In single-phase 2-pulse (or full-wave) converters, regulating the output voltage is achievable by modifying the firing angle delay of the thyristors.The conversion from AC to DC can utilize either mid-point or bridge-type circuits.This section covers discussions on both mid-point and bridge-type configurations, considering input from a single-phase source.
Single-phase Full-wave Mid-point Converter
The circuit diagram depicted in Figure show in above illustrates a single-phase full-wave converter utilizing a center-tapped transformer. When terminal a is positive relative to n, terminal n becomes positive in relation to b. Consequently, van equals vnb, or van is equals to minus vbn, given that n represents the midpoint of the secondary winding.The corresponding equivalent circuit is presented in Figure below. For this explanation, we assume a continuous load or output current and a unity turns ratio from the primary to each secondary.
.jpg "Single-phase full-wave mid-point converter (a) circuit diagram (b) equivalent circuit and (c) various voltage and current waveforms") |
| Single-phase full-wave mid-point converter (a) circuit diagram (b) equivalent circuit and (c) various voltage and current waveforms |
Thyristors T1 and T2 operate in forward bias during positive and negative half cycles, triggered accordingly. If T2 is already conducting, and after ωt = 0 with positive Vam, T1 becomes forward biased and is triggered at the delay angle α. This turn-on of T1 occurs at α, causing the supply voltage 2Vmsinα to reverse bias T2, leading to its turn-off. T1 is referred to as the incoming thyristor, and T2 as the outgoing thyristor.As T1 is triggered, the ac supply voltage induces reverse bias across T2, turning it off.This natural reversal of ac supply voltage, causing SCR turn-off, is known as natural or line commutation.
In the equivalent circuit shown in Figure, it is observed that if...
van =Vmsin ωt
then Vbn=-Vnb=-Vmsinωt
and vab =van +vnb =2Vm sinωt
At ωt = α, T1 initiates triggering. SCR T2 experiences a reverse voltage vab=2Vmsinα, as mentioned earlier.Current is shifted from T2 to T1, leading to the turn-off of T2. The reverse voltage across T2's magnitude can be determined by applying KVL to the loop efghe in the equivalent circuit of show in figure at the moment T1 is triggered.
vT2 -vbn +vT1 =0
or vT2 = vbn - van +vT1
while T1 is conducting, vT1 equals 0. Thus, the voltage across T2 at the moment ωt = α is expressed as follows:
vT2 = - Vm sinα - Vm sinα = - 2Vm sinα
This indicates that SCR T2 experiences a reverse bias of voltage 2Vmsinα, leading to its turn-off at ωt = α. Thyristor T1 conducts from α to π + α. Beyond ωt = π, although T1 is reverse biased, it continues to conduct since the forward-biased SCR T2 is not gated. At ωt = π + α, T2 is triggered, resulting in T1 being reverse biased by a voltage of magnitude 2Vmsinα. Current is transferred from T1 to T2, causing the turn-off of T1.
When ωt = α, T2 is turn-off and remains reverse biased from ωt = α to π, as illustrated in the various voltage and current waveforms shown in the figure.The turn-off time for SCR T2 in this circuit is consequently determined by...
\( t_c \) = \(\frac{\pi - \alpha}{\omega}\)sec .....equation(1)
At ωt = π + α, Thyristor T1 experiences turn-off.The figure, displaying various voltage and current waveforms, illustrates that T1 encounters a reverse voltage from ωt = π + α to ωt = 2π. Consequently, this circuit determines the turn-off time for thyristor T1 as...
\( t_c \) = \( \frac{2\pi - (\pi + \alpha)}{\omega} = \frac{\pi - \alpha}{\omega} \)
The turn-off time provided to thyristor T1 is the same as that provided to thyristor T2, as indicated by Equation (1).
Examining the voltage waveform v0, the average value of the output voltage is determined by.....
\[ V_0 = \frac{1}{\pi} \int_{\alpha}^{\alpha + \pi} V_m \sin \omega t \, d(\omega t) = \frac{2V_m}{\pi} \cos \alpha \] ..........equation(2)
This formula expresses the average value of the output voltage \( V_0 \) based on the provided integral expression.
