Power Electronics in Renewable Energy Systems Power Electronics in
POWER ELECTRONICS
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POWER ELECTRONICS Instructor: Eng. Jalal Al Roumy The Islamic University of Gaza Faculty of Engineering Electrical Engineering Department م ي ح ر ل ا ن م ح ر ل ه ا ل ل م ا س بEELE 5450 — Summer 2012 Lecture 17
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بسم الله الرحمن الرحيم. The Islamic University of Gaza Faculty of Engineering Electrical Engineering Department. POWER ELECTRONICS . EELE 5450 — Summer 2012. Instructor: Eng. Jalal Al Roumy. Lecture 17. HALF-BRIDGE WITH RESISTIVE LOAD. - PowerPoint PPT Presentation
Transcript of POWER ELECTRONICS
POWER ELECTRONICS
Instructor: Eng. Jalal Al Roumy
The Islamic University of GazaFaculty of EngineeringElectrical Engineering Department
الرحمن الله بسمالرحيم
EELE 5450 — Summer 2012
Lecture 17
HALF-BRIDGE WITH RESISTIVE LOAD
HALF-BRIDGE WITH RESISTIVE LOAD
Chapter seven
HALF-BRIDGE INVERTER WITH RESISTIVELOAD AND CAPACITIVE ELEMENTS
In the circuit shown, capacitors replace the centre-tapped battery, and a single battery acts as the d.c. source.
Chapter seven
HALF-BRIDGE INVERTER WITH RESISTIVELOAD AND CAPACITIVE ELEMENTS
C1 and C2 are each charged to Vb/2 prior to switching. When Q1 is switched on, C1 capacitor voltage is applied across the load, load current flows and exponentially decays until Q1 is switched off.
Chapter seven
HALF-BRIDGE INVERTER WITH RESISTIVELOAD AND CAPACITIVE ELEMENTS
Q2 is switched on to connect C2 as a reverse voltage across the load. Load current flows in reverse through the load, exponentially decaying until Q2 turns off and Q1 turns on to repeat the cycle.
Chapter seven
HALF-BRIDGE INVERTER WITH RESISTIVELOAD AND CAPACITIVE ELEMENTS
Load voltage
Chapter seven
HALF-BRIDGE INVERTER WITH RESISTIVELOAD AND CAPACITIVE ELEMENTS
Example Vb = 40V, C 1 - C 2 = 200μF, R1 = 10Ω and an inverter frequency
of 100Hz. Assuming ideal switches, determine maximum and minimum values of load voltage, assuming the waveform is symmetrical.
Chapter seven
HALF-BRIDGE INVERTER WITH RESISTIVELOAD AND CAPACITIVE ELEMENTS
Example Vb = 40V, C 1 - C 2 = 200μF, R1 = 10Ω and an inverter frequency
of 100Hz. Assuming ideal switches, determine maximum and minimum values of load voltage, assuming the waveform is symmetrical.
Chapter seven
Chapter seven
Chapter seven
HALF-BRIDGE WITH PURELY INDUCTIVELOAD
With resistive loads, at the half-cycle period Q1 is switched off and Q2 is switched on. The load current changes instantaneously from the maximum positive to the maximum negative value.
With inductive loads, instantaneous reversal is impossible. In fact, if an inductive load was turned off with maximum current and there was no alternative path for the current to flow, the current collapse would be very rapid, and the induced voltage, e = L di/dt, would be very large. This voltage would appear across the transistor switch in reverse and could destroy it.
Chapter seven
HALF-BRIDGE WITH PURELY INDUCTIVELOAD
To protect the transistor from this inductive surge, feedback diodes are connected in anti-parallel across the transistor to provide a path for the current to decay.
Chapter seven
HALF-BRIDGE WITH PURELY INDUCTIVELOAD
When Q1 switches off at maximum positive current, the inductive voltage VL reverses its polarity, the voltage rises above Vb2 and forward biases D2, allowing the decay of current until load current zero, Then Q2 will start the flow of current in the negative direction.
Chapter seven
HALF-BRIDGE WITH PURELY INDUCTIVELOAD
When negative current reaches its maximum value, Q2 switches off, VL reverses, rising above Vbl and forward biasing diode D1, until load current zero, Q1 will then allow current to grow in the positive direction and the cycle is repeated.
Chapter seven
HALF-BRIDGE WITH PURELY INDUCTIVELOAD
Chapter seven
HALF-BRIDGE WITH PURELY INDUCTIVELOAD
Chapter seven
HALF-BRIDGE WITH PURELY INDUCTIVELOAD
Chapter seven
HALF-BRIDGE WITH PURELY INDUCTIVELOAD
Chapter seven
End of Lecture
