Operation and analysis of single phase half controlled converters
Instructional Objectives
On completion the student will be able to
Introduction
Single phase fully controlled bridge converters are widely used in many industrial applications. They can supply unidirectional current with both positive and negative voltage polarity. Thus they can operate either as a controlled rectifier or an inverter. However, many of the industrial application do not utilize the inverter mode operation capability of the fully controlled converter. In such situations a fully controlled converter with four thyristors and their associated control and gate drive circuit is definitely a more complex and expensive proposition. Single phase fully controlled converters have other disadvantages as well such as relatively poor output voltage (and current for lightly inductive load) form factor and input power factor.
The inverter mode of operation of a single phase fully controlled converter is made possible by the forward voltage blocking capability of the thyristors which allows the output voltage to go negative. The disadvantages of the single phase fully controlled converter are also related to the same capability. In order to improve the output voltage and current form factor the negative excursion of the output voltage may be prevented by connecting a diode across the output as shown in Fig 11.1(a). Here as the output voltage tries to go negative the diode across the load becomes forward bias and clamp the load voltage to zero. Of course this circuit will not be able to operate in the inverter mode. The complexity of the circuit is not reduced, however. For that, two of the thyristors of a single phase fully controlled converter has to be replaced by two diodes as shown in Fig 11.1 (b) and (c) . The resulting converters are called single phase half controlled converters. As in the case of fully controlled converters, the devices T_{1} and D_{2} conducts in the positive input voltage half cycle after T_{1 }is turned on. As the input voltage passes through negative going zero crossing D_{4 }comes into conduction commutating D_{2} in Fig 11.1 (b) or T_{1} in Fig 11.1 (c). The load voltage is thus clamped to zero until T_{3} is fired in the negative half cycle. As far as the input and output behavior of the circuit is concerned the circuits in Fig 11.1 (b) and (c) are identical although the device designs differs. In Fig 11.1 (c) the diodes carry current for a considerably longer duration than the thyristors. However, in Fig 11.1 (b) both the thyristors and the diodes carry current for half the input cycle. In this lesson the operating principle and characteristics of a single phase half controlled converter will be presented with reference to the circuit in Fig 11.1 (b).
Operating principle of a single phase half controlled bridge converter
With reference of Fig 11.1 (b), it can be stated that for any load current to flow one device from the top group (T_{1} or T_{3}) and one device from the bottom group must conduct. However, T_{1} T_{3} or D_{2} D_{4} cannot conduct simultaneously. On the other hand T_{1} D_{4} and T_{3} D_{2} conducts simultaneously whenever T_{1} or T_{3} are on and the output voltage tends to go negative. Therefore, there are four operating modes of this converter when current flows through the load. Of course it is always possible that none of the four devices conduct. The load current during such periods will be zero. The operating modes of this converter and the voltage across different devices during these operating modes are shown in the conduction table of Fig 11.2. This table has been prepared with reference to Fig 11.1 (b).
It is observed that whenever D_{2} conducts the voltage across D_{4} is vi and whenever D_{4} conducts the voltage across D_{2} is vi. Since diodes can block only negative voltage it can be concluded that D_{2} and D_{4 }conducts in the positive and the negative half cycle of the input supply respectively. Similar conclusions can be drawn regarding the conduction of T_{1 }and T _{3}. The operation of the converter can be explained as follows when T_{1} is fired in the positive half cycle of the input voltage. Load current flows through T_{1 }and D_{2}. If at the negative going zero crossing of the input voltage load current is still positive it commutates from D_{2} to D_{4} and the load voltage becomes zero. If the load current further continuous till T_{3} is fired current commutates from T_{1} to T_{3}. This mode of conduction when the load current always remains above zero is called the continuous conduction mode. Otherwise the mode of conduction becomes discontinuous.
Exercise 11.1
Fill in the blanks(s) with the appropriate word(s)
Answer: (i) thyristors, diodes; (ii) diodes, two; (iii) same, different; (iv) form factor; (v) power factor.
2. Find out an expression of the ration of the thyristor to diode RMS current ratings in the single phase half controlled converter topologies of Fig. 11.1(b) & (c). Assume ripple free continuous output current.
Answer
In the first conduction diagram the diodes and the thyristors conduct for equal periods, since the load current is constant. The ration of the thyristors to the diode RMS current ratings will be unity for the circuit of Fig 11.1 (b).
From the second conduction diagram the thyristors conduct for π  α radians while the diodes conduct for π + α radians. Since the load current is constant.
in this case
Single phase half controlled converter in the continuous conduction mode
From the conduction table and the discussion in the previous section it can be concluded that the diode D_{2} and D_{4} conducts for the positive and negative half cycle of the input voltage waveform respectively. On the other hand T_{1} starts conduction when it is fired in the positive half cycle of the input voltage waveform and continuous conduction till T_{3} is fired in the negative half cycle. Fig. 11.3 shows the circuit diagram and the waveforms of a single phase half controlled converter supplying an R – L – E load.