The turn-off time (\(t_c\)) in Equation(1), depicted in show in figure single-phase full-wave mid-point converter with circuit diagram, should exceed the SCR turn-off time (\(t_q\)) specified in the data sheet. If \(t_c\) < \(t_q\), there's a risk of commutation failure, leading to a short circuit in the entire secondary winding. In the event of commutation failure, and a rapid rise in fault current, the incoming SCR might sustain damage if protective elements fail to clear the fault. Show in figure Single-phase full-wave mid-point converter with various voltage and current waveforms illustrates that each SCR experiences a peak voltage of 2\(V_m\).
The above study yields the following observations:
- For SCR commutation, the SCR to be commutated must be reverse-biased, while the incoming SCR needs to be forward-biased.
- Upon gating on the incoming SCR, there is a transfer of current from the outgoing SCR to the incoming one.
- The circuit turn-off time should exceed the SCR turn-off time.
It is evident that achieving thyristor commutation through the natural reversal of line voltage, known as line or natural commutation, is a straight forward method. This approach finds application in all phase-controlled rectifiers, AC voltage controllers, and cycloconverters.
Working Principle of Single-Phase Full-Wave Bridge Converter
Phase-controlled single-phase or three-phase full-wave converters can be categorized into three main types: uncontrolled converters, half-controlled converters, and fully-controlled converters.
- Uncontrolled converters, also known as rectifiers, rely solely on diodes, providing a DC output voltage that cannot be controlled.
- Half-controlled converters, or semiconverters, utilize a combination of diodes and thyristors, offering a limited degree of control over the DC output voltage.
- Fully-controlled converters, or full converters, exclusively employ thyristors, allowing for more extensive control over the level of the DC output voltage.
.jpg "One-quadrant converter and two-quadrant converter") |
| One-quadrant converter and two-quadrant converter |
A semiconverter operates as a one-quadrant converter, meaning it produces a dc output with a single polarity of voltage and current at its output terminals, as shown in figure. On the other hand, a two-quadrant converter allows the reversal of voltage polarity while maintaining unidirectional current flow due to the nature of thyristors, as depicted in Figure. this section focuses on a detailed study of single-phase bridge-type full converters and semiconverters.
Single-Phase Full Converter Bridge Analysis
Figure 1 illustrates a single-phase full converter bridge employing four SCRs and functioning as a one-quadrant converter. The load is assumed to be of RLE type, where E represents the load circuit emf, originating from a battery or the generated emf of a DC motor. The converter operates by triggering thyristor pairs T1 and T2 simultaneously, followed by pairs T3 and T4 with a delay of π radians. The supply voltage waveform, \(v_{ab}\), is depicted when a is positive relative to b, while \(v_{ba}\) is shown when b is positive relative to a. Notably, \(v_{ab} = -v_{ba}\), and the current directions and voltage polarities in the figure are treated as positive.
.jpg "Single-phase full converter bridge with RLE load (b) voltage and current waveforms for continuous load current.") |
| Single-phase full converter bridge with RLE load (b) voltage and current waveforms for continuous load current |
Assuming continuous load current (\(i_0\)) over the working range, the thyristors maintain a connection between the load and the AC voltage source. Between ωt = 0 and ωt = α, T1 and T2 are forward-biased through already conducting SCRs T3 and T4, blocking the forward voltage. For continuous current, T3 and T4 conduct even though reverse-biased after ωt = 0. When forward-biased T1 and T2 are triggered at ωt = α, they turn on, causing \(V_m\)sinα to appear across T3 and T4. These are turned off by natural (line) commutation. Simultaneously, load current \(i_0\) flows from T3 and T4 to T1 and T2 at ωt = α. Notably, T1 and T2 conduct from ωt = α to π + α, and at ωt = π + α, forward-biased T3 and T4 are triggered.The supply voltage turns off T1 and T2 by natural commutation, transferring load current from T1 and T2 to T3 and T4. It is essential to note that T1 and T2 conduct between ωt = α and π + α only if \(V_m\)sinα > E.
