Power, Electronic Systems, Applications and Resources on Electrical and Electronic Project-Thesis: Rectifiers

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THREE-PHASE DUAL CONVERTER

In many variable-speed drives, the four quadrant operation is generally required and three phase dual converters are extensively used in applications up to the 2000kW level.

THREEPHASE DUAL CONVERTER WAVE FORMS

previous Three-phase full-wave Controlled Rectifier

Three-phase full-wave Controlled Rectifier

Three-phase full-wave Controlled Rectifier with highly inductive load (Continuous load current)

Average Load/Output Voltage

${V_0} = \frac{{3\sqrt 3 {V_m}}}{\pi }\cos \alpha$

${V_0} = \frac{{3\sqrt 6 V}}{\pi }\cos \alpha$

V_m peak phase voltage

V rms phase voltage

Three-phase full-wave Controlled Rectifier with highly inductive load

Voltage and current waveforms of a three-phase full converter with a highly inductive load is shown in figure. This converter provides two quadrant operation and thyristors are fired at an interval of π/3 degrees. Since thyristors are fired every 60°, the frequency of the output ripple voltage is six times the frequency of the supply voltage. At ωt = π /6 + α, thyristor S₆ is already conducting and thyristor S₁ is turned on. For the interval ωt of π/6 to π/2 thyristors S₁ and S₆ conduct, and line to line voltage v_ab appears across the load. At ωt = π /2 + α, thyristor S₂ is turned on and thyristor S₆ is turned off due to natural commutation. This occurs because when thyristor S₂ is turned on, the line to line voltage across thyristor S₆ is the positive voltage v_bc from cathode to anode which reverse biases thyristor S₆. During the interval ωt of (π /2 + α) (5 π /6 + α), thyristors S₁ and S₂ conduct and line to line voltage appears across the load. The firing sequence of the thyristors is: 12, 23, 34, 45, 56 and 61.

The average output voltage is given by

${V_{dc}} = \frac{6}{{2\pi }}\int\limits_{\pi /6 + \alpha }^{\pi /2 + \alpha } {{V_{ab}}d\left( {\omega t} \right)}$

${V_{dc}} = \frac{3}{\pi }\int\limits_{\pi /6 + \alpha }^{\pi /2 + \alpha } {\sqrt 3 {V_m}\sin \left( {\omega t + \frac{\pi }{6}} \right)d\left( {\omega t} \right)}$

${V_{dc}} = \frac{{3\sqrt 3 {V_m}}}{\pi }\cos \alpha$

The maximum output dc voltage is given by

${V_{dm}} = \frac{{3\sqrt 3 {V_m}}}{\pi }$

The rms output voltage is given by

${V_{rms}} = {\left[ {\frac{3}{\pi }\int\limits_{\pi /6 + \alpha }^{\pi /2 + \alpha } {{{\left( {\sqrt 3 {V_m}\sin (\omega t + \frac{\pi }{6})} \right)}^2}d(\omega t)} } \right]^{\frac{1}{2}}}$

${V_{rms}} = \sqrt 3 {V_m}{\left( {\frac{1}{2} + \frac{{3\sqrt 3 }}{{4\pi }}\cos 2\alpha } \right)^{\frac{1}{2}}}$

Three-phase Converter Output Characteristics for continuous load current (Full Converter)

For fully controlled rectifier, The DC Motor operates in two modes.
Rectification [As Motoring]

V₀ = positive
E_a = Positive
I_o= positive
Power Flow (+ve) from input AC to DC machine

Inversion [As Regenerative Braking]

V₀ = negative
E_a = negative
I_o= positive
Power Flow (-ve) from DC machine to AC supply

Thyristor based Rectifiers (3-phase)

E_d becomes smaller as α increases, but still each thyristor conducts 120 deg. Power flow is from AC side to DC side. I_d=(E_d-E₀)/R

Thyristor based Line Commutated Inverter (3-phase)

I_d=(E_o-E_d)/R, real power flow is from DC to AC side, Polarity of E_d is reversed.
Triggering range:

Rectifier 15°-90°, inverter: 90°-165°. Thyristor may misfire for α less than 15° (def. 8°) for sudden change in line voltage and hence discontinuity in output current. If we go beyond 165°, the inverter may lose its ability to switch from one thyristor to the next. As a result currents build up very quickly until the CB trips. For safety margin max α is 150°.

previous Three-phase half-wave Controlled Rectifier

next THREE-PHASE DUAL CONVERTER

Three-phase half-wave Controlled Rectifier

Three-phase half-wave Controlled Rectifier circuit with R load

Average Load/Output Voltage
${V_{dc}} = \frac{3}{\pi }\int_{\pi /6 + \alpha }^\pi {\sqrt 2 V\sin \theta d\theta }$
$= \frac{{3\sqrt 2 }}{\pi }V\left( {1 + \cos \left( {\frac{\pi }{6} + \alpha } \right)} \right)$
Three-phase half-wave Controlled Rectifier circuit
The rms output voltage is obtained from with resistive load

