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Power Electronics 06EC73
SJBIT/Dept of ECE Page 1
UNIT 1
Introduction to Power Electronics
1. What are the important applications of power electronics?
(Dec 2011, Jun 2011, Jun 2010)
Solution:
Power Electronic Applications
1. COMMERCIAL APPLICATIONS
Heating Systems Ventilating, Air Conditioners, Central Refrigeration, Lighting,
Computers and Office equipments, Uninterruptible Power Supplies (UPS), Elevators,
and Emergency Lamps.
2. DOMESTIC APPLICATIONS
Cooking Equipments, Lighting, Heating, Air Conditioners, Refrigerators &
Freezers, Personal Computers, Entertainment Equipments, UPS.
3. INDUSTRIAL APPLICATIONS
Pumps, compressors, blowers and fans. Machine tools, arc furnaces, induction
furnaces, lighting control circuits, industrial lasers, induction heating, welding
equipments.
4. AEROSPACE APPLICATIONS
Space shuttle power supply systems, satellite power systems, aircraft power
systems.
5. TELECOMMUNICATIONS
Battery chargers, power supplies (DC and UPS), mobile cell phone battery
chargers.
6. TRANSPORTATION
Traction control of electric vehicles, battery chargers for electric vehicles, electric
locomotives, street cars, trolley buses, automobile electronics including engine
controls.
2. Mention and explain the different types of power electronics converter systems. Draw
there input/output characteristics. (Dec 2011, Jun 2011, Dec 2010)
Solution:
1. AC TO DC Converters (Rectifiers)
Line
CommutatedConverter
+
-
DC Output
V0(QC)
AC
Input
Voltage
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These are AC to DC converters. The line commutated converters are AC to DC power
converters. These are also referred to as controlled rectifiers. The line commutated converters
(controlled rectifiers) are used to convert a fixed voltage, fixed frequency AC power supply
to obtain a variable DC output voltage. They use natural or AC line commutation of the
Thyristors.
Fig1.4: A Single Phase Full Wave Uncontrolled Rectifier Circuit (Diode Full Wave Rectifier) using a
Center Tapped Transformer
Fig: 1.5 A Single Phase Full Wave Controlled Rectifier Circuit (using SCRs) using a Center Tapped
Transformer
Different types of line commutated AC to DC converters circuits are
Diode rectifiers – Uncontrolled Rectifiers
Controlled rectifiers using SCR’s.
o Single phase controlled rectifier.
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o Three phase controlled rectifiers.
Applications of Ac To Dc Converters
AC to DC power converters are widely used in
Speed control of DC motor in DC drives.
UPS.
HVDC transmission.
Battery Chargers.
2. a. AC TO AC Converters or AC regulators.
The AC voltage controllers convert the constant frequency, fixed voltage AC supply into
variable AC voltage at the same frequency using line commutation.
AC regulators (RMS voltage controllers) are mainly used for
Speed control of AC motor.
Speed control of fans (domestic and industrial fans).
AC pumps.
Fig.1.6: A Single Phase AC voltage Controller Circuit (AC-AC Converter using a TRIAC)
AC
VoltageController
V0(RMS)
fS
Variable AC
RMS O/P Voltage
AC
Input
Voltage
fs
Vs
fs
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2. b. AC TO AC Converters with Low Output Frequency or CYCLO CONVERTERS
The cyclo converters convert power from a fixed voltage fixed frequency AC supply to a
variable frequency and variable AC voltage at the output.
The cyclo converters generally produce output AC voltage at a lower output frequency. That
is output frequency of the AC output is less than input AC supply frequency.
Applications of cyclo converters are traction vehicles and gearless rotary kilns.
3. CHOPPERS or DC TO DC Converters
The choppers are power circuits which obtain power from a fixed voltage DC supply and
convert it into a variable DC voltage. They are also called as DC choppers or DC to DC
converters. Choppers employ forced commutation to turn off the Thyristors. DC choppers are
further classified into several types depending on the direction of power flow and the type of
commutation. DC choppers are widely used in
Speed control of DC motors from a DC supply.
DC drives for sub-urban traction.
Switching power supplies.
Cyclo
Converters
V , f0 0
f < f0 S
Variable Frequency
AC Output
Vs
fs
ACInput
Voltage
DC
Chopper
V0(dc)
-
Variable DC
Output VoltageVs
+
+
-
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Fig.1.7: A DC Chopper Circuit (DC-DC Converter) using IGBT
4. INVERTERS or DC TO AC Converters
The inverters are used for converting DC power from a fixed voltage DC supply into an AC
output voltage of variable frequency and fixed or variable output AC voltage. The inverters
also employ force commutation method to turn off the Thyristors.
Applications of inverters are in
Industrial AC drives using induction and synchronous motors.
Uninterrupted power supplies (UPS system) used for computers, computer labs.
Inverter
(ForcedCommutation)
AC
Output Voltage
+
-
DC
Supply
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Fig.1.8: Single Phase DC-AC Converter (Inverter) using MOSFETS
3. Mention and explain the classification of power semiconductor switching device, on
the bases of control characteristics. Give an examples. (Jun 2011, Dec 2010)
Solution:
The power semiconductor devices are used as switches. Depending on power
requirements, ratings, fastness & control circuits for different devices can be selected. The
required output is obtained by varying conduction time of these switching devices.
Control characteristics of Thyristors:
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Fig1.3: Control Characteristics of Power Switching Devices
4. Explain peripheral effects of power converter system. (Dec 2011)
Solution:
The power converter operations are based mainly on the switching of power
semiconductor devices and as a result the power converters introduce current and voltage
harmonics (unwanted AC signal components) into the supply system and on the output of the
converters.
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Fig.1.9: A General Power Converter System
These induced harmonics can cause problems of distortion of the output voltage,
harmonic generation into the supply system, and interference with the communication and
signaling circuits. It is normally necessary to introduce filters on the input side and output
side of a power converter system so as to reduce the harmonic level to an acceptable
magnitude. The figure below shows the block diagram of a generalized power converter with
filters added. The application of power electronics to supply the sensitive electronic loads
poses a challenge on the power quality issues and raises the problems and concerns to be
resolved by the researchers. The input and output quantities of power converters could be
either AC or DC. Factors such as total harmonic distortion (THD), displacement factor or
harmonic factor (HF), and input power factor (IPF), are measures of the quality of the
waveforms. To determine these factors it is required to find the harmonic content of the
waveforms. To evaluate the performance of a converter, the input and output
voltages/currents of a converter are expressed in Fourier series. The quality of a power
converter is judged by the quality of its voltage and current waveforms.
The control strategy for the power converters plays an important part on the harmonic
generation and the output waveform distortion and can be aimed to minimize or reduce these
problems. The power converters can cause radio frequency interference due to
electromagnetic radiation and the gating circuits may generate erroneous signals. This
interference can be avoided by proper grounding and shielding.
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UNIT 2
Power Transistor
1. With the necessary waveforms, explain the switching characteristics of a power
Transister. (Dec 2011, Dec 2010)
Solution: SWITCHING CHARACTERISTICS
A forward biased p-n junction exhibits two parallel capacitances; a depletion layer
capacitance and a diffusion capacitance. On the other hand, a reverse biased p-n junction has
only depletion capacitance. Under steady state the capacitances do not play any role.
However under transient conditions, they influence turn-on and turn-off behavior of the
transistor.
Transient Model Of BJT
Fig. 7: Transient Model of BJT
Fig. 8: Switching Times of BJT
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Due to internal capacitances, the transistor does not turn on instantly. As the voltage VB rises
from zero to V1 and the base current rises to IB1, the collector current does not respond
immediately. There is a delay known as delay time td, before any collector current flows. The
delay is due to the time required to charge up the BEJ to the forward bias voltage VBE(0.7V).
The collector current rises to the steady value of ICS and this time is called rise time tr.
The base current is normally more than that required to saturate the transistor. As a result
excess minority carrier charge is stored in the base region. The higher the ODF, the greater is
the amount of extra charge stored in the base. This extra charge which is called the saturating
charge is proportional to the excess base drive.
This extra charge which is called the saturating charge, is proportional to the excess base
drive and the corresponding current Ie.
. 1CSe B BS BS BS
II I ODF I I I ODF
Saturating charge ( 1)S s e s BSQ I I ODF where s is known as the storage time constant.
