This post aims to provide a comprehensive overview of the power circuit utilized in three-phase full-converter systems feeding RLE (Resistive-Inductive-Electronic) loads. It will delve into the components, operation, benefits, and applications of this advanced power electronics technology.
If all the diodes in the 3-phase full-converter bridge of Figure are substituted with thyristors, the resulting configuration, depicted in Fig(i), forms a three-phase full-converter bridge. This setup allows for the connection of a three-phase input at terminals A, B, C, with the RLE load linked across the output terminals. Similar to a single-phase full-converter, the thyristor power circuit in Fig. operates as a three-phase AC to DC converter when the firing angle delay falls within the range of 0°< α ≤ 90°. Beyond this range, between 90° < α < 180°, it functions as a three-phase line-commutated inverter. The preference for a three-phase full converter arises particularly in scenarios where power regeneration is necessary. The SCR numbering in Fig. follows the sequence of 1, 3, 5 for the positive group, and 4, 6, 2 for the negative group, aligning with the gating sequence of the six thyristors in a 3-phase full converter.
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| Fig(i) Power circuit for a 3-phase full-converter feeding RLE load |
For α = 0°, thyristors T1 to T6 function akin to diodes, illustrated in Fig(ii)(a). Their conduction sequence is depicted in the same figure. Notably, at α = 0°, T1 triggers at Ï€/6, T2 at 90°, T3 at 150°, and so forth. Consequently, the load voltage assumes the waveform depicted in Fig(Output voltage waveform of 3-phase six pulse diode bridge). At α = 60°, the conduction sequence of thyristors T1 to T6 alters, showcased in Fig(ii) (b). Here, T1 triggers at ωt = 30°+60°=90°, T2 at 90°+60°=150°, T3 at 150°+60°, and so on. Illustrating the conduction intervals of T1, T2,..., T6 upfront facilitates the drawing of voltage and current waveforms. Each SCR conducts for 120°; when T1 triggers, reverse-biased thyristor T5 deactivates, and T1 engages. As T1 connects to A and T6 to B, the load experiences voltage \(v_{ab}\), ranging from 1.5\(V_m\) to zero.Here \(V_{mp}\) is the maximum value of phase voltage. Continuing, when T2 activates, T6 commutates from the negative group, and T1 sustains conduction. As T1 links to A and T2 to C, voltage \(v_{ac}\) arises across the load, varying from 1.5\(V_m\) to zero. This sequence persists for the remaining SCRs.
Note that the positive group of SCRs triggers at intervals of 120°, mirroring the pattern for the negative group. However, both groups synchronize their firing at 60° intervals. This implies alternate commutation every 60° between upper and lower groups of SCRs. Each SCR in both groups conducts for 120°. Consequently, for the source to energize the load, two SCRs, one from the positive group and the other from the negative group, must conduct simultaneously. In the case of an ABC phase sequence in the three-phase supply, thyristors operate in pairs: T1 and T2, T2 and T3, T3 and T4, and so on. This synchronized operation ensures smooth power delivery and effective load management in the circuit.
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| Fig(ii) Voltage waveforms and conduction of thyristors for a 3-phase full converter. |
The sequence depicted in Fig(ii) can be better illustrated by considering line voltages instead of phase voltages. Fig( iii)(a) shown cases line voltages \(v_{ab}, v_{ac}, v_{bc}, v_{ba}\) etc. When α = 0°, SCRs T1 to T6 behave like diodes, producing an output voltage waveform illustrated in Fig(iii)(a) by \(v_{ab}, v_{ac},v_{bc}\) etc. Here, for α = 0°, T1 triggers at ωt = 60°, T2 at ωt = 120°, T3 at ωt = 180°, and onward. Consequently, in Fig(iii)(a), the firing angle is measured from 60° for T1, 120° for T2, 180° for T3, and so on, enhancing clarity in understanding the timing and operation of the SCRs in the circuit.
Readers may question why T1, at α = 0°, commences conduction from ωt = 60° and not form ωt = 0°. The use of subscripts like ab, ac, bc, ba aids in clarifying this confusion. Notably, subscripts appear twice in sequence. When the first subscript repeats, the SCR in the positive group corresponding to that line conducts for 120°. Similarly, when the second subscript repeats, the SCR in the negative group related to that line conducts for 120°. For instance, in \(v_{ab}\) and \(V_{ac}\), the first subscript 'a' repeats, indicating that SCR T1 from the positive group initiates conduction when vab appears, i.e., at ωt = 60° in Fig(iii) (a). Likewise, in \(v_{ac}\) and \(v_{bc}\), the second subscript 'C' repeats, signifying that SCR T2 from the negative group initiates conduction when \(v_{ac}\) appears, i.e., from ωt = 120° in Fig(iii)(a). This consistent pattern aids in understanding the triggering sequence of SCRs in the circuit. Similarly, the first subscript 'b' repeats in \(v_{bc}\) and \(v_{ba}\), indicating that SCR T3 from the positive group begins conduction when \(v_{bc}\) appears, i.e., at ωt = 180° in Fig(iii)(a).
