Full-Wave AC Voltage Controller Circuit Design Assignment Sample

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Introduction of Design Of Mechatronic Systems

Circuit Design and explanation

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Rather than only a single thyristor as well as a single diode, the full-wave or unidirectional ac voltage controller employs two thyristors in it. These 2 thyristors are coupled in such an anti-parallel configuration. Now it is shown below the working principle of this AC voltage controller with a R load situated in its terminal. This working principle is totally based on the full wave circuit.

The image underneath depicts the regulated voltage supply schematic for an ac voltage controller having a resistive load in its terminal and the circuit is in a single phase (Amin and Molinas, 2018). In the time of the +ve half-cycle, the i/p power supply which is provided in the direction of the load terminal via the 1st thyristor that is T1, as well as throughout the -ve half-cycle, the given i/p power supply which is transferred via the 2nd thyristor that is T2.

Circuit Diagram of full wave “AC volage controller” for 0 degrees firing angle

(Source: Self-made)

 Circuit Diagram for 120 degrees firing angle

(Source: PSpice)

 Circuit Diagram for 180 degrees firing angle

(Source: Self-made)

The 1st thyristor which is indicated as T1 is in forward biased as well as the 2nd thyristor which is indicated as T2 is in reverse biased in the time of the +ve half cycle of such supply potential VS. There is no current transmission seen in the load terminal till the 1st thyristor that is T1 is activated at a certain triggering angle (Qoriaet al. 2018). As a result, the full volt. source VS displays having the similar polarization alongside the 1st thyristor as well as inverted polarization throughout the 2nd thyristor. The 1st thyristor begins flowing as immediately as the 1st thyristor is activated at the time ωt = α, as well as the whole supplying volts VS, save for the fall over the 1st thyristor, emerges throughout the resistance in the load terminal.

Output waves and characteristics

 o/p waveform

(Source: https://electricalworkbook.com/ac-voltage-controller/)

The output waveforms which are shown above represent the input volt. source, gated impulses, o/p volt., o/p current, as well as the voltages alongside the Circuits. When the voltages in the load terminal reaches ωt = π, the current in the load terminal goes to 0, as well as the 1st thyristor ceases to flow. Following that, in the next stage the 2nd thyristor would be forward biased as well as the 1st thyristor would be reverse biased, with no conductivity till the 2nd thyristor is activated, that is till the (π + α).

As longer even though the 1st thyristor is in the ON condition, in other words, from t, the potential alongside the 2nd thyristor equals VT2 = - VT1. In this condition some volt. drop occurs in the 1st thyristor. The 2nd Thyristor is then activated as well as begins flowing at ωt = (π + α). Apart from a tiny power loss along the 2nd thyristor, the whole distribution voltage is developed throughout the resistance in the load terminal as well as the current in the load terminal passes (Sirisha and Nazeemuddin, 2022). As longer as the 2nd thyristor is turned on, that is from (π + α) <ωt< 2π, the voltages over the 1st thyristor would be VT1 = -VT2. This is just because some drop in the voltage of the 2nd thyristor.

Formulas

As demonstrated with in waveforms, this procedure is repeated itself for each round (Hanet al. 2019). For such a resistive load in the terminal, the supply signal would follow the waveforms of the given supply voltages, implying that the supply volt. as well as the current would be into phase with one another.

If the triggering angle of 1st thyristor as well as the 2nd thyristor are similar having such a resistive load in the terminal, then the value of the RMS of o/p volt is defined by,

The RMS value of the load current will be I rms = V rms / R L

So, the i/p power factor will be PF = I 2 rms * R L

The RMS value for the 3rd harmonics would be

E3 = V3rms(3*2πƒt) = V3rms(6πƒt) = V3rms(3ωt)

The RMS value for the 5th harmonics would be

E5 = V5rms(5*2πƒt) = V5rms(10πƒt) = V5rms(5ωt)

The RMS value for the 7th harmonics would be

E7 = V7rms(7*2πƒt) = V7rms(14πƒt) = V7rms(7ωt)

The RMS value for the 9th harmonics would be

E9 = V9rms(9*2πƒt) = V9rms(18πƒt) = V9rms(9ωt)

References

Journals

Amin, M. and Molinas, M., 2018. A gray-box method for stability and controller parameter estimation in HVDC-connected wind farms based on nonparametric impedance. IEEE Transactions on Industrial Electronics66(3), pp.1872-1882.

Han, D., Fang, J., Yu, J., Tang, Y. and Debusschere, V., 2019, September. Small-signal modeling, stability analysis, and controller design of grid-friendly power converters with virtual inertia and grid-forming capability. In 2019 IEEE Energy Conversion Congress and Exposition (ECCE) (pp. 27-33). IEEE.

Ji, K., Chen, W., Wu, X., Pang, H., Hu, J., Liu, S., Cheng, F. and Tang, G., 2021. High frequency stability constraints based MMC controller design applying NSGA-III algorithm. CSEE Journal of Power and Energy Systems.

Qoria, T., Gruson, F., Colas, F., Guillaud, X., Debry, M.S. and Prevost, T., 2018, June. Tuning of cascaded controllers for robust grid-forming voltage source converter. In 2018 Power Systems Computation Conference (PSCC) (pp. 1-7). IEEE.

Sirisha, B. and Nazeemuddin, M.A., 2022. A novel five-level voltage source inverter interconnected to grid with srf controller for voltage synchronization. Bulletin of Electrical Engineering and Informatics11(1), pp.50-58.

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