Sol: The thyristor will conduct when the instantaneous value of ac voltage is more than 50V or
100sin
ω
t = 50
Or
ω
t =
π
/6 and 5
π/6
i =
= 10sinωt
5
Average current =
=

10cosωt

5ωt
π/6
5π/6
=
(10cos
+ 10cos
 5x
+ 5x
) =
1.09A
7.
If a halfwave controlled rectifier has a purely resistive load of R and the delay angle is
α
=
π
/3.
Determine
(a) Rectification efficiency, (b) Form factor, (c) Ripple factor, (d) Transformer
utilization factor and (e) Peak inverse voltage for SCR
Solution:
(a)
Rectification efficiency,
η
=
where, Pdc = dc load power = V
2
dc/R and Pac = rms load
power = V
2
rms/R.
47
Vdc
=
(1+cos
α
) , since,
α
=
π
/3 therefore Vdc = 0.239Vm
Also Vrms = Vm[
+
]
0.5
For firing angle
α
=
π
/3 , Vrms = 0.485Vm
Therefore rectification efficiency
η
= (0.239Vm)
2
/(0.485Vm)
2
=
0 .2428 = 24.28%
(b)
Form factor, (ff) = Vrms/Vdc = 0.485Vm/0.239Vm =
2.033 = 203.3%
(c)
Ripple factor (Rf)
=
√
(ff
2
–
1) = 1.77 = 177%
(d)
Trasformer utilization factor (TUF) =
where Vs and Is are the rms secondary voltage and
current respectively.
Now Vs = Vm/
√
2 = 0.707Vm and Is = rms load current = Vrms/R = 0.485Vm/R
Therefore TUF =(V
2
dc/R)/Vs.Is = (0.239Vm)
2
/R/(0.707Vm x 0.485Vm/R)
=
0.166 or 16.6%
(e)
Peak inverse voltage =
Vm
8.
An SCR is used to control the power of 1kW, 230V, 50Hz heater.
Determine the heater power
for firing angles of 45° and 90°.
Solution
:
The heater resistance = R, and the rms current is the heat producing component of load
current.
Vrms = Vm[
+
]
0.5
(i)
At
α
=
π
/4, Vrms = 155V, Therefore heat power W = V
2
rms/R
R = 230
2
/1kW = 52.90
Ω
Therefore W = (155)
2
/52.90 =
454.15 watts
(ii)
At
α
=
π
/2, Vrms = 115V, Therefore W = (115)
2
/52.90 =
250watts
UJT Triggered SCR Phase Control
In SCR power control circuit a control circuit is needed to vary the firing or the phase angle at
which the SCR is triggered.
The value of phase angle controls the power delivered to the load.
If we want to deliver the maximum power to the load, then the SCR must be triggered as soon
as the a.c. voltage across it goes positive i.e., the firing angle must be zero.
Therefore, the SCR
will conduct for 180°after triggering.
If the power requirement is below the maximum value,
the firing angle must be between 0° and 180°.
In order to adjust the firing angle between 0°
and 180°, we need a control circuit. We have already seen that a 180° halfwave rectifier circuit
can be used to accomplish the control.
But this circuit is affected by loading and supply voltage
variation.
Therefore, this circuit does not give a better phase control.
However, a better
control cab be achieved by using a unijunction transistor (UJT) relaxation oscillator circuit.
48
This circuit shows a UJT oscillator controlling a S
CR circuit.
The SCR is in it ‘OFF’ state until the
UJT fires.
When the UJT fires, the capacitor will discharge quickly through the resistor R4.
This
positive pulse will turn ‘ON’ the SCR.
The SCR will remain ‘ON’ until the SCR line voltage
approaches z
ero.
At this point, the SCR will turn ‘OFF’ throughout the entire negative circle.
A
similar circuit can also be made for triac phase control.
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 Bipolar junction transistor, SCR