The voltage across thyristors T1 and T2 is denoted as \(V_{T1} = V_{T2}\), while that across T3 and T4 is represented as \(V_{T3} = V_{T4}\).The maximum reverse voltage across T1, T2, T3, or T4 is \(V_m\), and at the triggering instant with a firing angle α, each SCR faces a reverse voltage of \(V_m\)sinα.The source current \(i_s\) is considered positive in the arrow direction. Following this assumption, the source current is depicted as positive when T1 and T2 are conducting and negative when T3 and T4 are conducting. The figure illustrates voltage and current waveforms for continuous load current, maintaining these conventions.
From α to π, both \(v_s\) and \(i_s\) exhibit positivity, indicating power flowing from the AC source to the load. In the interval π to (π + α), \(v_s\) is negative, yet \(i_s\) remains positive, signifying that the load returns some energy to the supply system.However, the overall power flow remains from the AC source to the DC load since (π - α) > α in the figure depicting voltage and current waveforms for continuous load current.
The load terminal voltage, or the output voltage \(v_0\) of the full converter, is illustrated in the figure.The average value of the output voltage \(V_0\) is expressed as:
\[ V_0 = \frac{1}{\pi} \int_{\alpha}^{\alpha + \pi} V_m \sin \omega t \, d(\omega t) = \frac{2V_m}{\pi} \cos \alpha \] .......... equation(2)
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| Voltage and current waveform for single-phase full converter of figures for α > 90° |
Equation (3) indicates that when α exceeds 90°, \(V_0\) becomes negative.Refer to the Voltage and current waveform diagram for a single-phase full converter in the attached figure, specifically for α > 90°. The figure explicitly displays α greater than 90°, highlighting the scenario where the average terminal voltage \(V_0\) is negative. When the load circuit's electromotive force (EMF) E is reversed, this reversed source E transfers power back to the AC supply. This operational state of the full converter is termed the inverter operation of the converter. A full converter with a firing angle delay surpassing 90° is referred to as a line-commutated inverter. Such an operation finds application in the regenerative braking mode of a DC motor, wherein E represents the counter EMF of the DC motor.
In the interval 0 to α, the AC source voltage \(v_s\) is positive, while the AC source current \(i_s\) is negative, causing power to flow from the DC source to the AC source.Moving from α to π, both \(v_s\) and \(i_s\) are positive, resulting in power flowing from the AC source to the DC source.However, the overall power flow remains from the DC source to the AC source, as (π-α) < α in the figure.
The output voltage \(V_0\) must surpass the load circuit electromotive force (E) value. During inverter operation, the inverted load circuit electromotive force, when converted to AC, should exceed the AC supply voltage.Essentially, the DC source voltage (E) must be greater than the inverter voltage \(V_o\) for power to flow from the DC source to the AC supply system.In both converter and inverter modes, thyristors must be forward-biased, and the current through Silicon Controlled Rectifiers (SCRs) should flow in the same direction since these are unidirectional devices.This explains why the output current \(i_0\) is depicted as positive in the figure. Similarly, the source current \(i_s\) is positive when T1 and T2 are conducting.
The change in voltage across thyristors T1, T2, T3, or T4 indicates that the circuit turn-off time for both converter and inverter operations is determined by...
\( t_c \) = \(\frac{\pi - \alpha}{\omega}\)sec
After studying both types of phase-controlled converters, the advantages of the single-phase bridge converter over the single-phase mid-point converter can be highlighted:
- SCRs are subjected to a peak inverse voltage of 2 \(V_m\) in mid-point converter and \(V_m\) in full converter. Thus for the same voltage and current ratings of SCRs, power handled by mid-point configuration is about half of that handled by bridge configuration.
- In the mid-point converter, each secondary must be capable of supplying the load power, leading to a transformer rating double the load rating. This contrast does not apply to the single-phase bridge converter.
It may thus be inferred from above that bridge configuration is preferred over mid-point configuration. However, the choice between these two types depends primarily on cost of the various components, available source voltage and the load voltage required. Mid-point configuration is used in case the terminals on dc side human have to be grounded.