When the input voltage is reversed from V1 to -V2, the reverse current –IB2 helps to discharge
the base. Without –IB2 the saturating charge has to be removed entirely due to recombination
and the storage time ts would be longer.
Once the extra charge is removed, BEJ charges to the input voltage –V2 and the base current
falls to zero. tf depends on the time constant which is determined by the reverse biased BEJ
capacitance.
on d r
off s f
t t t
t t t
2. Compare the BJT and MOSFET. (Dec 2011)
Solution:
Comparison of MOSFET With BJT
Power MOSFETS have lower switching losses but its on-resistance and conduction
losses are more. A BJT has higher switching loss bit lower conduction loss. So at high
frequency applications power MOSFET is the obvious choice. But at lower operating
frequencies BJT is superior.
MOSFET has positive temperature coefficient for resistance. This makes parallel
operation of MOSFET’s easy. If a MOSFET shares increased current initially, it heats
up faster, its resistance increases and this increased resistance causes this current to
shift to other devices in parallel. A BJT is a negative temperature coefficient, so
current shaving resistors are necessary during parallel operation of BJT’s.
In MOSFET secondary breakdown does not occur because it have positive
temperature coefficient. But BJT exhibits negative temperature coefficient which
results in secondary breakdown.
Power MOSFET’s in higher voltage ratings have more conduction losses.
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Power MOSFET’s have lower ratings compared to BJT’s . Power MOSFET’s
500V to 140A, BJT 1200V, 800A.
3. Discuss the different methods of providing isolation of gate drive circuits from power
circuit. (Dec 2011)
Solution:
Necessity
Driver circuits are operated at very low power levels. Normally the gating circuit are digital in
nature which means the signal levels are 3 to 12 volts. The gate and base drives are connected
to power devices which operate at high power levels.
There are two ways of floating or isolating control or gate signal with respect to ground.
Pulse transformers
Optocouplers
Pulse Transformers
Pulse transformers have one primary winding and can have one or more secondary
windings.
Multiple secondary windings allow simultaneous gating signals to series and parallel
connected transistors. The transformer should have a very small leakage inductance and the
rise time of output should be very small.
The transformer would saturate at low switching frequency and output would be distorted.
V1
-V2
0
Logicdrive
circuit
Q1
IC
RC
+
-VCC
RB
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Optocouplers
Optocouplers combine infrared LED and a silicon photo transistor. The input signal is
applied to ILED and the output is taken from the photo transistor. The rise and fall times of
photo transistor are very small with typical values of turn on time = 2.5 s and turn off of
300ns. This limits the high frequency applications. The photo transistor could be a darlington
pair. The phototransistor requires separate power supply and adds to complexity and cost and
weight of driver circuits.
4. What is the necessity of base drive control in power transistors? Explain proportional
base control. (Jun 2011)
Solution:
This is required to optimize the base drive of transistor. Optimization is
required to increase switching speeds. ont can be reduced by allowing base current peaking
during turn-on, CSF
B
Iforced
I resulting in low forces at the beginning. After turn
on, F can be increased to a sufficiently high value to maintain the transistor in quasi-
saturation region. offt can be reduced by reversing base current and allowing base current
peaking during turn off since increasing 2BI decreases storage time.
A typical waveform for base current is shown.
Fig.2.10: Base Drive Current Waveform
Optocoupler
R
Vg1
1
+
-
RBR1
1Q1
0
R2
+VCC
Q3G
R3
ID
RGRD
S
M1
D
ID
G
VDD
+
-1
t0
-IB2
IBS
IBIB1
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Some common types of optimizing base drive of transistor are
Turn-on Control.
Turn-off Control.
Proportional Base Control.
Antisaturation Control
Proportional Base Control
This type of control has advantages over the constant drive circuit. If the collector
current changes due to change in load demand, the base drive current is changed in
proportion to collector current.
When switch 1S is turned on a pulse current of short duration would flow through the base of
transistor 1Q and 1Q is turned on into saturation. Once the collector current starts to flow, a
corresponding base current is induced due to transformer action. The transistor would latch
on itself and 1S can be turned off. The turns ratio is 2
1
C
B
INN I
. For proper operation
of the circuit, the magnetizing current which must be much smaller than the collector current
should be as small as possible. The switch 1S can be implemented by a small signal transistor
and additional arrangement is necessary to discharge capacitor 1C and reset the transformer
core during turn-off of the power transistor.
Fig.2.13: Proportional base drive circuit
5. For the transistor switch as shown in figure (Jun 2011)
a. Calculate forced beta, f
of transistor.
b. If the manufacturers specified is in the range of 8 to 40, calculate the
minimum overdrive factor (ODF).
c. Obtain power loss TP in the transistor.
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Solution
(i) 10 1.5
11.330.75
B BE sat
B
B
V VI A
R
200 1.0
18.0911
CC CE sat
CS
C
V VI A
R
Therefore min
18.092.26
8
CSBS
II A
18.09
1.611.33
CSf
B
I
I
(ii) 11.33
5.012.26
B
BS
IODF
I
(iii) 1.5 11.33 1.0 18.09 35.085T BE B CE CP V I V I W
6. For the transistor switch as shown in figure (Dec 2010)
d. Calculate forced beta, fof transistor.
e. If the manufacturers specified is in the range of 8 to 40, calculate the
minimum overdrive factor (ODF).
f. Obtain power loss TP in the transistor.
10 , 0.75 ,
1.5 , 11 ,
1 , 200
B B
CBE sat
CCCE sat
V V R
V V R
V V V V
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Solution
(i) 10 1.5
11.330.75
B BE sat
B
B
V VI A
R
200 1.0
18.0911
CC CE sat
CS
C
V VI A
R
Therefore min
18.092.26
8
CSBS
II A
18.09
1.611.33
CSf
B
I
I
(ii) 11.33
5.012.26
B
BS
IODF
I
(iii) 1.5 11.33 1.0 18.09 35.085T BE B CE CP V I V I W
7. The of a bipolar transistor varies from 12 to 75. The load resistance is 1.5CR .
The dc supply voltage is VCC=40V and the input voltage base circuit is VB=6V. If VCE (sat)
=1.2V, VBE (sat) =1.6V and RB=0.7 determine
a. The overdrive factor ODF.
b. The forced f.
c. Power loss in transistor PT (Jun 2010)
Solution
( ) 40 1.2
25.861.5
CC CE sat
CS
C
V VI A
R
min
25.862.15
12
CSBS
II A
10 , 0.75 ,
1.5 , 11 ,
1 , 200
B B
CBE sat
CCCE sat
V V R
V V R
V V V V
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Also ( ) 6 1.6
6.280.7
B BE sat
B
B
V VI A
R
(a) Therefore 6.28
2.922.15
B
BS
IODF
I
Forced 25.86
4.116.28
CSf
B
I
I
(c) T BE B CE CP V I V I
1.6 6.25 1.2 25.86
41.032
T
T
P
P Watts
8. Explain the steady state characteristics of Power MOSFET and compare this with
power BJT. (Jun 2010)
Solution:
Fig.2.15 Symbol of n-channel depletion type MOSFET
Operation
When 0GSV V and DSV is applied and current flows from drain to source similar to
JFET. When 1GSV V , the negative potential will tend to pressure electrons towards the p-
type substrate and attracts hole from p-type substrate. Therefore recombination occurs and
will reduce the number of free electrons in the n-channel for conduction. Therefore with
increased negative gate voltage DI reduces.
For positive values,gsV , additional electrons from p-substrate will flow into the channel and
establish new carriers which will result in an increase in drain current with positive gate
voltage.
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Drain Characteristics
Transfer Characteristics
9. With necessary sketches, explain briefly the switching characteristics of IGBT.
(Jun 2011)
Solution:
SWITCHING CHARACTERISTIC OF IGBT
Figure below shows the switching characteristic of an IGBT. Turn-on time consists of
delay time d on
t and rise time rt .
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Fig. : Switching Characteristics
The turn on delay time is the time required by the leakage current CEI to rise to 0.1 CI , where
CI is the final value of collector current. Rise time is the time required for collector current
to rise from 0.1 CI to its final value CI . After turn-on collector-emitter voltage CEV will be
very small during the steady state conduction of the device.
