Single Phase Inverter

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Single Phase Inverter

By | April 2011
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Power Electronic Systems 3
ECAD1 Exercise

John McSorley
Introduction
A single phase inverter circuit is supplied from a 250V DC source which also supplies and 8 Ohm resister in series with a 0.02H inductor. The inverter is configured to operate in quasi-square mode at 50Hz with a duty cycle of 0.6. Objectives

The main objectives of this report is to build and simulate a single phase inverter circuit with a a load consisting of a resistor and an inductor using the Pspice student software and using this software to obtain the load current and load voltage waveforms and also plot the RMS (Root Mean Squared) load current, the RMS diode current (D9), the RMS thyristor current (D14) and the mean power in the load based on the I2R calculation. Circuit Diagram

Figure 1: Single phase inverter schematic

Analysis
The waveform for the load voltage is 197V but shows a distorted square wave due to the load being inductive. The peak of the waveform slopes because the inductor is de-energizing and when the voltage passes 0V line and it takes 2ns to reach the zero line because the inductor takes a short time to become energized. The load current starts at 25A and because the load is inductive the current lags the voltage by 90 degrees. According to the Pspice program the RMS current through the diode (D14) is 0A which doesn’t seem correct. The RMS thyristor current starts at 25A and drops to 10A after 100ns and continues to operate at this current. The RMS load current begins at 25A and drops to 16A, this current is also DC. The power dissipated is 4.85kW and the beginning but then drops and varies between 2.1kW and 2.15kW after 100ns Results

Figure 2: Simulation graph
In Figure 2 shows:
* V(R1:1,L1:2) is the load voltage waveform
* I(R1) is the load current waveform
* RMS(I(D14) is the RMS diode current
* RMS(I(D9) is the RMS thyristor current
* RMS(I(R1) is the RMS load current
* RMS(I(R1) * RMS(I(R1) * 8 (resister value) is the mean...

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