Rightarrow \frac{{2{V_m}}}{\pi } = 400\)

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

⇒ V_s = 444.28 V

Load resistance (R) = 10 Ω

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

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

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

\({\left( {{I_{{s_1}}}} \right)_{rms}} = \frac{{2\sqrt 2 }}{\pi }{I_0}\cos \frac{\alpha }{2}\)

(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 \({\rm{\Delta }}{V_{d0}} = \frac{{{V_{d0}}}}{2}\;\left[ {\cos \alpha - \cos \left( {\alpha + \mu } \right)} \right]\)

μ is the overlap angle

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

\(= \frac{{3{V_{mL}}}}{{2\pi }}\) for 3 pulse

\(= \frac{{3{V_m}}}{\pi }\) 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)

\({V_{d0}} = \frac{{3{V_m}}}{{2\pi }}\)

\(\frac{{3{V_m}}}{{2\pi }}\cos \alpha - 3\;f\;{L_s}\;{I_0} = - 160\)

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

On solving we get cos α = - 0.57

⇒ α = 124.76°

For three pulse converters:

\(\frac{{{V_{d0}}}}{2}\left[ {\cos \alpha - \cos \left( {\alpha + \mu } \right)} \right] = 3f\;{L_s}{I_0}\)

\(\frac{{3{V_m}}}{{2\pi \cdot 2}}\;\left[ {\cos \left( {127.76} \right) - \cos \left( {124.76 + \mu } \right)} \right] = 3\;f\;{L_s}{I_0}\)

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

cos (α + μ) = -0.6145

μ = 3.15°   

Concept:

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

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

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

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

Calculation:

Supply voltage (V_s) = 230 V

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

(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

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