The turn-off time consists of delay off time d off
t and fall timeft . Off time delay is the time
during which collector current falls from CI to 0.9 CI and GEV falls to threshold voltage GETV
. During the fall time ft the collector current falls from 0.90 CI to 0.1 CI . During the turn-off
time interval collector-emitter voltage rises to its final value CEV .
IGBT’s are voltage controlled power transistor. They are faster than BJT’s, but still not quite
as fast as MOSFET’s. the IGBT’s offer for superior drive and output characteristics when
compared to BJT’s. IGBT’s are suitable for high voltage, high current and frequencies upto
20KHz. IGBT’s are available upto 1400V, 600A and 1200V, 1000A.
t
t
t
VGET
0.9 VCE0.9 VCE
0.9 ICE
0.1 VCE0.1 VCE
0.1 ICE
IC
VGE
VCE
td(on)td(off)
td(off)
tf
tf
tr
t = t +t
t = t +t(on) d(on) r
(off) d(off) f
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UNIT 3
Introduction to Thysistors
1. Draw the transistor model of a thyristor and derive an expression for the anode
current in terms of the common base current gain of the transirtor.
(Dec 2011, Dec 2010)
Solution:
Two Transistor Model
The general transistor equations are,
1
1
C B CBO
C E CBO
E C B
B E CBO
I I I
I I I
I I I
I I I
The SCR can be considered to be made up of two transistors as shown in above figure.
Considering PNP transistor of the equivalent circuit,
1 1 1 1
1 1
1
1
, , , ,
1 1
E A C C CBO CBO B B
B A CBO
I I I I I I I I
I I I
Considering NPN transistor of the equivalent circuit,
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2 2 2
2 2
2 2
2
2
, ,
2
C C B B E K A G
C k CBO
C A G CBO
I I I I I I I I
I I I
I I I I
From the equivalent circuit, we see that
2 1
2 1 2
1 21
C B
g CBO CBO
A
I I
I I II
Two transistors analog is valid only till SCR reaches ON state
Case 1: When 0gI ,
1 2
1 21
CBO CBO
A
I II
The gain 1 of transistor 1T varies with its emitter current E AI I . Similarly varies with
E A g KI I I I . In this case, with 0gI , 2 varies only with AI . Initially when the applied
forward voltage is small, 1 2 1 .
If however the reverse leakage current is increased by increasing the applied forward voltage,
the gains of the transistor increase, resulting in 1 2 1 .
From the equation, it is seen that when 1 2 1 , the anode current AI tends towards .
This explains the increase in anode current for the break over voltage 0BV .
Case 2: With gate current gI applied.
When sufficient gate drive is applied, we see that 2B gI I is established. This in turn results
in a current through transistor 2T , these increases 2 of 2T . But with the existence of
2 22 2C gI I I , a current through T, is established. Therefore,
1 1 21 1 2 1 2C B B gI I I I . This current in turn is connected to the base of 2T . Thus the
base drive of 2T is increased which in turn increases the base drive of 1T , therefore
regenerative feedback or positive feedback is established between the two transistors. This
causes 1 2 to tend to unity therefore the anode current begins to grow towards a large
value. This regeneration continues even if gI is removed this characteristic of SCR makes it
suitable for pulse triggering; SCR is also called a Lathing Device.
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2. Write a short note on dv/dt and di/dt protection. (Dec 2011)
Solution:
The dv
dt across the thyristor is limited by using snubber circuit as shown in figure (a)
below. If switch 1S is closed at 0t , the rate of rise of voltage across the thyristor is limited
by the capacitor SC . When thyristor 1T is turned on, the discharge current of the capacitor is
limited by the resistor SR as shown in figure (b) below.
Fig.3.18 (a)
Fig.3.18 (b)
Fig.3.18 (c)
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The voltage across the thyristor will rise exponentially as shown by fig (c) above. From fig.
(b) above, circuit we have (for SCR off)
t 0
10S S c for
V i t R i t dt VC
.
Therefore s
tS
S
Vi t e
R, where s S SR C
Also T S SV t V i t R
s
tS
T S S
S
VV t V e R
R
Therefore 1s s
t t
T S S SV t V V e V e
At t = 0, 0 0TV
At st , 0.632T s SV V
Therefore 0 0.632T s T S
s S S
V V Vdv
dt R C
And SS
TD
VR
I.
TDI is the discharge current of the capacitor.
It is possible to use more than one resistor for dv
dt and discharging as shown in the
figure (d) below. The dv
dt is limited by 1R and SC . 1 2R R limits the discharging current such
that 1 2
STD
VI
R R
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Fig.3.18 (d)
The load can form a series circuit with the snubber network as shown in figure (e) below.
The damping ratio of this second order system consisting RLC network is given as,
0 2
S S
S
R R C
L L, where SL stray inductance and L, R is is load inductance
and resistance respectively.
To limit the peak overshoot applied across the thyristor, the damping ratio should be in the
range of 0.5 to 1. If the load inductance is high, SR can be high and SC can be small to retain
the desired value of damping ratio. A high value of SR will reduce discharge current and a
low value of SC reduces snubber loss. The damping ratio is calculated for a particular circuit
SR and SC can be found.
Fig.3.18 (e)
di
dt PROTECTION
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Practical devices must be protected against high di
dt . As an example let us consider the
circuit shown above, under steady state operation mD conducts when thyristor 1T is off. If 1T
is fired when mD is still conducting di
dtcan be very high and limited only by the stray
inductance of the circuit. In practice the di
dt is limited by adding a series inductor SL as
shown in the circuit above. Then the forward S
S
Vdi
dt L.
3. Sketch the gate characteristics of SCR and explain different regions of gate
characteristics. Also indicate different regions, different voltages, and different currents
on gate characteristics. (Jun 2011)
Solution:
Thyristor Gate Characteristics
Fig 3.6 Gate Characteristics
Fig. 3.6 shows the gate trigger characteristics.The gate voltage is plotted with respect to
gate current in the above characteristics. Ig(max) is the maximum gate current that can flow
through the thyristor without damaging it Similarly Vg(max) is the maximum gate voltage to be
applied. Similarly Vg (min) and Ig(min) are minimum gate voltage and current, below which
thyristor will not be turned-on. Hence to turn-on the thyristor successfully the gate current
and voltage should be
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Ig(min) < Ig < Ig(max)
Vg (min) < Vg < Vg (max)
The characteristic of Fig. 3.6 also shows the curve for constant gate power (Pg). Thus for
reliable turn-on, the (Vg, Ig) point must lie in the shaded area in Fig. 3.6. It turns-on thyristor
successfully. Note that any spurious voltage/current spikes at the gate must be less than Vg
(min) and Ig(min) to avoid false triggering of the thyristor. The gate characteristics shown in Fig.
3.6 are for DC values of gate voltage and current.
4. Explain Resistance triggering. (Jun 2011)
Solution:
A simple resistance triggering circuit is as shown. The resistor 1R limits the current
through the gate of the SCR. 2R is the variable resistance added to the circuit to achieve
control over the triggering angle of SCR. Resistor ‘R’ is a stabilizing resistor. The diode D is
required to ensure that no negative voltage reaches the gate of the SCR.
Fig.3.4: Resistance firing circuit
LOAD
vOa b
i R1
R2
D
R Vg
VT
v =V sin tS m
VS
2
3 4
t
V sin tm
Vg Vgt
t
t
t
t
Vo
io
VT
VgpVgtVgp
(a)
t
t
t
t
t
t
t
t
t
t
2
3 4
2
3 4
VS
Vg
Vo
io
VT
VS
Vg
Vo
io
VT
V =Vgp gt
2700
2
3 4
900 =90
0
(c)(b)
<900
V >Vgp gt
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Fig.3.8: Resistance firing of an SCR in half wave circuit with dc load
(a) No triggering of SCR (b) = 900 (c) < 90
0
Design
With 2 0R , we need to ensure that 1
mgm
VI
R, where
gmI is the maximum or peak gate
current of the SCR. Therefore 1m
gm
VR
I.
Also with 2 0R , we need to ensure that the voltage drop across resistor ‘R’ does not exceed
gmV , the maximum gate voltage
1
1
1
1
mgm
gm gm m
gm m gm
gm
m gm
V RV
R R
V R V R V R
V R R V V
V RR
V V
Operation
Case 1: gp gtV V
gpV , the peak gate voltage is less then gtV since 2R is very large. Therefore, current ‘I’ flowing
through the gate is very small. SCR will not turn on and therefore the load voltage is zero and
scrv is equal to sV . This is because we are using only a resistive network. Therefore, output
will be in phase with input.
