Q3

Question 1

Q3. slide 1 main ideas

Answer 1

• 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

Question 2

Explain operation of 6-pulse rectifier

Answer 2

• 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

Question 3

Q3. slide 2 main ideas

Answer 3

• 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

Question 4

Q3. slide 2 explain top and bottom group operation

Answer 4

• 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

Question 5

Q3. slide 2 explain phase a I figure

Answer 5

• 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

Question 6

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

Answer 6

• 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

Question 7

Q3. slide 3 main ideas

Answer 7

• 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

Question 8

slide 3. explain input I waveform when Ls small

Answer 8

• 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.

Question 9

Q3. slide 4 explain effect of large Ls

Answer 9

• 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

Question 10

Explain general impact of grid-side inductance

Answer 10

• 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

Question 11

Q3. slide 6 main ideas

Answer 11

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

Question 12

Explain cause of inrush I and problems

Answer 12

• 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

Question 13

Describe solutions to limit inrush I

Answer 13

• 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.

Question 14

Main problems of passive rectifiers

Answer 14

• 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.

Question 15

List and explain solution for problems of passive rectifiers

Answer 15

• 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

Question 16

Characteristics and advantages of PFC boost

Answer 16

• 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.