Lecture 12 & 13
description
Transcript of Lecture 12 & 13
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18/04/2023Rijil Ramchand
Electrical Drives
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Lecture 12 & 13 (24/25-02-2015)
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18/04/2023Rijil Ramchand2
Controlled rectifiers produce variable DC output, whose magnitude is varied by Phase control.
Phase Control
DC output from rectifier is controlled by controlling duration of the conduction period by varying the point at which gate signal is applied to SCR.
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Controlled rectifiers are of two types,1- Fully Controlled rectifiers
DC current is unidirectional, but DC voltage has either polarity. With one polarity, flow of power is from AC source to DC load---Rectification.
With the reversal of DC voltage by the load, flow of power is from DC load to AC source---Inversion.
2- Half controlled rectifiersHalf of SCRs are replaced by diodes.
DC output current and voltage are unidirectional. i.e., flow of power is from AC source to DC load.
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Half-Wave Controlled Rectifiers
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With Resistive Load
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Control characteristics of half-wave rectifier
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With an Inductive (RL) Load
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With Inductive Load and Freewheeling Diode
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Full-Wave Controlled Center-Tap Rectifiers
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With Resistive Load
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With an Inductive (RL) Load
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Control Characteristics for center-tap rectifier
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With Freewheeling Diode
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Example 6.4
Explain with the help of waveforms the operation of a full-wave center-tap rectifier with RL load for the following firing angles:(a) 0°
(b) 45°
(c) 90°
(d) 135°
(e) 180°
Assume highly Inductive Load14
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During positive-half cycle of source voltage, SCR1 is forward biased and SCR2 is reverse biased. During negative half-cycle, SCR2 is forward biased and SCR1 is reverse biased. In either case voltage across the load is Vs.
Output is similar to uncontrolled rectifier.
Each SCR conducts for 180° and supplies current to the load for this period
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Voltage and current waveforms for α=0°
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Average DC output voltage decreases.
If SCR1 is triggered at 45°, SCR2 will conduct upto that point, even though the source voltage is zero, due to highly inductive nature of load.
When SCR1 is turned on, SCR2 is turned off.
Current to the load is supplied by SCR1 and SCR2, each conducting for 180°
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Voltage and current waveforms for α=45°
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Average DC voltage is zero, so there is no transfer of power from AC source to DC load.
Each SCR remains in conduction for 180°
As firing angle is increased from 0 to 90°, the power supplied to the DC load decreases, becoming zero at α=90°
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Voltage and current waveforms for α=90°
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Average DC voltage is negative.
Load current still flows in each SCR for 180° in its original direction.
Load voltage has changed polarity.
Power now flows from DC load to AC source .
Circuit acts like an inverter.
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Voltage and current waveforms for α=135°
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Average output DC voltage is at its maximum negative value.
SCRs remain in conduction for 180°
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Voltage and current waveforms for α=180°
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Example
Show direction of power flow and operating mode (rectifying or inversion) of center-tap rectifier circuit with following firing angles:
A) α > 0°
B) α < 90°
C) α > 90°
D) α < 180°
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Solution
For firing angle in the range 0° < α < 90°
1. Average output voltage is positive.
2. Converter operates in the rectifying mode.
3. Power to the load is positive
4. Power flow is from AC source to the DC load.
For firing angle in the range 90° < α < 180°
1. Average output voltage is negative
2. Converter operates in inversion mode
3. Power to the load is negative
4. Power flow is from DC load to AC source21
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Full-Wave Controlled Bridge Rectifier
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With Resistive Load
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18/04/2023Rijil Ramchand24
)cos1(
)t(d)tsin(V1
V mo
mV
angle delay
)cos1(R
V
R
VI mo
o
4
)2sin(
22
1
R
V
)t(d)tsinR
V(
1I
m
2mrms
The power delivered to the load RIP rms2
The rms current in source is the same as the rms current in the load.
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With an Inductive (RL) Load
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)cos(cos
)t(d)tsin(V1
V mo
mV
angle delay
)cos(cosR
V
R
VI mo
o
4
)2sin(
4
)2sin(
22R
V
)t(d)tsinR
V(
1I
m
2mrms
The power delivered to the load RIP rms2
The rms current in source is the same as the rms current in the load.
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For L >>> R
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cos
)t(d)tsin(V1
V mo
m2V
angle delay
)(cosR
V2
R
VI mo
o
)t(d)tsin
R
V(
1I 2m
rms
The power delivered to the load RIP rms2
The rms current in source is the same as the rms current in the load.
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Control characteristics for bridge rectifier
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With RL load and freewheeling diode
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Half-Controlled or
Semicontrolled Bridge Rectifiers
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In fully-controlled rectifier, only rectification can be obtained by connecting a freewheeling diode across the output terminals of the rectifier.
Another method of obtaining rectification in bridge rectifiers is replacing half of the SCRs with diodes. These circuits are called semicontrolled bridge rectifiers.
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Full-wave semicontrolled bridge rectifier circuit
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Semicontrolled bridge rectifier with FWD
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Dual Converter
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1. A purely resistive load is supplied with power from an ideal single-phase supply of fixed voltage and frequency, denoted by e =Em sinωt, through an arrangement of four ideal diodes and an ideal SCR switch (Figure below). The switch is gated during every half cycle of its positive anode voltage state. Sketch (1) the supply current waveform and (2) the load current waveform for a switching angle α of 600. Derive an expression for the average value of the load current in terms of Em and α. What is the per-unit value of this when α is 600 compared with α =0?
e =Em sinωt
R
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2. A single-phase full converter is designed to have an average output current of 10A through a load resistance of 20Ω when supplied from a 230V, 50Hz supply. Due to some problems in the triggering circuit there was a delay of 1ms in triggering from the desired triggering instant. Find the average value of the voltage across the load and the average current through the load under this condition.
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3. A rectifier circuit containing two ideal diodes D and an ideal semiconductor switch (shown as an SCR) is given in Figure below. Sketch waveforms for the supply voltage, the three resistor currents, the switch current iT, and the supply current is if the SCR firing angle α=600. Write a mathematical expression for is(ωt) to define the waveform of the supply current and calculate its average value over a supply cycle in terms of Vm.
V=Vm sin ωt
10Ω
20Ω
30Ω
D
D
is