Analysis of Single-Phase Semiconverter Bridge
Figure depicts a single-phase semiconverter bridge featuring two thyristors (T1, T2) and three diodes (D1, D2). Additionally, a freewheeling diode (FD) is connected across the load. Similar to the full converter bridge, the load is of the RLE type.The figure illustrates various voltage and current waveforms for this converter, assuming continuous load current throughout the operational range.
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| Single-phase semiconverter bridge (a) Power-circuit diagram with RLE load and (b) voltage and current waveforms for continuous load current |
After ωt=0, thyristor T1 becomes forward-biased only when the source voltage \(V_m\)sinωt exceeds E. Consequently, T1 is triggered with a firing angle delay α, ensuring \(V_m\)sinα > E. During the period ωt = α to π, with T1 conducting, the load connects to the source through T1 and D1. The load current \(i_0\) flows through RLE, D1, the source, and T1, resulting in the load terminal voltage \(v_0\) having the same waveshape as the AC source voltage \(v_s\).
As ωt approaches π, the load voltage \(v_o\) starts to reverse due to the changing polarity of the AC source voltage. At ωt = π+, the freewheeling diode (FD) becomes forward-biased and conducts. The load current \(i_0\) is then transferred from T1 and D1 to FD. With FD conducting, SCRT1 becomes reverse-biased at ωt = π+, leading to T1 turning off.This causes the load terminals to be short-circuited through FD, resulting in the load voltage \(v_o\) being zero during π < ωt < (π + α).
After ωt = π, during the negative half cycle, T2 becomes forward-biased when the source voltage exceeds E. At ωt = π + α, T2 is triggered, and FD, being reverse-biased, turns off. Load current shifts from FD to T2 and D2. At ωt = 2π, FD becomes forward-biased again, and the output current \(i_0\) is transferred from T2 and D2 to FD, following the explained sequence.
The source current \(i_s\) is positive from α to π when T1 and D1 conduct and negative from (π + α) to 2π when T1 and D2 conduct, as illustrated in the waveforms for continuous load current figure.
In the period from α to π, T1 and D1 conduct, allowing the AC source to supply energy to the load circuit. This energy is distributed, with a portion stored in the inductance (L), another portion converted into electric energy in the load-circuit electromotive force (E), and the rest dissipated as heat in resistor R. During the freewheeling interval from α to (π + α), the energy stored in the inductance is recovered. This recovered energy is then partially dissipated in resistor R and partially added to the energy stored in the load electromotive force (E). Importantly, no energy is fed back to the source during the freewheeling period.
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| Converter output voltage as a function of firing angle for semi- and full-converters |
The average output voltage \(V_0\) for the semiconverter, as depicted in the figure, is expressed as:
\[ V_0 = \frac{1}{\pi} \int_{\alpha}^{ \pi} V_m \sin \omega t \, d(\omega t) = \frac{V_m}{\pi} (1+\cos \alpha) \] ......... equation(3)
The variation in voltage across T1 and T2 is illustrated in the figure. Examination of these waveforms reveals that the circuit turn-off time for the semiconverter is...
\( t_c \) = \(\frac{\pi - \alpha}{\omega}\)sec
The figure illustrates the variation in the average value of the converter output voltage as a function of the firing angle (α) for both the semiconverter and full converter.
Continuous Conduction Analysis of Two-Pulse Bridge Converter
In this section, we present a steady-state analysis of a single-phase two-pulse converter for both types.
Semiconverter. In the conduction period, the voltage equation for the power circuit diagram with RLE load is as follow:
\(v_{0} = v_{s} = R \ i_{0} + L \frac{d i_0}{dt} + E\) ........... Equation(4)
for α < ωt ≤ π
In the freewheeling period, the voltage equation can be expressed as follows.
