Power Electronics 2 Course Syllabus

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

/







Theoretical syllabus
Week
Regenrating fully controlled interters , examples , dc motor speed control.
𝟏𝒔𝒕
Three face inverters , output voltage wave from with , triggering pulses and
equations.
𝟐𝒏𝒅
Thyristor protection from the high rate change in current and voltage ,
protection from the transient change in source voltage , fully protection circuit
from all possible faults due to current and voltage .
𝟑𝒓𝒅
Dc to ac inverters methods of forcing the thyristor to get off .
𝟒𝒕𝒉
Parallel and senies inverter, single and three phase , control mdthods in
charging frequency and voltage , output wave forms1 .
𝟓𝒕𝒉
Inverter application , emergency power supply , single phase dc motor speed
control.
𝟔𝒕𝒉
Three phase motor control by using a constant ratio of variation frequency and
voltage.
𝟕𝒕𝒉
Choppers , dc to dc inverter frequency constant , line constant .
𝟖𝒕𝒉
Types of choppers , dc motor speed control .
𝟗𝒕𝒉
Ac to ac inverter , single phase voltage requlator , three phase voltage requlator
.
𝟏𝟎𝒕𝒉
General application on single and three induction motor speed control due to
the change in stat or voltage , using the closed loop feedback circuit to control
the dlippery rings of ac motor .
𝟏𝟏𝒕𝒉
Cyclic inverter , ac to dc cyclic inverter , dc to dc cyclic inverter
𝟏𝟐𝒕𝒉
Ac to ac cyclic inverter control black diagram .
𝟏𝟑𝒕𝒉
Using amplitude modulation for speed control.
𝟏𝟒𝒕𝒉
Using polar transistor for ac motor speed control .
𝟏𝟓𝒕𝒉















Fully controlled full wave Rectifier
: -
2Vm
Vd.c = g cos
If α >= 90°
Vd.c = 2Vm cos 120= Vm
g g
EX:- if L =50 mH , R= 10 Ω , α=3 , V= 325 sin 100
π
t Calculate :-
1- Vd.c
2-
the angle of the load (Ф)
3-
time to commutate the SCR
4-
Discuss the waveform of (I load )
Solution :-
Vd.c = Vo = ∫α^+α) Vmsin(wt)d(wt) = (2 Vm)/π cosα
Vd.c = (2*325 ) cos30° = 179.2 v
tanФ =WL/R
Ф = tan- 1 (100π*50*10-3)/10 = 57.
Wtq = β = π – α
tq = - α)/w = (π- /6))/100π = 8.3 msec
ﺎﻣﺑﺔﯾ ﻣﺣﻟ(Ф) ﺑﻛﺔﯾ ﯾﺧﺄﺗ) ﺎﻓﺎﯾﺗﺣﻟﻛﯾﻣﺗﺳﻣ

Converter Control of DC Machines
The speed of a dc motor can be controlled by controlling the dc voltage across its
armature terminals. A
phase controlled thyristor converter can provide this dc voltage source. For a low- power drive, a single-
phase bridge converter can be used ,whereas for a high-power drive, a three- phase bridge circuit is
preferred. The machine can be a permanent magnet or wound field type. The wound field type permits
variation and
reversal
of field and is normally preferred in large power machines.
DC Motor Drives:-
The DC drive is relatively simple and cheap (compared to induction motor
drives). But DC
motor itself is more expensive.
Due to the numerous disadvantages of DC motor (esp. maintenance), it is
getting less
popular, particularly in high power applications.
For low power applications the cost of DC motor plus
drives is
still economical.
For servo application, DC drives is still popular because
of good
dynamic response and ease of control.
Future Trend? Not so bright prospect for DC, esp. in
high
power drives
.
Thyristor/SCR drives
Basic single-phase drive

Example
Consider a 500V, 10kW , 20A rated- DC motor with armature resistance of 1 ohm.
When supplied at 500V, the UNLOADED motor runs at 1040 rev/min, drawing a
current of 0.8A (ideally current is zero at no-load).
Estimate the full load speed at rated values
Estimate the no-load speed at 250V.
Solution:-
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