Case 2: gp gtV V , 2R optimum value.
When 2R is set to an optimum value such that gp gtV V , we see that the SCR is triggered at
090 (since gpV reaches its peak at 090 only). The waveforms shows that the load voltage is
zero till 090 and the voltage across the SCR is the same as input voltage till it is triggered at 090 .
Case 3: gp gtV V , 2R small value.
The triggering value gtV is reached much earlier than 090 . Hence the SCR turns on earlier
than SV reaches its peak value. The waveforms as shown with respect to sins mV V t .
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At , , sinS gt m gp gt gpt V V V V V V
Therefore 1sin
gt
gp
V
V
But 1 2
mgp
V RV
R R R
Therefore 1 21sin
gt
m
V R R R
V R
Since 1, ,gtV R R are constants
5. Distinguish between :
(i) Latching current and holding current.
(ii) Converter grade thyristor and inverter grade thyristor.
(iii) Thyristor turn off time and circuit turn off time. (Dec 2010, Jun 2010)
Solution:
Latching current and holding current.
Latching current is the minimum amount of current required to maintain the thyristor in on
state immediately after a thyristor is turned on and Holding current is a minimum current that
is required to maintain the thyristor in on-state not allowing it to turn off.
Converter grade or Phase Control thyristors: These devices are the work horses of
the Power Electronics. They are turned off by natural (line) commutation and are reverse
biased at least for a few milliseconds subsequent to a conduction period. No fast switching
feature is desired of these devices. They are available at voltage ratings in excess of 5 KV
starting from about 50 V and current ratings of about 5 KA. The largest converters for HVDC
transmission are built with series-parallel combination of these devices. Conduction voltages
are device voltage rating dependent and range between 1.5 V (600V) to about 3.0 V (+5 KV).
These devices are unsuitable for any 'forced-commutated' circuit requiring unwieldy large
commutation components.
The dynamic di/dt and dv/dt capabilities of the SCR have vastly improved over the years
borrowing emitter shorting and other techniques adopted for the faster variety. The
requirement for hard gate drives and di/dt limting inductors have been eliminated in the
process.
Inverter grade thyristors: Turn-off times of these thyristors range from about 5 to 50 μsecs
when hard switched. They are thus called fast or 'inverter grade' SCR's. The SCR's are mainly
used in circuits that are operated on DC supplies and no alternating voltage is available to
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turn them off. Commutation networks have to be added to the basic converter only to turn-
off the SCR's. The efficiency, size and weight of these networks are directly related to the
turn-off time, tq of the SCR. The commutation circuits utilised resonant networks or charged
capacitors. Quite a few commutation networks were designed and some like the McMurray-
Bedford became widely accepted.
(iii) Thyristor turn off time and circuit turn off time.
The turn off time of a thyristor is defined as the time between the instant anode current
becomes zero and the instant SCR regains forward blocking capability. During time tq, all the
excess carriers from the four layers of SCR must be removed. This removal of excess carriers
consists of sweeping out of holes from outer p layer and electrons from outer n layer.
Circuit turn off time is defined as the time during which a reverse voltage is applied across
the thyristor during its commutation process.
6. With the help of neat circuit diagram and waveforms, explain RC firing circuit used
with half controlled rectifier. (Jun 2010)
Solution:
RC Half Wave
Capacitor ‘C’ in the circuit is connected to shift the phase of the gate voltage. 1D is used to
prevent negative voltage from reaching the gate cathode of SCR.
In the negative half cycle, the capacitor charges to the peak negative voltage of the supply
mV through the diode 2D . The capacitor maintains this voltage across it, till the supply
voltage crosses zero. As the supply becomes positive, the capacitor charges through resistor
‘R’ from initial voltage of mV , to a positive value.
When the capacitor voltage is equal to the gate trigger voltage of the SCR, the SCR is fired
and the capacitor voltage is clamped to a small positive value.
Fig.: RC half-wave trigger circuit
LOAD
vO
R
C
VT
v =V sin tS m
D2
VC
+
-D1
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Fig.3.9: Waveforms for RC half-wave trigger circuit
(a) High value of R (b) Low value of R
Case 1: R Large.
When the resistor ‘R’ is large, the time taken for the capacitance to charge from mV to gtV is
large, resulting in larger firing angle and lower load voltage.
Case 2: R Small
When ‘R’ is set to a smaller value, the capacitor charges at a faster rate towards gtV resulting
in early triggering of SCR and hence LV is more. When the SCR triggers, the voltage drop
across it falls to 1 – 1.5V. This in turn lowers, the voltage across R & C. Low voltage across
the SCR during conduction period keeps the capacitor discharge during the positive half
cycle.
Design Equation
From the circuit 1C gt dV V V . Considering the source voltage and the gate circuit, we can
write s gt Cv I R V . SCR fires when
s gt Cv I R V that is 1S g gt dv I R V V . Therefore
1s gt d
gt
v V VR
I. The RC time constant for zero output voltage that is maximum firing angle
for power frequencies is empirically gives as 1.32
TRC .
vs
0
V sin tm
0 t
tt
avc
- /2
a
vc
Vgt
vo
vT
Vm
-Vm
vs
0
V sin tm
0 t
avc
- /2
a
vc
Vgt
0
0
vo
vT
Vm Vm
-Vm(2 + )
(a) (b)
t t
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UNIT 4
Controlled Rectifiers
1. With the necessary circuit and waveforms, explain the principle of operation of single
phase full converter with R-L load, Derive the expression for the average output
voltage. (Dec 2010, Jun 2011)
Solution:
Single Phase Full Wave Full Converter With R,L, & E Load
Waveforms of Single Phase Full Converter Assuming Continuous (Constant Load
Current) & Ripple Free Load Current
To Derive an Expression for the Average DC Output Voltage of a Single Phase Full
Converter assuming Continuous & Constant Load Current
iOConstant Load Current i =IO a
i
iT1
T2&
Ia
t
t
t
Iai
iT3
T4&
Ia
Ia
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2
0
The average dc output voltage
can be determined by using the expression
1. ;
2
The o/p voltage waveform consists of two o/p
pulses during the input supply time period of
0 to 2 r
dc OO dcV V v d t
adians. Hence the Average or dc
o/p voltage can be calculated as
2sin .
2
2cos
2
2cos
dc mO dc
mdcO dc
mdcO dc
V V V t d t
VV V t
VV V
0
max
max
Maximum average dc output voltage is
calculated for a trigger angle 0
and is obtained as
2 2cos 0
2
m mdmdc
mdmdc
V VV V
VV V
max
The normalized average output voltage is given by
2cos
cos2
O dc dcdcn n
dmdc
m
dcn nm
V VV V
V V
V
V VV
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2. With the circuit diagram, and waveform, explain the principle of operation of dual
converter, with and without circulating current. (Dec 2010)
Solution:
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3. What are the advantages and drawbacks of circulating current mode of operation of
a dual converter. (Dec 2010)
Solution:
Advantages of Circulating Current Mode Of Operation
• The circulating current maintains continuous conduction of both the converters over
the complete control range, independent of the load.
• One converter always operates as a rectifier and the other converter operates as an
inverter, the power flow in either direction at any time is possible.
• As both the converters are in continuous conduction we obtain faster dynamic
response. i.e., the time response for changing from one quadrant operation to another
is faster.
Disadvantages of Circulating Current Mode Of Operation
• There is always a circulating current flowing between the converters.
• When the load current falls to zero, there will be a circulating current flowing
between the converters so we need to connect circulating current reactors in order to
limit the peak circulating current to safe level.
The converter thyristors should be rated to carry a peak current much greater than the peak
load current.
4. Explain with the help of wave forms, 1-φ semi converter (half bridge converter) with
R load. Derive an expression for Vo(avg) and Vo(rms). (Dec 2011)
Solution: Single Phase Half-Wave Thyristor Converter with a Resistive Load
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Equations:
sin i/p ac supply voltage
max. value of i/p ac supply voltage
RMS value of i/p ac supply voltage2
output voltage across the load
s m
m
mS
O L
v V t
V
VV
v v
When the thyristor is triggered at
sin ; to
Load current; to
sinsin ; to
Where max. value of load current
O L m
OO L
mO L m
mm
t
v v V t t
vi i t
R
V ti i I t t
R
VI
R
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5. Explain the operation of single phase semi converter with highly inductive load.
