FYP2-2 Progress Evaluation

Final component: Inverter circuit

• Full bridge inverter is chosen to be the inverter circuit of the UPS system.
• The inverter is connected parallel with a battery.

• A Lead-Acid battery is used as it can provide high-current output when needed during power outages.

• A filter circuit is added at the end of the inverter to obtain a pure sinewave of the inverter output voltage.

• A unipolar PWM generator is used to control the power switches of the inverter.
• The main function of the inverter is to convert the DC output voltage of the buck converter to an AC voltage. 

Important note:

The simulation was run for 1 second where the ideal switch is ON during the first 0.5 seconds, and switched OFF on the next 0.5 seconds.

Full Bridge Inverter circuit

Result:

Discussion:

• The battery is charged with 60 VDC during ON state condition.
• The battery is no longer charged during the next 0.5s (OFF state) and only continues supplying the inverter.
• The charged battery will supply the load once a utility power supply fails.

• The inverter is able to convert 60 VDC to 48 VAC.

• The filter circuit is used to reduce the total harmonic distortion (THD) of the inverter output voltage.

Calculations:

1. Inverter

Modulation index, ma = 0.8

Reference Frequency = 50 Hz

Carrier Frequency = 2500 Hz

Frequency modulation ratio, mf = (carrier frequency)/(reference frequency) = 2500/50 = 50

Output voltage, V1 = ma × VDC = 0.8 × 60 = 48 V

2. Filter Circuit

Frequency = 50 Hz

Resistance = 100 Ω

Cut-off frequency, fc = √(50×2500) = 353.55 Hz

Capacitor, C = 1/2πRf = 1/2π(100)(353.55) =4.5 μF

Inductor, L = (1/C)x(1/2πf)^2 = (1/4.5μ)x(1/(2π×353.55))^2 = 90.03 mH