Download Q. 1 Two inverse parallel connected SCRs are used as a contactor

POWER ELECTRONICS
TUTORIAL
ER
PO W
POWER CONVERTERS
ICS
TRON
ELEC
AC-AC PHASE ANGLE CONTROL : R +L
VT1ak
iT1ak
iinin(t)
iout
iT2ak
vin (t )
VT2ak
VR R
Vout
VL
Q. 1
L
Two inverse parallel connected SCRs are used as a contactor to connect an inductive load to a
240 V rms, 50 Hz ac supply. The load is an 8  resistor in series with a 0.12 H inductor.
Determine the phase angle at which the SCRs should be first triggered in order that there is no
turn-on current transient.
(  = 78o)
Q. 2 An a.c. regulator is used to control the power delivered to a load consisting of a 100 resistor
in series with a 550 mH inductor. The regulator SCRs receive gate drive from the instant of
triggering until the following input-voltage zero-crossing.
Determine the minimum range of firing delay necessary to allow control of output power over the
maximum possible range.
(  = 60o to 180o )
Q. 3 A single-phase ac controller supplies a series R-L load with R = 5  and L = 5 mH. The input
voltage is 240 V rms, 50 Hz. The SCR firing delay angles   ) are symmetrical and equal to 60o.
Determine
EET307
a.
the scr conduction angle 
b.
the rms output voltage
c.
the rms output current
d.
the scr rms current
e.
the scr average current
f.
the input power factor
POWER ELECTRONICS
TUTORIAL PETac-ac2: R+L
1
Prof R T Kennedy
VT1ak
iT1ak
Q. 4
iinin(t)
iout
iT2ak
VT2ak
vin (t )
VR R
Vout
VL
L
Vin,rms
f
Z
cos 
240 V
50 Hz
2
0.5
(i) If the circuit shown is used as a contactor to connect the load to the supply determine the trigger delay
to minimize turn-on transient current.
(60o)
(ii) If the SCR conduction angle   120o in the AC regulator determine using performance curves
.
(a)
60o
 , the SCR trigger delay angle
(b)
rms output voltage
162.92 V
(c)
ac switch rms current
55.545 A
(d)
SCR rms current
38.94 A
(e)
input power factor
0.26
(iii) sketch the circuit waveforms using PEW: PAC:AC:R
VT1ak
Q. 5
iT1ak
iinin(t)
vin (t )
  90o
iout
iT2ak
  270o
VR R
VT2ak
VL
Vin,rms
f
L
240 V
50 Hz
18 mH
R
10

Vout
L
(i) The circuit shown is used as an AC regulator to control the load power.
Using performance curves determine
230 V rms!
(a)
 , the SCR conduction angle
120o
(b)
output power
1555 W
(c)
individual SCR average current
4.53 A
(d)
input current
12.47 A
(e)
input power factor
0.542
(ii) Show with the aid of circuit waveforms (PEW: PAC: AC : R+L) that there is a need to consider
snubber circuits in parallel with the SCRs.
(iii) Determine the permissible control range for 
EET307
POWER ELECTRONICS
TUTORIAL PETac-ac2: R+L
( 30o  180o )
2
Prof R T Kennedy
Q. 6
VT1ak
iT1ak
  90o
iout
iinin(t)
  270
VT2ak
o
vin (t )
iT2ak
VR R
Vin,rms
f
L
240 V
50 Hz
18 mH
R
10

Vout
VL
L
(a) The circuit shown is used as an AC regulator to control the load power.
Using performance curves determine
(i)
 the SCR conduction angle
140o
(ii)
the output power
290.95 W
(b) If the inductor is removed determine the new firing delay angle for the same load power
( 1300 )
EET307
POWER ELECTRONICS
TUTORIAL PETac-ac2: R+L
3
Prof R T Kennedy