Question

DC-AC inverters have many commercial and residential applications, including uninterruptible power supply (UPS) units, adjustable speed drives (ASD) for AC motors, and electronic frequency changer circuits. They are also used in solar electrical systems where batteries are charged from solar panels and used as the input DC voltage source. The inverter circuit converts the DC into AC voltage of desired frequency. Most of us are also familiar with HVDC transmission systems, where high voltages are first converted from AC to DC for transmission and then inverted back from DC to AC before being stepped down for commercial and residential use. Principle of operation As explained in sections 12.4 and 12.5 of the textbook, switching transistors connected in the "H shape can be used to form a basic single-phase DC-AC inverter, as shown below: D 01 031-D) ELL 8 EH Figure 4. Single-phase DC-AC inverter Here D is the duty ratio, and Q1, Q2, Q3, and Q4 represent the four switching transistors. Switches Q1 and Q2 open and close alternately, as do Q3 and Q4. From part one of the term paper: E DE and Eg (-D)E. Hence Eu = Er(1D-1) . If D=0.5 and the switching frequency is 100 kHz, the output waveform will be: AR 10 15 20 s Figure 5. Output waveform with D 0.5 Note that this voltage contains a zero DC output plus harmonics. Why is the DC output zero? If the DC supply voltage is 100 V, the output waveform contains a fundamental sinusoidal component (first harmonic) which peak amplitude is 127 V, see below: 127 V +100 - 100- Figure 6. Fundamental sinusoidal component from the square wave. Why is the peak amplitude 127 V? Theoretically, a low pass filter can be designed to keep the fundamental waveform and block the higher harmonics. Practically, however, it does not work well because the third harmonic is not that far apart from the first. DC-AC inverter with PWM Instead of designing a filter to obtain the sinusoidal output, what if we change the duty ratio D? Suppose now that D is varied periodically between 0.2 and 0.8 at a frequency that is much lower than the switching frequency. Now the output voltage ELL is no longer zero, see below: 0.8 0.2 0 AR 0.6 EH 气し -0.6 EH Figure 7. Output of inverter with a varying D One interesting observation from the above figure is that the output waveform follows exactly the waveform of D From equation ELL EH (2D-1) earlier, we can solve for D: Note that in the above equation both the duty ratio and the output are functions of time. Suppose we want to generate a sinusoidal output of E Esin(2ft+6 , then the duty ratio is given by IH The output waveform will look like the dotted line below Figure 8. Inverter output with a sinusoidal D This follows the waveform of the duty ratio Dit). The "varying rectangles" are the waveform of EAB as before. The varying duty ratio is provided by some control signal using PWM. Given a 200 V DC source with a switching frequency of 1 kHz, calculate DIt) if we want to generate a sinusoidal output of E-100sin(2833 using the circuit above. Obtain the data for one cycle (0 to 360 degrees with 30 degree increment) for D and ELL, and draw the waveform similar to the one above. What is the function of the four diodes? Please include all necessary circuit diagrams or block diagrams in your paper. Please also draw the necessary waveforms together with your comments or discussions.

Answer 1

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