Derive the expression for V(dc) and V(rms). (Jun 2010)
Solution:
Single Phase Full Wave Half Controlled Bridge Converter (Single Phase Semi
Converter)
Trigger Pattern of Thyristors
Waveforms of single phase semi-converter with general load & FWD for > 900
Single Quadrant Operation
1
2
0
1 2
, 2 ,...
, 3 ,...
& 180
Thyristor T is triggered at
t at t
Thyristor T is triggered at
t at t
The time delay between the gating
signals of T T radians or
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Thyristor T1 and D1 conduct from ωt = α to π
Thyristor T2 and D2 conduct from ωt = (π + α) to 2 π
FWD conducts during ωt = 0 to α, π to (π + α) , …..
Load Voltage & Load Current Waveform of Single Phase Semi Converter for < 900 &
Continuous load current operation
vOVm
0
iO
t
( )
t0
( )
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(i) To Derive an Expression for The DC Output Voltage of A Single Phase Semi
Converter with R, L, & E Load & FWD For Continuous, Ripple Free Load Current
Operation
(ii) RMS O/P Voltage VO(RMS)
0
1.
1sin .
cos
cos cos ; cos 1
1 cos
dc OO dc
t
dc mO dc
mdcO dc
mdcO dc
mdcO dc
V V v d t
V V V t d t
VV V t
VV V
VV V
max
can be varied from a max.
2value of 0 by varying from 0 to .
For 0, The max. dc o/p voltage obtained is
Normalized dc o/p voltage is
2
11 cos
2
dc
m
m
dc
mdn
mdmdc
dcn n
V
Vto
V
V
V V
VVV
V
V 1 cos2
1
22 2
1
2 2
1
2
2sin .
2
1 cos 2 .2
1 sin 2
22
mO RMS
m
O RMS
m
O RMS
V V t d t
VV t d t
VV
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UNIT 5
Commutation
1. What do mean by commutation? Explain briefly the different types of commutation
(Dec 2010)
Solution:
The process of turning off a conducting thyristor is called commutation.
Types:
Self Commutation
Resonant Pulse Commutation
Complementary Commutation
Impulse Commutation
External Pulse Commutation.
Load Side Commutation.
Line Side Commutation.
2. Explain the self commutation with the help of neat sketch and obtain the expression
for the capacitor voltage and current.
(Dec 2011, Dec 2010, Jun 2010)
Solution:
In this type of commutation the current through the SCR is reduced below the holding
current value by resonating the load. i.e., the load circuit is so designed that even though the
supply voltage is positive, an oscillating current tends to flow and when the current through
the SCR reaches zero, the device turns off. This is done by including an inductance and a
capacitor in series with the load and keeping the circuit under-damped. Figure 5.3 shows the
circuit.
This type of commutation is used in Series Inverter Circuit.
Fig. 5.3: Circuit for Self Commutation
(i) Expression for Current
V
R L V (0)c
C
T i
Load
+ -
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At 0t , when the SCR turns ON on the application of gate pulse assume the current
in the circuit is zero and the capacitor voltage is 0CV .
Writing the Laplace Transformation circuit of figure 5.3 the following circuit is obtained
when the SCR is conducting.
Fig.: 5.4.
0
1
C
S
V V
SI S
R sLC
2
0
1
S CC V V
S
RCs s LC
2
0
1
CC V V
RLC s s
L LC
2
0
1
CV V
LR
s sL LC
2 2
2
0
1
2 2
CV V
L
R R Rs s
L LC L L
VS
R sL
1CS
V (0)
SC
C
T I(S) + +- -
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22 2
0
1
2 2
CV V
L
R Rs
L LC L
2 2
A
s,
Where
20 1, ,
2 2
CV V R RA
L L LC L
is called the natural frequency
2 2
AI S
s
Taking inverse Laplace transforms
sintAi t e t
Therefore expression for current
20
sinR
tC L
V Vi t e t
L
Peak value of current 0CV V
L
(ii) Expression for voltage across capacitor at the time of turn off
Applying KVL to figure 1.3
c R Lv V v V
c
div V iR L
dt
Substituting for i,
sin sint t
c
A d Av V R e t L e t
dt
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sin cos sint t t
c
A Av V R e t L e t e t
sin cos sint
c
Av V e R t L t L t
sin cos sin2
t
c
A Rv V e R t L t L t
L
sin cos2
t
c
A Rv V e t L t
Substituting for A,
0
sin cos2
C t
c
V V Rv t V e t L t
L
0
sin cos2
C t
c
V V Rv t V e t t
L
SCR turns off when current goes to zero. i.e., at t .
Therefore at turn off
0
0 cosC
c
V Vv V e
0c Cv V V V e
Therefore 20R
Lc Cv V V V e
Note: For effective commutation the circuit should be under damped.
That is
21
2
R
L LC
With R = 0, and the capacitor initially uncharged that is 0 0CV
sinV t
iL LC
But 1
LC
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Therefore sin sinV t C t
i LC VL LLC LC
and capacitor voltage at turn off is equal to 2V.
Figure 5.5 shows the waveforms for the above conditions. Once the SCR turns off
voltage across it is negative voltage.
Conduction time of SCR .
Fig. 5.5: Self Commutation – Wave forms of Current and Capacitors Voltage
3. Calculate the values of LR and C to be used for commutating the main SCR in the
circuit shown in figure 1.24. When it is conducting a full load current of 25 A flows. The
minimum time for which the SCR has to be reverse biased for proper commutation is
40 sec. Also find 1R , given that the auxiliary SCR will undergo natural commutation
when its forward current falls below the holding current value of 2 mA. (Dec 2011)
Current i
Capacitor voltage
Gate pulse
Voltage across SCR
0 /2t
t
t
t
V
V
2V
CV
L
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Fig. 5.24
Solution
In this circuit only the main SCR carries the load and the auxiliary SCR is used to turn
off the main SCR. Once the main SCR turns off the current through the auxiliary SCR is the
sum of the capacitor charging current ci and the current 1i through 1R , ci reduces to zero after
a time ct and hence the auxiliary SCR turns off automatically after a time ct , 1i should be less
than the holding current.
Given 25LI A
That is 100
25L L
VA
R R
Therefore 4LR
40 sec 0.693c Lt R C
That is 640 10 0.693 4 C
Therefore 640 10
4 0.693C
14.43C F
1
1
Vi
R should be less than the holding current of auxiliary SCR.
Therefore 1
100
R should be < 2mA.
Therefore 1 3
100
2 10R
That is 1 50R K
V=100V
R1 RL
MainSCR
AuxiliarySCR
iC
C
ILi1
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4. An impulse commutated thyristor circuit is shown in figure 5.29. Determine the
available turn off time of the circuit if V = 200 V, R = 10 and C = 5 F. Voltage across
capacitor before 2T is fired is V volts with polarity as shown.
(Dec 2010, Jun 2010)
Fig. 5.29
Solution
When 2T is triggered the circuit is as shown in figure 5.30.
Fig. 5.30
Writing the transform circuit, we obtain
Fig. 5.31
We have to obtain an expression for capacitor voltage. It is done as follows:
C
T1
T2
V (0)C
V +
+
-
-
R
C
i(t)
T2V
++-
-
R
V (O)C
Vs
V (0)
sC
+
+
I(s)
1Cs
R
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10
1
CV VsI S
RCs
0
1
CC V VI S
RCs
0
1
CV VI S
R sRC
Voltage across capacitor 01 C
C
VV s I s
Cs s
0 01
1
C C
C
V V VV s
RCs ss
RC
0 0 0
1
C C C
C
V V V V VV s
s ss
RC
0
1 1C
C
VV VV s
ss s
RC RC
1 0t t
RC RCc Cv t V e V e
In the given problem 0CV V
Therefore 1 2tRC
cv t V e
The waveform of cv t is shown in figure 5.32.