${V_{rms}} = {\left[ {\frac{3}{{2\pi }}\int\limits_{\frac{\pi }{6} + \alpha }^\pi {V_m^2{{\sin }^2}\omega td\left( {\omega t} \right)} } \right]^{\frac{1}{2}}}$
${V_{rms}} = \sqrt 3 {V_m}{\left[ {\frac{5}{{24}} - \frac{\alpha }{{4\pi }} + \frac{1}{{8\pi }}\sin \left( {\frac{\pi }{3} + 2\alpha } \right)} \right]^{\frac{1}{2}}}$
For a continuous load current with highly inductive load, the average output voltage is given by
${V_{dc}} = \frac{3}{{2\pi }}\int\limits_{\pi /6 + \alpha }^{5\pi /6 + \alpha } {{V_m}} \sin \omega td\left( {\omega t} \right)$
${V_{dc}} = \frac{{3\sqrt 3 {V_m}}}{{2\pi }}\cos \alpha$
The rms output voltage is obtained from
${V_{rms}} = {\left[ {\frac{3}{{2\pi }}\int\limits_{\pi /6 + \alpha }^{5\pi /6 + \alpha } {V_m^2{{\sin }^2}\omega t\left( {\omega t} \right)} } \right]^{\frac{1}{2}}}$
${V_{rms}} = \sqrt 3 {V_m}{\left[ {\frac{1}{6} + \frac{{\sqrt 3 }}{{8\pi }}\cos 2\alpha } \right]^{\frac{1}{2}}}$
previous THREE-PHASE CONTROLLED RECTIFIER
next Three-phase full-wave Controlled Rectifier

SINGLE-PHASE CONVERTERs

SINGLE-PHASE SEMI-CONVERTER

${V_0} = \frac{{\sqrt 2 V}}{\pi }(1 + \cos \alpha )$
SINGLEPHASE DUAL CONVERTER

previous Single-Phase Full Converter with RL load
next THREE-PHASE CONTROLLED RECTIFIER

Single-Phase Full Converter with RL load

The output load current of the converter comprises of two components per cycle. One component flows when thyristors S1 and S2 are fired and connects the supply voltage to the load and the other component flows when thyristors S3 and S4 are turned on again connecting supply voltage to the load. Since both components of current are identical, only one component will be studied. For the interval α ≤ ωt ≤ (α + π) the load current is given by (at ωt = α, i_L = I_L0 )

$L\frac{{d{i_L}}}{{dt}} + R{i_L} + E = {V_m}\sin \omega t$

${i_L} = \frac{{{V_m}}}{Z}\sin (\omega t - \theta ) + \left[ {{I_{L0}} + \frac{E}{R} - \frac{{{V_m}}}{Z}\sin (\alpha - \theta )} \right]{e^{(R/L)(\alpha /\omega - t)}} - \frac{E}{R}$

Single-Phase Full Converter with RL load

The current magnitude at the end of the first component of load current is the same as that at the beginning of the second component of load current, that is at ωt = π+ α, i_L = I_L0 and the current I_L0 can be obtained by substituting this condition in the above equation which yields

${I_{L0}} = \frac{{{V_m}}}{Z}\frac{{\left[ { - \sin (\alpha - \theta ) - \sin (\alpha - \theta ){e^{ - (R/L)(\pi /\omega )}}} \right]}}{{1 - {e^{ - (R/L)(\pi /\omega )}}}} - \frac{E}{R}$

The value of firing angle α at which current I_L0 = 0 can be obtained for known values of the parameters in the above equation, using an iterative method. The rms value of thyristor current is given by

${I_R} = {\left[ {\frac{1}{{2\pi }}\int\limits_\alpha ^{\pi + \alpha } {i_L^2d(\omega t)} } \right]^{ - \frac{1}{2}}}$

The rms output current is given by

${I_{rms}} = {[I_R^2 + I_R^2]^{\frac{1}{2}}} = \sqrt 2 {I_R}$

The average output current is given by
I_dc = I_A + I_A = 2I_A
The average current of one thyristor is given by

${I_A} = \frac{1}{{2\pi }}\int\limits_\alpha ^{\pi + \alpha } {{i_L}d\left( {\omega t} \right)}$

previous Full Controlled Rectifier

next SINGLE-PHASE CONVERTERs

Full Controlled Rectifier

Full Controlled Rectifier with R Load

If α= 0° ,The output will be same as a Full Uncontrolled Rectifier that uses DIODE.

Average Load/Output Voltage

${V_0} = \frac{2}{{2\pi }}\int_\alpha ^\pi {\sqrt 2 V\sin \theta d\theta }$

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

Average Load/Output Current

${I_0} = \frac{{{V_0}}}{R}$

Full Controlled Rectifier with DC Motor Load
For 90° < α < 180°

Average Load/Output Voltage
${V_0} = \frac{2}{{2\pi }}\int_\alpha ^{\pi + \alpha } {{V_m}\sin \theta d\theta }$

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

Average Load/Output Current

V_o= I_oR_a+ E_a

E_a = armature back emf.

For 90° < α < 180°

Converter Output Characteristics for continuous load current

For fully controlled rectifier, the DC Motor operates in two modes.
1. Rectification [As Motoring]
V₀ = positive
E_a = Positive
I_o= positive
Power Flow (+ve) from input AC to DC machine
2. Inversion [As Regenerative Braking]
V₀ = negative
E_a = negative
I_o= positive
Power Flow (-ve) from DC machine to AC supply

The output voltage can be varied from a maximum of 2Vm/π to a minimum of zero as the firing angle a varies from zero to π. The rms output voltage is given by

${V_{rms}} = {\left[ {\frac{2}{{2\pi }}\int\limits_\alpha ^\pi {V_m^2{{\sin }^2}\omega td\left( {\omega t} \right)} } \right]^{\frac{1}{2}}}$

${V_{rms}} = \frac{{{V_m}}}{{\sqrt 2 }}{\left[ {\frac{1}{\pi }\left( {\pi - \alpha + \frac{{\sin 2\alpha }}{2}} \right)} \right]^{\frac{1}{2}}}$

previous Performance of Single-phase, half-wave controlled rectifiers

next Single-Phase Full Converter with RL load

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