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| Fig(iii) Voltage and current waveforms for a 3-phase full-converter for different firing angles. |
At α = 60°, T1 triggers at ωt = 60 + 60 = 120°, T2 at ωt = 180°, T3 at ωt = 240°, and so forth. When T1 activates at ωt = 120°, T5 deactivates, while T6 continues conduction. Given T1 and T6's connection to terminals A and B, the load voltage must be \(v_{ab}\), as depicted in Fig(iii)(b). Upon T2's activation, T6 commutates. With both T1 and T2 conducting, the load voltage becomes \(v_{ac}\), as shown in Fig(iii)(b). This sequential turning on or off of other SCRs facilitates drawing the load voltage waveform. At α = 90°, the load voltage becomes symmetrical about the reference line, resulting in an average value of zero. At α = 150°, T1 triggers at ωt = 210°, T2 at ωt = 270°, and so forth, producing an output voltage waveform as depicted in Fig(iii)(d). Notably, the average voltage reverses polarity, indicating power delivery from a DC source to an AC source, characteristic of line-commutated inverter operation. For α ranging from 0° to 90°, the power circuit in Fig(iii) functions as a 3-phase full converter, delivering power from an AC source to a DC load. Conversely, at α = 90° to 180°, it operates as a line-commutated inverter, supplying power from a DC source to an AC load. This inverter mode is feasible only if the load possesses a direct EMF due to a battery or a DC motor. Remarkably, the current direction remains fixed for both converter and inverter operations, while the output voltage polarity reverses.
In Fig(iii)(d), the source current \(i_A\) in phase A is depicted for α = 150°. As per the arrow direction indicated in Fig(i), \(i_A\) is considered positive. Hence, \(i_A\) is positive when T1 conducts, corresponding to the first subscript for voltages or currents being 'a'. Conversely, \(i_A\) is negative when T4 conducts, aligned with the second subscript for voltages or currents being 'a'. Source current waveforms for the remaining two phases can be similarly represented. For other firing angles, source currents can be drawn in a similar manner, ensuring consistency with the conduction sequence of the thyristors and the directional conventions established in the circuit diagram.
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| Fig(iv) Output voltage waveform for a 3-phase full converter. |
To determine the expression for the average output voltage \( v_{0} \), we can refer to Fig(iv), which depicts waveforms such as \(v_{ab}\) and \(v_{ac}\) from Fig(iii)(a) with a firing angle delay \( \alpha < 30° \). The periodicity of the output voltage waveform is \( \frac{\pi}{3} \) radians.
The average value of the output voltage \( v_{0} \) is calculated by first identifying the dashed area abcd over one complete cycle as shown in Fig(iv), and then dividing this area by the period of the waveform. Assuming OO' as the origin at the peak value of \(v_{ab}\), the expression for \( v_{0} \) is given by
\[ V_0 = \frac{3}{\pi} \int_{-(\frac{\pi}{6} - \alpha)}^{\frac{\pi}{6} + \alpha} v_{ml} \cos(\omega t) \, d(\omega t) \]
Integrating the cosine function with respect to \( \omega t \) over the given interval, we get:
\[ V_0 = \frac{3 v_{ml}}{\pi} [\sin(\alpha + \frac{\pi}{6}) - \sin(\alpha - \frac{\pi}{6})]\]
First, let's use the angle subtraction formula for sine:
\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]
Applying this formula to the equation:
\[ \sin(\alpha - \frac{\pi}{6}) = \sin(\alpha) \cos(\frac{\pi}{6}) - \cos(\alpha) \sin(\frac{\pi}{6}) \]
\[ \sin(\alpha + \frac{\pi}{6}) = \sin(\alpha) \cos(\frac{\pi}{6}) + \cos(\alpha) \sin(\frac{\pi}{6}) \]
Substituting these into the original equation:
\[ \frac{3 v_{ml}}{\pi} \left[ \sin(\alpha + \frac{\pi}{6}) - \sin(\alpha - \frac{\pi}{6}) \right] \]
\[ = \frac{3 v_{ml}}{\pi} \left[ \left( \sin(\alpha) \cos(\frac{\pi}{6}) + \cos(\alpha) \sin(\frac{\pi}{6}) \right) - \left( \sin(\alpha) \cos(\frac{\pi}{6}) - \cos(\alpha) \sin(\frac{\pi}{6}) \right) \right] \]
Simplify the expression:
\[ = \frac{3 v_{ml}}{\pi} \left[ \sin(\alpha) \cos(\frac{\pi}{6}) + \cos(\alpha) \sin(\frac{\pi}{6}) - \sin(\alpha) \cos(\frac{\pi}{6}) + \cos(\alpha) \sin(\frac{\pi}{6}) \right] \]
\[ = \frac{3 v_{ml}}{\pi} \left[ 2 \cos(\alpha) \sin(\frac{\pi}{6}) \right] \]
Now, using the fact that \( \sin(\frac{\pi}{6}) = \frac{1}{2} \):
\[ = \frac{3 v_{ml}}{\pi} \left[ 2 \cos(\alpha) \times \frac{1}{2} \right] \]
\[ = \frac{3 v_{ml}}{\pi} \cos(\alpha) \]...........equestion(1)
\( V_{ml} \) represents the peak value of the line voltage.