\(0 = R \ i_{0} + L \frac{d i_0}{dt} + E\) ........equation(5)
or π < ωt ≤ (π+α)
Integrating the aforementioned voltage equations across their corresponding time intervals yields,
form equation(4), \(\int_{\frac{a}{\omega}}^{\frac{\pi}{\omega}} v_{0} \, dt\) = R\(\int_{\frac{a}{\omega}}^{\frac{\pi}{\omega}} i_{0} \, dt\) + \(L \int_{i_1}^{i_2} \, di_{0}\) +\(\int_{\frac{a}{\omega}}^{\frac{\pi}{\omega}} E \, dt\)
form equation(5), \(\int_{\frac{a}{\omega}}^{\frac{\pi}{\omega}} 0 . \, dt\) = R\(\int_{\frac{a}{\omega}}^{\frac{\pi}{\omega}} i_{0} \, dt\) + \(L \int_{i_1}^{i_2} \, di_{0}\) +\(\int_{\frac{a}{\omega}}^{\frac{\pi}{\omega}} E \, dt\)
In the above equations, \(i_1\) = output current at ωt = α, π+α etc. and \(i_2\) = output current at ωt= π, 2π etc.
Combining the terms from the aforementioned equations, and then dividing the sum by the duration of one-half cycle (i.e., \(\frac{\pi}{\omega}\)).
\(\frac{\omega}{\pi} \int_{\frac{a}{\omega}}^{\frac{\pi}{\omega}} V_{m} \sin(\omega t) \, dt\) = \(\frac{\omega \cdot R}{\pi} \int_{\frac{a}{\omega}}^{\frac{\pi + a}{\omega}} i_{0} \, dt\) + L[(\(i_2\) - \(i_1\)) + (\(i_1\) - \(i_2\))] + \(\frac{\omega \cdot E}{\pi} \left[\pi - \alpha + \pi + \alpha - \pi\right]\)
or \(V_{0} = R \ I_{0} + E\) ....... equation (6)
where \(V_{0} = \frac{V_{m}}{\pi} \cdot (1 + \cos \alpha)\) = average voltage applied to the load
\(i_0\) = average load current; E = Load circuit emf
In the case where the load is a DC motor, the equations are:
\(E = K - \omega_{m}\) (Back electromotive force)
\(R = r_{a}\) (Armature-circuit resistance)
\(I_{0} = I_{a}\) (Armature current)
\(T_{c} = K_{m} I_{a}\) (Electromagnetic torque)
Where Tc= ectromagnetic torque in Nm.
\(K_m\) = torque constant in Nm/A, or emf constant in V-sec/rad
\(V_{0} = r_{a} \ I_{a} + K_{m} \omega_{m}\)
or \(\omega_{m} = \frac{{V_{0} - r_{a} \cdot I_{0}}}{{K_{m}}}\)
\(\omega_{m} = \frac{\left(\frac{V_{m}}{\pi}\right)(1 + \cos \alpha)}{K_{m}} - \frac{r_{a}}{K_{m}^{2}} T_{c}\) ........ equation(7)
Full-converter.The voltage equation, for the circuit of show in figure, is
\(v_{0} = v_{s} = R \ i_{0} + L \frac{d i_0}{dt}+ E\)
Its average value, as in a semi-converter, is
\(V_{0} = R \ I_{0} + E\) ..... equation(8)
Where \(V_{0} = \frac{2V_{m}}{\pi} \cos \alpha\)
In case load is a dc motor, \(V_{0} = r_{a} \ I_{a} + K_{m} \omega_{m}\)
Or \(\omega_{m} = \frac{\left(\frac{V_{m}}{\pi}\right) \cos \alpha}{K_{m}} - \frac{r_{a}}{K_{m}^{2}} T_{c}\) .........equation(9)
Comparison of Semi-Converters and Full-Converters
| Feature | Semi-Converter | Full Converter |
|---|---|---|
| Number of Controlled Phases | Single-phase | Single-phase or Three-phase |
| Output Voltage Waveform | Unidirectional | Bidirectional (AC to DC or DC to AC) |
| Device Utilization | Utilizes only one half of the AC waveform | Utilizes the entire AC waveform |
| Ripple Factor | Higher ripple in output | Lower ripple in output |
| Transformer Utilization | 50% (for single-phase) | 100% (for full-wave) |
| Circuit Complexity | Simpler design | More complex design |
| Applications | Limited applications | Wider range of applications |