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Fig. 5.32
At ct t , 0cv t
Therefore 0 1 2ct
RCV e
1 2ct
RCe
1
2
ctRCe
Taking natural logarithms
1
log2
ce
t
RC
ln 2ct RC
610 10 10 ln 2ct
69.3 secct .
t
tC
V
V (0)C
v (t)C
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UNIT 6
AC Voltage Controller
1. Draw the circuit diagram of a single phase AC voltage controller and explain the
principle of ON-OFF control, with the help of relevant wave forms. Derive the
expression for rms output voltage in terms of rms supply voltage. ( Dec 2011)
Solution:
Principle of On-Off Control Technique (Integral Cycle Control)
The basic principle of on-off control technique is explained with reference to a single phase
full wave ac voltage controller circuit shown below. The thyristor switches 1T and 2T are
turned on by applying appropriate gate trigger pulses to connect the input ac supply to the
load for ‘n’ number of input cycles during the time interval ONt . The thyristor switches 1T and
2T are turned off by blocking the gate trigger pulses for ‘m’ number of input cycles during the
time interval OFFt . The ac controller ON time ONt usually consists of an integral number of
input cycles.
Fig.: Single phase full wave AC voltage controller
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Fig.: Waveforms
LR R = Load Resistance
To derive an expression for the rms value of output voltage, for on-off control method.
Output RMS voltage 2 2
0
1.
ONt
mO RMS
O t
V V Sin t d tT
22
0
.ONt
m
O RMS
O
VV Sin t d t
T
Substituting for 2 1 2
2
CosSin
2
0
1 2
2
ONt
m
O RMS
O
V Cos tV d t
T
2
0 0
2 .2
ON ONt t
m
O RMS
O
VV d t Cos t d t
T
2
0 0
2
22
ON ONt t
m
O RMS
O
V Sin tV t
T
Vs
Vo
io
ig1
ig2
wt
wt
wt
wt
Gate pulse of T1
Gate pulse of T2
n m
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2 sin 2 sin 00
2 2
m ONONO RMS
O
V tV t
T
Now ONt = an integral number of input cycles; Hence
, 2 ,3 ,4 ,5 ,.....ONt T T T T T & 2 ,4 ,6 ,8 ,10 ,......ONt
Where T is the input supply time period (T = input cycle time period). Thus we note that
sin 2 0ONt
2
2 2
m ON m ON
O RMS
O O
V t V tV
T T
ON ONSO RMS i RMS
O O
t tV V V
T T
Where 2
mSi RMS
VV V = RMS value of input supply voltage;
ON ON
O ON OFF
t t nT nk
T t t nT mT n m= duty cycle (d).
S SO RMS
nV V V k
m n
2. Distinguish between ON-OFF control and phase control of AC voltage controller.
(Jun 2010)
Solution:
ON/OFF
Advantages:
1. As the SCRs turn on and off at the zero crossing instants of ac supply, the supply current
does not get distorted. Therefore no additional harmonic components are introduced due to
switching of SCRs.
2. Disturbance to the neighbouring electronic circuits due to EMI is minimum. This is
because the switching takes place at zero crossing instants.
3. External commutation components are not required.
Disadvantages:
1. The SCRs and triac are turned on at α=0. Therefore these devices are being underutilized,
because SCRs or triac can be turned on at any value from 00 to 180
0. This virtue of these
devices is not being exploited.
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2. Precise control of heat or intensity cannot be achieved.
3. Explain the single phase bidirectional AC voltage controller with resistive load with
waveform. (Dec 2011)
Solution:
Single phase full wave ac voltage controller circuit using two SCRs or a single triac is
generally used in most of the ac control applications. The ac power flow to the load can be
controlled in both the half cycles by varying the trigger angle ' ' .
The RMS value of load voltage can be varied by varying the trigger angle ' ' . The
input supply current is alternating in the case of a full wave ac voltage controller and due to
the symmetrical nature of the input supply current waveform there is no dc component of
input supply current i.e., the average value of the input supply current is zero.
A single phase full wave ac voltage controller with a resistive load is shown in the
figure below. It is possible to control the ac power flow to the load in both the half cycles by
adjusting the trigger angle ' ' . Hence the full wave ac voltage controller is also referred to as
to a bi-directional controller.
Fig 6.7: Single phase full wave ac voltage controller (Bi-directional Controller) using SCRs
The thyristor 1T is forward biased during the positive half cycle of the input supply
voltage. The thyristor 1T is triggered at a delay angle of ' ' 0 radians .
Considering the ON thyristor 1T as an ideal closed switch the input supply voltage appears
across the load resistor LR and the output voltage O Sv v during t to radians. The
load current flows through the ON thyristor 1T and through the load resistor LR in the
downward direction during the conduction time of 1T from t to radians.
At t , when the input voltage falls to zero the thyristor current (which is
flowing through the load resistor LR ) falls to zero and hence 1T naturally turns off . No
current flows in the circuit during t to .
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The thyristor 2T is forward biased during the negative cycle of input supply and when
thyristor 2T is triggered at a delay angle , the output voltage follows the negative
halfcycle of input from t to 2 . When 2T is ON, the load current flows in the
reverse direction (upward direction) through 2T during t to 2 radians. The time
interval (spacing) between the gate trigger pulses of 1T and 2T is kept at radians or 1800. At
2t the input supply voltage falls to zero and hence the load current also falls to zero and
thyristor 2T turn off naturally.
Instead of using two SCR’s in parallel, a Triac can be used for full wave ac voltage
control.
Fig 6.8: Single phase full wave ac voltage controller (Bi-directional Controller) using TRIAC
Fig 6.9: Waveforms of single phase full wave ac voltage controller
Equations
Input supply voltage
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sin 2 sinS m Sv V t V t ;
Output voltage across the load resistor LR ;
sinO L mv v V t ;
for to t and to 2t
Output load current
sinsinO m
O m
L L
v V ti I t
R R ;
for to t and to 2t
4. A single phase full wave controller has an input voltage of 230 V (RMS) and a load
resistance of 10 ohm. The firing angle of thyristor is 2 . Find
a. Average output voltage.
b. RMS output voltage.
c. The Average thyristor current.
d. The rms current value of the thyristor.
e. Diode average current.
f. Diode rms current. (Dec 2010)
Solution
090 , 120 V, 62
SV R
RMS Value of Output Voltage
1
21 sin 2
2O SV V
1
21 sin180120
2 2OV
84.85 VoltsOV
RMS Output Current
84.85
14.14 A6
OO
VI
R
Load Power
2
O OP I R
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2
14.14 6 1200 wattsOP
Input Current is same as Load Current
Therefore 14.14 AmpsS OI I
Input Supply Volt-Amp 120 14.14 1696.8 S SV I VA
Therefore
Input Power Factor = Load Power 1200
0.707Input Volt-Amp 1696.8
lag
Each Thyristor Conducts only for half a cycle
Average thyristor current T AvgI
1
sin .2
mT AvgI V t d t
R
m1 cos ; V 2
2
mS
VV
R
2 120
1 cos90 4.5 A2 6
RMS thyristor current T RMSI
2 2
2
sin1
2
m
T RMS
V tI d t
R
2
2
1 cos 2
2 2
mtV
d tR
1
21 sin 2
2 2
mV
R
1
22 1 sin 2
2 2
SV
R
1
22 120 1 sin18010 Amps
2 6 2 2
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5. What is an ac voltage controller? With the help of circuit diagram and waveform,
explain the principle of phase control. (Jun 2011)
Solution:
Principle of AC Phase Control
The basic principle of ac phase control technique is explained with reference to a
single phase half wave ac voltage controller (unidirectional controller) circuit shown in the
below figure.
The half wave ac controller uses one thyristor and one diode connected in parallel
across each other in opposite direction that is anode of thyristor 1T is connected to the
cathode of diode 1D and the cathode of 1T is connected to the anode of 1D . The output
voltage across the load resistor ‘R’ and hence the ac power flow to the load is controlled by
varying the trigger angle ‘ ’.
The trigger angle or the delay angle ‘ ’ refers to the value of t or the instant at
which the thyristor 1T is triggered to turn it ON, by applying a suitable gate trigger pulse
between the gate and cathode lead.
The thyristor 1T is forward biased during the positive half cycle of input ac supply. It
can be triggered and made to conduct by applying a suitable gate trigger pulse only during the
positive half cycle of input supply. When 1T is triggered it conducts and the load current
flows through the thyristor 1T , the load and through the transformer secondary winding.
