Lecture01 EE 4109 Dept of EEE KUET This

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Lecture-01 EE 4109 Dept. of EEE, KUET

Lecture-01 EE 4109 Dept. of EEE, KUET

This lecture is for Learning Outcome 3 (LO 3): At the end of the

This lecture is for Learning Outcome 3 (LO 3): At the end of the course students will be able to identify and design DC-AC converters (inverters) and their gate driver circuits q Basic of inverter q Principle of operation of inverters q Performance parameters of inverters

Inverter basic DC to AC converters are known as inverters The function of an

Inverter basic DC to AC converters are known as inverters The function of an inverter is to change a dc input voltage to a symmetric ac output voltage of desired magnitude and frequency The output voltage could be fixed or variable at a fixed or variable frequency. Variable output voltage can be obtained by varying the gain of the inverter.

Inverter basic Inverter gain= ac output/dc input (which is controlled by using PWM) Applications:

Inverter basic Inverter gain= ac output/dc input (which is controlled by using PWM) Applications: variable speed ac motor, induction heating, IPS, and UPS Classification: single-phase and three-phase

Single-phase half-bridge inverter

Single-phase half-bridge inverter

Operational Details 3 -wire DC source • • Consists of 2 choppers, 3 -wire

Operational Details 3 -wire DC source • • Consists of 2 choppers, 3 -wire DC source Transistors switched on and off alternately Need to isolate the gate signal for Q 1 (upper device) Each provides opposite polarity of Vs/2 across the load

Q 1 on, Q 2 off, vo = Vs/2 Peak Reverse Voltage of Q

Q 1 on, Q 2 off, vo = Vs/2 Peak Reverse Voltage of Q 2 = Vs

Q 1 off, Q 2 on, vo = -Vs/2

Q 1 off, Q 2 on, vo = -Vs/2

Voltage and current Waveforms

Voltage and current Waveforms

Look at the output voltage rms value of the output voltage, Vo

Look at the output voltage rms value of the output voltage, Vo

rms value of the fundamental component

rms value of the fundamental component

When the load is highly inductive

When the load is highly inductive

Turn off Q 1 at t = To/2 Current falls to 0 via D

Turn off Q 1 at t = To/2 Current falls to 0 via D 2, L, Vs/2 lower + Vs/2 -

Turn off Q 2 at t = To Current falls to 0 via D

Turn off Q 2 at t = To Current falls to 0 via D 1, L, Vs/2 upper + Vs/2 -

Load Current for a highly inductive load Transistors are only switched on for a

Load Current for a highly inductive load Transistors are only switched on for a quarter-cycle, or 90

Fundamental Output Power In most cases, the useful power

Fundamental Output Power In most cases, the useful power

Performance Parameters • Harmonic factor of the nth harmonic (HFn) for n>1 Von =

Performance Parameters • Harmonic factor of the nth harmonic (HFn) for n>1 Von = rms value of the nth harmonic component V 01 = rms value of the fundamental component

Performance Parameters (continued) • Total Harmonic Distortion (THD) • Measures the “closeness” in shape

Performance Parameters (continued) • Total Harmonic Distortion (THD) • Measures the “closeness” in shape between a waveform and its fundamental component

Performance Parameters (continued) • Distortion Factor (DF) • Indicates the amount of HD that

Performance Parameters (continued) • Distortion Factor (DF) • Indicates the amount of HD that remains in a particular waveform after the harmonics have been subjected to second-order attenuation. for n>1

Performance Parameters (continued) • Lowest order harmonic (LOH) • The harmonic component whose frequency

Performance Parameters (continued) • Lowest order harmonic (LOH) • The harmonic component whose frequency is closest to the fundamental, and its amplitude is greater than or equal to 3% of the amplitude of the fundamental component.

Single-phase full-bridge inverter

Single-phase full-bridge inverter

Operational Details • • Consists of 4 choppers and a 3 -wire DC source

Operational Details • • Consists of 4 choppers and a 3 -wire DC source Q 1 -Q 2 and Q 3 -Q 4 switched on and off alternately Need to isolate the gate signal for Q 1 and Q 3 (upper) Each pair provide opposite polarity of Vsacross the load

Q 1 -Q 2 on, Q 3 -Q 4 off, vo = Vs +

Q 1 -Q 2 on, Q 3 -Q 4 off, vo = Vs + Vs -

Q 3 -Q 4 on, Q 1 -Q 2 off, vo = -Vs -

Q 3 -Q 4 on, Q 1 -Q 2 off, vo = -Vs - Vs +

When the load is highly inductive Turn Q 1 -Q 2 off – Q

When the load is highly inductive Turn Q 1 -Q 2 off – Q 3 -Q 4 off

Turn Q 3 -Q 4 off – Q 1 -Q 2 off

Turn Q 3 -Q 4 off – Q 1 -Q 2 off

Load current for a highly inductive load Problems 6. 1, 6. 2, and 6.

Load current for a highly inductive load Problems 6. 1, 6. 2, and 6. 3 from M. H. Rashid’s book

End of Lecture 1

End of Lecture 1