Referring to Fig 11.3 (b) T_{1} D_{2} starts conduction at ωt = α. Output voltage during this period becomes equal to v_{i}. At ωt = π as v_{i} tends to go negative D_{4} is forward biased and the load current commutates from D_{2} to D_{4} and freewheels through D_{4} and T_{1}. The output voltage remains clamped to zero till T_{3 }is fired at ωt = π + α. The T_{3} D_{4} conduction mode continues upto ωt = 2π. Where upon load current again free wheels through T_{3} and D_{2} while the load voltage is clamped to zero.
From the discussion in the previous paragraph it can be concluded that the output voltage (hence the output current) is periodic over half the input cycle. Hence
Clearly in addition to the average component, the output voltage (and current) contains a large number of harmonic components. The minimum harmonic voltage frequency is twice the input supply frequency. Magnitude of the harmonic voltages can be found by Fourier series analysis of the load voltage and is left as an exercise.
The Fourier series representation of the load current can be obtained from the load voltage by applying superposition principle in the same way as in the case of a fully controlled converter.
However, the closed form expression of i_{o} can be found as explained next.
In the period α ≤ ω t ≤ π
The input current ii is given by
ii = i0 for α ≤ ω t ≤ π
ii =  i0 for π + α ≤ ωt ≤ 2π
i_{i} = 0 otherwise (11.14)
However, it will be very difficult to find out the characteristic parameters of i_{i} using equation 11.14 since the expression of i_{0} is considerably complex. Considerable simplification can however be obtained if the actual i_{i} waveform is replaced by a quasisquare wave current waveform with an amplitude of I_{oav }as shown in Fig 11.5.
Single phase half controlled converter in the discontinuous conduction mode.
So far we have discussed the operating characteristics of a single phase half controlled converter in the continuous conduction mode without identifying the condition required to achieve it. Such a condition exists however and can be found by carefully examining the way this converter works.
Referring to Fig 11.3 (b), when T_{1} is fired at ωt = α the output voltage (instantaneous value) is larger than the back emf. Therefore, the load current increases till v_{o} becomes equal to E again at ωt = π – θ. There, onwards the load current starts decreasing. Now if i_{o} becomes zero before T_{3} is fired at ωt = π + α the conduction becomes discontinuous. So clearly the condition for continuous conduction will be
If the condition in Eq. 11.22 is violated the conduction will become discontinuous. Clearly, two possibilities exist. In the first case the load current becomes zero before ωt = π. In the second case i_{o} continuous beyond ωt = π but becomes zero before ωt = π + α. In both cases however, i_{o} starts from zero at ωt = α.
Fig. 11.6 shows the wave forms in these two cases
Of these two cases the second one will be analyzed in detail here. The analysis of the first case is left as an exercise.
For this case
However IORMS cannot be computed directly from VORMS. For this the closed form expression for io has to be obtained. This will also help to find out an expression for the conduction angle β.
Fill in the blank(s) with the appropriate word(s).
(iv) At the boundary between continuous and discontinuous conduction the value of the output current at ωt = α is ___________________.
(v) The output voltage and current waveform of a single phase fully controlled and half controlled converter will be same provided the extinction angle β is less than
___________________.
(vi) For the same value of the firing angle the average output voltage of a single phase half controlled converter is ___________________ in the discontinuous conduction mode compared to the continuous conduction mode.
(vii) Single phase half controlled converters are most suitable for loads requiring
___________________ voltage and current.
Answer: (i) zero; (ii) π; (iii) higher; (iv) unidirectional.
2. A single phase half controlled converter charges a 48v 50Ah battery from a 50v, 50Hz single phase supply through a 50mH line inductor. The battery has on interval resistance of 0.1Ω. The
firing angle of the converter is adjusted such that the battery is charged at C/5 rate when it is fully discharged at 42 volts. Find out whether the conduction will be continuous or discontinuous at this condition. Up to what battery voltage will the conduction remain continuous? If the charging current of the battery is to become zero when it is fully charged at 52 volts what should be the value of the firing angle.
Lesson Summary
Practice Problems and Answers
Q1. The thyristor T_{3} of Fig 1.1(b) fails to turn on at the desired instant. Describe how this circuit will work in the presence of the fault.
Q2. A single phase half controlled converter is used to boost the no load speed of a separately excited dc machine by weakening its field supply. At α = 0° the half controlled converter produces the rated field voltage. If the field inductance is large enough to make the field current almost ripple face what will be the input power factor when the dc motor no load speed is bossed to 150%?
Q3. A single phase half controlled converter supplies a 220V, 1500rpm, 20A dc motor from a 230V 50HZ single phase supply. The motor has a armature resistance of 1.0Ω and inductance of 50mH. What will be the operating modes and torques for α = 30°; and speed of 1400 RPM.
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