By assuming 1T as an ideal thyristor switch it can be considered as a closed switch
when it is ON during the period t to radians. The output voltage across the load
follows the input supply voltage when the thyristor 1T is turned-on and when it conducts from
t to radians. When the input supply voltage decreases to zero at t , for a
resistive load the load current also falls to zero at t and hence the thyristor 1T turns off
at t . Between the time period t to 2 , when the supply voltage reverses and
becomes negative the diode 1D becomes forward biased and hence turns ON and conducts.
The load current flows in the opposite direction during t to 2 radians when 1D is ON
and the output voltage follows the negative half cycle of input supply.
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Fig 6.4: Halfwave AC phase controller (Unidirectional Controller)
6. With the necessary waveforms explain the single phase Full wave controller with
inductive load. Derive the expression for the rms output voltage. (Jun 2011)
Solution:
The operation and performance of a single phase full wave ac voltage controller with RL
load. In practice most of the loads are of RL type. For example if we consider a single phase
full wave ac voltage controller controlling the speed of a single phase ac induction motor, the
load which is the induction motor winding is an RL type of load, where R represents the
motor winding resistance and L represents the motor winding inductance.
A single phase full wave ac voltage controller circuit (bidirectional controller) with an
RL load using two thyristors 1T and 2T ( 1T and 2T are two SCRs) connected in parallel is
shown in the figure below. In place of two thyristors a single Triac can be used to implement
a full wave ac controller, if a suitable Traic is available for the desired RMS load current and
the RMS output voltage ratings.
Fig 6.11: Single phase full wave ac voltage controller with RL load
The thyristor 1T is forward biased during the positive half cycle of input supply. Let
us assume that 1T is triggered at t , by applying a suitable gate trigger pulse to 1T during
the positive half cycle of input supply. The output voltage across the load follows the input
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supply voltage when 1T is ON. The load current Oi flows through the thyristor 1T and through
the load in the downward direction. This load current pulse flowing through 1T can be
considered as the positive current pulse. Due to the inductance in the load, the load current Oi
flowing through 1T would not fall to zero at t , when the input supply voltage starts to
become negative.
The thyristor 1T will continue to conduct the load current until all the inductive energy
stored in the load inductor L is completely utilized and the load current through 1T falls to
zero at t , where is referred to as the Extinction angle, (the value of t ) at which the
load current falls to zero. The extinction angle is measured from the point of the beginning
of the positive half cycle of input supply to the point where the load current falls to zero.
The thyristor 1T conducts from t to . The conduction angle of 1T is
, which depends on the delay angle and the load impedance angle . The
waveforms of the input supply voltage, the gate trigger pulses of 1T and 2T , the thyristor
current, the load current and the load voltage waveforms appear as shown in the figure below.
Fig 6.12: Input supply voltage & Thyristor current waveforms
is the extinction angle which depends upon the load inductance value.
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Fig 6.13: Gating Signals
Waveforms of single phase full wave ac voltage controller with RL load for .
Discontinuous load current operation occurs for and ;
i.e., , conduction angle .
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Fig 6.14: Waveforms of Input supply voltage, Load Current, Load Voltage and Thyristor Voltage across 1T
To Derive an Expression For rms Output Voltage
O RMSV of a Single Phase Full-Wave Ac
Voltage Controller with RL Load.
When O , the load current and load voltage waveforms become discontinuous as
shown in the figure above. 1
22 21sin .mO RMS
V V t d t
Output sino mv V t , for to t , when 1T is ON.
122 1 cos 2
2
m
O RMS
tVV d t
1
22
cos 2 .2
m
O RMS
VV d t t d t
1
22sin 2
22
m
O RMS
V tV t
12 2sin 2 sin 2
2 2 2
m
O RMS
VV
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121 sin 2 sin 2
2 2 2mO RMS
V V
1
21 sin 2 sin 2
2 22
m
O RMS
VV
The RMS output voltage across the load can be varied by changing the trigger angle
.
For a purely resistive load 0L , therefore load power factor angle 0 .
1tan 0L
R ;
Extinction angle 0 radians 180
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UNIT 7
DC Chopper
1. Give the classification of choppers. Explain class E chopper with circuit and quadrant
diagram. (Dec 2010, Jun 2010)
Solution:
Classification Of Choppers
Choppers are classified as
• Class A Chopper
• Class B Chopper
• Class C Chopper
• Class D Chopper
• Class E Chopper
Class E Chopper
Four Quadrant Operation
• Class E is a four quadrant chopper
V
v0
i0L ER
CH2 CH4D2 D4
D1 D3CH1 CH3
+
v0
i0
CH - CH ON
CH - D Conducts1 4
4 2
D D2 3 - Conducts
CH - D Conducts4 2
CH - CH ON
CH - D Conducts3 2
2 4
CH - D Conducts
D - D Conducts2 4
1 4
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• When CH1 and CH4 are triggered, output current iO flows in positive direction
through CH1 and CH4, and with output voltage Vo = V.
• This gives the first quadrant operation.
• When both CH1 and CH4 are OFF, the energy stored in the inductor L drives io
through D2 and D3 in the same direction, but output voltage Vo = -V.
• Therefore the chopper operates in the fourth quadrant.
• When CH2 and CH3 are triggered, the load current io flows in opposite direction &
output voltage Vo = -V.
• Since both io and VO are negative, the chopper operates in third quadrant.
• When both CH2 and CH3 are OFF, the load current iO continues to flow in the same
direction D1 and D4 and the output voltage VO = V.
Therefore the chopper operates in second quadrant as VO is positive but io is negative.
2. Explain the principle of operation of a step up chopper with suitable circuit diagram
and waveforms. Derive the expression for average output voltage of step up chopper.
(Dec 2011, Dec 2010, Jun 2010)
Solution:
Principle of Step-up Chopper
Step-up chopper is used to obtain a load voltage higher than the input voltage V.
The values of L and C are chosen depending upon the requirement of output
voltage and current.
When the chopper is ON, the inductor L is connected across the supply.
The inductor current ‘I’ rises and the inductor stores energy during the ON time of
the chopper, tON.
When the chopper is off, the inductor current I is forced to flow through the diode
D and load for a period, tOFF.
The current tends to decrease resulting in reversing the polarity of induced EMF
in L.
Therefore voltage across load is given by
• A large capacitor ‘C’ connected across the load, will provide a continuous output
voltage .
• Diode D prevents any current flow from capacitor to the source.
• Step up choppers are used for regenerative braking of dc motors.
+
VOV
Chopper
CLOAD
DLI
+
. ., O O
dIV V L i e V V
dt
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Expression For Output Voltage
3. A Chopper circuit is operating on TRC at a frequency of 2 kHz on a 460 V supply. If
the load voltage is 350 volts, calculate the conduction period of the thyristor in each
cycle. (Dec 2011)
Solution:
(energy is supplied by inductor to load)
Voltage across
Energy supplied by inductor
where period of Chopper.
Neg
When Chopper
lecting losses, energy stored in inductor
is OFF
O
O OFF
OFF
L V V
L V V It
t OFF
L = energy supplied by inductor L
Assume the average inductor current to be
during ON and OFF time of Chopper.
Voltage across inductor
Therefore energy stored in inductor
= . .
Where
When Chopper
period of chopper.
is ON
ON
ON
I
L V
V I t
t ON
Where
T = Chopping period or period
of switching.
ON O OFF
ON OFF
O
OFF
O
ON
VIt V V It
V t tV
t
TV V
T t
1
1
1
1
Where duty cyle
ON OFF
OON
O
ON
T t t
V Vt
T
V Vd
td
T
3
460 V, = 350 V, f = 2 kHz
1Chopping period
10.5 sec
2 10
Output voltage
dc
ONdc
V V
Tf
T m
tV V
T
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4. With neat circuit diagram and waveform, explain the operation of Jone’s chopper
(Jun 2011)
Solution:
Fig. 2.35: Jone’s Chopper
Figure 2.35 shows a Jone’s Chopper circuit for an inductive load with free wheeling
diode. Jone’s Chopper is an example of class D commutation. Two thyristors are used, T1 is
the main thyristor and T2 is the auxiliary thyristor. Commutating circuit for T1 consists of
thyristor T2, capacitor C, diode D and autotransformer (L1 and L2).
