Q3 Flashcards

1
Q

Q3. slide 1 main ideas

A
  • six-pulse rectifier most common topology for 3 phase ac voltages
  • simplify analysis, redraw circuit
  • Consists of 2 group of diodes, top and bottom group
  • Top group diode operation
  • Bottom group diode operatioin
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2
Q

Explain operation of 6-pulse rectifier

A
  • In top group of diodes, cathode terminals at same potential
  • diode with its anode at highest potential will conduct, other 2 will be reversed biased
  • in bottom group, anode terminals at same potential
  • diode with its cathode at the lowest potential will conduct, other 2 will be reversed biased
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3
Q

Q3. slide 2 main ideas

A
  • see voltage waveforms 3 phase voltages, output voltage of rectifier, phase a input I assuming constant output I
  • Voltages Vpn and Vnn, are voltages at points p and n with respect to ac voltage neutral point
  • Output voltage equal to difference of Vpn and Vnn
  • Top group and bottom group operation
  • figure of phase a
  • calculate average output voltage
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4
Q

Q3. slide 2 explain top and bottom group operation

A
  • In top group diodes with its anode at highest potential will conduct
  • Vpn will be equal to phase voltage with highest potential
  • in bottom group same principle applies
  • Vnn equal to phase voltage with lowest potential
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5
Q

Q3. slide 2 explain phase a I figure

A
  • phase a I equal to output I when D1 conducts, that is when phase a voltage is the highest
  • phase a I equal to - output I when D4 conducts, that0s when phase a voltage is the lowest
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6
Q

How to obtain average output voltage of six-pulse rectifier diode

A
  • To obtain average output voltage, we consider one of 6 segments of output voltage
  • each segment of output voltage equals L-L voltage
  • L-L voltage can be expressed as a cosine function from -pi/6 to pi/6
  • we integrate this expression and divide by pi/3
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7
Q

Q3. slide 3 main ideas

A
  • If we replace constant output I for and C and R we can have two cases: Ls small, Ls large
  • Input I when Ls small
  • Explain phase a I with figure
  • Input I goes to zero
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8
Q

slide 3. explain input I waveform when Ls small

A
  • Input I in each half-cycle consists of 2 pulses occurring when L-L voltage is higher than output voltage
  • for phase a I, first pulse occurs when Vab is higher than output voltage, and D1 and D6 are conducting
  • second pulse occurs when Vac is higher than output voltage, and D1 and D2 are conducting.
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9
Q

Q3. slide 4 explain effect of large Ls

A
  • input I in each half-cycle does not go to zero
  • it increases conduction time, reduces peak I, and reduces THD
  • We obtain smoother current rise
  • Impedance of Ls is usually selected as 2% or more than base impedance
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10
Q

Explain general impact of grid-side inductance

A
  • Increases conduction time and reduces peak I
  • I waveform is improved imrpoving PF to an acceptable level
  • THD is reduced
  • Output voltage depends on output load, lower compared to case with no L (due to voltage drop in L and varying output I)
  • L and output C form a LPF, reducing peak-peak voltage ripple at the output
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11
Q

Q3. slide 6 main ideas

A
  • Necessary to control large inrush I that occurs when rectifier is connected to voltage source
  • Cause of inrush I
  • Problem of inrush I
  • Solutions
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12
Q

Explain cause of inrush I and problems

A
  • caused by the very large output C (to keep voltage nearly constant)
  • Initially voltage across C is zero, thus act as a short circuit and large I flows through diode to charge dc-capacitor
  • Large I might destroy the diode
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13
Q

Describe solutions to limit inrush I

A
  • current-limiting resistor or NTC allows to charge the C with a small controlled I. When C reaches certain voltage, R is bypassed by mechanical contactors or thyristor to reduce power dissipation.
  • Replace diode by thyristors to control start of conduction. Thyristors turned on at instant close to the end of half-wave so that source voltage is small and I to charge capacitor is limited. When C is charge to 80% thyristors are turned as diodes.
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14
Q

Main problems of passive rectifiers

A
  • Input I are not pure sinusoidal, leading to high THD and low PF.
  • Injection of harmonic I impacts quality of AC voltage waveforms.
  • Harmonic distortion and low PF overloads circuit wiring (if PF is low, I to supply given load increases).
  • There is no control on the output voltage.
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15
Q

List and explain solution for problems of passive rectifiers

A
  • Add regulator circuits to provide stable output voltage regardless of input voltage.
  • THD, PF and I waveform can be improved by used large grid side L or filters.
  • PFC converter to shape the input I
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16
Q

Characteristics and advantages of PFC boost

A
  • Allows to shape I current to a sinusoidal waveform and improving THD and PF
  • No isolated
  • Regulated voltage will be higher than maximum input voltage
  • CCM or DCM operation modes
  • Nearly unity PF (CCM)
  • Higher cost, higher power losses, more complex.