If a sine function is employed for the source voltage, then the expression for \( v_{ab} \) can be given as \( v_{ab} = V_{ml} \sin(\omega t) \) because \( v_{ab} \) equals zero when \( \omega t = 0 \).
\[ V_0 = \frac{3}{\pi} \int_{\frac{\pi}{3} + \alpha}^{\frac{2\pi}{3} + \alpha} v_{ml} \sin(\omega t) \, d(\omega t) \]
\[ V_0 = \frac{3 v_{ml}}{\pi} \int_{\frac{\pi}{3} + \alpha}^{\frac{2\pi}{3} + \alpha} \sin(\omega t) \, d(\omega t) \]
Evaluating the integral of \( \sin(\omega t) \) over the interval \( [\frac{\pi}{3} + \alpha, \frac{2\pi}{3} + \alpha] \):
\[ V_0 = \frac{3 v_{ml}}{\pi} \left[ -\cos(\omega t) \right]_{\frac{\pi}{3} + \alpha}^{\frac{2\pi}{3} + \alpha} \]
\[ V_0 = \frac{3 v_{ml}}{\pi} \left[ -\cos\left(\omega t \bigg|_{\frac{2\pi}{3} + \alpha}\right) + \cos\left(\omega t \bigg|_{\frac{\pi}{3} + \alpha}\right) \right] \]
\[ V_0 = \frac{3 v_{ml}}{\pi} \left[ -\cos\left(\frac{2\pi}{3} + \alpha\right) + \cos\left(\frac{\pi}{3} + \alpha\right) \right] \]
Using the angle addition formula for cosine:
\[ \cos\left(\frac{2\pi}{3} + \alpha\right) = \cos\left(\frac{2\pi}{3}\right)\cos(\alpha) - \sin\left(\frac{2\pi}{3}\right)\sin(\alpha) \]
Knowing that \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \) and \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \), substitute these values:
\[ \cos\left(\frac{2\pi}{3} + \alpha\right) = -\frac{1}{2}\cos(\alpha) - \frac{\sqrt{3}}{2}\sin(\alpha) \]
Similarly, using the angle addition formula for cosine:
\[ \cos\left(\frac{\pi}{3} + \alpha\right) = \cos\left(\frac{\pi}{3}\right)\cos(\alpha) - \sin\left(\frac{\pi}{3}\right)\sin(\alpha) \]
Knowing that \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \), substitute these values:
\[ \cos\left(\frac{\pi}{3} + \alpha\right) = \frac{1}{2}\cos(\alpha) - \frac{\sqrt{3}}{2}\sin(\alpha) \]
Substitute these results back into the expression for \( V_0 \):
\[ V_0 = \frac{3 v_{ml}}{\pi} \left[ -\left( -\frac{1}{2}\cos(\alpha) - \frac{\sqrt{3}}{2}\sin(\alpha) \right) + \left( \frac{1}{2}\cos(\alpha) - \frac{\sqrt{3}}{2}\sin(\alpha) \right) \right] \]
\[ V_0 = \frac{3 v_{ml}}{\pi} \left[ \frac{1}{2}\cos(\alpha) + \frac{\sqrt{3}}{2}\sin(\alpha) + \frac{1}{2}\cos(\alpha) - \frac{\sqrt{3}}{2}\sin(\alpha) \right] \]
\[ V_0 = \frac{3 v_{ml}}{\pi} \left[ \cos(\alpha) \right] \]
\[ V_0 = \frac{3 v_{ml}}{\pi} \cos(\alpha) \]....... equation(1)
From Fig(iii), it can be observed that the source current for phase A (i.e., \( i_A \) or for any other phase) flows for 120° out of every 180°. Therefore, if the output current is assumed to be constant at \( I_{0n} \), the root mean square (rms) value of the source current is calculated as follows:
The source current flows for \( \frac{120°}{180°} = \frac{2}{3} \) of the total time period. If \( I_{0n} \) is constant, the rms value \( I_{\text{rms}} \) can be calculated based on the duty cycle:
\[ I_{\text{rms}} = I_{0n} \times \sqrt{\frac{2}{3}} \]
This formula accounts for the fact that the current is flowing for 2/3 of the time during each cycle. Therefore, the rms value of the source current can be determined using this relationship.
Each Silicon-Controlled Rectifier (SCR) conducts for 120° out of every 360° cycle. Therefore, to calculate the root mean square (rms) value of the thyristor current, we can use the following relationship:
The conducting angle for each SCR is \( \frac{120°}{360°} = \frac{1}{3} \) of the total cycle. If the thyristor current \( I_{\text{thyristor}} \) is assumed constant during its conduction period, the rms value \( I_{\text{rms}} \) can be calculated as:
\[ I_{\text{rms}} = I_{\text{thyristor}} \times \sqrt{\frac{1}{3}} \]
This formula reflects that the SCR conducts for 1/3 of the total cycle time, leading to the calculation of its rms current value based on this duty cycle.
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