In this part four of the term paper, design a single phase DC-AC inverter with p

Question

In this part four of the term paper, design a single phase DC-AC inverter with pulse width modulation (PWM) and provide answers lo the questions below. 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 lo DC for transmission and then inverted back from DC to AC before being stepped dawn 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: 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_A = DE_H and E_B = (1 - D)E_H Hence E_LL = E_H(2D - 1). If D = 0.5 and the switching frequency is 100 kHz, the output waveform will be: 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: 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. Practical/however, it does not work well because the third harmonic is not that far apart from the first. DO-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 E_LL is no longer zero, see below: One interesting observation from live above figure is that the output waveform follows exactly the waveform of D. From equation E_LL = E_H(2D-1) earlier, we can solve for D: D(t) = 0.5[1 + E_LL(t)/E_H] Note that in the above equation both the duty ratio and the output are functions of lime. Suppose we want to generate a sinusoidal output of E = E_n sin(2 pi ft + theta), then the duty ratio is given by D(t) = 0.5[1 + E_ /E_H sin(2 pi ft + theta)] The output waveform will look like the dotted line below: This follows the waveform of the duty ratio D(t). The "varying rectangles" are the waveform of E_AB 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 D(t) if we want to generate a sinusoidal output of E = 100 sin(2 pi 83.3 3t) using the circuit above. Obtain the data for one cycle (0 to 360 degrees with 30 degree increment) for D and E_LL, and draw the waveform similar to the one above. What is the function of the lour 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. Your paper should be at least two pages, typed single-spaced, using 12-point font and 1-inch margins on the top, bottom, right, and left. You should use the APA guidelines for writing and citations.

Explanation / Answer

Ea = DEh;

Eb = (1-D)Eh;

Ell = (2D-1)Eh;

Why DC output is zero?

A: It is because the Dc averages out and gives output as zero.

Why peak amplitude 127V?

A: It is because harmonic peak voltage is 1.27 times of DC voltage. Here DC voltage is 100V, hence peak voltage is 127V.

What is the function of 4 diodes?

A: The function of 4 diodes is to convert DC voltage to AC voltage. These 4 diodes together act as inverter.

Ell = Emsin(2pift + theta)

Ell = 100 sin(2pi83.33t + 30) (as theta is increment by 30deg and t will be 1/1kHz)

Ell = -77.8548V

D = 0.5(1+Ell(t)/Eh);

D = 0.5(1+(-77.8548/200));

D = 0.305363;

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