Initially thyristor T2 is turned ON and capacitor C is charged to a voltage V with a
polarity as shown in figure 2.35. As C charges, the charging current through thyristor T2
decays exponentially and when current falls below holding current level, thyristor T2 turns
OFF by itself. When thyristor T1 is triggered, load current flows through thyristor T1, L2 and
load. The capacitor discharges through thyristor T1, L1 and diode D. Due to resonant action
C
D
+
V
+
LFWD
R
T1
T2
L2
L1
v0
3
Conduction period of thyristor
0.5 10 350
460
0.38 msec
dcON
ON
ON
T Vt
V
t
t
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of the auto transformer inductance L2 and capacitance C, the voltage across the capacitor
reverses after some time.
It is to be noted that the load current in L1 induces a voltage in L2 due to
autotransformer action. Due to this voltage in L2 in the reverse direction, the capacitor
charges to a voltage greater than the supply voltage. (The capacitor now tries to discharge in
opposite direction but it is blocked by diode D and hence capacitor maintains the reverse
voltage across it). When thyristor T1 is to be commutated, thyristor T2 is turned ON resulting
in connecting capacitor C directly across thyristor T1. Capacitor voltage reverse biases
thyristor T1 and turns it off. The capacitor again begins to charge through thyristor T2 and
the load for the next cycle of operation.
The various waveforms are shown in figure 2.36
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5. A dc chopper in figure 2.25 has a resistive load of 10R and input voltage of V =
200 V. When chopper is ON, its voltage drop is 2 V and the chopping frequency is 1
kHz. If the duty cycle is 60%, determine
Average output voltage
RMS value of output voltage
Effective input resistance of chopper
Chopper efficiency.
(Jun 2011)
Gate pulse of T2 Gate pulse of T1 Gate pulse of T2
Capacitor Voltage
Capacitordischarge current
Current of T1
Voltage across T1
Auto transformer action
Resonant action
Ig
IL
IL
VC
+V
V
t
t
t
t
tC
tC
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Fig. 2.25
Solution
V = 200 V, 10R , Chopper voltage drop, 2chV V , d = 0.60, f = 1 kHz.
Average output voltage
dc chV d V V
0.60 200 2 118.8 VoltsdcV
RMS value of output voltage
O chV d V V
0.6 200 2 153.37 VoltsOV
Effective input resistance of chopper is
i
S dc
V VR
I I
118.811.88 Amps
10
dcdc
VI
R
20016.83
11.88i
S dc
V VR
I I
Output power is
2
0
0
1dT
O
vP dt
T R
2
0
1dT
ch
O
V VP dt
T R
V
i0
Chopper
+
R v0
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2
ch
O
d V VP
R
20.6 200 2
2352.24 watts10
OP
Input power, 0
1dT
i OP Vi dtT
0
1dT
ch
O
V V VP dt
T R
0.6 200 200 22376 watts
10
ch
O
dV V VP
R
Chopper efficiency,
100O
i
P
P
2352.24100 99%
2376
6. Explain the different control strategies used in choppers. (Jun 2011)
Solution:
Methods of Control
The output dc voltage can be varied by the following methods.
Pulse width modulation control or constant frequency operation.
Variable frequency control.
Pulse Width Modulation
In pulse width modulation the pulse width ONt of the output waveform is varied
keeping chopping frequency ‘f’ and hence chopping period ‘T’ constant. Therefore output
voltage is varied by varying the ON time, ONt . Figure 2.3 shows the output voltage
waveforms for different ON times.
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Fig. 2.3: Pulse Width Modulation Control
Variable Frequency Control
In this method of control, chopping frequency f is varied keeping either ONt or OFFt
constant. This method is also known as frequency modulation.
Figure 2.4 shows the output voltage waveforms for a constant ONt and variable
chopping period T.
In frequency modulation to obtain full output voltage, range frequency has to be
varied over a wide range. This method produces harmonics in the output and for large OFFt
load current may become discontinuous.
Fig. 2.4: Output Voltage Waveforms for Time Ratio Control
V0
V
V
V0
t
ttON
tON tOFF
tOFF
T
v0
V
V
v0
t
t
tON
tON
T
T
tOFF
tOFF
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UNIT 8
Inverters
1. Explain the performance of inverters. (Dec 2011)
The converters which converts the power into ac power popularly known as the
inverters,. The application areas for the inverters include the uninterrupted power supply
(UPS), the ac motor speed controllers, etc.
Fig.8.1 Block diagram of an inverter.
The inverters can be classified based on a number of factors like, the nature of output
waveform (sine, square, quasi square, PWM etc), the power devices being used (thyristor
transistor, MOSFETs IGBTs), the configuration being used, (series. parallel, half bridge, Full
bridge), the type of commutation circuit that is being employed and Voltage source and
current source inverters.
The thyristorised inverters use SCRs as power switches. Because the input source of power is
pure de in nature, forced commutation circuit is an essential part of thyristorised inverters.
The commutation circuits must be carefully designed to ensure a successful commutation of
SCRs. The addition of the commutation circuit makes the thyristorised inverters bulky and
costly. The size and the cost of the circuit can be reduced to some extent if the operating
frequency is increased but then the inverter grade thyristors which are special thyristors
manufactured to operate at a higher frequency must be used, which are costly.
2. With a neat circuit diagram, explain the principle of variable DC link.
(Dec 2011, Dec 2010)
Solution:
The circuit diagram of a variable DC-link inverter is shown in Fig.8.18. This circuit
can be divided into two parts namely a block giving a variable DC voltage and the second
part being the bridge inverter itself.
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Fig.8.18. Variable DC link Inverter
The components Q, Dm, Land C give out a variable DC output. L and C are the filter
components. This variable DC voltage acts as the supply voltage for the bridge inverter.
Fig.8.19. Output voltage Waveforms for different DC input voltages
The pulse width (conduction period) of the transistors is maintained constant and the
variation in output voltage is obtained by varying the DC voltage.
The output voltage waveforms with a resistive load for different dc input voltages are shown
in Fig. 8.19.
We know that for a square wave inverter, the rms value of output voltage is given by,
V0 (rms) = Vdc volts
Hence by varying Vdc, we can vary V0 (rms)
One important advantage of variable DC link inverters is that it is possible to eliminate or
reduce certain harmonic components from the output voltage waveform.
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The disadvantage is that an extra converter stage is required to obtain a variable DC voltage
from a fixed DC. This converter can be a chopper.
3. Write a short note on CSI (current source inverter). (Dec 2011, Jun 2011)
The circuit diagram of current source inverter is shown in Fig. 8.14. The variable dc voltage
source is converted into variable current source by using inductance L.
Fig.8.14. CSI using Thyristor
The current IL supplied to the single phase transistorised inverter is adjusted by the
combination of variable dc voltage and inductance L.
The waveforms of base currents and output current io are as shown in Fig. 8.15. When
transistors Q1 and Q2 conduct simultaneously, the output current are positive and equal to +
IL. When transistors Q3 and Q4 conduct simultaneously the output current io = - IL.
But io = 0 when the transistors from same arm i.e. Q( Q4 or Q2 Q3 conduct simultaneously.
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Fig.8.15: Waveforms for single phase current source
The output current waveform of Fig. 8.15 is a quasi-square waveform. But it is possible to
Obtain a square wave load current by changing the pattern of base driving signals. Such
waveforms are shown in Fig. 8.16.
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5. Explain the principle of operation of single phase half bridge inverter. (Dec 2010)
Solution:
Principle of Operation:
1. The principle of single phase transistorised inverters can be explained with the help of Fig.
8.2. The configuration is known as the half bridge configuration.
2. The transistor Q1 is turned on for a time T0/2, which makes the instantaneous voltage
across the load Vo = V12.
3. If transistor Q2 is turned on at the instant T0/2 by turning Q1 off then -V/2 appears across
the load.
Fig.8.2 Half bridge inverter
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Fig. Load voltage and current waveforms with resistive load for half bridge inverter.
6. Write and explain the performance parameters of an inverter.
(Dec 2010,Jun 2010)
Solution:
Performance parameters of inverters
The output of practical inverters contains harmonics and the quality of an inverter is normally
evaluated in terms of following performance parameters:
• Harmonic factor of nth
harmonic.
• Total harmonic distortion.
• Distortion factor.
• Lowest order harmonic.
Harmonic factor of nth
harmonics HFn:
The harmonic factor is a measure of contribution of indivisual harmonics. It is defined as the
ratio of the rms voltage of a particular harmonic component to the rms value of fundamental
component.
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Total Harmonic Distortion
Distortion Factor DF
Lowest order Harmonic