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POWER ELECTRONICS DC-AC CONVERTERS (INVERTERS) PART 1

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Presentation on theme: "POWER ELECTRONICS DC-AC CONVERTERS (INVERTERS) PART 1"— Presentation transcript:

1 POWER ELECTRONICS DC-AC CONVERTERS (INVERTERS) PART 1
SITI ZARINA BINTI MD NAZIRI TMT 404 ADVANCED ENGINEERING 4

2 Content Content Concept Full-bridge Converters Half-bridge Inverters
Square-wave Inverters Half-bridge Inverters Introductory notes. SOURCES: “Power Electronics”, Daniel W. Hart, Mc-Graw Hill, “Power Electronics & Drives”, Lecture Notes, Dr. Zainal Salam, UTM

3 DC-AC Converter (Inverter)
DEFINITION: Converts DC to AC power by switching the DC input voltage (or current) in a pre-determined sequence so as to generate AC voltage (or current) output. TYPICAL APPLICATIONS: Un-interruptible power supply (UPS), Industrial (induction motor) drives, Traction, HVDC General block diagram

4 Full-Bridge Converter
Basic circuit used to convert dc to ac ac output is synthesized from a dc input by closing & opening switches in appropriate sequence List of vo according to switch combinations: Overlap of switch “on” time will result short-circuit, across dc voltage source The time allowed for switching is called blanking time Switches Closed Output Voltage, vo S1 and S2 +Vdc S3 and S4 -Vdc S1 and S3 S2 and S4

5 Full-Bridge Converter
S1, S2 closed Vo = -Vdc Vo = +Vdc S3, S4 closed S2, S4 closed Vo = 0 S1, S3 closed Vo = 0

6 Square-wave Inverter Simplest switching scheme for full bridge converter produce a square wave output voltage For RL load Sum of force resources Sum of natural resources (S1, S2 closed) (S3, S4 closed) Note: A and B are constants evaluated from initial condition m

7 Square-wave Inverter Steady-state operation

8 Square-wave Inverter Steady-state operation
Since the square each of the current half-periods is identical, only the 1st half period need to be evaluated: Power absorbed by load: If switches are ideal, the power supplied must be the same as absorbed by load. Power from a dc source:

9 Square-wave Inverter Waveforms

10 Full-Bridge Inverter With insulated gate bipolar transistors (IGBT) switches & feedback diodes

11 Full-Bridge Inverter With IGBT switches & feedback diodes – Operation
When IGBTs Q1 & Q2 are turned off, the load must be continuous, transferred to D3 and D4, Vo = -Vdc; effectively turning the switch path 3 & 4 before Q3 and Q4 are turned on. IGBTs of Q3 & Q4 must be turned on before the load current decays to 0.

12 Half-bridge Inverter The number of switches is reduce to 2, by dividing the dc source voltage into 2 parts with capacitors Capacitors are same in value, voltage = Vdc/2 S1 closed, Vo = -Vdc/2 S2 closed, Vo = +Vdc/2

13 Questions/Discussions
Square-wave Inverter with RL load Given f = 60 Hz, Vdc = 100 V, R = 10 Ohm, L = 25 mH. Determine: An expression for load current The power absorbed by load The average current in dc source


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