ee366-chap-3-3
TRANSCRIPT
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Mode 2
At the time of turn-on, none of the
devices is conducting.
Just after turn-on, two diodes will
conduct in addition to the SCRSubsequently one diode will turn off
Followed by the SCR and the otherdiode.
The expressions for the instantaneous
out ut volta e are as follows:
322
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(a) 65 (Dand)T2t0For :conductingD,T 65 0van
(b) 621 (Dand)D(),Tt2For :conductingNone 0van
(c) 6D32tFor :conductingD,D,T 621 tV
m sinvan
(d) 21 D(and)T67t32For :conductingD,T 21
6
sin2
3van
tV
m
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(e) 243 D(and)D(),T32t67For
:conductingNone 0van
(f) 2D34t32For :conductingD,T,D 432 tV
m sinvan
(g)
43
D(and)T611t34For
:conductingD,T 43
6
sin2
3van
tV
m
(h) 465
(Dand)D(,)T34t611For :conductingNone 0van
(i) :D2t34For 4 :conductingD,TD 654,
tVm sinvan
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Mode 3At the time of turn-on, none of the
devices is conducting.
Just after turn-on, only one SCR and
one diode will conduct.
At zero power is delivered to the
load.The expressions for the instantaneous
output voltage are as follows:
6732
67
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(a) 65 (Dand)T3/2t0For
:conductingNone 0van
(b) 65 (Dand)T2
t3/2For
:conductingD,T 65 0van
(c) 21 (Dand)Tt2For :conductingNone 0van
(d) 21
(Dand)T67tFor
:conductingD,T 21
6
sin2
3van
tV
m
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(e) 43 (Dand)T32t67For :conductingNone 0van
(f) 43 (Dand)T611t32For :conductingD,T 43
6
sin2
3van
tV
m
(g) 2t611For :conductingisNone 0van
T1, D2, T3, D4, T5, D6, T1, D2, T3,.....
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The rms output voltage can be obtained asfollows
Mode 1 20
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Mode 2 322
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Mode 3 6732
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Example 3The 3-phase half-controlled regulator
supplies a star connected resistive load of
R= 10 per phase and the line-to-line
voltage is 208 V, 60 Hz. = 60o. Find
(a) The rms output phase voltage
(b) The input power factor and
(c) Expressions for the instantaneous
output voltage of phase a
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Solution(a)
(b) RMS phase currentOutput power
Input volt-amperes
The power factor PF
A094.111094.110 oI W3692.311011.09433 22 RI
o
VA3996.80094.1120833 oLIV
lagging0.923996.89
3692.31
VA
P
o
8
2sin
43
13
21
VV
o
21
8
32sin
123
1208
V110.94
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The instantaneous output phase voltage,which depends on the number of conducting
devices are as follows:
anv
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3.3 Other Regulator CircuitsDelta-connected ac regulator
It is used for delta-connected loads
where each end of each phase is
accessible.
It consists of three single-phase acregulators operating independently of
each other as shown below.
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Delta-connected ac regulator contd
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Delta-connected ac regulator contdThe SCRs should be capable of
carrying the phase currents and
withstand the line voltages.
The firing angle control range is 0 to180ofor resistance load.
The firing angle is measured from the
zero crossing of the line-to-line voltage.
This can be employed to reduce SCR
current ratings.
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Open-Star 3-Phase AC Regulator withSix SCRs
They are used for star-connected load
where neutral exists and can be
accessed and opened.
In this regulator, shown on the next
slide, current can flow between twolines even if one SCR is conducting, so
each SCR requires one firing pulse per
cycle.
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Ia
Ib
Ic
ZL
ZL
ZL
T2
T3
T6
T4
T1
T5
vc
vb
va
a
b
c
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Open-Star 3-Phase AC Regulator withSix SCRs Contd
The voltage and current ratings of the
SCRs are nearly the same as those of
the delta-connected ac regulator.
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Open-Star 3-Phase AC Regulator withThree SCRs
It is possible to operate the open-star 3-
phase ac regulator with three SCRs as
shown on the next slide.
Each SCR is provided with gate pulses
in each cycle spaced 120o
apart.Though, the number of devices is
fewer, their current ratings must be
higher.
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Integral Cycle ControlIn SCR heating applications, load
harmonics are unimportant and integral
cycle control, or burst firing, can be
employed.
Basic triac burst-firing controller circuitis shhown on the next slide.
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I l C l C l C d
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Integral Cycle Control ContdThe output voltage waveforms:
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Integral Cycle Control ContdCommercially produced gate drivers may be
used to trigger the Triacs.
The lowest order harmonics in the load is 1/Tp.
The rms output voltage isThe output power is
The supply displacement factor is unity
The input power factor where
n = the number of oncycles and
N = number of cycles in the period Tp
21
NnVVo
NnRVP 2
NnPF
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Example 4A burst-firing ac voltage regulator
supplies power to a 100-ohm resistance
from a 250 V, 50 Hz supply. Determine
the values of (a) load voltage (b) load
Current (c) output power and (d) input
power when n=2andN = 8.
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Solution(a)
(b)
(c) Output power
(d)Inputand hence input
V12582250 21
21
NnVVo
A25.1100125 RVIoo
W3.15610025.122 RIP
o
VA5.31225.1250 oVIS
50.05.3123.156 SPPF
A 3 h h lf l f d b d li
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A 3-phase half-wave ac regulator fed by a star-connected source supplies power to a star-
connected resistive load ofR = 10 per phase. The phase voltages are va= 170 sint, vb= 170
sin(t120o) and vc= 170 sin(t + 120
o). The circuit operates with a delay angle =135
o.
(a)Draw the circuit diagram of the regulator with T1andD4connected anti-parallel in phasea, T3andD6anti-parallel in phase b and T5andD2anti-parallel in phase c. [4marks]
(b)Given that for a delay angle =135o, all switches are off before an SCR is fired and an SCRwhen fired switches on with only one diode, determine the following:
(i) The instant in degrees (= t) at which T1is fired, the diode that switches on with T1andthe instant (= t) at which T1 turns off. [3 marks]
(ii) The instant in degrees (= t) at which T3is fired, the diode that switches on with T3andthe instant (= t) when T3turns off. [3 marks]
(iii)The instant in degrees (= t) at which T5is fired, the diode that switches on with T5andthe instant (= t) at which T5turns off. [3 marks]
(c)Obtain expressions for the instantaneous output voltage of phase a over a period. Givenon zeroexpressions only. [4
marks]
(d)Using the expressions obtained in part (c), calculate the rms output phase voltage. [4 marks](e)Calculate the input power factor. [4 marks]