Discontinuous load current of single phase to single phase cyclo converter Inductive Load
Discontinuous Load Current of Single Phase to Single Phase Cyclo Converter Inductive Load:
Fig: Single phase to single phase cyclo-converter (using thyristor bridge) Fig: Input (a) and output (b) voltage, and current (c) waveforms for a cyclo-converter with discontinuous
- The load current in this case is discontinuous, as the inductance, L in series with the resistance, R, is low. Here, also non-circulating mode of operation takes place, with only one of the bridges − #1 (positive), or #2 (negative), conducting at a time, but two bridges do not conduct at the same time, as this will result in a short circuit.
- The output frequency is assumed as (12.5 Hz), the input frequency being same as (f1 = 50 Hz), i.e.,f1 = 4.f2, or f2 = f1/4 . So, four positive half cycles, or two full cycles of the input to the full-wave bridge converter (#1), are required to produce one positive half cycle of the output waveform, as the output frequency is one-fourth of the input frequency.
- With resistive load, taking bridge 1, and assuming the top point of the ac supply as positive, in the positive half cycle of ac input, the odd-numbered thyristor pair, P1 & P3, is triggered after phase delay (θ = ωt = α1), such that current starts flowing the inductive load in this half cycle.
- But, the current flows even after the input voltage has reversed (after θ = π) till it reaches zero (θ = β1) with π α1 > β1 >π, due to inductance being present in series with resistance, its value being low.
- The thyristor pair is naturally commutated. In the next (negative) half cycle, the other thyristor pair (even-numbered), P2 & P4, is triggered at (π α2).
- The current flows through the load in the same direction, with the output voltage also remaining positive. The current goes to zero at (π β2), with π α3 > β2 >π . This procedure continues for the next two half cycles, making a total of four positive half cycles.
- From these four waveforms, one combined positive half cycle of output voltage is produced across the inductive load. The firing angle (α) of the converter is first decreased, in this case for second half cycle only, kept nearly same in the third one, and finally increased in the last (fourth) one.
- To obtain negative output voltage, in the next four half cycles of output voltage, bridge 2 is used. Following same logic, if the bottom point of the ac supply is taken as positive in the negative half of ac input, the odd-numbered thyristor pair, N1 & N3 conducts, by triggering them after phase delay (θ = 4 π α1).
- The current flows now in the opposite (negative) direction through the inductive load, with the output voltage being also negative. The current goes to zero at (4 π β1), due to load being inductive as given earlier. Similarly, the even-numbered thyristor pair, N2 & N4 conducts in the next half cycle, after they are triggered at (5 π α2).
- The current goes to zero at (5 π β2).
- Both the output voltage and current are now negative. The above process also continues for two more half cycles of input voltage, making a total of four.
- From these four waveforms, one combined negative half cycle of output voltage is produced with same output frequency. The pattern of firing angle − first decreasing and then increasing, is also followed in the negative half cycle.
- The load (output) current is discontinuous, as also load (output) voltage. One positive half cycle, along with one negative half cycle, constitute one complete cycle of output (load) voltage waveform, its frequency being 12.5 Hz as stated earlier. The ripple frequency remains also same at 100 Hz, with the ripple in load current being filtered by the inductance present in the load.